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Contents
CHAPTER 4 The Chain Rule
4.1 Derivatives by the Chain Rule
4.2 Implicit Differentiation and Related Rates
4.3 Inverse Functions and Their Derivatives
4.4 Inverses of Trigonometric Functions
CHAPTER 5 Integrals
5.1 The Idea of the Integral 177
5.2 Antiderivatives 182
5.3 Summation vs. Integration 187
5.4 Indefinite Integrals and Substitutions 195
5.5 The Definite Integral 201
5.6 Properties of the Integral and the Average Value 206
5.7 The Fundamental Theorem and Its Consequences 213
5.8 Numerical Integration 220
CHAPTER 6 Exponentials and Logarithms
6.1 An Overview 228
6.2 The Exponential ex 236
6.3 Growth and Decay in Science and Economics 242
6.4 Logarithms 252
6.5 Separable Equations Including the Logistic Equation 259
6.6 Powers Instead of Exponentials 267
6.7 Hyperbolic Functions 277
CHAPTER 7 Techniques of Integration
7.1 Integration by Parts
7.2 Trigonometric Integrals
7.3 Trigonometric Substitutions
7.4 Partial Fractions
7.5 Improper Integrals
CHAPTER 8 Applications of the Integral
8.1 Areas and Volumes by Slices
8.2 Length of a Plane Curve
8.3 Area of a Surface of Revolution
8.4 Probability and Calculus
8.5 Masses and Moments
8.6 Force, Work, and Energy
CHAPTER 6
Exponentials and Logarithms
This chapter is devoted to exponentials like 2" and 10" and above all ex. The goal is
to understand them, differentiate them, integrate them, solve equations with them,
and invert them (to reach the logarithm). The overwhelming importance of ex makes
this a crucial chapter in pure and applied mathematics.
In the traditional order of calculus books, ex waits until other applications of the .
integral are complete. I would like to explain why it is placed earlier here. I believe
that the equation dyldx = y has to be emphasized above techniques of integration.
The laws of nature are expressed by drflerential equations, and at the center is ex. Its
applications are to life sciences and physical sciences and economics and engineering
(and more-wherever change is influenced by the present state). The model produces
a differential equation and I want to show what calculus can do.
The key is always bm+" (bm)(b3.
= Section 6.1 applies that rule in three ways:
1. to understand the logarithm as the exponent;
2. to draw graphs on ordinary and semilog and log-log paper;
3. to find derivatives. The slope of b" will use bX+*" (bx)(bh").
=
hAn Overview
6.1
There is a good chance you have met logarithms. They turn multiplication into
addition, which is a lot simpler. They are the basis for slide rules (not so important)
and for graphs on log paper (very important). Logarithms are mirror images of
exponentials-and those I know you have met.
Start with exponentials. The numbers 10 and lo2 and lo3 are basic to the decimal
system. For completeness I also include lo0, which is "ten to the zeroth power" or
1. The logarithms of those numbers are the exponents. The logarithms of 1 and 10 and
100 and 1000 are 0 and 1 and 2 and 3. These are logarithms "to base 1 , because
0"
the powers are powers of 10.
Question When the base changes from 10 to b, what is the logarithm of l ?
Answer Since b0 = 1, logJ is always zero. To base b, the logarithm of bn is n.
6.1 An Overview 229
Negative powers are also needed. The number 10x is positive, but its exponent x can
be negative. The first examples are 1/10 and 1/100, which are the same as 10-' and
10- 2 . The logarithms are the exponents -1 and -2:
1000 = 103 and log 1000 = 3
1/1000 = 10- 3
and log 1/1000 = - 3.
Multiplying 1000 times 1/1000 gives 1 = 100. Adding logarithms gives 3 + (- 3) = 0.
m "
Always 10 times 10" equals 10 +" .In particular 103 times 102 produces five tens:
(10)(10)(10) times (10)(10) equals (10)(10)(10)(10)(10) = 105.
The law for b" times b" extends to all exponents, as in 104.6 times 10'. Furthermore
the law applies to all bases (we restrict the base to b > 0 and b - 1). In every case
multiplication of numbers is addition of exponents.
6A bm times b" equals b'",so logarithms (exponents) add
b' divided by b" equals b", so logarithms (exponents) subtract
logb(yZ) = lOgby + lOgbz and logb(Y/Z) = lOgby - lOgbz. (1)
Historical note In the days of slide rules, 1.2 and 1.3 were multiplied by sliding
one edge across to 1.2 and reading the answer under 1.3. A slide rule made in
Germany would give the third digit in 1.56. Its photograph shows the numbers on a
log scale. The distance from 1 to 2 equals the distance from 2 to 4 and from 4 to 8.
By sliding the edges, you add distances and multiply numbers.
Division goes the other way. Notice how 1000/10 = 100 matches 3 - 1 = 2. To divide
1.56 by 1.3, look back along line D for the answer 1.2.
The second figure, though smaller, is the important one. When x increases by 1, 2 x
is multiplied by 2. Adding to x multiplies y. This rule easily gives y = 1, 2, 4, 8, but
look ahead to calculus-which doesn't stay with whole numbers.
Calculus will add Ax. Then y is multiplied by 2ax. This number is near 1. If
ax
Ax = A then 2" 1.07-the tenth root of 2. To find the slope, we have to consider
(2 ax - 1)/Ax. The limit is near (1.07 - 1)/- = .7, but the exact number will take time.
^ ^
2>
1 1+1 1+1+1
Fig. 6.1 An ancient relic (the slide rule). When exponents x add, powers 2x multiply.
6 Exponentials and Logarithms
Base Change Bases other than 10 and exponents other than 1,2,3, ... are needed
for applications. The population of the world x years from now is predicted to grow
by a factor close to 1.02". Certainly x does not need to be a whole number of years.
And certainly the base 1.02 should not be 10 (or we are in real trouble). This prediction
will be refined as we study the differential equations for growth. It can be rewritten
to base 10 if that is preferred (but look at the exponent):
1.02" is the same as 10('Og .02)".
When the base changes from 1.02 to 10, the exponent is multiplied-as we now see.
For practice, start with base b and change to base a. The logarithm to base a will
be written "log." Everything comes from the rule that logarithm = exponent:
base change for numbers: b = d o g b .
Now raise both sides to the power x. You see the change in the exponent:
base change for exponentials: bx = a('0g
,Ix.
Finally set y = bX.Its logarithm to base b is x. Its logarithm to base a is the exponent
on the right hand side: logay = (log,b)x. Now replace x by logby:
base change for logarithms: log, y = (log, b) (log, y ).
We absolutely need this ability to change the base. An example with a = 2 is
b = 8 = Z3 g2 = (z3), = 26 log, 64 = 3 2 = (log28)(log864).
The rule behind base changes is (am)"= am". When the mth power is raised to the
xth power, the exponents multiply. The square of the cube is the sixth power:
(a)(a)(a)times (a)(a)(a) equals (a)(a)(a)(a)(a)(a): (a3),=a6.
Another base will soon be more important than 10-here are the rules for base
changes:
The first is the definition. The second is the xth power of the first. The third is the
logarithm of the second (remember y is bx). An important case is y = a:
log, a = (log, b)(logba) = 1 so log, b = 1/log, a. (3)
EXAMPLE 8 = 23 means 8lI3 = 2. Then (10g28)(l0g82) (3)(1/3) = 1.
=
This completes the algebra of logarithms. The addition rules 6A came from
(bm)(b") bm+".The multiplication rule 68 came from (am)"= am". We still need to
=
deJine b" and ax for all real numbers x. When x is a fraction, the definition is easy.
The square root of a8 is a4 (m = 8 times x = 112). When x is not a fraction, as in 2",
the graph suggests one way to fill in the hole.
.
23141100,.. . As the fractions r approach
We could defne 2" as the limit of 23, 231110,
7t, the powers 2' approach 2". This makes y = 2" into a continuous function, with the
desired properties (2")(2") = 2"'" and (2")" = 2""-whether m and n and x are inte-
gers or not. But the E'S and 6's of continuity are not attractive, and we eventually
choose (in Section 6.4) a smoother approach based on integrals.
GRAPHS O b" AND logby
F
It is time to draw graphs. In principle one graph should do the job for both functions,
because y = bx means the same as x = logby. These are inverse functions. What one
function does, its inverse undoes. The logarithm of g(x) = bXis x:
In the opposite direction, the exponential of the logarithm of y is y:
g(g - = b('08b~)=
Y. (9
This holds for every base b, and it is valuable to see b = 2 and b = 4 on the same
graph. Figure 6.2a shows y = 2" and y = 4". Their mirror images in the 45" line give
the logarithms to base 2 and base 4, which are in the right graph.
When x is negative, y = bx is still positive. If the first graph is extended to the left,
it stays above the x axis. Sketch it in with your pencil. Also extend the second graph
down, to be the mirror image. Don't cross the vertical axis.
Fig. 6.2 Exponentials and mirror images (logarithms). Different scales for x and y.
There are interesting relations within the left figure. All exponentials start at 1,
because b0 is always 1. At the height y = 16, one graph is above x = 2 (because 4' =
16). The other graph is above x = 4 (because 24 = 16). Why does 4" in one graph equal
2," in the other? This is the base change for powers, since 4 = 2,.
The figure on the right shows the mirror image-the logarithm. All logarithms
start from zero at y = 1. The graphs go down to - co at y = 0. (Roughly speaking
2-" is zero.) Again x in one graph corresponds to 2x in the other (base change for
logarithms). Both logarithms climb slowly, since the exponentials climb so fast.
The number log, 10 is between 3 and 4, because 10 is between 23 and 24. The slope
of 2" is proportional to 2"-which never happened for xn. But there are two practical
difficulties with those graphs:
1. 2" and 4" increase too fast. The curves turn virtually straight up.
2. The most important fact about Ab" is the value of 6-and the base
doesn't stand out in the graph.
There is also another point. In many problems we don't know the function y =
f(x). We are looking for it! All we have are measured values of y (with errors mixed
in). When the values are plotted on a graph, we want to discover f(x).
Fortunately there is a solution. Scale the y axis dfferently. On ordinary graphs,
each unit upward adds a fixed amount to y. On a log scale each unit multiplies y by
6 Exponentials and Logarithms
aJixed amount. The step from y = 1 to y = 2 is the same length as the step from 3 to
6 or 10 to 20.
On a log scale, y = 11 is not halfway between 10 and 12. And y = 0 is not there at
all. Each step down divides by a fixed amount-we never reach zero. This is com-
pletely satisfactory for Abx, which also never reaches zero.
Figure 6.3 is on semilog paper (also known as log-linear), with an ordinary x axis.
The graph of y = Abx is a straight line. To see why, take logarithms of that equation:
log y = log A + x log b. (6)
The relation between x and log y is linear. It is really log y that is plotted, so the graph
is straight. The markings on the y axis allow you to enter y without looking up its
logarithm-you get an ordinary graph of log y against x.
Figure 6.3 shows two examples. One graph is an exact plot of y = 2 loX.It goes
upward with slope 1, because a unit across has the same length as multiplication by
10 going up. lox has slope 1 and 10("gb)" (which is bx) will have slope log b. The
crucial number log b can be measured directly as the slope.
Fig. 6.3 2 = 10" and 4 10-"I2 on semilog paper. Fig. 6.4 Graphs of AX^ on log-log paper.
The second graph in Figure 6.3 is more typical of actual practice, in which we start
with measurements and look for f(x). Here are the data points:
We don't know in advance whether these values fit the model y = Abx. The graph is
strong evidence that they do. The points lie close to a line with negative slope-
indicating log b < 0 and b < 1. The slope down is half of the earlier slope up, so the
6.1 An Overview
model is consistent with
y = Ado-X12 or log y = l o g A - f x . (7)
When x reaches 2, y drops by a factor of 10. At x = 0 we see A z 4.
Another model-a power y = Axk instead of an exponential-also stands out with
logarithmic scaling. This time we use log-log paper, with both axes scaled. The
logarithm of y = Axk gives a linear relation between log y and log x:
log y = log A + k log x. (8)
The exponent k becomes the slope on log-log paper. The base b makes no difference.
We just measure the slope, and a straight line is a lot more attractive than a power
curve.
4
The graphs in Figure 6.4 have slopes 3 and and - 1. They represent Ax3 and
A& and Alx. To find the A's, look at one point on the line. At x = 4 the height is
8, so adjust the A's to make this happen: The functions are x3/8 and 4& and 32/x.
On semilog paper those graphs would not be straight!
You can buy log paper or create it with computer graphics.
THE DERIVATIVES OF y = bxAND x= log,y
This is a calculus book. We have to ask about slopes. The algebra of exponents is
done, the rules are set, and on log paper the graphs are straight. Now come limits.
The central question is the derivative. What is dyldx when y = bx? What is dxldy
when x is the logarithm logby? Thpse questions are closely related, because bx and
logby are inverse functions. If one slope can be found, the other is known from
dxldy = l/(dy/dx). The problem is to find one of them, and the exponential comes
first.
You will now see that those questions have quick (and beautiful) answers, except
for a mysterious constant. There is a multiplying factor c which needs more time. I
think it is worth separating out the part that can be done immediately, leaving c in
dyldx and llc in dxldy. Then Section 6.2 discovers c by studying the special number
called e (but c # e).
I 6C The derivative of bX is a multiple ebx. The number c depends on the
base b. I
The product and power and chain rules do not yield this derivative. We are pushed
all the way back to the original definition, the limit of AylAx:
Key idea: Split bx+hinto bXtimes bh. Then the crucial quantity bx factors out. More
than that, bx comes outside the limit because it does not depend on h. The remaining
limit, inside the brackets, is the number c that we don't yet know:
This equation is central to the whole chapter: dyldx equals cbx which equals cy. The
rate of change of y is proportional to y. The slope increases in the same way that bx
increases (except for the factor c). A typical example is money in a bank, where
6 Exponentials and Logarithms
interest is proportional to the principal. The rich get richer, and the poor get slightly
richer. We will come back to compound interest, and identify b and c.
The inverse function is x = logby. Now the unknown factor is l/c:
I 6D The slope of logby is llcy with the same e (depending on b). I
Proof If dy/dx = cbx then dxldy = l/cbx = llcy. (11)
That proof was like a Russian toast, powerful but too quick! We go more carefully:
f(bx) = x (logarithm of exponential)
f '(bx)(cbx)= 1 (x derivative by chain rule)
f '(bx) = l/cbx (divide by cbx)
f '(y) = l/cy (identify bx as y)
The logarithm gives another way to find c. From its slope we can discover l/c. This
is the way that finally works (next section).
-1 0 1
Fig. 6.5 The slope of 2" is about .7 2". The slope of log2y is about 11.7 ~.
Final remark It is extremely satisfying to meet an f(y) whose derivative is llcy.
At last the " - 1 power" has an antiderivative. Remember that j'xndx = xn+'/(n 1) +
is a failure when n = - 1. The derivative of x0 (a constant) does not produce x-'.
'
We had no integral for x - , and the logarithm fills that gap. If y is replaced by x or t
(all dummy variables) then
d 1 d 1
-log,x=- and -log,t=-.
dx cx dt ct
The base b can be chosen so that c = 1. Then the derivative is llx. This final touch
comes from the magic choice b = e-the highlight of Section 6.2.
6.1 EXERCISES
Read-through questions On ordinary paper the graph of y = I is a straight line.
Its slope is m . On semilog paper the graph of y = n
In lo4 = 10,000, the exponent 4 is the a of 10,000. The is a straight line. Its slope is 0 . On log-log paper the
base is b = b . The logarithm of 10" times 10" is c . graph of y = p is a straight line. Its slope is 9 .
The logarithm of 10m/lOn is d . The logarithm of 10,000"
is e . If y = bX then x = f . Here x is any number, The slope of y = b" is dyldx = r , where c depends on
b. The number c is the limit as h - 0 of s . Since x =
,
and y is always s .
k logby is the inverse, (dx/dy)(dy/dx)= t . Knowing
A base change gives b = a -and b" = a - . Then dyldx = cb" yields dxldy = u . Substituting b" for y, the
8' is 2". In other words log2y is i times log8y. When slope of log,?; is v . With a change of letters, the slope of
y = 2 it follows that log28 times log82 equals k . log,x is w .
6.1 An Overview
Problems 1-10 use the rules for logarithms. 14 Draw semilog graphs of y = lo1-' and y = ~fi)".
1 Find these logarithms (or exponents): 15 The Richter scale measures earthquakes by loglo(I/Io)=
(a)log232 (b) logz(1/32) ( 4 log32(1/32) R. What is R for the standard earthquake of intensity I,? If
the 1989 San Francisco earthquake measured R = 7, how did
(d) (e) log, dl0-) (f) log2(l0g216) its intensity I compare to I,? The 1906 San Francisco quake
2 Without a calculator find the values of had R = 8.3. The record quake was four times as intense with
(a)310g35 (b) 3210835 R= .
(c) log, 05 + log1o2 (d) (l0g3~)(logbg) 16 The frequency of A above middle C is 440/second. The
(e) 10510-4103 (f) log256- log27 frequency of the next higher A is . Since 2'/l2 x 1.5,
the note with frequency 660/sec is
3 Sketch y = 2-" and y = g4") from -1 to 1 on the same
graph. Put their mirror images x = - log2y and x = log42y 17 Draw your own semilog paper and plot the data
on a second graph.
4 Following Figure 6.2 sketch the graphs of y = (iy and x =
Estimate A and b in y = Abx.
logl12y.What are loglI22and loglI24?
5 Compute without a computer:
18 Sketch log-log graphs of y = x2 and y = &.
(a)log23 + log2 3 (b) log2(i)10 19 On log-log paper, printed or homemade, plot y = 4, 11,
(c) log,010040 21, 32, 45 at x = 1, 2, 3, 4, 5. Estimate A and k in y = AX^.
( 4 (log1 0 4(loge10)
(e) 223/(22)3 (f logdlle)
Questions 20-29 are about the derivative dyldx = cbx.
6 Solve the following equations for x:
20 g(x) = bx has slope g' = cg. Apply the chain rule to
(a)log10(10")= 7 (b) log 4x - log 4 = log 3
g f (y))= y to prove that dfldy = llcy.
(
(c) logXlO 2
= (d) 10g2(l/x) 2
,=
(e) log x + log x = log 8 (f) logx(xx) 5
=
21 If the slope of log x is llcx, find the slopes of log (2x) and
log (x2)and log (2").
7 The logarithm of y = xn is logby= . 22 What is the equation (including c) for the tangent line to
*8 Prove that (1ogba)(logdc) (logda)(logbc).
= ?
y = 10" at x = O Find also the equation at x = 1.
9 2'' is close to lo3 (1024 versus 1000). If they were equal 23 What is the equation for the tangent line to x = log, ,y at
then log,lO would be . Also logl02 would be y = l? Find also the equation at y = 10.
instead of 0.301.
24 With b = 10, the slope of 10" is c10". Use a calculator for
10 The number 21°00has approximately how many (decimal) small h to estimate c = lim (loh- l)/h.
digits?
25 The unknown constant in the slope of y = (.l)" is
Questions 11-19 are about the graphs of y = bx and x = logby. L =lim (. l h- l)/h. (a) Estimate L by choosing a small h.
(b) Change h to -h to show that L = - c from Problem 24.
11 By hand draw the axes for semilog paper and the graphs
of y = l.lXand y = lq1.1)". 26 Find a base b for which (bh- l)/h x 1. Use h = 114 by hand
or h = 1/10 and 1/100 by calculator.
12 Display a set of axes on which the graph of y = loglox is
a straight line. What other equations give straight lines on 27 Find the second derivative of y = bx and also of x = logby.
those axes? 28 Show that C = lim (lWh- l)/h is twice as large as c =
13 When noise is measured in decibels, amplifying by a factor lim (10" - l)/h. (Replace the last h's by 2h.)
A increases the decibel level by 10 log A. If a whisper is 20db 29 In 28, the limit for b = 100 is twice as large as for b = 10.
and a shout is 70db then 10 log A = 50 and A = . So c probably involves the of b.
236 6 Exponentials and Logarithms
h
6.2 T e Exponential eX
The last section discussed bx and logby. The base b was arbitrary-it could be 2 or 6
or 9.3 or any positive number except 1. But in practice, only a few bases are used.
I have never met a logarithm to base 6 or 9.3. Realistically there are two leading
candidates for b, and 10 is one of them. This section is about the other one, which is
an extremely remarkable number. This number is not seen in arithmetic or algebra
or geometry, where it looks totally clumsy and out of place. In calculus it comes into
its own.
The number is e. That symbol was chosen by Euler (initially in a fit of selfishness,
but he was a wonderful mathematician). It is the base of the natural logarithm.
It also controls the exponential ex, which is much more important than In x.
Euler also chose 7c to stand for perimeter-anyway, our first goal is to find e.
Remember that the derivatives of bx and logby include a constant c that depends
on b. Equations (10) and (1 1) in the previous section were
d d 1
-b" = cb"
dx
and - logby = -. (1)
d~ CY
At x = 0, the graph of bx starts from b0 = 1. The slope is c. At y = 1, the graph of
logby starts from logbl = 0. The logarithm has slope llc. With the right choice of the
base b those slopes will equal 1 (because c will equal 1).
For y = 2" the slope c is near .7. We already tried Ax = .1 and found Ay z -07. The
base has to be larger than 2, for a starting slope of c = 1.
We begin with a direct computation of the slope of logby at y = 1:
1 1
- = slope
C
at 1 = lim - [logb(l
h+O h
+ h) - logbl] = hlim logb[(l + h)'lh].
-0
Always logbl = 0. The fraction in the middle is logb(l + h) times the number l/h. This
number can go up into the exponent, and it did.
The quantity (1 + h)'Ih is unusual, to put it mildly. As h + 0, the number 1 h is +
approaching 1. At the same time, l/h is approaching infinity. In the limit we have
1". But that expression is meaningless (like 010). Everything depends on the
balance bet.ween "nearly 1" and "nearly GO." This balance produces the extraordinary
number e:
DEFINITION The number e is equal to lim (1 +'h)'lh. Equivalently e = lim
h+O n+ c
o
Before computing e, look again at the slope llc. At the end of equation (2) is the
logarithm of e:
When the base is b = e, the slope is logee = 1. That base e has c = 1 as desired
1
The derivative of ex is 1 ex and the derivative of log,y is -
1 my' (4)
This is why the base e is all-important in calculus. It makes c = 1.
To compute the actual number e from (1 + h)'lh, choose h = 1, 1/10, 1/100, ... . Then
the exponents l/h are n = 1, 10, 100, .... (All limits and derivatives will become official
in Section 6.4.) The table shows (1 + h)lih approaching e as h - 0 and n - oo:
, ,
6 2 The Exponential eX
.
The last column is converging to e (not quickly). There is an infinite series that
converges much faster. We know 125,000 digits of e (and a billion digits of n). There
are no definite patterns, although you might think so from the first sixteen digits:
e = 2.7 1828 1828 45 90 45 .-. (and lle z .37).
The powers of e produce y = ex. At x = 2.3 and 5, we are close to y = 10 and 150.
The logarithm is the inversefunction. The logarithms of 150 and 10, to the base e,
are close to x = 5 and x = 2.3. There is a special name for this logarithm--the natural
logarithm. There is also a special notation "ln" to show that the base is e:
In y means the same as log,y. The natural logarithm is the exponent in ex = y.
The notation In y (or In x-it is the function that matters, not the variable) is standard
in calculus courses. After calculus, the base is generally assumed to be e. In most of
science and engineering, the natural logarithm is the automatic choice. The symbol
"exp (x)" means ex, and the truth is that the symbol "log x" generally means In x.
Base e is understood even without the letters In. But in any case of doubt-on a
calculator key for example-the symbol "ln x" emphasizes that the base is e.
THE DERIVATIVES OF ex AND In x
Come back to derivatives and slopes. The derivative of bx is cbx, and the derivative
of log, y is llcy. If b = e then c = 1 . For all bases, equation (3) is llc = logbe.
This gives c-the slope of bx at x = 0:
c = In b is the mysterious constant that was not available earlier. The slope of 2" is
In 2 times 2". The slope of ex is In e times ex (but In e = 1). We have the derivatives
on which this chapter depends:
6F The derivatives of ex and In y are ex and 1fy. For other bases
d d 1
- bx = (In b)bx and - logby= --- (6)
dx d~ (in b ) ~ '
To make clear that those derivatives come from the functions (and not at all from
the dummy variables), we rewrite them using t and x:
d d 1
-e'=ef and -lnx=-.
dt dx x
6 Exponentials and Logarithms
Remark on slopes at x = 0: It would be satisfying to see directly that the slope of 2"
is below 1, and the slope of 4" is above 1. Quick proof: e is between 2 and 4.
But the idea is to see the slopes graphically. This is a small puzzle, which is fun to
solve but can be skipped.
2" rises from 1 at x = 0 to 2 at x = 1. On that interval its average slope is 1. Its
slope at the beginning is smaller than average, so it must be less than 1-as desired.
:
On the other hand 4" rises from at x = - to 1 at x = 0. Again the average slope
,,
is L/L = 1. Since x = 0 comes at the end of this new interval, the slope of 4" at that
point exceeds 1. Somewhere between 2" and 4" is ex, which starts out with slope 1.
This is the graphical approach to e. There is also the infinite series, and a fifth
definition through integrals which is written here for the record:
1. e is the number such that ex has slope 1 at x = 0
2. e is the base for which In y = log,y has slope 1 at y = 1
:r
3. e is the limit of 1 + - as n - co
( ,
5. the area 5; x - l dx equals 1.
The connections between 1, 2, and 3 have been made. The slopes are 1 when e is the
limit of (1 + lln)". Multiplying this out wlll lead to 4, the infinite series in Section 6.6.
The official definition of in x comes from 1dxlx, and then 5 says that in e = 1. This
approach to e (Section 6.4) seems less intuitive than the others.
Figure 6.6b shows the graph of e-". It is the mirror image of ex across the vertical
axis. Their product is eXe-" = 1. Where ex grows exponentially, e-" decays
exponentially-or it grows as x approaches - co. Their growth and decay are faster
than any power of x. Exponential growth is more rapid than polynomial growth, so
that e"/xn goes to infinity (Problem 59). It is the fact that ex has slope ex which keeps
the function climbing so fast.
Fig. 6.6 ex grows between 2" and 4". Decay of e-", faster decay of e-"'I2.
The other curve is y = e-"'I2. This is the famous "bell-shaped curve" of probability
theory. After dividing by fi,
it gives the normal distribution, which applies to so
many averages and so many experiments. The Gallup Poll will be an example in
Section 8.4. The curve is symmetric around its mean value x = 0, since changing x to
- x has no effect on x2.
About two thirds of the area under this curve is between x = - 1 and x = 1. If you
pick points at random below the graph, 213 of all samples are expected in that
interval. The points x = - 2 and x = 2 are "two standard deviations" from the center,
6.2 The Exponential ex 239
enclosing 95% of the area. There is only a 5% chance of landing beyond. The decay
is even faster than an ordinary exponential, because -ix2 has replaced x.
THE DERIVATIVES OF eX AND eu x)
The slope of ex is ex. This opens up a whole world of functions that calculus can deal
with. The chain rule gives the slope of e3 x and esinx and every e"(x):
6G The derivative of euix) is eu(x) times du/dx. (8)
Special case u = cx: The derivative of e" is cecx. (9)
3 3
EXAMPLE 1 The derivative of e x is 3e x (here c = 3). The derivative of esinx is
esin x cos x (here u = sin x). The derivative of f(u(x)) is df/du times du/dx. Here
f= e"so df/du = e". The chain rule demands that secondfactor du/dx.
EXAMPLE 2 e(In 2 is the same as 2x. Its derivative is In times 2x. The chain rule
2)x 2
rediscovers our constant c = In 2. In the slope of bx it rediscovers the factor c = Inb.
Generally ecx is preferred to the original bx. The derivative just brings down the
constant c. It is better to agree on e as the base, and put all complications (like c =
b)
In up in the exponent. The second derivative of ecx is c2ecx.
EXAMPLE 3 The derivative of e-x2/2 is - xe -x 2/ 2 (here u = - x 2/2 so du/dx= - x).
EXAMPLE 4 The second derivative off= e - x2/2, by the chain rule and product rule,
is
2 - 2/2
f" (-1)
= e- 2/2 +
x (
2 2
x) 2 e-x / = ( - l)e x . (10)
Notice how the exponential survives. With every derivative it is multiplied by more
factors, but it is still there to dominate growth or decay. The points of inflection,
where the bell-shaped curve hasf" = 0 in equation (10), are x = 1 and x = - 1.
"n
EXAMPLE 5 (u = n Inx). Since en is x"in disguise, its slope must be nx -1:
slope = e""nx (n In x)= x(n) = nx (11)
This slope is correctfor all n, integer or not. Chapter 2 produced 3x2 and 4x 3 from
the binomial theorem. Now nx"- 1 comes from In and exp and the chain rule.
EXAMPLE 6 An extreme case is xx = (eInx)x. Here u = x In and we need du/dx:
x
d (x) = exnxIn x+ x- = xx(ln x + 1).
dx x)
INTEGRALS OF e" AND e" du/dx
The integral of ex is ex. The integral of ecx is not ecx. The derivative multiplies by c so
the integral divides by c. The integralof ecx is ecx/c (plus a constant).
+
EXAMPLES e2xdx - e2x + C bxdx = C
2 f Inb
6 Exponentiais and Logarithms
The first one has c = 2. The second has c = In b-remember again that bx = e('nb)x.
The integral divides by In b. In the third one, e3("+')is e3" times the number e3 and
that number is carried along. Or more likely we see e3'"+'I as eu. The missing du/dx =
3 is fixed by dividing by 3. The last example fails because duldx is not there. We
cannot integrate without duldx:
Here are three examples with du/dx and one without it:
The first is a pure eudu. So is the second. The third has u = and du/dx = l/2&,
so only the factor 2 had to be fixed. The fourth example does not belong with the
others. It is the integral of du/u2, not the integral of eudu. I don't know any way to
tell you which substitution is best-except that the complicated part is 1 + ex and it
is natural to substitute u. If it works, good.
5
Without an extra ex for duldx, the integral dx/(l + looks bad. But u = 1 + ex
is still worth trying. It has du = exdx = (u - 1)dx:
That last step is "partial fractions.'' The integral splits into simpler pieces (explained
in Section 7.4) and we integrate each piece. Here are three other integrals:
5
The first can change to - eudu/u2, hich is not much better. (It is just as impossible.)
w
The second is actually J u d u , but I prefer a split: 54ex and 5e2" are safer to do
5
separately. The third is (4e-" + l)dx, which also separates. The exercises offer prac-
tice in reaching eudu/dx - ready to be integrated.
Warning about dejinite integrals When the lower limit is x = 0, there is a natural
tendency to expect f(0) = 0-in which case the lower limit contributes nothing. For
a power f = x3 that is true. For an exponential f = e3" it is definitely not true, because
f(0) = 1:
6 2 The Exponential eX
. 241
6.2 EXERCISES
Read-through questions 24 The function that solves dyldx = - y starting from y = 1
The number e is approximately a . It is the limit of (1 + h)
at x = 0 is . Approximate by Y(x h) - Y(x)= +
to the power b . This gives l.O1lOOwhen h = c . An
- hY(x). If h = what is Y(h)after one step and what is Y ( l )
after four steps?
equivalent form is e = lim ( d )".
25 Invent three functions f, g, h such that for x > 10
When the base is b = e, the constant c in Section 6.1 is e .
Therefore the derivative of y = ex is dyldx = f . The deriv-
+
(1 llx)" <f ( x )< e" < g(x)< e2" < h(x)< xx.
ative of x = logey is dxldy = g . The slopes at x = 0 and 26 Graph ex and #
at x = - 2, -1, 0, 1, 2. Another form
y = 1 are both h . The notation for log,y is I , which offiis .
is the I logarithm of y.
Find antiderivatives for the functions in 27-36.
The constant c in the slope of bx is c = k . The function
bx can be rewritten as I . Its derivative is m . The
derivative of eU(") n . The derivative of ednX
is is 0 .
The derivative of ecxbrings down a factor P .
The integral of ex is q . The integral of ecxis r .
The integral of eU(")du/dx s . In general the integral of
is +
33 xeX2 xe-x2 34 (sin x)ecO" + (cos x)e"'""
eU(") itself is t to find.
by +
35 @ (ex)' 36 xe" (trial and error)
37 Compare e-" with e-X2.Which one decreases faster near
Find the derivatives of the functions in 1-18. x = O Where do the graphs meet again? When is the ratio of
?
e-x2 to e-X less than 1/100?
38 Compare ex with xX:Where do the graphs meet? What
are their slopes at that point? Divide xx by ex and show that
the ratio approaches infinity.
39 Find the tangent line to y = ex at x = a. From which point
on the graph does the tangent line pass through the origin?
40 By comparing slopes, prove that if x > 0 then
(a)ex> 1 + x (b)e-"> 1 - x .
41 Find the minimum value of y = xx for x >0.Show from
dZy/dx2that the curve is concave upward.
42 Find the slope of y = x1lXand the point where dy/dx = 0.
17 esinx
+ sin ex 18 x- 'Ix (which is e-) Check d2y/dx2to show that the maximum of xllx is
19 The difference between e and (1 + l/n)" is approximately 43 If dyldx = y find the derivative of e-"y by the product
Celn. Subtract the calculated values for n = 10, 100, 1000 from rule. Deduce that y(x) = Cex for some constant C.
2.7183 to discover the number C.
44 Prove that xe = ex has only one positive solution.
20 By algebra or a calculator find the limits of ( 1 + l/n)2n
and
+
(1 l / n ) 4 Evaluate the integrals in 45-54. With infinite limits, 49-50 are
...
21 The limit of (11/10)1°,(101/100)100, is e. So the limit of "improper."
(1 - l/ny.
...
(10111)1°, (100/101)100, is
(lO/ll)ll , (100/101)101,. is
..
. So the limit of
. The last sequence is 46 Jb" sin x ecoSx
dx
22 Compare the number of correct decimals of e for
(l.OO1)lOOO (l.OOO1)lOOOO if possible (l.OOOO1)lOOOOO.
and and
48 Sl 2-. dx
Which power n would give all the decimals in 2.71828?
23 The function y = ex solves dyldx = y. Approximate this
50 J; xe-.. dx
equation by A Y A x = Y; which is Y(x+ h) - Y(x)= h Y(x).
With h = & find Y(h) after one step starting from Y(0)= 1.
What is Y ( l )after ten steps?
242 6 Exponentials and Logarithms
53 :
1 cos
2sinx x dx 54 1'' (1 -ex)'' ex dx
59 This exercise shows that F(x) = x"/ex - 0 as x + m.
,
(a) Find dF/dx. Notice that F(x) decreases for x > n > 0.
The maximum of xn/e", at x = n, is nn/en.
55 Integrate the integrals that can be integrated: (b) F(2x) = (2x)"/ezx= 2"xn/eXex < 2"n"/en ex.
Deduce that F(2x) + 0 as x + bo. Thus F(x) + 0.
60 With n = 6, graph F(x) = x6/ex on a calculator or com-
puter. Estimate its maximum. Estimate x when you reach
F(x) = 1. Estimate x when you reach F(x) = 4.
61 Stirling's formula says that n! z @JZn. it to esti-
Use
56 Find a function that solves yl(x) = 5y(x) with y(0) = 2. mate 66/e6 to the nearest whole number. Is it correct? How
57 Find a function that solves yl(x) = l/y(x) with y(0) = 2.
many decimal digits in lo!?
62 x6/ex - 0 is also proved by l'H6pital's rule (at x = m):
58 With electronic help graph the function (1 + llx)". What ,
are its asymptotes? Why? lim x6/ex= lim 6xs/ex = fill this in = 0.
6.3 Growth and Decay in Science and Economics
The derivative of y = e" has taken time and effort. The result was y' = cecx, which
means that y' = cy. That computation brought others with it, virtually for free-the
derivatives of bx and x x and eu(x). I want to stay with y' = cy-which is the most
But
important differential equatibn in applied mathematics.
Compare y' = x with y' = y. The first only asks for an antiderivative of x . We quickly
find y = i x 2 + C. The second has dyldx equal to y itself-which we rewrite as dy/y =
d x . The integral is in y = x + C. Then y itself is exec. Notice that the first solution is
$x2 plus a constant, and the second solution is ex times a constant.
There is a way to graph slope x versus slope y. Figure 6.7 shows "tangent arrows,"
which give the slope at each x and y. For parabolas, the arrows grow steeper as x
1 2 1
Fig. 6.7 The slopes are y' =x and y' = y. The solution curves fit those slopes.
6.3 Growth and Decay in Science and Economics 243
grows-because y' = slope = x. For exponentials, the arrows grow steeper as y
grows-the equation is y'= slope = y. Now the arrows are connected by y = Aex.
A differential equation gives afield of arrows (slopes). Its solution is a curve that stays
tangent to the arrows - then the curve has the right slope.
A field of arrows can show many solutions at once (this comes in a differential
equations course). Usually a single Yo is not sacred. To understand the equation we
start from many yo-on the left the parabolas stay parallel, on the right the heights
stay proportional. For y' = - y all solution curves go to zero.
From y' = y it is a short step to y' = cy. To make c appear in the derivative, put c
into the exponent. The derivative of y = ecx is cecx, which is c times y. We have reached
the key equation, which comes with an initial condition-a starting value yo:
dy/dt = cy with y = Yo at t = 0. (1)
A small change: x has switched to t. In most applications time is the natural variable,
rather than space. The factor c becomes the "growth rate" or "decay rate"-and ecx
converts to ect.
The last step is to match the initial condition. The problem requires y = Yo at
t = 0. Our ec' starts from ecO = 1. The constant of integration is needed now-the
solutions are y = Ae". By choosing A = Yo, we match the initial condition and solve
equation (1). The formula to remember is yoec'.
61 The exponential law y = yoec' solves y' = cy starting from yo.
The rate of growth or decay is c. May I call your attention to a basic fact? The
formula yoec' contains three quantities Yo, c, t. If two of them are given, plus one
additional piece of information, the third is determined. Many applications have one
of these three forms: find t, find c, find yo.
1. Find the doubling time T if c = 1/10. At that time yoecT equals 2yo:
In 2 .7
e T = 2 yields cT= In 2 so that T= I --. (2)
c .1
The question asks for an exponent T The answer involves logarithms. If a cell grows
at a continuous rate of c = 10% per day, it takes about .7/.1 = 7 days to double in
size. (Note that .7 is close to In 2.) If a savings account earns 10% continuous interest,
it doubles in 7 years.
In this problem we knew c. In the next problem we know T
2. Find the decay constant c for carbon-14 if y = ½yo in T= 5568 years.
ecr = 4 yields cT= In I so that c (In 5)/5568. (3)
After the half-life T= 5568, the factor e T equals 4. Now c is negative (In = - In 2).
c
Question 1 was about growth. Question 2 was about decay. Both answers found
ecT as the ratio y(T)/y(O). Then cT is its logarithm. Note how c sticks to T.
T has the units of time, c has the units of "1/time."
Main point: The doubling time is (In 2)/c, because cT= In 2. The time to multiply
by e is 1/c. The time to multiply by 10 is (In 10)/c. The time to divide by e is - 1/c,
when a negative c brings decay.
3. Find the initial value Yo if c = 2 and y(l) = 5:
y(t) = yoec' yields Yo = y(t)e - c = 5e-2
6 Exponentials and Logarithms
.
(1.05 13)20
(1 .05l2O
2
simple interest
f
cT=ln2 5 10 15 20 years
Fig. 6.8 Growth (c > 0) and decay (c < 0. Doubling time T = (In 2)lc. Future value at 5%.
)
All we do is run the process backward. Start from 5 and go back to yo. With time
reversed, ect becomes e-". The product of e2 and e-2 is 1-growth forward and
decay backward.
Equally important is T + t. Go forward to time Tand go on to T + t:
which is (yoecT)ect.
y(T+ t) is yoec(T+t) (4)
Every step t, at the start or later, multiplies by the same ect.This uses the fundamental
property of exponentials, that eT+'= eTet.
EXAMPLE 1 Population growth from birth rate b and death rate d (both constant):
dyldt = by - dy = cy (the net rate is c = b - d).
The population in this model is yoect= yoebte-dt.It grows when b > d (which makes
c > 0). One estimate of the growth rate is c = 0.02/year:
In2 .7
The earth's population doubles in about T = -x - = 35 years.
c .02
First comment: We predict the future based on c. We count the past population
to find c. Changes in c are a serious problem for this model.
Second comment: yoectis not a whole number. You may prefer to think of bacteria
instead of people. (This section begins a major application of mathematics to economics
and the life sciences.) Malthus based his theory of human population on this equation
y' = cy-and with large numbers a fraction of a person doesn't matter so much. To
use calculus we go from discrete to continuous. The theory must fail when t is very
large, since populations cannot grow exponentially forever. Section 6.5 introduces the
logistic equation y' = cy - by2, with a competition term - by2 to slow the growth.
Third comment: The dimensions of b, c, d are "l/time." The dictionary gives birth
rate = number of births per person in a unit of time. It is a relative rate-people
divided by people and time. The product ct is dimensionless and ectmakes sense (also
-
dimensionless). Some texts replace c by 1 (lambda). Then 1/A is the growth time or
decay time or drug elimination time or diffusion time.
EXAMPLE 2 Radioactive dating A gram of charcoal from the cave paintings in
France gives 0.97 disintegrations per minute. A gram of living wood gives 6.68 disin-
tegrations per minute. Find the age of those Lascaux paintings.
The charcoal stopped adding radiocarbon when it was burned (at t = 0). The
amount has decayed to yoect.In living wood this amount is still yo, because cosmic
6.3 Growth and ÿ gay in Science and Economics
rays maintain the balance. Their ratio is ect= 0.97/6.68. Knowing the decay rate c
from Question 2 above, we know the present time t:
ct = ln (~3 5568 0.97
yields t = -in -
-.7 (6.68)
= 14,400 years.
f
Here is a related problem-the age o uranium. Right now there is 140 times as much
U-238 as U-235. Nearly equal amounts were created, with half-lives of (4.5)109 and
(0.7)109 years. Question: How long since uranium was created? Answer: Find t by
sybstituting c = (In $)/(4.5)109and C = (ln ;)/(0.7)109:
In 140
ect/ect=140 * -
ct - Ct = In 140 =. t = - 6(109) years.
c-C
EXAMPLE 3 Calculus in Economics: price inflation and the value o money
f
We begin with two inflation rates - a continuous rate and an annual rate. For the
price change Ay over a year, use the annual rate:
Ay = (annual rate) times (y) times (At). (5)
Calculus applies the continuous rate to each instant dt. The price change is dy:
k
dy = (continuous rate) times (y) times (dt). (6)
Dividing by dt, this is a differential equation for the price:
dyldt = (continuous rate) times (y) = .05y.
The solution is yoe.05'.Set t = 1. Then emo5= 1.0513 and the annual rate is 5.13%.
When you ask a bank what interest they pay, they give both rates: 8% and 8.33%.
The higher one they call the "effective rate." It comes from compounding (and depends
how often they do it). If the compounding is continuous, every dt brings an increase
of dy-and eeo8is near 1.0833.
Section 6.6 returns to compound interest. The interval drops from a month to a
day to a second. That leads to (1 + lln)", and in the limit to e. Here we compute the
effect of 5% continuous interest:
Future value A dollar now has the same value as esoST dollars in T years.
Present value A dollar in T years has the same value as e--OSTdollars now.
Doubling time Prices double (emosT= 2) in T= In 21.05 x 14 years.
With no compounding, the doubling time is 20 years. Simple interest adds on 20
times 5% = 100%. With continuous compounding the time is reduced by the factor
In 2 z -7, regardless of the interest rate.
EXAMPLE 4 In 1626 the Indians sold Manhattan for $24. Our calculations indicate
that they knew what they were doing. Assuming 8% compound interest, the original
$24 is multiplied by e.08'. After t = 365 years the multiplier is e29.2and the $24 has
grown to 115 trillion dollars. With that much money they could buy back the land
and pay off the national debt.
This seems farfetched. Possibly there is a big flaw in the model. It is absolutely
true that Ben Franklin left money to Boston and Philadelphia, to be invested for 200
years. In 1990 it yielded millions (not trillions, that takes longer). Our next step is a
new model.
6 Exponentlals and Logarithms
Question How can you estimate e2'm2 with a $24 calculator (log but not In)?
Answer Multiply 29.2 by loglo e = .434 to get 12.7. This is the exponent to base 10.
After that base change, we have or more than a trillion.
GROWTH OR DECAY WlTH A SOURCE TERM
The equation y' = y will be given a new term. Up to now, all growth or decay has
started from yo. No deposit or withdrawal was made later. The investment grew by
itself-a pure exponential. The new term s allows you to add or subtract from the
account. It is a "source"-or a "sink" if s is negative. The source s = 5 adds 5dt,
proportional to dt but not to y:
Constant source: dyldt = y + 5 starting from y = yo.
Notice y on both sides! My first guess y = et+' failed completely. Its derivative is et+'
+
again, which is not y + 5. The class suggested y = et 5t. But its derivative et + 5 is
still not y + 5. We tried other ways to produce 5 in dyldt. This idea is doomed to
failure. Finally we thought o y = Aet - 5. That has y' = Aet = y + 5 as required.
f
Important: A is not yo. Set t = 0 to find yo = A - 5. The source contributes 5et - 5:
+
The solution is (yo+ 5)e' - 5. That is the same as yOef 5(et- 1).
s = 5 multiplies the growth term ef - 1 that starts at zero. yoefgrows as before.
EXAMPLE 5 dyldt = - y + 5 has y = (yo- 5)e-' + 5. This is y0e-' + 5(1 - e-'). 7 ,lOet-5
That final term from the soul-ce is still positive. The other term yoe-' decays to zero.
The limit as t + is y, = 5 . A negative c leads to a steady state y,.
Based on these examples with c = 1 and c = -- 1, we can find y for any c and s.
Oet -5
EQUATION WlTH SOURCE 2 = cy + s starts from y = yo at t = 0.
dt (7)
5e&+5
The source could be a deposit of s = $1000/year, after an initial investment of yo = 5 =Y,
$8000. Or we can withdraw funds at s = - $200/year. The units are "dollars per year"
to match dyldt. The equation feeds in $1000 or removes $200 continuously-not all 0 -5e-'+5
at once. 1
Note again that y = e(c+s)t not a solution. Its derivative is (c + sly. The combina-
is Rgmdm9
tion y = ect+ s is also not a solution (but closer). The analysis of y' = cy + s will be
our main achievement for dzrerential equations (in this section). The equation is not
restricted to finance-far from it-but that produces excellent examples.
I propose to find y in four ways. You may feel that one way is enough.? The first
way is the fastest-only three lines-but please give the others a chance. There is no
point in preparing for real problems if we don't solve them.
Solution by Method 1 (fast way) Substitute the combination y = Aec' + B. The solu-
tion has this form-exponential plus constant. From two facts we find A and B:
the equation y' = cy + s gives cAect= c(Aect+ B) + s
the initial value at t = 0 gives A + B = yo.
tMy class says one way is more than enough. They just want the answer. Sometimes I cave
in and write down the formula: y is y,ect plus s(e" - l)/c from the source term.
6.3 Growth and Decay in Science and Economics
The first line has cAect on both sides. Subtraction leaves cB + s = 0 or B = - SIC.
,
Then the second line becomes A = yo - B = yo + (slc):
y = yoect+ -(ect - 1).
S
KEY FORMULA y = or
C
With s = 0 this is the old solution yoect (no source). The example with c = 1 and
s = 5 produced ( y o + 5)ef - 5. Separating the source term gives yo& + 5(et - 1).
Solution by Method 2 (slow way) The input yo produces the output yo@. After t
years any deposit is multiplied by ea. That also applies to deposits made after the
account is opened. If the deposit enters at time 'IS the growing time is only t - T
-
Therefore the multiplying factor is only ec(t This growth factor applies to the small
deposit (amount s d T ) made between time T and T + dT.
Now add up all outputs at time t. The output from yo is yoea. The small deposit
dT.
s dTnear time T grows to ec('-T)s The total is an integral:
This principle of Duhamel would still apply when the source s varies with time.
Here s is constant, and the integral divides by c:
That agrees with the source term from Method 1, at the end of equation (8). There
we looked for "exponential plus constant," here we added up outputs.
Method 1 was easier. It succeeded because we knew the form A&'+ B-with
"undetermined coefficients." Method 2 is more complete. The form for y is part of
the output, not the input. The source s is a continuous supply of new deposits, all
growing separately. Section 6.5 starts from scratch, by directly integrating y' = cy + s.
Remark Method 2 is often described in terms of an integrating factor. First write
the equation as y' - cy = s. Then multiply by a magic factor that makes integration
possible:
( y r - cy)e-ct = se-c' multiply by the factor e-"
S
ye-"]: = - - e - ~ t $ integrate both sides
C
S
ye - C t - yo = - - (e- C f - 1) substitute 0 and t
C
y = ectyo+ - (ect- 1 )
S
isolate y to reach formula (8)
C
The integrating factor produced a perfect derivative in line 1. I prefer Duhamel's idea,
that all inputs yo and s grow the same way. Either method gives formula (8) for y.
H
T E MATHEMATICS OF FINANCE (AT A CONTINUOUS RATE)
The question from finance is this: What inputs give what outputs? The inputs can
come at the start by yo, or continuously by s. The output can be paid at the end or
continuously. There are six basic questions, two of which are already answered.
The future value is yoect from a deposit of yo. To produce y in the future, deposit
the present value ye-". Questions 3-6 involve the source term s. We fix the continuous
6 Exponentlab and Logarithms
rate at 5% per year (c = .05), and start the account from yo = 0. The answers come
fast from equation (8).
Question 3 With deposits of s = $1000/year, how large is y after 20 years?
One big deposit yields 20,000e z $54,000. The same 20,000 via s yields $34,400.
Notice a small by-product (for mathematicians). When the interest rate is c = 0,
our formula s(ec'- l)/c turns into 010. We are absolutely sure that depositing
$1000/year with no interest produces $20,000 after 20 years. But this is not obvious
from 010. By l'H6pital's rule we take c-derivatives in the fraction:
s(ec'- 1) steC'
lim -= lim - = st. This is (1000)(20)= 20,000.
c+O C c-ro 1 (11)
Question 4 What continuous deposit of s per year yields $20,000 after 20 years?
S
.05
1000
20,000 = -(e(.0"(20) 1) requires s = - 582.
-
e- 1
-
Deposits of $582 over 20 years total $11,640. A single deposit of yo = 20,00O/e =
$7,360 produces the same $20,000 at the end. Better to be rich at t = 0.
Questions 1and 2 had s = 0 (no source). Questions 3 and 4 had yo = 0 (no initial
deposit). Now we come to y = 0. In 5, everything is paid out by an annuity. In 6,
everything is paid up on a loan.
Question 5 What deposit yo provides $1000/year for 20 years? End with y = 0.
y = yoec' + - (ec'- 1) = 0 requires yo = -(1 - e-").
S -S
C C
Substituting s = - 1000, c = .05, t = 20 gives yo x 12,640. If you win $20,000 in a
lottery, and it is paid over 20 years, the lottery only has to put in $12,640. Even less
if the interest rate is above 5%.
Question 6 What payments s will clear a loan of yo = $20,000 in 20 years?
Unfortunately, s exceeds $1000 per year. The bank gives up more than the $20,000
to buy your car (and pay tuition). It also gives up the interest on that money. You pay
that back too, but you don't have to stay even at every moment. Instead you repay
at a constant rate for 20 years. Your payments mostly cover interest at the start and
principal at the end. After t = 20 years you are even and your debt is y = 0.
This is like Question 5 (also y = O), but now we know yo and we want s:
y = yoec'+ - (ec' - 1)= 0 requires s = - cyoec'/(ec'- 1).
S
C
The loan is yo = $20,000, the rate is c = .05/year, the time is t = 20 years. Substituting
in the formula for s, your payments are $1582 per year.
Puzzle How is s = $1582 for loan payments related to s = $582 for deposits?
0 -+ $582 per year + $20,000 and $20,000 + - $1582 per year + 0.
6.3 Growth and Decay in Science and Economics 249
That difference of exactly 1000 cannot be an accident. 1582 and 582 came from
e 1 e-1
1000 • and 1000 with difference 1000 - 1000.
e-1 e-1 e-1
Why? Here is the real reason. Instead of repaying 1582 we can pay only 1000 (to
keep even with the interest on 20,000). The other 582 goes into a separate account.
After 20 years the continuous 582 has built up to 20,000 (including interest as in
Question 4). From that account we pay back the loan.
Section 6.6 deals with daily compounding-which differs from continuous com-
pounding by only a few cents. Yearly compounding differs by a few dollars.
34400
s = 1000 y'= - 3y + 6
+
20000 - 20000 s =-1582 6 2
12640 Yoo - 3 - 1
s= 582 +2
20 s =-1000 20
Fig. 6.10 Questions 3-4 deposit s. Questions 5-6 repay loan or annuity. Steady state -s/c.
TRANSIENTS VS. STEADY STATE
Suppose there is decay instead of growth. The constant c is negative and yoec" dies
out. That is the "transient" term, which disappears as t -+ co. What is left is the
"steady state." We denote that limit by y.
Without a source, y, is zero (total decay). When s is present, y, = - s/c:
6J The solution y = Yo + - e" - - approaches y, =- - when ec -*0.
At this steady state, the source s exactly balances the decay cy. In other words
cy + s = 0. From the left side of the differential equation, this means dy/dt = 0. There
is no change. That is why y, is steady.
Notice that y. depends on the source and on c-but not on yo.
EXAMPLE 6 Suppose Bermuda has a birth rate b = .02 and death rate d = .03. The
net decay rate is c = - .01. There is also immigration from outside, of s = 1200/year.
The initial population might be Yo = 5 thousand or Yo = 5 million, but that number
has no effect on yo. The steady state is independent of yo.
In this case y. = - s/c = 1200/.01 = 120,000. The population grows to 120,000 if
Yo is smaller. It decays to 120,000 if Yo is larger.
EXAMPLE 7 Newton's Law of Cooling: dy/dt = c(y - y.). (12)
This is back to physics. The temperature of a body is y. The temperature around it
is y.. Then y starts at Yo and approaches y,, following Newton's rule: The rate is
proportionalto y - y. The bigger the difference, the faster heat flows.
The equation has - cy. where before we had s. That fits with y. = - s/c. For the
solution, replace s by - cy. in formula (8). Or use this new method:
6 Exponentlab and bgariihms
Solution by Method 3 The new idea is to look at the dzrerence y - y, . Its derivative
is dy/dt, since y, is constant. But dy/dt is c(y - y,)- this is our equation. The differ-
ence starts from yo - y,, and grows or decays as a pure exponential:
d
-(y-y,)=c(y-y,) hasthesolution (y-y,)=(yo-y,)e". (13).
dt
This solves the law of cooling. We repeat Method 3 using the letters s and c:
(y + :) = c(y + :) has the solution (y + f) = (yo + :)ect. (14)
Moving s/c to the right side recovers formula (8). There is a constant term and an
exponential term. In a differential equations course, those are the "particularsolution"
and the "homogeneous solution." In a calculus course, it's time to stop.
EXAMPLE 8 In a 70" room, Newton's corpse is found with a temperature of 90". A
day later the body registers 80". When did he stop integrating (at 98.6")?
Solution Here y, = 70 and yo = 90. Newton's equation (13) is y = 20ec' 70. Then +
y = 80 at t = 1 gives 206 = 10. The rate of cooling is c = In ). Death occurred when
+
2 0 8 70 = 98.6 or ect= 1.43. The time was t = In 1.43/ln ) = half a day earlier.
6.3 EXERCISES
Read-through exercises Solve 5-8 starting from yo = 10. At what time does y increase
to 100 or drop to l?
If y' = cy then At) = a . If dyldt = 7y and yo = 4 then
y(t) = b . This solution reaches 8 at t = c . If the dou-
bling time is Tthen c = d . If y' = 3y and y(1) = 9 then yo
was e . When c is negative, the solution approaches
f astjoo. 9 Draw a field of "tangent arrows" for y' = -y, with the
solution curves y = e-" and y = - e-".
The constant solution to dyldt = y + 6 is y = g . The
general solution is y = Aet - 6. If yo = 4 then A = h . The 10 Draw a direction field of arrows for y' = y - 1, with solu-
solution of dyldt = cy + s starting from yo is y = Ae" + B = tion curves y = eX + 1 and y = 1.
i . The output from the source s is i . An input at
time T grows by the factor k at time t. Problems 11-27 involve yoect. They ask for c or t or yo.
At c = lo%, the interest in time dt is dy = 1 . This 11 If a culture of bacteria doubles in two hours, how many
equation yields At) = m . With a source term instead of hours to multiply by lo? First find c.
yo, a continuous deposit of s = 4000/year yields y = n 12 If bacteria increase by factor of ten in ten hours, how
after 10 years. The deposit required to produce 10,000 in 10 many hours to increase by 100? What is c?
years is s = 0 (exactly or approximately). An income of
4000/year forever (!) comes from yo = P . The deposit to 13 How old is a skull that contains 3 as much radiocarbon
give 4OOOIyear for 20 years is yo = 9 . The payment rate as a modern skull?
s to clear a loan of 10,000 in 10 years is r . 14 If a relic contains 90% as much radiocarbon as new mate-
The solution to y' = - 3y + s approaches y, = s . rial, could it come from the time of Christ?
15 The population of Cairo grew from 5 million to 10 million
Solve 1-4 starting from yo = 1 and from yo = - 1. Draw both in 20 years. From y' = cy find c. When was y = 8 million?
solutions on the same graph.
16 The populations of New York and Los Angeles are grow-
ing at 1% and 1.4% a year. Starting from 8 million (NY) and
6 million (LA), when will they be equal?
6.3 Growth and Decay in Sclenco and Economics 251
17 Suppose the value of $1 in Japanese yen decreases at 2% 30 Solve y' = 8 - y starting from yo and y = Ae-' + B.
per year. Starting from $1 = Y240, when will 1 dollar equal 1
yen?
18 The effect of advertising decays exponentially. If 40% Solve 31-34 with yo = 0 and graph the solution.
remember a new product after three days, find c. How long
will 20% remember it?
19 If y = 1000 at t = 3 and y = 3000 at t = 4 (exponential
growth), what was yo at t = O?
20 If y = 100 at t = 4 and y = 10 at t = 8 (exponential decay)
when will y = l? What was yo?
35 (a) What value y = constant solves dy/dt = - 2y + 12?
(b) Find the solution with an arbitrary constant A.
21 Atmospheric pressure decreases with height according to (c) What solutions start from yo = 0 and yo = lo?
dpldh = cp. The pressures at h = 0 (sea level) and h = 20 km (d) What is the steady state y,?
are 1013 and 50 millibars. Find c. Explain why p =
halfway up at h = 10. 36 Choose + +
signs in dyldt = 3y f 6 to achieve the
following results starting from yo = 1. Draw graphs.
22 For exponential decay show that y(t) is the square root of
y(0) times y(2t). How could you find y(3t) from y(t) and y(2t)? (a) y increases to GO (b) y increases to 2
(c) y decreases to -2 (d) y decreases to - GO
23 Most drugs in the bloodstream decay by y' = cy @st-
order kinetics). (a) The half-life of morphine is 3 hours. Find 37 What value y = constant solves dyldt = 4 - y? Show that
its decay constant c (with units). (b) The half-life of nicotine +
y(t) = Ae-' 4 is also a solution. Find y(1) and y, if yo = 3.
is 2 hours. After a six-hour flight what fraction remains? +
38 Solve y' = y e' from yo = 0 by Method 2, where the
24 How often should a drug be taken if its dose is 3 mg, it is deposit eT at time Tis multiplied by e'-T. The total output
cleared at c =.Ol/hour, and 1 mg is required in the blood- ',
at time t is y(t) = j eTe' - d ~ = . Substitute back to
stream at all times? (The doctor decides this level based on check y' = y + et.
body size.) 39 Rewrite y' = y + et as y' - y = et. Multiplying by e-', the
25 The antiseizure drug dilantin has constant clearance rate left side is the derivative of . Integrate both sides
,
y' = - a until y = yl . Then y' = - ayly . Solve for y(t) in two from yo = 0 to find y(t).
pieces from yo. When does y reach y,? 40 Solve y' = - y + 1 from yo = 0 by rewriting as y' + y = 1,
26 The actual elimination of nicotine is multiexponential:y = multiplying by et, and integrating both sides.
+
Aect ~ e ~The first-order equation (dldt - c)y = 0 changes
' . 41 Solve y' = y + t from yo = 0 by assuming y = Aet + Bt + C.
to the second-order equation (dldt - c)(d/dt - C)y = 0. Write
out this equation starting with y", and show that it is satisfied
by the given y.
Problems 42-57 are about the mathematics of finance.
27 True or false. If false, say what's true.
42 Dollar bills decrease in value at c = - .04 per year because
(a) The time for y = ec' to double is (In 2)/(ln c). of inflation. If you hold $1000, what is the decrease in dt
(b) If y' = cy and z' = cz then (y + 2)' = 2c(y + z). years? At what rate s should you print money to keep even?
(c) If y' = cy and z' = cz then (ylz)' = 0.
43 If a bank offers annual interest of 74% or continuous
(d)If y' = cy and z' = Cz then (yz)' = (c + C)yz. interest of 74%, which is better?
m
28 A rocket has velocity u. Burnt fuel of mass A leaves at 44 What continuous interest rate is equivalent to an annual
velocity v - 7. Total momentum is constant: rate of 9%? Extra credit: Telephone a bank for both rates
and check their calculation.
+
m = (m - Am)(v Av) + Am(u - 7).
u
45 At 100% interest (c = 1)how much is a continuous deposit
What differential equation connects m to v? Solve for v(m) not of s per year worth after one year? What initial deposit yo
v(t), starting from vo = 20 and mo = 4. would have produced the same output?
46 To have $50,000 for college tuition in 20 years, what gift
Problems 29-36 are about solutions of y' = cy + s. yo should a grandparent make now? Assume c = 10%. What
continuous deposit should a parent make during 20 years? If
29 Solve y' = 3y+ 1 with yo = 0 by assuming y = Ae3' + B the parent saves s = $1000 per year, when does he or she reach
and determining A and B. $50,000 arid retire?
252 6 Exponentials and Logarithms
47 Income per person grows 3%, the population grows 2%, Problems 58-65 approach a steady state y, as t -+ m.
the total income grows . Answer if these are (a)
58 If dyldt =- y + 7 what is y,? What is the derivative of
annual rates (b) continuous rates.
y - y,? Then y - y, equals yo - y , times .
48 When dyldt = cy + 4, how much is the deposit of 4dT at
time T worth at the later time t? What is the value at t = 2 of 59 Graph y(t) when y' = 3y - 12 and yo is
deposits 4dTfrom T= 0 to T= I? (a)below 4 (b) equal to 4 (c) above 4
49 Depositing s = $1000 per year leads to $34,400 after 20 60 The solutions to dyldt = c(y - 12) converge to y , =
years (Question 3). To reach the same result, when should you provided c is .
deposit $20,000 all at once?
61 Suppose the time unit in dyldt = cy changes from minutes
50 For how long can you withdraw s = $500/year after to hours. How does the equation change? How does dyldt =
depositing yo = $5000 at 8%, before you run dry? +
- y 5 change? How does y , change?
51 What continuous payment s clears a $1000 loan in 60
days, if a loan shark charges 1% per day continuously?
62 True or false, when y, and y, both satisfy y' = cy + s.
52 You are the loan shark. What is $1 worth after a year of
(a)The sum y = y, + y, also satisfies this equation.
continuous compounding at 1% per day? (b)The average y = $(yl + y2) satisfies the same equation.
53 You can afford payments of s = $100 per month for 48 (c) The derivative y = y; satisfies the same equation.
months. If the dealer charges c = 6%, how much can you 63 If Newton's coffee cools from 80" to 60" in 12 minutes
borrow? (room temperature 20G),find c. When was the coffee at 100G?
54 Your income is Ioe2" per year. Your expenses are Eoect
64 If yo = 100 and y(1) = 90 and y(2) = 84, what is y,?
per year. (a) At what future time are they equal? (b) If you
borrow the difference until then, how much money have you 65 If yo = 100 and y(1) = 90 and y(2) = 81, what is yr?
borrowed?
66 To cool down coffee, should you add milk now or later?
55 If a student loan in your freshman year is repaid plus 20% The coffee is at 70°C, the milk is at lo0, the room is at 20".
four years later, what was the effective interest rate?
(a) Adding 1 part milk to 5 parts coffee makes it 60". With
56 Is a variable rate mortgage with c = .09 + .001t for 20 y, = 20", the white coffee cools to y(t) = .
years better or worse than a fixed rate of lo%?
(b)The black coffee cools to y,(t) = . The milk
57 At 10% instead of 8%, the $24 paid for Manhattan is warms to y,(t) = . Mixing at time t gives
worth after 365 years. (5yc + y J 6 =-
-
6.4 Logarithms
We have given first place to ex and a lower place to In x. In applications that is
absolutely correct. But logarithms have one important theoretical advantage (plus
many applications of their own). The advantage is that the derivative of In x is l/x,
whereas the derivative of ex is ex. We can't define ex as its own integral, without
circular reasoning. But we can and do define In x (the natural logarithm) as the
integral of the " - 1 power" which is llx:
Note the dummy variables, first x then u. Note also the live variables, first x then y.
Especially note the lower limit of integration, which is 1 and not 0. The logarithm is
the area measured from 1. Therefore In 1 = 0 at that starting point-as required.
6.4 Logarithms 253
Earlier chapters integrated all powers except this "-1 power." The logarithm is
that missing integral. The curve in Figure 6.11 has height y = 1/x-it is a hyperbola.
At x = 0 the height goes to infinity and the area becomes infinite: log 0 = - 00.
The minus sign is because the integral goes backward from 1 to 0. The integral
does not extend past zero to negative x. We are defining In x only for x > O.t
1I
1 1
Fig. 6.11
x 1 a ab
Logarithm as area. Neighbors In a + In b = In ab. Equal areas: -In
In2-
1/2 1 2 4
= In 2 = In 4.
With this new approach, In x has a direct definition. It is an integral (or an area).
Its two key properties must follow from this definition. That step is a beautiful
application of the theory behind integrals.
Property 1: In ab = In a + In b. The areas from 1 to a and from a to ab combine into
a single area (1 to ab in the middle figure):
a 1 ab fab
Neighboring areas: dx + - dx - dx. (2)
x x x
The right side is In ab, from definition (1). The first term on the left is In a. The
problem is to show that the second integral (a to ab) is In b:
- d x
du = In b. (3)
We need u = 1 when x = a (the lower limit) and u = b when x = ab (the upper limit).
The choice u = x/a satisfies these requirements. Substituting x = au and dx = a du
yields dx/x = du/u. Equation (3) gives In b, and equation (2) is In a + In b = In ab.
Property2: In b" = n In b. These are the left and right sides of
{b"1 dx (?) n -Jdu. (4)
This comes from the substitution x = u". The lower limit x = 1 corresponds to u = 1,
and x = b" corresponds to u = b. The differential dx is nu"-ldu. Dividing by x = u"
leaves dx/x = n du/u. Then equation (4) becomes In b" = n In b.
Everything comes logically from the definition as an area. Also definite integrals:
3x3x
EXAMPLE I Compute - dt. Solution: In 3x - In x = In - In 3.
EXAMPLE 2 Compute
11
- dx. Solution: In 1 - In .1 = In 10. (Why?)
tThe logarithm of -1 is 7ni (an imaginary number). That is because e"'= -1. The logarithm
of i is also imaginary-it is ½7i. In general, logarithms are complex numbers.
254 6 Exponentials and Logarithms
EXAMPLE 3 Compute ' du. Solution: In e2 = 2. The area from 1 to e2 is 2.
Remark While working on the theory this is a chance to straighten out old debts.
The book has discussed and computed (and even differentiated) the functions ex and
bx and x", without defining them properly. When the exponent is an irrational number
how
like rt, do we multiply e by itself i times? One approach (not taken) is to come
closer and closer to it by rational exponents like 22/7. Another approach (taken now)
is to determine the number e' = 23.1 ... its logarithm.t Start with e itself:
by
e is (by definition) the number whose logarithm is 1
e"is (by definition) the number whose logarithm is 7r.
When the area in Figure 6.12 reaches 1, the basepoint is e. When the area reaches 7E,
the basepoint is e'. We are constructing the inverse function (which is ex). But how
do we know that the area reaches 7t or 1000 or -1000 at exactly one point? (The
area is 1000 far out at e1000 . The area is -1000 very near zero at e-100ooo0.) To define
e we have to know that somewhere the area equals 1!
For a proof in two steps, go back to Figure 6.11c. The area from 1 to 2 is more
than 1 (because 1/x is more than - on that interval of length one). The combined area
from 1 to 4 is more than 1. We come to area = 1 before reaching 4. (Actually at
e = 2.718....) Since 1/x is positive, the area is increasing and never comes back to 1.
To double the area we have to square the distance. The logarithm creeps upwards:
Inx
In x -+ oo but --*0. (5)
x
The logarithm grows slowly because ex grows so fast (and vice versa-they are
inverses). Remember that ex goes past every power x". Therefore In x is passed by
every root x'l". Problems 60 and 61 give two proofs that (In x)/xl"I approaches zero.
We might compare In x with x/. x = 10 they are close (2.3 versus 3.2). But out
At
at x = e'o the comparison is 10 against e5, and In x loses to x.
I
e
e
e 1 ex e
Fig. 6.12 Area is logarithm of basepoint. Fig. 6.13 In x grows more slowly
than x.
tChapter 9 goes on to imaginary exponents, and proves the remarkable formula e"' = - 1.
6.4 Logarithms 255
APPROXIMATION OF LOGARITHMS
The limiting cases In 0 = - co and In oo = + co are important. More important are
1=
- logarithms near the starting point In 1 = 0. Our question is: What is In (1 + x) for x
T
x near zero? The exact answer is an area. The approximate answer is much simpler.
area x If x (positive or negative) is small, then
minus
area x2/2 In (1 +x) x and ex ;1 + x.
1 1+x
The calculator gives In 1.01 = .0099503. This is close to x = .01. Between 1 and 1 + x
S= ex the area under the graph of 1/x is nearly a rectangle. Its base is x and its height is 1.
I areax2/2 So the curved area In (1 + x) is close to the rectangular area x. Figure 6.14 shows
area x how a small triangle is chopped off at the top.
The difference between .0099503 (actual) and .01 (linear approximation) is
Ox
-. 0000497. That is predicted almost exactly by the second derivative: ½ times (Ax)2
Rg. 6.14 times (In x)" is (.01)2( - 1)= - .00005. This is the area of the small triangle!
In(1 + x) . rectangular area minus triangular area = x - Ix 2.
3
The remaining mistake of .0000003 is close to x (Problem 65).
May I switch to ex? Its slope starts at eo = 1, so its linear approximation is 1 + x.
Then In (ex) %In(1 + x) x x. Two wrongs do make a right: In (ex) = x exactly.
0"1
The calculator gives e as 1.0100502 (actual) instead of 1.01 (approximation). The
second-order correction is again a small triangle: ix 2 = .00005. The complete series
for In (1 + x) and ex are in Sections 10.1 and 6.6:
In (1+x)= x- x 2 /2 + x 3 /3- ... ex = 1 + x + x 2/2+ x 3/6 + ....
DERIVATIVES BASED ON LOGARITHMS
Logarithms turn up as antiderivatives very often. To build up a collection of integrals,
we now differentiate In u(x) by the chain rule.
6K The derivative of In x is -.
1 The derivative of In u(x) is
du
x u .:x
The slope of In x was hard work in Section 6.2. With its new definition (the integral
of 1/x) the work is gone. By the Fundamental Theorem, the slope must be 1/x.
For In u(x) the derivative comes from the chain rule. The inside function is u, the
outside function is In. (Keep u > 0 to define In u.) The chain rule gives
d 1 1 ( !) d 3
dIn cx= -c- In X 3 = 3x 2
/x 3 =3
dx cx x dx x
d
d -sin x
d In (x 2 + 1)= 2x/(x 2 + 1) in cos x - tan x
dx dx cos x
d 11
In ex = exlex = 1 In (In x)= I
dx dx In x x
Those are worth another look, especially the first. Any reasonable person would
expect the slope of In 3x to be 3/x. Not so. The 3 cancels, and In 3x has the same
slope as In x. (The real reason is that In 3x = In 3 + In x.) The antiderivative of 3/x is
not In 3x but 3 In x, which is In x 3.
6 Exponentials and Logarithms
Before moving to integrals, here is a new method for derivatives: logarithmic dzreren-
tiation or LD. It applies to products and powers. The product and power rules are
always available, but sometimes there is an easier way.
Main idea: The logarithm of a product p(x) is a sum of logarithms. Switching to
In p, the sum rule just adds up the derivatives. But there is a catch at the end, as you
see in the example.
EXAMPLE 4 Find dpldx if p(x) = xxJx - 1. Here ln p(x) = x in x + f ln(x - 1).
1 1
ld
Take the derivative of In p: --p = x . - + l n x + -
pdx x 2(x - 1)'
Now multiply by p(x):
The catch is that last step. Multiplying by p complicates the answer. This can't be
helped-logarithmic differentiation contains no magic. The derivative of p =fg is the
same as from the product rule: In p = l n f + In g gives
For p = xex sin x, with three factors, the sum has three terms:
In p = l n x + x + l n sin x and p l = p
L
We multiply p times pl/p (the derivative of In p). Do the same for powers:
AE
INTEGRALS B S D ON LOGARITHMS
Now comes an important step. Many integrals produce logarithms. The foremost
example is llx, whose integral is In x. In a certain way that is the only example, but
its range is enormously extended by the chain rule. The derivative of In u(x) is uf/u,
so the integral goes from ul/u back to In u:
dx = ln u(x) or equivalently = In u.
Try to choose u(x) so that the integral contains duldx divided by u.
EXAMPLES
6.4 Logarithms
Final remark When u is negative, In u cannot be the integral of llu. The logarithm
is not defined when u < 0. But the integral can go forward by switching to - u:
jdu? I-du/dx
d x = - = In(- u).
dx
-U
Thus In(- u) succeeds when In u fails.? The forbidden case is u = 0. The integrals In u
and In(- u), on the plus and minus sides of zero, can be combined as lnlul. Every
integral that gives a logarithm allows u < 0 by changing to the absolute value lul:
The areas are -1 and -In 3. The graphs of llx and l/(x - 5) are below the x axis.
We do not have logarithms of negative numbers, and we will not integrate l/(x - 5)
from 2 to 6. That crosses the forbidden point x = 5, with infinite area on both sides.
The ratio dulu leads to important integrals. When u = cos x or u = sin x, we are
integrating the tangent and cotangent. When there is a possibility that u < 0, write
the integral as In lul.
Now we report on the secant and cosecant. The integrals of llcos x and llsin x
also surrender to an attack by logarithms - based on a crazy trick:
1 sec dx = 1 GeC + sec x
+
tan x
tan x) dx = In isec x + tan XI. (9)
1 CSC j
x dx = csc x
csc x - cot x
(CSC - X) dx = ln csc x - cot xi. (10)
+
Here u = sec x + tan x is in the denominator; duldx = sec x tan x sec2 x is above it.
The integral is In lul. Similarly (10) contains duldx over u = csc x - cot x.
In closing we integrate In x itself. The derivative of x In x is In x + 1. To remove
I
the extra 1, subtract x from the integral: ln x dx = x in x -x.
In contrast, the area under l/(ln x) has no elementary formula. Nevertheless it is
the key to the greatest approximation in mathematics-the prime number theorem.
The area J: dxlln x is approximately the number o primes between a and b. Near eloo0,
f
about 1/1000 of the integers are prime.
6.4 EXERCISES
Read-through questions e . As x + GO, In x approaches f . But the ratio
The natural logarithm of x is a . This definition leads
(ln x)/& approaches g . The domain and range of in x
are h .
to In xy = b and In xn = c . Then e is the number
whose logarithm (area under llx curve) is d . Similarly
ex is now defined as the number whose natural logarithm is The derivative of In x is I . The derivative of ln(1 + x)
x
?The integral of llx (odd function) is In 11 (even function). Stay clear of x = 0.
258 6 Exponentials and logarithms
is I . The tangent approximation to ln(1 + x) at x = 0 is
k . The quadratic approximation is I . The quadratic
approximation to ex is m .
The derivative of In u(x) by the chain rule is n . Thus
(ln cos x)' = 0 . An antiderivative of tan x is P . The
Evaluate 37-42 by any method.
product p = x e5" has In p = q . The derivative of this equ-
ation is r . Multiplying by p gives p' = s , which is
LD or logarithmic differentiation.
The integral of ul(x)/u(x) is t . The integral of
2x/(x2+ 4) is u . The integral of llcx is v . The integ-
+
ral of l/(ct s)is w . The integral of l/cos x, after a trick,
is x . We should write In 1 1 for the antiderivative of llx,
x
since this allows Y . Similarly Idu/u should be written
d
41 - ln(sec x + tan x) +
42 lsec2x sec x tan x
dx
2 .
dx sec x + tan x
Find the derivative dyldx in 1-10. Verify the derivatives 43-46, which give useful antiderivatives:
3 y=(ln x)-' 4 y = (ln x)/x
5 y = x ln x - x d x-a 2a
6 y=loglox 44 -In - --
dx (x + a) - (X2 a')
-
Find the indefinite (or definite) integral in 11-24.
Estimate 47-50 to linear accuracy, then quadratic accuracy,
by ex x 1 + x + ix2. Then use a calculator.
In(' ex- 1
51 Compute lim - 52 Compute lim -
+
x+O x x-ro x
bX- 1
53 Compute lim logdl x, 9 Compute lim -
x x
+
x+O x-ro
19 1- cos x dx
sin x
55 Find the area of the "hyperbolic quarter-circle" enclosed
byx=2andy=2abovey=l/x.
56 Estimate the area under y = l/x from 4 to 8 by four upper
21
I tan 3x dx 22
I
cot 3x dx
rectangles and four lower rectangles. Then average the
answers (trapezoidal rule). What is the exact area?
1
57 Why is - + - +
1
--•
1
+ - near In n? Is it above or below?
2 3 n
58 Prove that ln x < 2(& - 1)for x > 1. Compare the integ-
rals of l/t and 1 1 4 , from 1 to x.
25 Graph y = ln (1 x) + 26 Graph y = In (sin x)
59 Dividing by x in Problem 58 gives (In x)/x < 2(&
- l)/x.
Deduce that (In x)/x - 0 as x - co.Where is the maximum
, ,
Compute dyldx by differentiating In y. This is LD:
of (In x)/x?
27 y=,/m 28 Y=,/m
Jn 60 Prove that (In x)/xlln also approaches zero. (Start with
29 y = esinx 30 =x-llx (In xlln)/xlln- 0 )Where is its maximum?
, .
6.5 Separable Equations Including the Logistic Equation 259
61 For any power n, Problem 6.2.59 proved ex > xnfor large 70 The slope of p = xx comes two ways from In p = x In x:
x. Then by logarithms, x > n In x. Since (In x)/x goes below 1 Logarithmic differentiation (LD): Compute (In p)' and
l/n and stays below, it converges to . multiply by p.
62 Prove that y In y approaches zero as y -+ 0, by changing 2 Exponential differentiation (ED): Write xX as eXlnX,
y to llx. Find the limit of yY(take its logarithm as y + 0). take its derivative, and put back xx.
What is .I.' on your calculator? 71 If p = 2" then In p = . LD gives p' = (p)(lnp)' =
63 Find the limit of In x/log,,x as x + co. . ED gives p = e and then p' = .
64 We know the integral th-' dt = [th/h]Z = (xh- l)/h. 72 Compute In 2 by the trapezoidal rule and/or Simpson's
Its limit as h + 0 is . rule, to get five correct decimals.
65 Find linear approximations near x = 0 for e-" and 2". 73 Compute In 10 by either rule with Ax = 1, and compare
with the value on your calculator.
66 The x3 correction to ln(1 + x) yields x - i x 2 + ix3. Check
that In 1.01 x -0099503and find In 1.02. 74 Estimate l/ln 90,000, the fraction of numbers near 90,000
that are prime. (879 of the next 10,000 numbers are actually
67 An ant crawls at 1foot/second along a rubber band whose prime.)
original length is 2 feet. The band is being stretched at 1
footlsecond by pulling the other end. At what time T, ifever, 75 Find a pair of positive integers for which xY= yx. Show
does the ant reach the other end? how to change this equation to (In x)/x = (In y)/y. So look for
One approach: The band's length at time t is t + 2. Let y(t) two points at the same height in Figure 6.13. Prove that you
be the fraction of that length which the ant has covered, and have discovered all the integer solutions.
explain *76 Show that (In x)/x = (In y)/y is satisfied by
(a) y' = 1/(t + 2) (b)y = ln(t + 2) - ln 2 (c) T = 2e - 2.
68 If the rubber band is stretched at 8 feetlsecond, when if
ever does the same ant reach the other end?
+
69 A weaker ant slows down to 2/(t 2) feetlsecond, so y' = with t # 0. Graph those points to show the curve xY= y. It
'
+
2/(t 2)2. Show that the other end is never reached. crosses the line y = x at x = , where t + co.
6.5 Separable Equations Including the Logistic Equation
This section begins with the integrals that solve two basic differential equations:
dy -
- - CY and dy
- - cy + s.
dt dt
We already know the solutions. What we don't know is how to discover those solu-
tions, when a suggestion "try eC"' has not been made. Many important equations,
including these, separate into a y-integral and a t-integral. The answer comes directly
from the two separate integrations. When a differential equation is reduced that far-
to integrals that we know or can look up-it is solved.
One particular equation will be emphasized. The logistic equation describes the
speedup and slowdown of growth. Its solution is an S-curve, which starts slowly,
rises quickly, and levels off. (The 1990's are near the middle of the S, if the
prediction is correct for the world population.) S-curves are solutions to nonlinear
equations, and we will be solving our first nonlinear model. It is highly important
in biology and all life sciences.
6 Exponeniials and Logarithms
SEPARABLE EQUNIONS
The equations dyldt = cy and dyldt = cy + s (with constant source s) can be solved
by a direct method. The idea is to separate y from t:
9= c dt
Y
and - c dt.
dy -
Y + (sld
All y's are on the left side. All t's are on the right side (and c can be on either side).
This separation would not be possible for dyldt = y + t.
Equation (2) contains differentials. They suggest integrals. The t-integrals give ct
and the y-integrals give logarithms:
In y = ct + constant and In (3)
The constant is determined by the initial condition. At t = 0 we require y = yo, and the
right constant will make that happen:
lny=ct+lnyo and
( 3 (
In y + - = c t + l n y o + - .
3
Then the final step isolates y. The goal is a formula for y itself, not its logarithm, so
take the exponential of both sides (elnyis y):
y = yoeC' and y +: + = (yo :)ec'.
It is wise to substitute y back into the differential equation, as a check.
This is our fourth method for y' = cy + s. Method 1 assumed from the start that
+
y = Aect B. Method 2 multiplied all inputs by their growth factors ec('- )' and added
up outputs. Method 3 solved for y - y,. Method 4 is separation of variables (and all
methods give the same answer). This separation method is so useful that we repeat
its main idea, and then explain it by using it.
To solve dyldt = u(y)v(t), separate dy/u(y)from v(t)dt and integrate both sides:
Then substitute the initial condition to determine C, and solve for y(t).
EXAMPLE I dyldt = y2 separates into dyly2 = dt. Integrate to reach - l/y = t + C.
Substitute t = 0 and y = yo to find C = - l/yo. Now solve for y:
1
--=
1 Yo
t-- and y=-.
Y Yo 1 - tYo
This solution blows up (Figure 6.15a) when t reaches lly,. If the bank pays interest
on your deposit squared (y' = y2), you soon have all the money in the world.
EXAMPLE 2 dyldt = ty separates into dy/y = t dt. Then by integration in y = f t2 + C.
Substitute t = 0 and y = yo to find C = In yo. The exponential of *t2 + In yo gives
y = yoe'2'2. When the interest rate is c = t, the exponent is t2/2.
EXAMPLE 3 dyldt = y + t is not separable. Method 1 survives by assuming y =
6.5 Separable Equations Including the Logistic Equation
I I
I
blowup times r =
I
l
Yo
0 1 2 0 1
dy dy dy
Fig. 6.15 The solutions to separable equations - = y2 and - = n- or -= n-.dt
Y
dt d t t y t
+
Ae' B + Dt-with an extra coefficient D in Problem 23. Method 2 also succeeds-
but not the separation method.
E A P E 4 Separate dyldt = nylt into dyly = n dtlt. By integration In y = n In t + C.
XML
Substituting t = 0 produces In 0 and disaster. This equation cannot start from time
zero (it divides by t). However y can start from y, at t = 1, which gives C = In y, . The
,
solution is a power function y = y t ".
This was the first differential equation in the book (Section 2.2). The ratio of dyly
to dtlt is the "elasticity" in economics. These relative changes have units like
dollars/dollars-they are dimensionless, and y = tn has constant elasticity n.
On log-log paper the graph of In y = n In t + C is a straight line with slope n.
THE LOGISTIC EQUATION
The simplest model of population growth is dyldt = cy. The growth rate c is the birth
rate minus the death rate. If c is constant the growth goes on forever-beyond the
point where the model is reasonable. A population can't grow all the way to infinity!
Eventually there is competition for food and space, and y = ectmust slow down.
The true rate c depends on the population size y. It is a function c(y) not a constant.
The choice of the model is at least half the problem:
Problem in biology or ecology: Discover c(y).
Problem in mathematics: Solve dyldt = c(y)y.
Every model looks linear over a small range of y's-but not forever. When the rate
drops off, two models are of the greatest importance. The Michaelis-Menten equation
has c(y) = c/(y + K). The logistic equation has c(y) = c - by. It comes first.
The nonlinear effect is from "interaction." For two populations of size y and z, the
number of interactions is proportional to y times z. The Law of Mass Action produces
a quadratic term byz. It is the basic model for interactions and competition. Here we
have one population competing within itself, so z is the same as y. This competition
slows down the growth, because - by2 goes into the equation.
The basic model of growth versus competition is known as the logistic equation:
Normally b is very small compared to c. The growth begins as usual (close to ect).
The competition term by2 is much smaller than cy, until y itselfgets large. Then by2
6 Exponentlals and Logarithms
(with its minus sign) slows the growth down. The solution follows an S-curve that
we can compute exactly.
What are the numbers b and c for human population? Ecologists estimate the
natural growth rate as c = .029/year. That is not the actual rate, because of b. About
1930, the world population was 3 billion. The cy term predicts a yearly increase of
(.029)(3billion) = 87 million. The actual growth was more like dyldt = 60 millionlyear.
That difference of 27 millionlyear was by2:
27 millionlyear = b(3 b i l l i ~ n leads to b = 3 10- 12/year.
)~
Certainly b is a small number (three trillionths) but its effect is not small. It reduces
87 to 60. What is fascinating is to calculate the steady state, when the new term by2
equals the old term cy. When these terms cancel each other, dyldt = cy - by2 is zero.
The loss from competition balances the gain from new growth: cy = by2 and y = c/b.
The growth stops at this equilibrium point-the top of the S-curve:
c .029
Y,=T;= -1012= 10 billion people.
3
According to Verhulst's logistic equation, the world population is converging to 10
billion. That is from the model. From present indications we are growing much faster.
We will very probably go beyond 10 billion. The United Nations report in Section 3.3
predicts 11 billion to 14 billion.
Notice a special point halfway to y, = clb. (In the model this point is at 5 billion.)
It is the inflection point where the S-curve begins to bend down. The second derivative
d2y/dt2is zero. The slope dyldt is a maximum. It is easier to find this point from the
differential equation (which gives dyldt) than from y. Take one more derivative:
y" = (cy - by2)' = cy' - 2byy' = (c - 2by)y'. (8)
The factor c - 2by is zero at the inflection point y = c/2b, halfway up the S-curve.
THE S-CURVE
The logistic equation is solved by separating variables y and t:
J
dyldt = cy - by2 becomes dy/(cy - by2)= dt. )
The first question is whether we recognize this y-integral. No. The second question
is whether it is listed in the cover of the book. No. The nearest is Idx/(a2 - x2),which
can be reached with considerable manipulation (Problem 21). The third question is
whether a general method is available. Yes. "Partial fractions" is perfectly suited to
l/(cy - by2), and Section 7.4 gives the following integral of equation (9):
Y
In-=ct+C andthen Yo
In-=C.
c - by (10)
c - YO
That constant C makes the solution correct at t = 0. The logistic equation is integ-
rated, but the solution can be improved. Take exponentials of both sides to remove
the logarithms:
-- - ect Yo
y
c-by c-byo'
This contains the same growth factor ec' as in linear equations. But the logistic
6.5 Separable Equations Including the Logistic Equation 263
equation is not linear-it is not y that increases so fast. According to (ll), it is
y/(c - by) that grows to infinity. This happens when c - by approaches zero.
The growth stops at y = clb. That is the final population of the world (10 billion?).
We still need a formula for y. The perfect S-curve is the graph of y = 1/(1 + e-'). It
equals 1 when t = oo, it equals 4 when t = 0, it equals 0 when t = - co. It satisfies
y' = y - y2, with c = b = 1. The general formula cannot be so beautiful, because it
allows any c, b, and yo. To find the S-curve, multiply equation (11) by c - by and
solve for y:
When t approaches infinity, e-" approaches zero. The complicated part of the for-
mula disappears. Then y approaches its steady state clb, the asymptote in Figure 6.16.
The S-shape comes from the inflection point halfway up.
1 2 3 4 1988
Fig. 6.16 The standard S-curve y = 1/(1 + e - ' ) . The population S-curve (with prediction).
Surprising observation: z = l/y satisjes a linear equation. By calculus z' = - y'/y2. So
This equation z' = - cz + b is solved by an exponential e-" plus a constant:
Year US Model
Population
1790 3.9 = 3.9
1800 5.3 5.3 Turned upside down, y = l/z is the S-curve (12). As z approaches blc, the S-curve
1810 7.2 7.2 approaches clb. Notice that z starts at l/yo.
1820 9.6 9.8
1830 12.9 13.1 EXAMPLE 1 (United States population) The table shows the actual population and
1840 17.1 17.5
the model. Pearl and Reed used census figures for 1790, 1850, and 1910 to compute
1850 23.2 = 23.2
c and b. In between, the fit is good but not fantastic. One reason is war-another is
1860 31.4 30.4
1870 38.6 39.4 depression. Probably more important is immigration."fn fact the Pearl-Reed steady
1880 50.2 50.2 state c/b is below 200 million, which the US has already passed. Certainly their model
1890 62.9 62.8 can be and has been improved. The 1990 census predicted a stop before 300 million.
1900 76.0 76.9 For constant immigration s we could still solve y' = cy - by2 + s by partial fractions-
1910 92.0 = 92.0 but in practice the computer has taken over. The table comes from Braun's book
1920 105.7 107.6 DifSerentiaE Equations (Springer 1975).
1930 122.8 123.1
1940 131.7 # 136.7
1950 150.7 149.1 ?Immigration does not enter for the world population model (at least not yet).
6 Exponentials and Logarithms
Remark For good science the y2 term should be explained and justified. It gave a
nonlinear model that could be completely solved, but simplicity is not necessarily
truth. The basic justification is this: In a population of size y, the number of encounters
is proportional to y2. If those encounters are fights, the term is - by2. If those
encounters increase the population, as some like to think, the sign is changed. There
is a cooperation term + by2, and the population increases very fast.
EXAMPLE 5 y' = cy + by2: y goes to infinity in afinite time.
EXAMPLE 6 y' = - dy + by2: y dies to zero if yo < dlb.
In Example 6 death wins. A small population dies out before the cooperation by2
can save it. A population below dlb is an endangered species.
The logistic equation can't predict oscillations-those go beyond dyldt =f(y).
The y line Here is a way to understand every nonlinear equation y' =f(y). Draw a
" y line." Add arrows to show the sign of f(y). When y' =f ( y ) is positive, y is increasing
(itfollows the arrow to the right). When f is negative, y goes to the left. When f is zero,
the equation is y' = 0 and y is stationary:
y' = cy - by2 (this is f(y)) y' = - dy + by2 (this is f(y))
The arrows take you left or right, to the steady state or to infinity. Arrows go toward
stable steady states. The arrows go away, when the stationary point is unstable. The
y line shows which way y moves and where it stops.
The terminal velocity of a falling body is v, = &
in Problem 6.7.54. For f ( y ) =
sin y there are several steady states:
falling body: dvldt = g - v2 dyldt = sin y
EXAMPLE 7 Kinetics of a chemical reaction mA + nB -+ pC.
The reaction combines m molecules of A with n molecules of B to produce p
molecules of C. The numbers m, n, p are 1, 1,2 for hydrogen chloride: H, + C1, =
2 HCl. The Law of Mass Action says that the reaction rate is proportional to the
product of the concentrations [ A ] and [ B ] . Then [ A ] decays as [ C ] grows:
d[A]/dt= - r [ A ][ B ] and d [Clldt = + k [ A ][ B ] .
(15)
Chemistry measures r and k. Mathematics solves for [ A ] and [ C ] . Write y for the
concentration [ C ] , the number of molecules in a unit volume. Forming those y
molecules drops the concentration [ A ] from a, to a, - (m/p)y. Similarly [B] drops
from b, to b, - (n/p)y.The mass action law (15)contains y2:
6.5 Separable Equations Including the laglttlc Equation
This fits our nonlinear model (Problem 33-34). We now find this same mass action
in biology. You recognize it whenever there is a product of two concentrations.
THE MM EQUATION wdt=- cy/(y+ K)
Biochemical reactions are the keys to life. They take place continually in every living
organism. Their mathematical description is not easy! Engineering and physics go
far with linear models, while biology is quickly nonlinear. It is true that y' = cy is
extremely effective in first-order kinetics (Section 6.3), but nature builds in a nonlinear
regulator.
It is enzymes that speed up a reaction. Without them, your life would be in slow
motion. Blood would take years to clot. Steaks would take decades to digest. Calculus
would take centuries to learn. The whole system is awesomely beautiful-DNA tells
amino acids how to combine into useful proteins, and we get enzymes and elephants
and Isaac Newton.
Briefly, the enzyme enters the reaction and comes out again. It is the catalyst. Its
combination with the substrate is an unstable intermediate, which breaks up into a
new product and the enzyme (which is ready to start over).
Here are examples of catalysts, some good and some bad.
The platinum in a catalytic converter reacts with pollutants from the car engine.
(But platinum also reacts with lead-ten gallons of leaded gasoline and you
can forget the platinum.)
Spray propellants (CFC's) catalyze the change from ozone (03) into ordinary
oxygen (0J. This wipes out the ozone layer-our shield in the atmosphere.
Milk becomes yoghurt and grape juice becomes wine.
Blood clotting needs a whole cascade of enzymes, amplifying the reaction at
every step. In hemophilia-the "Czar's diseasew-the enzyme called Factor VIII
is missing. A small accident is disaster; the bleeding won't stop.
Adolph's Meat Tenderizer is a protein from papayas. It predigests the steak.
The same enzyme (chymopapain) is injected to soften herniated disks.
Yeast makes bread rise. Enzymes put the sour in sourdough.
Of course, it takes enzymes to make enzymes. The maternal egg contains the material
for a cell, and also half of the DNA. The fertilized egg contains the full instructions.
We now look at the Michaelis-Menten (MM) equation, to describe these reactions.
It is based on the Law o Mass Action. An enzyme in concentration z converts a
f
substrate in concentration y by dyldt = - byz. The rate constant is 6, and you see
the product of "enzyme times substrate." A similar law governs the other reactions
(some go backwards). The equations are nonlinear, with no exact solution. It is
typical of applied mathematics (and nature) that a pattern can still be found.
What happens is that the enzyme concentration z(t) quickly drops to z, K/(y + K).
The Michaelis constant K depends on the rates (like 6) in the mass action laws.
Later the enzyme reappears (z, = 2,). But by then the first reaction is over. Its law
of mass action is effectively
with c =.bz,K. This is the Michaelis-Menten equation-basic to biochemistry.
The rate dyldt is all-important in biology. Look at the function cy/(y + K):
when y is large, dyldt x -c when y is small, dyldt x - cylK.
6 Exponentials and Logarithms
The start and the finish operate at different rates, depending whether y dominates K
or K dominates y. The fastest rate is c.
A biochemist solves the MM equation by separating variables:
S y d y =
Set t = 0 as usual. Then C = yo
-
S+
c dt gives y + K In y = - ct + C.
K In yo. The exponentials of the two sides are
We don't have a simple formula for y. We are lucky to get this close. A computer
can quickly graph y(t)-and we see the dynamics of enzymes.
Problems 27-32 follow up the Michaelis-Menten theory. In science, concentrations
and rate constants come with units. In mathematics, variables can be made dimen-
sionless and constants become 1. We solve d v d T = Y/(Y + 1) and then $witch back
to y, t, c, K. This idea applies to other equations too.
Essential point: Most applications of calculus come through dzrerential equations.
That is the language of mathematics-with populations and chemicals and epidemics
obeying the same equation. Running parallel to dyldt = cy are the difference equations
that come next.
6.5 EXERCISES
Read-through questions
The equations dy/dt = cy and dyldt = cy + s and dyldt =
u(y)v(t) are called a because we can separate y from t.
1
Integration of idyly = c dt gives b . Integration of
6 dy/dx=tan ycos x, y o = 1
1 +
dy/(y sjc) = i c dt gives c . The equation dyldx = 7 dyldt = y sin t, yo = 1
- xly leads to d . Then y2 + x2 = e and the solution
stays on a circle. 8 dyldt = et-Y, yo = e
9 Suppose the rate of rowth is proportional to & instead
The logistic equation is dyldt = f . The new term - by2 of y. Solve dyldt = c&starting from yo.
represents g when cy represents growth. Separation gives
i dy/(cy - by2)= [ dt, and the y-integral is l/c times In h . 10 The equation dyjdx = nylx for constant elasticity is the
Substituting yo at t = 0 and taking exponentials produces same as d(ln y)/d(ln x) = . The solution is In y =
y/(c - by) = ect( i ). As t + co,y approaches i . That
is the steady state where cy - by2 = k . The graph of y 11 When c = 0 in the logistic equation, the only term is y' =
looks like an I , because it has an inflection point at - by2. What is the steady state y,? How long until y drops
y= m . from yo to iyo?
In biology and chemistry, concentrations y and z react at 12 Reversing signs in Problem 11, suppose y' = + by2. At
a rate proportional to y times n . This is the Law of
what time does the population explode to y = co, starting
o . In a model equation dyldt = c(y)y, the rate c depends from yo = 2 (Adam + Eve)?
on P . The M M equation is dyldt = q . Separating
variables yields j r dy = s = - ct + C. Problems 13-26 deal with logistic equations y' = cy - by2.
13 Show that y = 1/(1+ e-') solves the equation y' = y - y2.
Draw the graph of y from starting values 3 and 3.
Separate, integrate, and solve equations 1-8.
14 (a) What logistic equation is solved by y = 2/(1 + e-')?
(b) Find c and b in the equation solved by y = 1/(1 + e-3t).
15 Solve z' = - z + 1 with zo = 2. Turned upside down as in
3 dyjdx = xly2, yo = 1 (1 3), what is y = l/z?
6.6 Powers Instead of Exponential6 267
16 By algebra find the S-curve (12) from y = l/z in (14). aspirin follows the MM equation. With c = K = yo = 1, does
17 How many years to grow from yo = $c/b to y = #c/b? Use
aspirin decay faster?
equation (10) for the time t since the inflection point in 1988. 28 If you take aspirin at a constant rate d (the maintenance
When does y reach 9 billion = .9c/b? dose), find the steady state level where d = cy/(y + K). Then
18 Show by differentiating u = y/(c - by) that if y' = cy - by2 y' = 0.
then u' = cu. This explains the logistic solution (11) - it is 29 Show that the rate R = cy/(y + K) in the MM equation
u = uoect. increases as y increases, and find the maximum as y -* a.
19 Suppose Pittsburgh grows from yo = 1 million people in 30 Graph the rate R as a function of y for K = 1 and K =
1900 to y = 3 million in the year 2000. If the growth rate is 10. (Take c = 1.) As the Michaelis constant increases, the rate
y' = 12,00O/year in 1900 and y' = 30,00O/year in 2000, substi- . At what value of y is R = *c?
tute in the logistic equation to find c and b. What is the steady
31 With y = KY and ct = KT, find the "nondimensional"
state? Extra credit: When does y = y, /2 = c/2b?
MM equation for dY/dT. From the solution erY=
20 Suppose c = 1 but b = - 1, giving cooperation y' = y + y2. e-= eroYorecover the y, t solution (19).
Solve for fit) if yo = 1. When does y become infinite?
32 Graph fit) in (19) for different c and K (by computer).
21 Draw an S-curve through (0,O) with horizontal asymp-
33 The Law of Mass Action for A + B + C is y' =
+
totes y = - 1 and y = 1. Show that y = (et- e-')/(et e-') has
k(ao- y)(bo- y). Suppose yo = 0, a. = bo = 3, k = 1. Solve for
those three properties. The graph of y2 is shaped like
y and find the time when y = 2.
34 In addition to the equation for d[C]/dt, the mass action
22 To solve y' = cy - by3 change to u = l/y2. Substitute for
law gives d[A]/dt =
y' in u' = - 2y'/y3 to find a linear equation for u. Solve it as
in (14) but with uo = ljy;. Then y = I/&. 35 Solve y' = y + t from yo = 0 by assuming y = Aet + B + Dt.
Find A, B, D.
23 With y = rY and t = ST, the equation dyldt = cy - by2
changes to d Y/d T = Y- Y '. Find r and s. 36 Rewrite cy - by2 as a2 - x2, with x = Gy - c/2$ and
24 In a change to y = rY and t = ST,how are the initial values
a= . Substitute for a and x in the integral taken
from tables, to obtain the y-integral in the text:
yo and yb related to Yo and G?
25 A rumor spreads according to y' = y(N - y). If y people --In- {A=-ln-
1 Y
know, then N - y don't know. The product y(N - y) measures -a2-x2 2a x
a '-x cy-by2 c c-by
the number of meetings (to pass on the rumor). 37 (Important) Draw the y-lines (with arrows as in the text)
(a) Solve dyldt = y(N - y) starting from yo = 1. for y' = y/(l - y) and y' = y - y3. Which steady states are
(b) At what time T have N/2 people heard the rumor? approached from which initial values yo?
(c) This model is terrible because T goes to as 38 Explain in your own words how the y-line works.
N + GO. A better model is y' = by(N - y).
39 (a) Solve yl= tan y starting from yo = n / 6 to find
26 Suppose b and c are bcth multiplied by 10. Does the sin y = $et.
middle of the S-curve get steeper or flatter? (b)Explain why t = 1 is never reached.
(c) Draw arrows on the y-line to show that y approaches
71 - when does it get there?
12
Problems 27-34 deal with mass action and the MM equation
40 Write the logistic equation as y' = cy(1 - y/K). As y'
+
y' = - cy/(y K).
approaches zero, y approaches . Find y, y', y" at the
27 Most drugs are eliminated acording to y' = - cy but inflection point.
6.6 Powers lnstead of Exponentials
You may remember our first look at e. It is the special base for which ex has slope 1
at x = 0 That led to the great equation of exponential growth: The derivative of
.
ex equals ex. But our look at the actual number e = 2.71828 ... was very short.
6 Exponentlals and Logarithms
It appeared as the limit of (1 + lln)". This seems an unnatural way to write down
such an important number.
+
I want to show how (1 lln)" and (1 + xln)" arise naturally. They give discrete
growth infinite steps-with applications to compound interest. Loans and life insur-
ance and money market funds use the discrete form of yf = cy + s. (We include extra
information about bank rates, hoping this may be useful some day.) The applications
in science and engineering are equally important. Scientific computing, like account-
ing, has diflerence equations in parallel with differential equations.
Knowing that this section will be full of formulas, I would like to jump ahead and
tell you the best one. It is an infinite series for ex. What makes the series beautiful is
that its derivative is itself:
Start with y = 1 + x. This has y = 1 and yt = 1 at x = 0. But y" is zero, not one.
Such a simple function doesn't stand a chance! No polynomial can be its own deriva-
tive, because the highest power xn drops down to nxn-l. The only way is to have no
highest power. We are forced to consider infinitely many terms-a power series-to
achieve "derivative equals function.''
+
To produce the derivative 1 + x, we need 1 x + ix2. Then i x 2 is the derivative
of Ax3, which is the derivative of &x4. The best way is to write the whole series at
once:
+ + i x 2 + 4x3 + &x4 + -.
Infinite series ex = 1 x (1)
This must be the greatest power series ever discovered. Its derivative is itself:
The derivative of each term is the term before it. The integral of each term is the one
after it (so jexdx = ex + C). The approximation ex = 1 + x appears in the first two
are
terms. Other properties like (ex)(ex) eZX not so obvious. (Multiplying series is
=
hard but interesting.) It is not even clear why the sum is 2.718 ... when x = 1.
Somehow 1 + 1 + f + & + equals e. That is where (1 + lln)" will come in.
Notice that xn is divided by the product 1 2 3 * - . - n. This is "n factorial." Thus
-
x4 is divided by 1 2 3 4 = 4! = 24, and xS is divided by 5! = 120. The derivative of .
x5/120 is x4/24, because 5 from the derivative cancels 5 from the factorial. In general
xn/n! has derivative xn- '/(n - l)! Surprisingly O! is 1.
Chapter 10 emphasizes that xn/n! becomes extremely small as n increases. The
infinite series adds up to a finite number-which is ex. We turn now to discrete
growth, which produces the same series in the limit.
This headline was on page one of the New York Times for May 27, 1990.
213 Years After Loan, Uncle Sam is Dunned
San Antonio, May 26-More than 200 years ago, a wealthy Pennsylvania
merchant named Jacob DeHaven lent $450,000 to the Continental Congress to
rescue the troops at Valley Forge. That loan was apparently never repaid.
So Mr. DeHaven's descendants are taking the United States Government to
court to collect what they believe they are owed. The total: $141 billion if the
interest is compounded daily at 6 percent, the going rate at the time. If com-
pounded yearly, the bill is only $98 billion.
The thousands of family members scattered around the country say they are
not being greedy. "It's not the money-it's the principle of the thing," said
Carolyn Cokerham, a DeHaven on her father's side who lives in San Antonio.
6.6 Powen Instead of Exponentlals
"You have to wonder whether there would even be a United States if this man
had not made the sacrifice that he did. He gave everything he had."
The descendants say that they are willing to be flexible about the amount of
settlement. But they also note that interest is accumulating at $190 a second.
"None of these people have any intention of bankrupting the Government,"
said Jo Beth Kloecker, a lawyer from Stafford, Texas. Fresh out of law school,
Ms. Kloecker accepted the case for less than the customary 30 percent
contingency.
It is unclear how many descendants there are. Ms. Kloecker estimates that
based on 10 generations with four children in each generation, there could be as
many as half a million.
The initial suit was dismissed on the ground that the statute of limitations is
six years for a suit against the Federal Government. The family's appeal asserts
that this violates Article 6 of the Constitution, which declares as valid all debts
owed by the Government before the Constitution was adopted.
Mr. DeHaven died penniless in 1812. He had no children.
C O M P O U N D INTEREST
The idea of compound interest can be applied right away. Suppose you invest $1000
at a rate of 100% (hard to do). If this is the annual rate, the interest after a year is
another $1000. You receive $2000 in all. But if the interest is compounded you receive
more:
after six months: Interest of $500 is reinvested to give $1500
end of year: New interest of $750 (50% of 1500) gives $2250 total.
The bank multiplied twice by 1.5 (1000 to 1500 to 2250). Compounding quarterly
multiplies four times by 1.25 (1 for principal, .25 for interest):
after one quarter the total is 1000 + (.25)(1000) = 1250
after two quarters the total is 1250 + (.25)(1250)= 1562.50
after nine months the total is 1562.50 + (.25)(1562.50)= 1953.12
after a full year the total is 1953.12 + (.25)(@53.12) = 2441.41
Each step multiplies by 1 + (l/n), to add one nth of a year's interest-still at 100%:
quarterly conversion: (1 + 1/4)4x low = 2441.41
monthly conversion: (1 + 1/12)" x 1 Q h 2613.04
=
daily conversion: (1 + 1/365)36% 1000 = 2714.57.
Many banks use 360 days in a year, although computers have made that obsolete.
Very few banks use minutes (525,600 per year). Nobody compounds every second
(n = 31,536,000). But some banks offer continuous compounding. This is the limiting
case (n -+ GO) that produces e:
x 1000 approaches e x 1000 = 2718.28.
(1+
1
1. Quick method for (1 + lln)": Take its logarithm. Use ln(1 + x) x x with x = -:
n
6 Exponentlals and Logartthms
As l/n gets smaller, this approximation gets better. The limit is 1. Conclusion:
+
(1 l/n)" approaches the number whose logarithm is 1. Sections 6.2 and 6.4 define
the same number (which is e).
2. Slow method for (1 + l/n)": Multiply out all the terms. Then let n + a.
This is a brutal use of the binomial theorem. It involves nothing smart like logarithms,
but the result is a fantastic new formula for e.
Practice for n = 3:
Binomial theorem for any positive integer n:
Each term in equation (4) approaches a limit as n + a.Typical terms are
Next comes 111 2 3 4. The sum of all those limits in (4) is our new formula for e:
In summation notation this is Z,"=, l/k! = e. The factorials give fast convergence:
Those nine terms give an accuracy that was not reached by n = 365 compoundings.
A limit is still involved (to add up the whole series). You never see e without a limit!
It can be defined by derivatives or integrals or powers (1 + l/n)" or by an infinite
series. Something goes to zero or infinity, and care is required.
All terms in equation (4) are below (or equal to) the corresponding terms in (5).
The power (1 + l/n)" approaches efrom below. There is a steady increase with n. Faster
compounding yields more interest. Continuous compounding at 100% yields e, as
each term in (4) moves up to its limit in (5).
Remark Change (1 + lln)" to (1 + xln)". Now the binomial theorem produces ex:
Please recognize ex on the right side! It is the infinite power series in equation (1).
The next term is x3/6 (x can be positive or negative). This is a final formula for ex:
The logarithm of that power is n In(1 + x/n) x n(x/n) = x. The power approaches ex.
To summarize: The quick method proves (1 + lln)" + e by logarithms. The slow
method (multiplying out every term) led to the infinite series. Together they show the
agreement of all our definitions of e.
DIFFERENCE EQUATIONS VS. DIFFERENTIAL EQUATIONS
We have the chance to see an important part of applied mathematics. This is not a
course on differential equations, and it cannot become a course on difference equ-
ations. But it is a course with a purpose-we aim to use what we know. Our main
application of e was to solve y' = cy and y' = cy + s. Now we solve the corresponding
difference equations.
f
Above all, the goal is to see the connections. The purpose o mathematics is to
understand and explain patterns. The path from "discrete to continuous" is beautifully
illustrated by these equations. Not every class will pursue them to the end, but I
cannot fail to show the pattern in a difference equation:
Each step multiplies by the same number a. The starting value yo is followed by ay,,
a2yo,and a3y0. The solution at discrete times t = 0, 1,2, ... is y(t) = atyo.
f
This formula atyo replaces the continuous solution ectyoo the differential equation.
decaying
Fig. 6.17 Growth for la1 > 1, decay for la1 < 1. Growth factor a compares to ec.
A source or sink (birth or death, deposit or withdrawal) is like y' = cy + s:
y(t + 1)= ay(t) + s.
Each step multiplies by a and adds s. The first outputs are
We saw this pattern for differential equations-every input s becomes a new starting
point. It is multiplied by powers of a. Since s enters later than yo, the powers stop at
t - 1. Algebra turns the sum into a clean formula by adding the geometric series:
y(t)= atyo+ s[at-' +at-' + +
+ a + 1]= atyo s(at- l)/(a- 1). (9)
EXAMPLE 1 Interest at 8% from annual IRA deposits of s = $2000 (here yo = 0).
The first deposit is at year t = 1. In a year it is multiplied by a = 1.08, because 8% is
added. At the same time a new s = 2000 goes in. At t = 3 the first deposit has been
multiplied by (1.08)2,the second by 1.08, and there is another s = 2000. After year t,
y(t) = 2000(1.08' - 1)/(1.08 - 1). (10)
+
With t = 1 this is 2000. With t = 2 it is 2000 (1.08 1)-two deposits. Notice how
a - 1 (the interest rate .08) appears in the denominator.
EXAMPLE 2 Approach to steady state when la1 < 1. Compare with c < 0.
With a > 1, everything has been increasing. That corresponds to c > 0 in the
differential equation (which is growth). But things die, and money is spent, so a can
be smaller than one. In that case atyo approaches zero-the starting balance disap-
f
pears. What happens if there is also a source? Every year half o the balance y(t) is
6 Exponentials and Logartthms
spent and a new $2000 is deposited. Now a = +:
y(t + 1) = $y(t) + 2000 yields y(t) = (f)ty, + 2000[((+)' - I)/(+- I)].
The limit as t - co is an equilibrium point. As (fy goes to zero, y(t) stabilizes to
,
y, = 200qO - I)/($ - 1) = 4000 = steady state. (11)
Why is 4000 steady? Because half is lost and the new 2000 makes it up again. The
, +
iteration is y,,, = fy,, 2000. Ztsfied point is where y, = fy, + 2000.
In general the steady equation is y, = ay, +
s. Solving for y, gives s/(l - a).
Compare with the steady differential equation y' = cy + s = 0:
S S
y, = - - (differential equation) us. y, =-(difference equation). (12)
c 1-a
EXAMPLE 3 Demand equals supply when the price is right.
Difference equations are basic to economics. Decisions are made every year (by a
farmer) or every day (by a bank) or every minute (by the stock market). There are
three assumptions:
1. Supply next time depends on price this time: S(t + 1) = cP(t).
+
2. Demand next time depends on price next time: D(t 1) = - dP(t + 1) + b.
3. Demand next time equals supply next time: D(t + 1) = S(t + 1).
Comment on 3: the price sets itself to make demand = supply. The demand slope - d
is negative. The supply slope c is positive. Those lines intersect at the competitive
price, where supply equals demand. To find the difference equation, substitute 1 and
2 into 3:
+ +
Difference equation: - dP(t 1) b = cP(t)
Steady state price: - dP, + b = cP,. Thus P, = b/(c + d).
If the price starts above P,, the difference equation brings it down. If below, the
price goes up. When the price is P,, it stays there. This is not news-economic
theory depends on approach to a steady state. But convergence only occurs if c < d.
f
I supply is less sensitive than demand, the economy is stable.
+
Blow-up example: c = 2, b = d = 1. The difference equation is - P(t 1) + 1 = 2P(t).
From P(0) = 1 the price oscillates as it grows: P = - 1, 3, - 5, 11, ... .
Stable example: c = 112, b = d = 1. The price moves from P(0) = 1 to P(m) = 213:
1 1 3 5 2
- P(t + 1) + 1 = - P(t) yields
2
P = 1' - - - "" approaching - .
2' 4' 8'
3
Increasing d gives greater stability. That is the effect of price supports. For d = 0
(fixed demand regardless of price) the economy is out of control.
H
T E MATHEMATICS OF FINANCE
It would be a pleasure to make this supply-demand model more realistic-with
curves, not straight lines. Stability depends on the slope-calculus enters. But we
also have to be realistic about class time. I believe the most practical application is
to solve the fundamentalproblems offinance. Section 6.3 answered six questions about
continuous interest. We now answer the same six questions when the annual rate is
x = .05 = 5% and interest is compounded n times a year.
6.6 Powers Instead of Exponentials
First we compute eflective rates, higher than .05 because of compounding:
( T:.
compounded quarterly 1 + - = 1.0509 [effective rate .0509 = 5.09%]
compounded continuously eno5= 1.O513 [effective rate 5.13%]
Now come the six questions. Next to the new answer (discrete) we write the old
answer (continuous). One is algebra, the other is calculus. The time period is 20 years,
so simple interest on yo would produce (.05)(20)(yo).That equals yo -money doubles
in 20 years at 5% simple interest.
Questions 1and 2 ask for the future value y and present value yo with compound
interest n times a year:
1. y growing from yo: y = (1 + yonyo y = e(~OS,(20)yo
yo = e-(-05)(20)y
2. deposit yo to reach y: yo = (1 + :F20ny
Each step multiplies by a = (1 + .05/n). There are 20n steps in 20 years. Time goes
backward in Question 2. We divide by the growth factor instead of multiplying. The
future value is greater than the present value (unless the interest rate is negative!). As
n + GO the discrete y on the left approaches the continuous y on the right.
Questions 3 and 4 connect y to s (with yo = 0 at the start). As soon as each s is
deposited, it starts growing. Then y = s + as + a2s + --.
(1 + .05/n)20n I]
- y = s [e(.05)(20) I]
-
3. y growing from deposits s: y = s[ .05/n .05
4. deposits s to reach y:
Questions 5 and 6 connect yo to s. This time y is zero-there is nothing left at the
end. Everything is paid. The deposit yo is just enough to allow payments of s. This
is an annuity, where the bank earns interest on your yo while it pays you s (n times
a year for 20 years). So your deposit in Question 5 is less than 20ns.
Question 6 is the opposite-a loan. At the start you borrow yo (instead of giving
the bank yo). You can earn interest on it as you pay it back. Therefore your payments
have to total more than yo. This is the calculation for car loans and mortgages.
5. Annuity: Deposit yo to receive 20n payments of s:
6. Loan:. Repay yo with 20n payments of s:
Questions 2 , 4 , 6 are the inverses of 1,3,5. Notice the pattern: There are three num-
f
bers y, yo, and s. One o them-is zero each time. If all three are present, go back to
equation (9).
The algebra for these lines is in the exercises. I t is not calculus because At is not dt.
All factors in brackets [ 1 are listed in tables, and the banks keep copies. It might
6 Exponenlials and Logartthms
also be helpful to know their symbols. If a bank has interest rate i per period over
N periods, then in our notation a = 1 + i = 1 + .05/n and t = N = 20n:
future value of yo = $1 (line 1):y(N) = (1 + i)N
present value of y = $1 (line 2): yo = (1 + i)-N
future value of s = $1 (line 3): y(N) = s~~= [(I + i)N- l]/i
present value of s = $1 (line 5): yo = a~~= [ l - (1 + i)-']/i
To tell the truth, I never knew the last two formulas until writing this book. The
mortgage on my home has N = (12)(25) monthly payments with interest rate i =
.07/12. In 1972 the present value was $42,000 = amount borrowed. I am now going
to see if the bank is honest.?
Remark In many loans, the bank computes interest on the amount paid back
instead of the amount received. This is called discounting. A loan of $1000 at 5%
for one year costs $50 interest. Normally you receive $1000 and pay back $1050.
With discounting you receive $950 (called the proceeds) and you pay back $1000.
The true interest rate is higher than 5%-because the $50 interest is paid on the
smaller amount $950. In this case the "discount rate" is 501950 = 5.26%.
IFRNI L IFRN E
SCIENTIFIC COMPUTING: DF E E TA EQUATIONS BY DF E E C EQUATIONS
In biology and business, most events are discrete. In engineering and physics, time
and space are continuous. Maybe at some quantum level it's all the same, but the
equations of physics (starting with Newton's law F = ma) are differential equations.
The great contribution of calculus is to model the rates of change we see in nature.
But to solve that model with a computer, it needs to be made digital and discrete.
These paragraphs work with dyldt = cy. It is the test equation that all analysts use,
as soon as a new computing method is proposed. Its solution is y = ect,starting from
yo = 1. Here we test Euler's method (nearly ancient, and not well thought of). He
replaced dyldt by AylAt:
The left side is dyldt, in the limit At + 0. We stop earlier, when At > 0.
The problem is to solve (13). Multiplying by At, the equation is
y(t + At) = (1 + cAt)y(t) (with y(0) = 1).
Each step multiplies by a = 1 + cAt, so n steps multiply by an:
y = an= (1 + cAt)" at time nAt. (14)
This is growth or decay, depending on a. The correct ectis growth or decay, depending
on c. The question is whether an and eczstay close. Can one of them grow while the
other decays? We expect the difference equation to copy y' = cy, but we might be
wrong.
A good example is y' = - y. Then c = - 1 and y = e-'-the true solution decays.
?It's not. s is too big. I knew it.
The calculator gives the following answers an for n = 2, 10,20:
The big step At = 3 shows total instability (top row). The numbers blow up when
they should decay. The row with At = 1 is equally useless (all zeros). In practice the
magnitude of cAt must come down to .10 or .05. For accurate calculations it would
have to be even smaller, unless we change to a better difference equation. That is the
right thing to do.
Notice the two reasonable numbers. They are .35 and .36, approaching e- = .37. '
They come from n = 10 (with At = 1/10) and n = 20 (with At = 1/20). Those have the
same clock time nAt = 1:
The main diagonal of the table is executing (1 + xln)" - e" in the case x = - 1.
,
Final question: How quickly are .35 and .36 converging to e-' = .37? With At = .10
the error is .02. With At = .05 the error is .01. Cutting the time step in half cuts the
error in half. We are not keeping enough digits to be sure, but the error seems close
to *At. To test that, apply the "quick method" and estimate an= (1 - Atr from its
logarithm:
ln(1- Atr = n ln(1- At) z n[- At - + ( ~ t ) = ] 1 - f At.
~-
The clock time is nAt = 1. Now take exponentials of the far left and right:
' '.
The differencebetween an and e- is the last term *Ate- Everything comes down
to one question: Is that error the same as *At? The answer is yes, because e-'12 is
115. If we keep only one digit, the prediction is perfect!
That took an hour to work out, and I hope it takes longer than At to read. I wanted
you to see in use the properties of In x and e". The exact property In an= n In a came
first. In the middle of (15) was the key approximation ln(1 + x) z x - f x2, with x =
- At. That x2 term uses the second derivative (Section 6.4). At the very end came
e"xl+x.
+
A linear approximation shows convergence: (1 x/n)" - ex. A quadratic shows the
,
error: proportional to At = l/n. It is like using rectangles for areas, with error propor-
tional to Ax. This minimal accuracy was enough to define the integral, and here it is
enough to define e. It is completely unacceptable for scientific computing.
The trapezoidal rule, for integrals or for y' = cy, has errors of order (Ax)2and (At)2.
All good software goes further than that. Euler's first-order method could not predict
the weather before it happens.
dy
Euler's Method for - = F(y, t): Y(' + At) - y(t) = ~ ( ~ ( t). ,
t)
dt At
276 6 Exponentials and Logarithms
6.6 EXERCISES
Read-through questions limit of (1 - l/n)". What is the sum of this infinite series -
the exact sum and the sum after five terms?
The infinite series for e" is a . Its derivative is b . The
denominator n! is called " c " and it equals d . At x = 9 Knowing that (1 + l/n)" -+ e, explain (1 + l/n)2n e2 and
-+
1 the series for e is e . (1 + 2/N)N-+e2.
To match the original definition of e, multiply out 10 What are the limits of (1 + l/n2)" and (1 + l/n)"*?
(1 + l/n)" = f (first three terms). As n + co those terms OK to use a calculator to guess these limits.
approach Q in agreement with e. The first three terms of
11 (a) The power (1 + l/n)" (decreases) (increases) with n, as
(1 + xln)" are h . As n + co they approach 1 in
we compound more often. (b) The derivative of f(x)=
agreement with ex. Thus (1 + xln)" approaches I .A
x ln(1 + llx), which is , should be (<0)(> 0). This is
+
quicker method computes ln(1 xln)" x k (first term
confirmed by Problem 12.
only) and takes the exponential.
Compound interest (n times in one year at annual rate x)
12 Show that ln(1 + l/x) > l/(x + 1) by drawing the graph of
llt. The area from t = 1 to 1 + l/x is . The rectangle
multiplies by ( I )". As n -+ co, continuous compounding
inside it has area .
multiplies by m . At x = 10% with continuous compound-
ing, $1 grows to n in a year. 13 Take three steps of y(t + 1) = 2y(t) from yo = 1.
The difference equation y(t + 1) = ay(t) yields fit) = o 14 Take three steps of y(t + 1) = 2y(t) + 1 from yo = 0.
times yo. The equation y(t + 1) = ay(t) + s is solved by y =
+ +
atyo+ $1 a + -.- at-']. The sum in brackets is P .
Solve the difference equations 15-22.
When a = 1.08 and yo = 0, annual deposits of s = 1 produce
y = q after t years. If a = 9 and yo = 0, annual deposits
of s = 6 leave r after t years, approaching y, = s .
The steady equation y, = ay, + s gives y, = t .
When i = interest rate per period, the value of yo = $1 after
N periods is y(N) = u . The deposit to produce y(N) = 1
is yo = v . The value of s = $1 deposited after each period
grows to y(N) = w . The deposit to reach y(N) = 1 is s =
In 23-26, which initial value produces y, = yo (steady state)?
x .
Euler's method replaces y' = cy by Ay = cyAt. Each step 23 y(t + 1) = 2y(t) - 6 24 y(t + 1) = iy(t) - 6
multiplies y by Y . Therefore y at t = 1 is (1 + cAt)ll'yo, 25 y(t + 1)= - y(t) + 6 26 y(t + 1)= - $y(t) + 6
which converges to as At -+ 0. The error is proportional
27 In Problems 23 and 24, start from yo = 2 and take three
to A , which is too B for scientific computing.
steps to reach y,. Is this approaching a steady state?
1 Write down a power series y = 1 - x + .-.whose derivative 28 For which numbers a does (1 - at)/(l - a) approach a limit
is -y. as t -+ oo and what is the limit?
2 Write down a power series y = 1 + 2x + .--whose deriva- 29 The price P is determined by supply =demand or
tive is 2y. -dP(t + +
1) b = cP(t). Which price P is not changed from
one year to the next?
3 Find two series that are equal to their second derivatives.
30 Find P(t) from the supply-demand equation with c = 1,
4 By comparing e = 1 + 1 + 9 + 4 + + -.. with a larger
d = 2, b = 8, P(0) = 0. What is the steady state as t -+ co?
series (whose sum is easier) show that e < 3.
5 At 5% interest compute the output from $1000 in a year
Assume 10% interest (so a = 1 + i = 1.1) in Problems 31-38.
with 6-month and 3-month and weekly compounding.
31 At 10% interest compounded quarterly, what is the effec-
6 With the quick method ln(1 + x) z x, estimate ln(1- lln)"
tive rate?
and ln(1 + 2/n)". Then take exponentials to find the two limits.
32 At 10% interest compounded daily, what is the effective
7 With the slow method multiply out the three terms of
rate?
(1 - $)2 and the five terms of (1 - $I4.
What are the first three
terms of (1 - l/n)", and what are their limits as n -+ oo? 33 Find the future value in 20 years of $100 deposited now.
8 The slow method leads to 1 - 1 + 1/2! - 1/3! + -.-for the 34 Find the present value of $1000 promised in twenty years.
6.7 Hyperbolic Functions 277
35 For a mortgage of $100,000 over 20 years, what is the do you still owe after one month (and after a year)?
monthly payment?
41 Euler charges c = 100% interest on his $1 fee for discover-
36 For a car loan of $10,000 over 6 years, what is the monthly ing e. What do you owe (including the $1) after a year with
payment? (a) no compounding; (b) compounding every week; (c) con-
tinuous compounding?
37 With annual compounding of deposits s = $1000, what is
the balance in 20 years? 42 Approximate (1 + 1/n)" as in (15) and (16) to show that
you owe Euler about e - e/2n. Compare Problem 6.2.5.
38 If you repay s = $1000 annually on a loan of $8000, when
are you paid up? (Remember interest.) 43 My Visa statement says monthly rate = 1.42% and yearly
rate = 17%. What is the true yearly rate, since Visa com-
39 Every year two thirds of the available houses are sold, and
pounds the interest? Give a formula or a number.
1000 new houses are built. What is the steady state of the
housing market - how many are available? 44 You borrow yo = $80,000 at 9% to buy a house.
40 If a loan shark charges 5% interest a month on the $1000 (a) What are your monthly payments s over 30 years?
you need for blackmail, and you pay $60 a month, how much (b) How much do you pay altogether?
I 6.7 Hyperbolic Functions
This section combines ex with e - x. Up to now those functions have gone separate
ways-one increasing, the other decreasing. But two particular combinations have
earned names of their own (cosh x and sinh x):
ex + e - x ex - e-x
hyperbolic cosine cosh x-= -
hyperbolic sine sinh x =
2 2
The first name rhymes with "gosh". The second is usually pronounced "cinch".
The graphs in Figure 6.18 show that cosh x > sinh x. For large x both hyperbolic
functions come extremely close to ½ex. When x is large and negative, it is e- x that
dominates. Cosh x still goes up to + 00 while sinh x goes down to - co (because
sinh x has a minus sign in front of e-x).
1 1 1 1
cosh x = eX+ e-x sinh x = -ex e
2 2 2 2
\ /I
1 1
e-X 1 ex
2 2
-1 1
Fig. 6.18 Cosh x and sinh x. The hyperbolic Fig. 6.19 Gateway Arch courtesy of the St.
functions combine 'ex and ½e- x . Louis Visitors Commission.
The following facts come directly from ((ex + e - x) and ½(ex - e-X):
cosh(- x) = cosh x and cosh 0 = 1 (cosh is even like the cosine)
sinh(- x) = - sinh x and sinh 0 = 0 (sinh is odd like the sine)
6 Exponentials and Logarithms
The graph of cosh x corresponds to a hanging cable (hanging under its weight).
Turned upside down, it has the shape of the Gateway Arch in St. Louis. That must
be the largest upside-down cosh function ever built. A cable is easier to construct
than an arch, because gravity does the work. With the right axes in Problem 55, the
height of the cable is a stretched-out cosh function called a catenary:
y = a cosh (x/a) (cable tension/cable density = a).
Busch Stadium in St. Louis has 96 catenary curves, to match the Arch.
The properties of the hyperbolic functions come directly from the definitions. There
are too many properties to memorize-and no reason to do it! One rule is the most
important. Every fact about sines and cosines is reflected in a correspondingfact about
sinh x and cosh x. Often the only difference is a minus sign. Here are four properties:
1. (cosh x)2 - (sinh x)2 = 1 instead of (cos x) 2 + (sin x)2 = 1]
- 2
e 2 x+2+e-2x e2 x+2 -e x
ex e-x 2 x e- 2 =
Check:
2. d (cosh x) = sinh x instead of d (cos x) - sin x
dx dx
3. d (sinh x) = cosh x like d sin x = cos x
4. f sinh x dx = cosh x + C and f cosh x dx = sinh x + C
t, sinh t)
t)
Fig. 6.20 The unit circle cos 2 t + sin 2 t = 1 and the unit hyperbola cosh 2 t - sinh 2 t = 1.
Property 1 is the connection to hyperbolas. It is responsible for the "h" in cosh and
sinh. Remember that (cos x)2 + (sin x)2 = 1 puts the point (cos x, sin x) onto a unit
circle. As x varies, the point goes around the circle. The ordinary sine and cosine are
"circular functions." Now look at (cosh x, sinh x). Property 1 is (cosh x) 2 - (sinh x) 2 =
1, so this point travels on the unit hyperbola in Figure 6.20.
You will guess the definitions of the other four hyperbolic functions:
sinh x ex - e-x cosh x ex + e-x
tanh x - - coth x - - -
cosh x ex + e - x sinh x ex - e - x
1 2 1 2
sech x cosh x ex + e-x
csch x sinh x ex - e-x
I think "tanh" is pronounceable, and "sech" is easy. The others are harder. Their
6.7 Hyperbolic Functions
properties come directly from cosh2x- sinh2x = 1. Divide by cosh2x and sinh2x:
1 - tanh 2x = sech2x and coth2x - 1 = csch2x
(tanh x)' = sech2x and (sech x)' = -sech x tanh x
1 sinh x
tanh x dx = S=dx = ln(cosh x) + C.
N E S Y E B LC
I V R E H P R O I FUNCTIONS
You remember the angles sin-'x and tan-'x and sec-'x. In Section 4.4 we
differentiated those inverse functions by the chain rule. The main application was to
integrals. If we happen to meet jdx/(l+ x2), it is tan-'x + C. The situation for
sinh- 'x and tanh- 'x and sech- 'x is the same except for sign changes - which are
expected for hyperbolic functions. We write down the three new derivatives:
y = sinh-'x (meaning x = sinh y) has 9= J 21T i
dx
1
y = tanh-'x (meaning x = tanh y) has 9= -
dx 1 - x2
-1
'
y = sech - x (meaning x = sech y) has dy =
dx X J i 7
Problems 44-46 compute dyldx from l/(dx/dy). The alternative is to use logarithms.
Since In x is the inverse of ex, we can express sinh-'x and tanh-'x and sech-'x as
logarithms. Here is y = tanh- 'x:
The last step is an ordinary derivative of 4 ln(1 + x) - ln(1 - x). Nothing is new
except the answer. But where did the logarithms come from? In the middle of the
following identity, multiply above and below by cosh y:
-- x - 1 + tanh y - cosh y + sinh y - - - - e2y.
1+ - eY
1 - x 1- tanh y cosh y - sinh y e-y
Then 2y is the logarithm of the left side. This is the first equation in (4), and it is the
third formula in the following list:
Remark 1 Those are listed onlyfor reference. If possible do not memorize them. The
derivatives in equations (I), (2), (3) offer a choice of antiderivatives - either inverse
functions or logarithms (most tables prefer logarithms). The inside cover of the book
has
1% = fln[E] +C (in place of tanh- 'x + C).
Remark 2 Logarithms were not seen for sin- 'x and tan- 'x and sec - 'x. You might
6 Exponentials and Logarithms
wonder why. How does it happen that tanh-'x is expressed by logarithms, when the
parallel formula for tan-lx was missing? Answer: There must be a parallel formula.
To display it I have to reveal a secret that has been hidden throughout this section.
The secret is one of the great equations of mathematics. What formulas for cos x
and sin x correspond to &ex + e-x) and &ex- e-x)? With so many analogies
(circular vs. hyperbolic) you would expect to find something. The formulas do exist,
but they involve imaginary numbers. Fortunately they are very simple and there is
no reason to withhold the truth any longer:
1 1 .
cosx=-(eix+eix) and sin~=-(e'~--e-'~). (5)
2 2i
It is the imaginary exponents that kept those identities hidden. Multiplying sin x by
i and adding to cos x gives Euler's unbelievably beautiful equation
cos x + i sin x = eiX. (6)
That is parallel to the non-beautiful hyperbolic equation cosh x + sinh x = ex.
I have to say that (6) is infinitely more important than anything hyperbolic will
ever be. The sine and cosine are far more useful than the sinh and cosh. So we end
our record of the main properties, with exercises to bring out their applications.
Read-through questions Find the derivatives of the functions 9-18:
Cosh x = a and sinh x = b and cosh2x - sinh2x = 9 cosh(3x + 1) 10 sinh x2
c . Their derivatives are d and e and f .
The point (x, y) = (cosh t , sinh t ) travels on the hyperbola 11 l/cosh x 12 sinh(1n x)
- -
g . A cable hangs in the shape of a catenary y = h . 13 cosh2x + sinh2x 14 cosh2x - sinh2x
The inverse functions sinh-'x and t a n h l x are equal to 15 tanh , =
, / 16 (1 + tanh x)/(l - tanh x)
ln[x + ,/x2 + 11 and 4ln I . Their derivatives are i
17 sinh6x 18 ln(sech x + tanh x)
and k . So we have two ways to write the anti I . The
parallel to cosh x + sinh x = ex is Euler's formula m . 19 Find the minimum value of cosh(1n x) for x > 0.
The formula cos x = $(eix+ ePix)involves n exponents.
20 From tanh x = +find sech x, cosh x, sinh x, coth x, csch x.
The parallel formula for sin x is o .
21 Do the same if tanh x = - 12/13.
1 Find cosh x + sinh x, cosh x - sinh x, and cosh x sinh x.
22 Find the other five values if sinh x = 2.
2 From the definitions of cosh x and sinh x, find their deriv-
atives. 23 Find the other five values if cosh x = 1.
3 Show that both functions satisfy y" = y. 24 Compute sinh(1n 5) and tanh(2 In 4).
4 By the quotient rule, verify (tanh x)' = sech2x.
5 Derive cosh2x + sinh2x = cosh 2x, from the definitions. Find antiderivatives for the functions in 25-32:
6 From the derivative of Problem 5 find sinh 2x. 25 cosh(2x + 1) 26 x cosh(x2)
7 The parallel to (cos x + i sin x r = cos nx + i sin nx is a 27 cosh2x sinh
hyperbolic formula (cosh x + sinh x)" = cosh nx + .
sinh x ex + e P x
8 Prove sinh(x + y) = sinh x cosh y + cosh x sinh y by 30 ~ 0 t h = ex ---
x -
- e-"
changing to exponentials. Then the x-derivative gives 29 1 +cosh x
cosh(x + y) = 31 sinh x + cosh x 32 (sinh x + cosh x)"
6.7 Hyperbolic Functions 281
33 The triangle in Figure 6.20 has area 3 cosh t sinh t.
(a) Integrate to find the shaded area below the hyperbola
(b)For the area A in red verify that dA/dt = 4
(c) Conclude that A = it + C and show C = 0.
Sketch graphs of the functions in 34-40.
34 y = tanh x (with inflection point)
35 y = coth x (in the limit as x 4 GO) 54 A falling body with friction equal to velocity squared
obeys dvldt = g - v2.
36 y = sech x
(a) Show that v(t) = & tanh &t satisfies the equation.
(b)Derive this v yourself, by integrating dv/(g - v2)= dt.
38 y=cosh-lx for x 3 1 (c) Integrate v(t) to find the distance f(t).
39 y = sech- 'x for 0 c x d 1 55 A cable hanging under its own weight has slope S = dyldx
40 : (i':)
= tanh-'x = - In - for lxlc 1
that satisfiesdS/dx = c d m . The constant c is the ratio of
cable density to tension.
(a) Show that S = sinh cx satisfies the equation.
41 (a) Multiplying x = sinh y = b(ey - e - Y by 2Y gives
) e (b)Integrate dyldx = sinh cx to find the cable height y(x),
(eq2- 2 4 8 ) - 1 = 0. Solve as a quadratic equation for eY. if y 0 = llc.
()
(b)Take logarithms to find y = sinh - 'x and compare with (c) Sketch the cable hanging between x = - L and x = L
the text. and find how far it sags down at x = 0.
42 (a) Multiplying x = cosh y = i ( 8 + ebY) by 2ey gives 56 The simplest nonlinear wave equation (Burgers' equation)
( e ~- 2x(e") + 1 = 0. Solve for eY.
)~ yields a waveform W(x) that satisfies W" = WW' - W'. One
(b)Take logarithms to find y = cosh- 'x and compare with integration gives W' = 3w2- W .
the text. (a) Separate variables and integrate:
43 Turn (4) upside down to prove y' = - l/(l - x2), if y = dx=dw/(3w2- W)=-dW/(2- W)-dW/W.
coth- 'x. (b) Check W' = 3W2- W.
44 Compute dy/dx = I/,/= by differentiating x = sinh y 57 A solitary water wave has a shape satisfying the KdV
and using cosh2y - sinh2y= 1. equation y" = y' - 6yy'.
(a) Integrate once to find y". Multiply the answer by y'.
45 Compute dy/dx = l/(l - x2) if y = tanh- 'x by differen-
tiating x = tanh y and using sech2y+ tanh2y = 1. (b) Integrate again to find y' (all constants of integration
are zero).
46 Compute dyldx = -l / x J E ? for y = sech- 'x, by (c) Show that y = 4 sech2(x/2) gives the shape of the
differentiating x = sech y. "soliton."
From formulas (I), (2), (3) or otherwise, find antiderivatives in 58 Derive cos ix = cosh x from equation (5). What is the
47-52: cosine of the imaginary angle i =
59 Derive sin ix = i sinh x from (5). What is sin i?
60 The derivative of eix= cos x + i sin x is
MIT OpenCourseWare
Resource: Calculus Online Textbook
Gilbert Strang
The following may not correspond to a particular course on MIT OpenCourseWare, but has been
provided by the author as an individual learning resource. |
Intermediate Algebra, with Applications
One of a series of developmental maths textbooks, this volume focuses on intermediate algebra. It provides a learning system organized by objectives, ...Show synopsisOne of a series of developmental maths textbooks, this volume focuses on intermediate algebra. It provides a learning system organized by objectives, around which all lessons, exercises, end-of-chapter review tests and ancillaries are arranged. The last objective in every section is, where applicable, devoted to applications, and a specific strategy is suggested for each major application problem, encouraging students to plan problem-solving strategies before addressing the problems.Hide synopsis
Description:New. Based on the best-selling series by the Aufmann team, this...New. Based on the best-selling series by the Aufmann team, this hardcover text for the intermediate algebra course adheres to the formula that has made the Aufmann developmental texts so reliable for both students and instructors. The text's clear writing.
Description:New. 0547197977 Brand-New, Unread Copy in Perfect Condition. To...New. 0547197977 |
Mapping geometry for 2 by 2 matrices - Editorial Review
Dan Kalman's Mapping Geometry module allows linear algebra students to visualize linear transformations of the form Ax = y, where A is a 2 by 2 matrix. The module is based on Mathwright Microworld technology, an HTML document format that allows embedding of portals used to display "story pages." To use it, users must first download and install MathwrightWeb ActiveX Control (which is available for free) for Internet Explorer. There is also a stand-alone (Windows only) version available on the site. Mapping Geometry consists of two web pages that are very similar in form. They differ only in that the first page contains two boxes of Cartesian planes, one for drawing domain points and second for drawing image points, whereas the second page displays both domain and image points inside one box. Each page allows the user to define a matrix A and then to select one of four possible modes by clicking on buttons: Click Points, Circle, Polygon, and Drag Points. Editing the entries of A can be done easily by using the keyboard and clicking on the "Change A" button. Each mode allows a different representation of domain vectors so that the resulting image vectors can be studied. This tool will help students visualize linear transformations in R2, being particularly helpful in illustrating the geometric interpretation of eigenvectors. |
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Introduction to Numerical Methods and Analysis
9780470049631
ISBN:
0470049634
Pub Date: 2007 Publisher: John Wiley & Sons Inc
Summary: Praise for the First Edition". . . outstandingly appealing with regard to its style, contents, considerations of requirements of practice, choice of examples, and exercises." -Zentrablatt Math". . . carefully structured with many detailed worked examples . . ." -The Mathematical Gazette". . . an up-to-date and user-friendly account . . ." -MathematikaAn Introduction to Numerical Methods and Analysis addresses the mat...hematics underlying approximation and scientific computing and successfully explains where approximation methods come from, why they sometimes work (or don't work), and when to use one of the many techniques that are available. Written in a style that emphasizes readability and usefulness for the numerical methods novice, the book begins with basic, elementary material and gradually builds up to more advanced topics.A selection of concepts required for the study of computational mathematics is introduced, and simple approximations using Taylor's Theorem are also treated in some depth.The text includes exercises that run the gamut from simple hand computations, to challenging derivations and minor proofs, to programming exercises. A greater emphasis on applied exercises as well as the cause and effect associated with numerical mathematics is featured throughout the book. An Introduction to Numerical Methods and Analysis is the ideal text for students in advanced undergraduate mathematics and engineering courses who are interested in gaining an understanding of numerical methods and numerical analysis.
Epperson, James F. is the author of Introduction to Numerical Methods and Analysis, published 2007 under ISBN 9780470049631 and 0470049634. Twenty nine Introduction to Numerical Methods and Analysis textbooks are available for sale on ValoreBooks.com, twenty used from the cheapest price of $50.69, or buy new starting at $70 |
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Radar Workbook - Problems and Answers in Marine Radar Operations Workbook is designed to be used in classroom or online courses in radar, or for individual study outside of the classroom. The lesson structure follows that used by several schools in the US, based on the background reader Radar for Mariners by David Burch. The Appendix on advanced radar plotting is included for professional mariners who seek more practice on interpreting ARPA output by working out the vector solutions themselves. The ability to manually interpret the radar interactions seen on the screen, independent of electronic solutions, is in keeping with the fundamental tenet of good navigation and seamanship that we should not rely on any one aid alone. |
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Prerequisites: MAT-152 or equivalent course with a minimum grade of C.
A continuation of MAT-152. A study of real functions of several real variables. Topics include differentiability and continuity, differential geometry, extrema, Lagrange multipliers, multiple integration, line and surface integrals, and the theorems of Green, Gauss and Stokes. |
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This dynamic new edition of this proven series adds cutting edge print and media resources. An emphasis on the practical applications of algebra motivates learners and encourages them to see algebra as an important part of their daily lives. The reader-friendly writing style uses short, clear sentences and easy-to-understand language, and the outstanding pedagogical program makes the material easy to follow and comprehend.KEY TOPICSChapter topics cover basic concepts; equations and inequalities; graphs and functions; systems of equations and inequalities; polynomials and polynomial functions; rational expressions and equations; roots, radicals, and complex numbers; quadratic functions; exponential and logarithmic functions; conic sections; and sequences, series and the binomial theorem.For the study of Algebra. |
Calculus II For Dummies, 2nd Edition
Calculus II is a prerequisite for many popular college majors, including pre-med, engineering, and physics. Calculus II For Dummies offers expert instruction, advice, and tips to help second semester calculus students get a handle on the subject and ace their exams.
It covers intermediate calculus topics in plain English, featuring in-depth coverage of integration, including substitution, integration techniques and when to use them, approximate integration, and improper integrals. This hands-on guide also covers sequences and series, with introductions to multivariable calculus, differential equations, and numerical analysis. Best of all, it includes practical exercises designed to simplify and enhance understanding of this complex subject.
Introduction to integration
Indefinite integrals
Intermediate Integration topics
Infinite series
Advanced topics
Practice exercises
Confounded by curves? Perplexed by polynomials? This plain-English guide to Calculus II will set you straight!
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Overview
The demand for more reliable geometric computing in robotics, computer vision and graphics has revitalized many venerable algebraic subjects in mathematics - among them, Grassmann-Cayley algebra and Geometric Algebra. Nowadays, they are used as powerful languages for projective, Euclidean and other classical geometries.
This book contains the author and his collaborators' most recent, original development of Grassmann-Cayley algebra and Geometric Algebra and their applications in automated reasoning of classical geometries. It includes two of the three advanced invariant algebras - Cayley bracket algebra, conformal geometric algebra, and null bracket algebra - for highly efficient geometric computing. They form the theory of advanced invariants, and capture the intrinsic beauty of geometric languages and geometric computing. Apart from their applications in discrete and computational geometry, the new languages are currently being used in computer vision, graphics and robotics by many researchers |
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In my experience, many introductory engineering mathematics textbooks these days tend to skip proofs and discuss logic only in the context of digital electronics. On the other hand, I can imagine that ...
I am a college professor in the American education system and find that the major concern of my students is trying to determine the specific techniques or problems which I will ask on the exam. This ...
As a student in high school, I never bothered to memorize equations or methods of solving, rather I would try to identify the logic behind the operations and apply them. However, now that I've begun ...
My youngest brother had a pre-algebra test yesterday and he was asked to tell if two expressions are equal or not.
We agreed on most of the things but on this one I find it hard to make him accept my ...
As a graduate student at my university, I have the option many times of deciding what type of work I do for support. The two basic options are to teach either a calculus or college algebra course, or ...
If this isn't the place, I apologize:
At the end of my calculus class, we asked the students (among other things) what some applications of calculus methods are. Disappointingly, many focused on theMy daughter's elementary school is currently using Saxon curriculum, and I'd like to figure out what other sources, if any, should be age appropriate for elementary school children to enhance their ...
I am supposed to give a 15-20 minutes math lecture, where I am expecting around 20-30 people. The lecture is about derivative. Since this would be my first "class", I would appreciate any suggestions ...
I'm helping out my former calculus teacher as a volunteer calculus advisor, and I have under my supervision 5 students. They've already had an exam and... well, they failed.
I read their exams and I ...
In my country in $3^{rd}$ grade in math kids learn the four basic arithmetic operation (addition, subtraction, multiplication and divison) up to $10 000$.
My sister this year goes to $3^{rd}$ grade ...
We're going to hold a three-hour math workshop for some middle school students. It'll about the Pascal's triangle.
Well, we can ask the students to find patterns in the triangle, or try to prove some ...
I am supposed to teach undergraduate students who do not major in mathematics and I would like to give them a short introduction to mathematical reasoning and to the concept of proof. I am looking for ... |
Tag: mathml
LaTeX is document markup system (similar to the popular Markdown) that's popular among high-level academia and mathematics authors. iBooks Author 2, which was released during Apple's big event, now supports the LaTeX protocol. It works through the MathML markup language, which a lot of... |
Each volume is broken into individual chapters. And each volume has an associated solutions manual (the last item under each...
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Each volume is broken into individual chapters. And each volume has an associated solutions manual (the last item under each volume). There is a file with corrections to both the text and the answer manual. Also note a file with the diagrammatic summary of the rules that appeared on the inside cover of the published version of the primmer. All files are in Adobe Acrobat PDF 6 format. – they require version 6, or newer, of Adobe, which can be down loaded here. All files are fully searchable.A Problem Course in Mathematical Logic is intended to serve as the text for an introduction to mathematical logic for...
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A Problem Course in Mathematical Logic is intended to serve as the text for an introduction to mathematical logic for undergraduates with some mathematical sophistication. It supplies definitions, statements of results, and problems, along with some explanations, examples, and hints. The idea is for the students, individually or in groups, to learn the material by solving the problems and proving the results for themselves. The book should do as the text for a course taught using the modified Moore-method.The material and its presentation are pretty stripped-down and it will probably be desirable for the instructor to supply further hints from time to time or to let the students consult other sources. Various concepts and and topics that are often covered in introductory mathematical logic or computability courses are given very short shrift or omitted entirely, among them normal forms, definability, and model theory.
This text has developed out of an alternate beginning physics course designed for those students with a strong interest in...
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This text has developed out of an alternate beginning physics course designed for those students with a strong interest in physics. The course includes students intending to major in physics, but is not limited to them. The idea for a "radically modern" course arose out of frustration with the standard two-semester treatment. It is basically impossible to incorporate a significant amount of "modern physics" (meaning post-19th century!) in that format. This text uses optics and waves together with relativity as a foundation for all of physics instead of classical mechanics.GNU Free Documentation LicenseAcceso is a complete, interactive curriculum for intermediate-level learners of Spanish. The materials on the site are...
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This is a free, online textbook that is part of WikiBooks. Therefore, it is continually being updated. According to the...
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This is a free, online textbook that is part of WikiBooks. Therefore, it is continually being updated. According to the site, "Accounting is the process by which financial information about a business is recorded, classified, summarized, interpreted, and communicated.״ The WikiBook covers a wide variety of topics.
״This text undertakes to provide a first-year course in accounting, with accepted principles of accounting arranged in an...
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This is a free, online textbook offered by the Global Text Project at University of Georgia. "The Global Text Project's...
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This is a free, online textbook offered by the Global Text Project at University of Georgia. "The Global Text Project's international version of the 8th edition of the well-known accounting text. well-known accounting text.״
Acid-Base Physiology was designed by Dr. Kerry Brandis, a clinical professor of anesthesiology at Gold Coast Hospital in...
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Acid-Base Physiology was designed by Dr. Kerry Brandis, a clinical professor of anesthesiology at Gold Coast Hospital in Queensland, Australia. The learning resources is an open textbook for medical students studying anesthesia. It would be useful to any learner who needs a comprehensive understanding of acid-base balance, for example, pulmonary medicine and nurse anesthetist students. The textbook includes eleven chapters: 1- Introduction 2 - Control of Acid-Base Balance 3 - Acid-Base Disorders 4 - Respiratory Acidosis 5 - Metabolic Acidosis 6 - Respiratory Alkalosis7 - Metabolic Alkalosis 8 - Major Types of Megabolic Acidosis 9 - Assessment of Acid-Base Disorders 10 - Quantitative Acid-Base Analysis 11 - Special Aspects of Acid-Base Physiology. It includes 30 different case studies, as well as references for each chapter. The learning resource was recognized in 2006 by the "American Thoracic Best of Web" series. |
Web Site Dave's Short Trig Course Check out the short trigonometry course and learn the new way of learning trig. This short course breaks into sections and allows user to learn at his/her o... Curriculum: Mathematics Grades: 9, 10, 11, 12
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Web Site S.O.S. Mathematics - Calculus Check out a good list of calculus problems with solutions. This is a free resource for math review material from Algebra to Differential Equations!
Web Site Order of Operations When a numerical expression involves two or more operations, there is a specific order in which these operations must be performed. The phrase PEMDA (Parenth... Curriculum: Mathematics Grades: 5, 6, 7, 8
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Web Site Intermediate Algebra Tutorials 42 Tutorials that math teachers can use with student or students can work on their own to reinforce skills, as homework, or review during class. Tutorials in... Curriculum: Mathematics Grades: 6, 7, 8, 9, 10, 11, 12, Junior/Community College, University
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Web Site Variables This site covers symbol variables and substitution of symbols to discover unknown values. In simple terms it shows you how a box is waiting for a value. (Key... Curriculum: Mathematics Grades: 6, 7, 8
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Web Site Introduction to Algebra Think Algebra is hard? Think again - this site explains the history along with simple equations. Each paragraph scaffolds skills until you get it. Than at th... Curriculum: Mathematics Grades: 3, 4, 5, 6
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"I used My eCoach for a few years as my classroom website and for surveys and quizzes.
Two years ago, the district asked us to learn to use moodle because it is linked to student's grades, etc... At first, the moodle class page was separate, but needed to be accessed by student username and password. Many students forgot passwords or changed them or couldn't get on because of other reasons. This year there is a link within the students portal site to a moodle classroom page. (They still need their username and password.) This link was not to the webpage I had made earlier, but to a new one. I switched all the information over to the new because I was told the parents would be able to access it from their portal site, also. This was not true. I then switched back to My eCoach and found it so easy to use!
The parents and students both can keep up with assignments because they do not need to remember passwords! This past week, I put together an online test in portal-moodle because I was told it would grade the test and link the grade to my gradebook. What a disaster! The transferred grade was not correct and I am dealing with many e-mails from students who could not submit their finished tests. I believe I will try the on-line quiz in My eCoach next time and just copy over the grade.
I am still only using a small portion of what is available in ecoach, but am looking forward to trying all the new things that Barbara and her team have developed over the past year or so."
Susan Brown
High School Science Teacher
Osceola High School - Pinellas, Florida |
At what age should mathematical proofs be taught to students
I think college is way tooo late to learn how to write mathematical proofs! Proof writing should begin at least either in elementary school or middle school. Proof writing is just as important to a students education as learning how to write sentences and learn how to combined those sentences to create paragraghs and learn how to combined paragraphs together properly to write a decent term paper, because PRoof writing will improve your deductive and reasoning skills. I think a lot of people hate mathematics because they don't understand how the equations were derived . In high school, math was just memorizing formulas and algorithms . When I got to college, They just threw proofs right at me, and now my system that I have been using all my life to passed mathematics failed because you had to apply systematic methodology for writing proofs and so sadly I dropped my math major.pentazoid
#3
Feb24-09, 10:36 AM
P: 185
Quote by SticksandStonesI hope 9 year olds aren't just learning how to add 3 digit numbers. IF they are, then the kids who had the luxury of being taught in the public education system, are really screwed.
sure they have the intellectual capacity to write proofs just like they have the intellectual capacity to to write sentences and paragraphs. I think they lacked mathematical sophistication because we are not taught how to write proofs. Students should learned how mathematics equations and diagrams like Pascal's Triangle and N-factorial are derived
because right now they really don't have a strong grasp on understanding what those equations mean.
Tobias Funke
#4
Feb24-09, 05:15 PM
P: 138
At what age should mathematical proofs be taught to students
I can't say what age exactly, but certainly sooner than they are now. Dophysics girl phd
#5
Feb24-09, 08:04 PM
P: 936
Quote by pentazoid
I hope 9 year olds aren't just learning how to add 3 digit numbers. IF they are, then the kids who had the luxury of being taught in the public education system, are really screwed.pentazoid
#6
Feb24-09, 08:33 PM
P: 185
Quote by physics girl phdWell I didn't encountered proofs until sophomore year of college and I did terrible in my proof class, because up until then , I just wasn't used to proof writing. At least in elementary school, they should teach you what deductive reasoning is and introduce you to logic, and they should demonstrate to the student how proofs are written. It just doesn't make sense to me to have study math for 13 years and just begin encountering proofs in your sophomore year just after you studied calculus. I supposed when you are only studying arithmetic, writing proofs are not necessary, but students should still be taught deductive reasoning. I think you should begin proof writing when the student begins to study geometry and algebra.
csprof2000
#7
Feb24-09, 09:59 PM
P: 287
It should be available at the advanced middle school level for good students, and at the early to intermediate level of high school for everybody.
Elementary school? Maybe some simple proof system... like propositional calculus. I bet a 10-year-old could prove "If A then B means NOT(A) or B", "If A then B and if B then C then if A then C", etc. This would have the added benefit of giving them exposure to things like set and logic notation. Proofs are also easy for logic.
Thoughts?
Dr.D
#8
Feb28-09, 08:40 PM
P: 619
Long, long time ago, I first encountered formal proofs as a high school junior in plane geometry, and I thought it was just wonderful. It was the thing that has been missing in all my math classes up to that point. It could have come earlier.
buffordboy23
#9
Mar2-09, 11:21 AM
P: 540
The first time I saw a proof was in my 8th grade algebra course. I don't recall the teacher going over them or assigning them as homework, but the text, Algebra I by Dolciani, offered numerous examples. I really didn't understand what a proof actually is and why they are necessary until college, which is really sad.
What's worse is that the middle school algebra texts of recent publication appear to offer no example proofs and are not very rigorous from what I have seen in the school district I used to work at. Worse still is the fact that mathematical rigor is being replaced by "gadget tricks", that are supposed to aid in student understanding but is probably detrimental over the long hall.
I think formal proofs can begin as early as the 6th or 7th grade. However, many students will need heavy scaffolding on the teacher's part to be successful at first. I think activities can be designed for K-6 that nurtures mathematical discovery and teaches one to think mathematically, but when it comes down to it, it's a lot easier for a teacher to run a game of "Around the World" to teach students their times tables for 2-3 weeks straight.
Bourbaki1123
#10
Mar2-09, 02:57 PM
P: 326
I know that when I was 7 or 8 I could probably have handled some simple proofs. I did math quite a bit and learned basic algebra skills by figuring out how to play the math games you got to play if you finished your work early. I really enjoyed algebra and worked on it(with my teachers and parents and books) until 6th grade when I had to repeat the same material in a so-called 'advanced' math class. I got bored and stopped until college only going up to algebra 2 in high school, and now I will be taking my first graduate course next year as a junior.
That being said, I know that not everyone has an affinity for math and that some people are even put off by proofs after they learn how to do them and drop the major; so, how certain are you that younger kids who have questionable enthusiasm for mathematics would latch onto proofs? I think that if we could have more extracurricular math besides competitions, where interested students can learn a more rigorous version of what is presented in class and not be scared away by the competition aspect(anyone should be allowed to join, no classroom competitions, save that for the math team), then we could certainly bring in those with inclination to see what mathematics is really about.
Wellesley
#11
Mar3-09, 07:09 PM
P: 276
Quote by Tobias Funke
DoMy first experience with formal proofs was 7th grade geometry....I hated it. Maybe it was how it was taught, or maybe it was the book, but when I got to high school AP Calculus (BC), proofs took on a whole new meaning. I couldn't necessarily write them, but I could interpret them. I went from no grasp of proofs, to applying them in one year. This probably had something to do with me being motivated and wanting to learn. But my point is that it can be done.
I am saddened that middle school's have "Advanced Algebra" that is really only doing the kids a disservice. Not being able to work with fractions?! In High School?!
I found this article on the Web. I'm sure some people have already read this, but I think it fits well with what is being talked about, so I'm going to post the link. Lockhart's Lament
It's kind of long, but it is the best article I've ever read that describes the current condition of the American Math system in public schools.
Redbelly98
#12
Mar3-09, 08:11 PM
Mentor
P: 12,070
It's somewhat standard to get proofs in h.s. geometry (9th or 10th grade). However, 2 years ago I tutored a kid in this subject and his teacher never had them do proofs.
So I guess it depends on what school system you're in, and maybe on whether you are in the "honors track" for math.
Astronuc
#13
Mar9-09, 08:42 PM
Admin
P: 21,827
IIRC, my first encounter with proofs was probably 7th grade in introductory algebra.
In 10th grade, geometry and trigonometry included many proofs, but the methods were based on what was studied in 7-9th grades.
I would have like to learn more about analytical geometry and linear algebra early. I was introduced to matrices as early as 6th grade.
I found the flow of mathematics and science was sporatic and disjoint. I would have preferred to be allowed to learn when I was ready, but the educational system wasn't structured for me.
khemix
#14
Mar13-09, 04:29 AM
P: 117
I believe they used to encounter proofs in grade 10, in a geometry course that lasted a year. Most schools don't do that anymore, and proofs are usually only briefly seen in a grade 11-12 geometry course that lasts a semester.buffordboy23
#15
Mar13-09, 08:56 AM
P: 540
Quote by khemixkhemix
#16
Mar13-09, 10:46 AM
P: 117
Quote by buffordboy23I'm surprised you'd defend science and not history, as I was expecting more criticism for removing a history course which at least gives young students culture.
A lot of this is from personal experience, but any science I did in elementary school hardly inspired curiosity. What I remember doing was some very basic chemistry and physics, which was both too shallow to be of any taste, and the teacher was not qualified to answer our questions. Astronomy on the other hand I really enjoyed, and the study of rocks, but that was only grades 3 and 4. Highschool grade 9 is when I got a real taste for science.
thrill3rnit3
#17
Apr4-09, 01:44 AM
PF Gold
P: 712
Quote by pentazoid
I think a lot of people hate mathematics because they don't understand how the equations were derived .
I think it's more like people forgot the equations they memorized the night before, and they don't know how to derive it.
buffordboy23
#18
Apr4-09, 10:10 AM
P: 540
Quote by khemix
I'm surprised you'd defend science and not history, as I was expecting more criticism for removing a history course which at least gives young students culture.
I just now had a chance to read your response. Don't get me wrong. I think history is important too, because of the reasons you mentioned and more. My window of perspective in the educational world was science, so I know now how important the structuring of the K-12 science curriculum is rather than the history curriculum.
The OP's post falls under the general category of how we can improve mathematics education. It's funny that you mentioned the removal of certain elementary school subjects to achieve this end, since I know that some school districts in the U.S. have previously done so to focus on mathematics. Why did they do this? Because of their students' poor performance on high-stakes standardized tests and the negative consequences that would follow.
In my opinion, this is not the answer. The thinking is somewhat analogous to throwing more and more money into education in the hopes that achievement scores will rise. Elementary subjects are the foundation of higher learning, so we must not sacrifice them. It's unfortunate that your science teacher was not qualified to teach science (although they may or may not have their teaching certificate), but this is one part of the problem, which also pertains to mathematics education. From my experience as a student and as a teacher, I see that much of mathematics education is based upon memorization. This does not lead to true understanding of the subject. Moreover, not exposing students to proofs and similar methods of mathematical thinking in their K-12 education is a great disservice to our students, because this is the central force that drives our discoveries in mathematics and students don't recognize and appreciate that. |
Essentials of College Mathematics - 2nd edition
Summary: With applications to fourteen different career categories, The Essentials of College Mathematics, Second Edition is designed for students from a wide range of technical and career fields that require a solid understanding of basic math, elementary algebra, trigonometry, and geometry. Economical and easy-to-use, it is written in a step-by-step format, incorporates a ''spiral learning'' approach, and is supported by numerous examples, exercises and practice tests. Thro...show moreughout the text, examples are presented in both symbolic and narrative form and all concepts are applied directly to careers and professions. Clear explanations and real-world applications make this the most relevant and student-friendly textbook for today's students. ...show less
2005 Paperback Fair THE SPINE IS IN ROUGH SHAPE. This item may not include any CDs, Infotracs, Access cards or other supplementary material.
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College Algebra, Hybrid
Reflecting Cengage Learning's commitment to offering value for students, this new hybrid edition features the instructional presentation found in the ...Show synopsisReflecting Cengage Learning's commitment to offering value for students, this new hybrid edition features the instructional presentation found in the full text while delivering end-of-section exercises online in Enhanced WebAssign. Your instructor has seen the benefits of Enhanced WebAssign, which is included with this text, and has integrated its use into your course giving you an interactive learning experience with the convenience of a text that is both brief and affordable. Learn to think mathematically and develop genuine problem-solving skills with Stewart, Redlin, and Watson's COLLEGE ALGEBRA, Sixth Edition. This straightforward and easy-to-use algebra book will help you learn the fundamentals of algebra in a variety of practical ways. The book features new tools to help you succeed, such as learning objectives before each section to prepare you for what you're about to learn, and a list of formulas and key concepts after each section that help reinforce what you've learned. In addition, the book includes many real-world examples that show you how mathematics is used to model in fields like engineering, business, physics, chemistry, and biology.Hide synopsis
Description:Very good. No Supplement Edition. All orders ship SAME or NEXT...Very good 6th Edition. Used books are NOT guaranteed to contain...Good. 6 |
More About the Author
Dexter James Booth Dexter was educated in Wales and Canada being awarded a PhD in Theoretical Physics in 1970 by the University of Victoria in British Columbia. Since that time he has taught in a variety of institutions, finally retiring from the University of Huddersfield as a Principal Lecturer in 2007. His abiding interest in the teaching of mathematics has spurred him to produce over thirty books in mathematics, statistics and computing and this latest Think of a number . . . is his first venture into eBook publishing. It was inspired by a desire to expand his annual first lecture to his engineering students who were exposed to a rather wildly extravagant description of the place of numbers in the general scheme of things. The whole point of the book is to display the elements of the structure of the number system in a fairly unintimidating way. Whilst the approach is unconventional and not in the least bit rigorous it does expose Dexter's approach to education in general - at least the students seemed to appreciate it.
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'For several decades Stroud has been the favourite text for teaching mathematics to first year engineering students.' -- EPSRC Biochemical Engineering Research Network
About the Author
KEN STROUD was Principal Lecturer in the Department of Mathematics at the University of Coventry. He is also the author of the companion volume Further Engineering Mathematics. - DEXTER BOOTH is Senior Lecturer in the School of Computing and Mathematics at the University of Huddersfield. He has previously written several mathematics textbooks, including Foundation Mathematics (AWL)Without doubt the best textbook that I have ever read. I first came across this book whilst studying for a Physics degree and since then I have used the book on countless occasions. K A Stroud has managed to create a book that teaches you as well as if you were actually in a lecture. Split into Programmes of study, each is sub-divided into Frames. The beginning of each Frame is almost a Dummies guide to... and yet by the time you reach Frame 15 you're strolling through some complex problems, as if you had been familiar with the subject for years. If all Physics and Engineering books were like this I'd have breezed through all of my exams. Unfortunately I only breezed through maths.
I know a lot of UK universities recommend this but if your university advises anything else, ignore them. Engineering Mathematics is the best text book I used in the course of my degree and I still use it today. I was fortunate to have a good maths lecturer but on the odd occasions I got lost during a lecture (or if I was unable to attend), following the relevant areas in Engineering Mathematics soon got me up to speed. The book is split into a series of 28 programmes taking you from the simple stuff like Complex Numbers through more difficult subjects like Calculus. There is some statistics in there two. Each programme is broken out into a series of exercises which if you follow through sequentially take you from the basics in such a well devised step by step way that you never feel there is too big of a leap being made which is a problem with some other maths books. Thank you K. A. Stroud for helping me through my degree!
Three things to start off with: 1) I very much doubt whether I would have passed my first year in electronic engineering without this book. 2) This is not a reference textbook. It is a mathematics course. 3)This book (5th edition) does contain laplace transforms, partial differentiation and multiple integrals. It doesn't contain Taylor series in two variables, which is slightly annoying but minor. The great thing about this book is that though not a book you can use to quickly refer to to find out some equations (I find Engineering Mathematics by Croft, Davidson and Hargreaves useful for that.) but for when you're actually learning the stuff, ie you don't understand things before you open the book. If you've gone back to your room afte a few weeks of lectures on a subject you really don't understand, open this book, find the relevant chapter and work your way through it. At the end of this, unlessyou're on an engineering course by complete mistake, you WILL have a good grasp of that subject. I has this book on my shelf for a year at uni and occasionally used it as reference (possible if hard), and failed my maths exam. Over the summer, I worked my way systematically through most of the book, and at the end of August, passed my resit exam. I don't think I could have survived without it. The book is divided up into chapters with about 50 frames in each which you work through, lots of examples and worked solutions given, and some test questions at the end (plus loads of 'further questions' for each chapter.) You could probably work through this in parallel with a lecture course if you so wished, it may give you a different method of explaining things to what your lecturerer says, which can be useful. At no point is it overly complex, and it even has twelve Foundation chapters, which are all pretty basic A level stuff, to pick up on any obvious things you may have missed out on. Though it may not cover every single thing you'll meet in your engineering first year, you can survive on this book if nothing else. It definitely goes further than further maths A level courses, and if you buy one book before starting an Electronics degree, let it be this one!Read more ›
K A Strouds books; Engineering Mathematics and Further Engineering Mathematics are quite simply the finest texts of its kind. K A Stroud clarifies any difficulties or lack of understanding with upmost ability. In a nutshell any mathematics is made easy. All students with engineering maths/pure maths within their course should buy this, confusion will be a thing of the past, these books got me from a fail at A-Level maths to a 2:1 in Engineering. Brilliant
Never a great fan of mathematics and only got through maths A level by cramming and learning a load of stuff by memory without really understanding it. When I went on to do a degree in electronics maths was an integral part of the syllabus and it was the one part of the course I was dreading. Then somebody recommended this book to me and I bought it out of desperation and hope. My God! What a revelation, for the first time here was a maths text book that helped me to understand exactly what was going on and took me step by step through increasingly difficult problems. I actually enjoyed using it, and that's something I never thought I'd say about a maths text book. Once I understood what I was supposed to be doing I started to enjoy doing it and my confidence grew. At my first year exams I actually got my best marks in maths. I never would have believed it possible at the beginning of the year. It's not a heavyweight maths text book but it's ideal for people who need to do maths as part of a technical degree course.
I studied this book as part of my degree several years ago (several chapters a week no less) and I'm still lending this battered book to other people who require an understanding of Engineering Mathematics. I reccommend this book to anybody who wishes to understand Mathematics quickly, or who want a quick refresher. |
Find a Island, TXThe Socratic Method is a "guided question" methodology that allows students to see how a problem is solved rather then just mindlessly applying a formula without a conceptual understanding of the problem. A lot of students have difficulties with math and the math found in other subjects such as physics. Math is an interesting subject with a myriad of techniques for finding an answer |
Semester Projects
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Project Goals
The goal of these projects is to give you the opportunity to make your own connections between mathematics and modern society by considering a wide variety of problems ranging from economic and environmental issues to social and political situations that can be modeled and solved by mathematical means. They will help you establish connections between Math 10260 and your other courses, and they will allow you to make contributions in areas of your interest and expertise. In addition, they will provide you with an opportunity to interact and collaborate with your classmates.
Rules for the Project
You can work in groups of size 1-4 students (from any section of Math 10260).
Each group submits one (typed) paper (and an electronic copy, if possible).
Each member of the group receives the same score - a number between 0 and 10 - which will count toward your 20 participation points.
Final version due by December 7. If you choose option d above, a first draft for approval is due by November 9. You must include: Project title, the names of your team members and the class section each member belongs to.
Alternative Project Topics
The Social Security. Some experts project that the Social Security shortfall over the next 75 years will be about four trillion dollars. Is that true? How do they know? Make your contribution in the national debate about saving Social Security using ideas and techniques you learned in Math 10260 (for example, income streams).
The Deficit. Visit the Webpage of the Congressional Budget Office (CBO) at and try to make sense of the numbers you will find in "Current Budget Projections." Note that income streams are useful in making projections.
Sub-prime Loans. What are sub-prime loans and what they have to do with the current housing and banking crisis?
Ponzi Scheme. Currently the Securities and Exchange Commission (SEC) is investigating an alleged $50 billion fraud, a Ponzi scheme, perpetrated by Bernard Madoff and the asset management company that he ran. Explain what is a Ponzi scheme and why in the Madoff's case it went unnoticed by the SEC for so long that it became so massive.
Arctic National Wildlife Refuge: To Drill or not to drill? A question for public debate these days is whether the Arctic National Wildlife Refuge (ANWR) contains enough oil to make its extraction worth both the economic cost and the environmental risk. Make your contribution by doing the numbers.
Energy Conservation. Some claim that there are ways for saving about 20% of the energy we consume today. Examine this claim by doing some quantitative analysis on energy consumption and energy wasted.
Oil Price. Is the current oil price the result of world demand & supply or/and market manipulation? Draw your own conclusions by collecting data from reliable sources and analyzing them using the mathematics you learned in Math 10260.
Wind Energy. Collect data about wind energy production in the U.S. since 2000 and draw a curve that fits these data. Also, draw the oil-price curve using data from reliable sources. Furthermore, compare the shape of these curves and make sense of the current projections of wind energy productions for the next 10-20 years. Finally, find out for which country in the world the percentage of the energy it uses from wind is maximum.
Solar Energy. Collect data about solar energy production in the U.S. since 2000 and draw a curve that fits these data. Also, draw the oil-price curve using data from reliable sources. Furthermore, compare the shape of these curves and make sense of the current projections of solar energy production for the next 10-20 years. Finally, find out for which country in the world the percentage of the energy it uses from the sun is maximum.
How much renewable energy do we use? Visit the Energy Information Administration to find U.S energy statistics (e.g. see and provide an answer to this question. Use the quantitative skills you acquired in Math 10260 to make your answer clear and informative.
Sustainable Development. How would you define the concept of sustainable development? Is the current socioeconomic system consistent with such a concept? If not, then provide some explanation.
The nearly $800 billion stimulus package. The following is from a speech given by President-elect Barack Obama on January 11, 2009, at George Mason University, to defend his stimulus package (text is taken from New York Times.): "I don't believe it's too late to change course, but it will be if we don't take dramatic action as soon as possible. If nothing is done, this recession could linger for years. The unemployment rate could reach double digits. Our economy could fall $1 trillion short of its full capacity, which translates into more than $12,000 in lost income for a family of four. We could lose a generation of potential and promise, as more young Americans are forced to forgo dreams of college or the chance to train for the jobs of the future. And our nation could lose the competitive edge that has served as a foundation for our strength and standing in the world...
I understand that some might be skeptical of this plan. Our government has already spent a good deal of money, but we haven't yet seen that translate into more jobs or higher incomes or renewed confidence in our economy. That's why the American Recovery and Reinvestment Plan won't just throw money at our problems - we'll invest in what works. The true test of the policies we'll pursue won't be whether they're Democratic or Republican ideas, but whether they create jobs, grow our economy, and put the American Dream within reach of the American people."
Using the materials we learned in Math10260 and your business knowledge, make a quantitative analysis of the proposed economic stimulus package. You may wish to explore its impact to the Deficit problem the country faces.
Mountains Beyond Mountains (preview). In this inspiring book. Tracey Kidder describes the quest of Dr. Paul Farmer, a man who would cure the world. Curing infectious diseases and bringing the lifesaving tools of modern medicine to those who need them most is his life calling. Read this book and use the mathematics you have learned in Math10260 to try to understand, analyze and propose possible solutions to the global health problem.
Universal Health Care. What are the benefits and problems of a universal health care system? Examine and compare the health care system of the U.S. and one or two from other developed countries like the U.K., Germany, France, Japan, etc.
The End of Poverty (preview). In the preface of this book its author Dr. Jeffrey Sachs (Quetelet Professor of Sustainable Development at Columbia University, Direct of the Earth Institute, and Director of the United Nations Millennium Project) writes: "When the end of poverty arrives, as it can and should in our generation, it will be citizens in a million communities in rich and poor countries alike, rather than a handful of political leaders, who will have turned the tide. The fight for the end of poverty is a fight that all of us must join in our own way." Read this very interesting book and use the mathematics you have learned in Math 10260 to try to understand (quantify, analyze) poverty as a world problem, and propose possible solutions that our generation can realize.
Top Ten. What are the top 10 major challenges for your generation? Provide some numbers to justify your choices.
The Paradox of Choice (preview). In this book, Barry Schwartz, among many other things, claims that freedom of choice can turn into a tyranny of choice. He even uses some math to make his point. For example, in pages 67-73 he uses familiar curves to give a general explanation of how we go about evaluating options and making decisions. Write a report on this very interesting book and try to relate it to ideas you learned in Math 10260.
Demand and Supply. Read carefully section 6.1 on consumer and producer surplus, compare it with writings in economics' literature, and explain how demand and supply are curves determined.
Flatland (read online). Imagine that you live in a plane (a 2D-space) and that you are not able to see 3D shapes. Then, think of ways for visualizing such shapes. A good source of ideas is the book "Flatland" by Edwin Abbott. Read this book and extend its ideas to describe how inhabitants of 3D-space (i.e., humans) could visualize 4D shapes.
A) Income distribution and Lorentz curves. The way that income is distributed throughout a given society is an important object of study for economists. The U.S. Census Bureau collects and analyzes income data, which it makes available at its website, In 2001, for instance, the poorest 20% of the U.S. population received 3.5% of the money income, while the richest 20% received 50.1% The cumulative proportions of population and income are shown in the following table:
Proportion of Population
Proportion of Income
0
0
0.20
0.035
0.40
0.123
0.60
0.268
0.80
0.499
1.00
1.00
For instance, the table shows that the lowest 40% of the population received 12.3% of the total income. We can think of the data in this table as being given by a functional equation y = f(x), where x is the cumulative proportion of the population and y is the cumulative proportion of income. For instance, f(0.60) = 0.268 and f(0.80) = 0.499. Such a function (or, more properly speaking, its graph) is called a Lorentz curve.
Show that f(x) = 0.1x + 0.9x2 is a possible Lorentz curve. Also, compute the income received by the lowest 0%, 50%, and 100% of the population.
Show that f(x) = 0.3x + 0.9x2 is not a Lorentz curve.
For the Lorentz curve in (i) show the following properties:
f(0) = 0, f(1) = 1, and 0 ≤ f(x) ≤ 1 for all 0 ≤ x ≤ 1,
f(x) is an increasing function,
f(x) ≤ x for all x, 0 ≤ x ≤ 1
Explain why properties (a) - (c) hold for every Lorentz curve.
Write many other different formulas for Lorentz curves.
Using real data, produce Lorentz curves for the U.S. in 2006
Sketch the graph of a Lorentz curve and compare it with the line y = z.
B) Coefficient of Inequality. If the Lorentz curve of a country is given by f(x) = x, then its total income is distributed equally. Otherwise, there are inequalities present in the distribution of income which are measured by the following number:
coefficient of inequality =
which is also called the Gini Index.
Compute the coefficient of inequality when f(x) = 0.1x + 0.9x2.
Show that the Gini Index is the ratio of the area of the region between y = f(x) and y = x to the area of the region under y = x, and provide an economic interpretation of this ratio.
Using real data estimate the Gini Index of the U.S in 2006.
A) The Cobb-Douglas Production Function. Show that the production function Q(K,L) having the properties:
(Marginal Product of Capital) * (Capital) = α * (output)
(Marginal Product of Labor) * (Labor) = (1-α) * (output)
for some constant α, 0 < α < 1, must be of the form Q(K,L)=KαL1-α, for some constant A. B) Read and understand the Solow Growth Model (Section 9.3) and do exercise 1, or 2, or 3 on page 607.
A) You are 35 years old and your company offers you the following three retirement plans:
At the beginning it deposits $50,000 into an account A and nothing more during the next 30 years.
For the next 30 years it deposits money continuously into an account B at a rate of $10,000 per year
At the age of 65, you will receive $1,2000,000 and nothing more during the next 30 years you will be working there
If the accounts A and B yield 8% interest, compounded continuously, which option will you choose? Explain your answer. B) Do part A again with interest rate at 10% compounded monthly. For Plan 2, assume that money will be deposited monthly into account B. To complete this part, you will have to set up a geometric series that gives the value of your retirement account. Go to your notes for continuous compounding and modify the setup for discrete compounding. Explain what each of the terms in the geometric series means. You should state clearly the first term, common ratio, and the formula you use to obtain the value of your retirement account from the geometric series. How would this change your decision in part A?
A) A homeowner takes out a 20-year mortgage with an interest rate of 5% compounded continuously. The homeowner plans to make payments totaling $1,500 per month. Let M(t) be the account owed after t years. Write an initial value problem modeling this situation. Then find the maximum amount of mortgage that the homeowner can afford. B) Do part A again with interest rate at 5% compounded quarterly. To complete this part, you will have to set up a geometric series that gives the value of the mortgage. Go to your notes for continuous compounding and modify the set up for discrete compounding. Explain what each of the terms in the geometric series means. You should state clearly the first term, common ratio, and the formula you use to obtain the mortgage value from the geometric series. (Hint: You should prorate the interest because you are paying monthly.)
Read carefully section 6.4 on population models and then do exercise 27 and 28 on page 445.
What does calculus have to do with change? The two central concepts in calculus are the derivative (instantaneous rate of change) and the integral (total change). Both are based on the fundamental calculus idea of "using elementary concepts (like slope of a line and area of a rectangle) to approximate advanced concepts (like slope of a curve and area enclosed by a curve)." Write in your own words the way you understand these concepts. Give examples from mathematics and its applications to demonstrate them. |
Calcula = THE CALCULATOR ... but not limited to the calculator. Calcula is not a scientific calculator. Calcula is a tool 'all-in-one': instead of having 1, 2, 5, 10, 20 applications that serve as 'technical means' we have only one: Calcula, indeed!So, what is and what makes Calcula?. calculator with the 4 operations, percentage, square root, exponentiation of x, a fraction of 1, form, and factor accumulation and subtraction in memory ... what they do all the calculators, some (not quite all, actually!). storing the list of all the transactions like a roll of paper with its zoom. button to cancel last input CI, C key to cancel the entire operation and key 'tearing paper' to delete all memorized transactions. selection of the number of decimal places, from 0 to 5, with which to develop. currency conversion online, in real time and then, leaning on a free service of common good (the result can be integrated in the operation in progress). conversion between many units of measurement: length, weight, volume, area, etc.. (The result can be integrated in the operation in progress). conversion between different number systems: decimal, octal, hexadecimal and binary (the result, of course decimal) can be integrated in the operation in progress). calculating perimeter, area and volume of many geometric shapes with a list of requests for images and input context to the figure (within the perimeter of the circle or to calculate the volume of the cylinder, or more or less according to your traps, etc..)(The result can be integrated in the operation in progress). expression processing up to 26 variables and many functions available, such as cos, sin, tan, etc.. (The result can be integrated in the operation in progress). development of algebraic proportions of the type: b = x: c-fit of the 3 known values and the processing of the result in 4 combinations (the result can be integrated in the operation in progress). generation of random numbers indicating the amount of numbers to be generated and the minimum and maximum limits (ability to select whether the numbers generated should all be different or with repetitions). elaborations of summations, differences between dates with even numbers add or subtract days. elaboration of summations, differences between zones with even add or subtract a preset time. stopwatch with lap times list the possibility of. flashlight (beam) with a selection of different colors. in cm and inch ruler, and color-changing background and calibration lines for even better viewing of the backlit. compass needle or rotary dial with digital indication of the degree. level graphics and digital indication of the degree of vertical tilt and horizontal. selection if it beeps when you press any buttons or voice with repetition of numbers typed and conducting operations in. ability to change the background color. appropriate option for the configuration settings. Detailed help on all aspects. Calcula the program is released with 2 screens, others are making and will be issued free of charge even after the purchase) to have more or fewer buttons then more or less the same size buttons. In version 1.1.00 there are 2 screens: the no. 0 with all the buttons available, some with 2 or 3 functions enabled via special button shift, the no. 1 instead of the calculator and all transactions with a button that serves as a menu to call up all the other functions.. all the screens are operated by the minimum resolution is 320x480 portrait or landscape (480x320). ON / OFF switch!The program is released in Italian, |
Understanding Intermediate Algebra - With CD - 6th edition
Summary: Lewis Hirsch and Alan Goodman strongly believe that students can understand what they are learning in algebra and why. The authors meticulously explain why things are done in a certain way, illustrate how and why concepts are related and demonstrate how 'new' topics are actually new applications of concepts already learned. The authors introduce topics at an elementary level and return to them at increasing levels of complexity. Their gradual introduction of concepts...show more, rules, and definitions through a wealth of illustrative examples - both numerical and algebraic-helps students compare and contrast related ideas and understand the sometimes subtle distinctions among a variety of situations. Through this learning this author team carefully prepares students to succeed in higher-level mathematics. ...show less
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$198.34 |
Introductory Algebra for College Students
The Blitzer Algebra Series combines mathematical accuracy with an engaging, friendly, and often fun presentation for maximum appeal. Blitzer's ...Show synopsisThe321759825-55982559825Fine. Hardcover. Instructor Edition: Same as student edition...Fine. Hardcover. Instructor Edition: Same as student edition with additional notes or answers. Almost new condition. SKU: 9780321759825Hardcover. Instructor Edition: Same as student edition with...Hardcover. Instructor Edition: Same as student edition with additional notes or answers. New Condition. SKU: 978032175982558958 May include moderately worn cover, writing,...Good. Hardcover. May include moderately worn cover, writing, markings or slight discoloration. SKU: 9780321758958Reviews of Introductory Algebra for College Students
tHIS IS NOT THE COVER Of the book. Its orange and has a bottle cap on the cover. BUUT this book is great. It clearly lists the steps and reasons for the math eq. and such. Although I DESPISE the ANSWER KEY in the back because it only lists the answers for ODD NUMBERS. Other than that, the condition is great and the text is as well
Unfortunately, this text was required for my class. I got to use the Lial series for PreAlgebra, and I will get to for Intermediate as well. They are much better for those who need more examples, description, worked problems, etc. This one assumes you know |
Elementary and Intermediate Algebra - 6th edition
Summary: Larson IS student success. ELEMENTARY AND INTERMEDIATE ALGEBRA: ALGEBRA WITHIN REACH owes its success to the hallmark features for which the Larson team is known: learning by example, a straightforward and accessible writing style, emphasis on visualization through the use of graphs to reinforce algebraic and numeric solutions and to interpret data, and comprehensive exercise sets. These pedagogical features are carefully coordinated to ensure that students are better able to make co...show morennections between mathematical concepts and understand the content. With a bright, appealing design, the new Sixth Edition builds on the Larson tradition of guided learning by incorporating a comprehensive range of student success materials to help develop students' proficiency and conceptual understanding of algebra. The text also continues coverage and integration of geometry in examples63.99 +$3.99 s/h
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128507467X used book - book appears to be recovered - has some used book stickers - free tracking number with every order. book may have some writing or highlighting, or used book stickers on front o...show morer back |
Engineering Computation with MATLAB
This textbook is ideal for MATLAB/Introduction to Programming courses in both Engineering and Computer Science departments. It is also a suitable ...Show synopsisThis textbook is ideal for MATLAB/Introduction to Programming courses in both Engineering and Computer Science departments. It is also a suitable introduction for beginning programmers. Engineering Computation with MATLAB introduces the power of computing to engineering students who have no programming experience. The book places the fundamental tenets of computer programming into the context of MATLAB, employing hands-on exercises, examples from the engineering industry, and a variety of core tools to increase programming proficiency and capability. With this knowledge, students are prepared to adapt learned concepts to other programming languages.Hide synopsis
Description:Good. Paperback. Missing components. May include moderately...Good. Paperback. Missing components. May include moderately worn cover, writing, markings or slight discoloration. SKU: 9780132568708-4-1-3 Orders ship the same or next business day. Expedited shipping within U.S. will arrive in 3-5 days. Hassle free 14 day return policy. Contact Customer Service for questions. ISBN: 9780132568708 0273769138 Brand new book. International Edition. Ship...New. 0273769138 |
Course Description: Algebraic geometry is the study of solutions to systems of polynomial equations. The algebra/geometry dictionary establishes a bijection between such solution sets and certain ideals in a polynomial ring. Hence, techniques from commutative algebra can be used to analyze these geometric objects. The course will provide an introduction to algebraic geometry. We will emphasize not only the theory, but also practical aspects of how to compute with polynomial ideals using Groebner bases. We will follow the book closely and discuss the following topics: Groebner bases, affine varieties, morphisms and rational maps, elimination theory, the Nullstellensatz, primary decomposition, projective varieties, Grassmannians, and Hilbert Functions.
Homework: Homework will be assigned regularly at the beginning of the semester and is due in class on Fridays. Students must write up their own solutions. Please indicate on your homework any sources that you used in preparing solutions (e.g. if another student helped with a solution, or you found the solution in a book). Students are encouraged to prepare homework solutions in LateX. A file explaining how to prepare your homework can be found here. Homework assignments can be found here.
Final Project: Students will be expected to prepare a final project. This will consist of critically reading a paper on a topic related to the material from this course or working on a research problem in algebraic geometry, preparing a paper describing your project, and giving a twenty minute presentation to the class on your project during the final weeks of class. Project topics will be chosen halfway through the course, based on some suggested topics on the course website and in discussion with the instructor. Suggested topics can be found here.
Exams: There will be no exams. However, students will be expected to attend the final exam period as this will be used for student final presentations (Friday, 12/10 8–11 AM).
Grades: Grades will be based on Homework (50%) and a Final Project and Presentation (50%).
Attendance: Students are expected to arrive on time, to contribute to group work and class discussions, and to stay until the class ends. Attendance at all meetings of the class is expected. Occasional absences will be approved if they meet University policies.
Adverse Weather: Announcements regarding scheduled delays or the closing of the University due to adverse weather conditions will be broadcast on local radio and television stations and posted on the University homepage.
H1N1 Policy:
If you are ill with symptoms of H1N1 influenza (i.e. fever over 100, sore throat, cough, stuffy or runny nose, fatigue, headache, body aches, vomiting and diarrhea) please do not come to class. Instead, immediately contact your medical provider or Student Health Services (515-7107) for advice or to arrange an appointment. If you are diagnosed with H1N1, please inform me immediately. You will be required to be isolated away from class for up to 7 days or 24 hours after symptoms subside, whichever is longest.
Cell Phones: Pagers, cellular phones and other types of telecommunication equipment are prohibited from use during class. Make sure that any pagers, phones or other equipment are turned off during the class period. If you have a special need to have your pager or phone on during class, please let me know.
Academic Integrity Statement: Students are required to follow the NCSU policy . "Academic dishonesty is the giving, taking, or presenting of information or material by a student that unethically or fraudulently aids oneself or another on any work which is to be considered in the determination of a grade or the completion of academic requirements or the enhancement of that student's record or academic career.'' (NCSU Code of Student Conduct). The Student Affairs website has more information.
Students with Disabilities: Reasonable accommodations will be made for students with verifiable disabilities. In order to take advantage of available accommodations, students must register with Disabilities Services for Students.
Class Evaluations: Online class evaluations will be available for students to complete during the last two weeks of class. Students will receive an email message directing them to a website where they can login using their Unity ID and complete evaluations. All evaluations are confidential; instructors will never know how any one student responded to any question, and students will never know the ratings for any particular instructors. |
Geometry, Probability and Statistics from an Elementary Point of View
General preparation in mathematics for elementary educators. Geometry with emphasis on shapes and measurement, transformational geometry and symmetry, logic, counting methods, permutations and combinations, an introduction to probability and statistics. Prerequisite: MA 110 or equivalent. (Placement test required) |
9780739885Core Skills: Math Review and Algebra prepare students to master the skills and concepts found on most state standardized tests. This book provides plenty of examples and practice for reinforcement. A preview test determines what the student does or does not know, unit reviews at the end of each unit test what was learned, and a final review compares against the preview test to measure improvement. Bound-in answer key |
Introduction to Topology We will study the basic theory and the topological properties of the Möbius bands, the torus and the Klein bottles to explain why a topologist cannot distinguish between a doughnut and a tea cup.
You Can Count on It - Maths in Finance In this brief course we shall look at how mathematics contributes to finance and business. Our course is suitable for people with previous experience of mathematics at the sixth-form level and aims to provide an elementary introduction to the mathematics.
Puzzles and Pastimes Puzzles and pastimes can be both amazing and amusing. Mathematical puzzles often seem like magic. Mechanical examples include Rubik's cube. Other topics include board and card games, game theory (hawks and doves) and Sudoku..
Alternatively you can perform a keyword search on all our courses using the 'Find courses' box on this page. |
has evolved to address the needs of today's student. While maintaining its unique table of contents and functions-based approach, the text now includes additional components to build skill, address critical thinking, solve applications, and apply technology to support traditional algebraic solutions. It continues to incorporate an open design, helpful features, careful explanations of topics, and a comprehensive package of supplements and study aids to provide new and relevant opportunities for learning and teaching. |
Comments
Can you be more specific about which form of math you are questioning? If you just mean math in general, there are countless practical reasons for improving your math skills, many involving money. Arithmetic is vital and algebra is really just a symbolic
form of arithmetic. The most important math skill that you will use--assuming you aren't going into a science/medical/engineering field--is the ability to visualize a story problem (a real life situation) algebraically so it can be reduce to simple arithmetic.
If you could give a specific example, I'm sure someone can counter with an explanation of a practical use for that type of math. I know when I was a kid and asked, "What's the point of all this?" it only frustrated me further when I was given a pep talk
about building character and what not. (The pep talks are all still true, but don't answer the fundamental question of what it all means.) What I wanted to know was, "How is this skill applied in the real world?" However, without knowing which specific skill
you are asking about, we cannot be more specific in our answers.
6 Answers
Melissa G.
If your teachers have not yet shown you how what you are learning is likely to be relevant to your life in the future, then it might be worth asking them directly - probably best done in office hours instead of putting them on the spot in the classroom.
As educators, we each have our own reasons for believing what we are teaching is really worth learning. In my classes, I focus a lot on helping make the learning activities and materials personally relevant for my students. As a student, when I was taking
a class I truly thought was a waste of time, I would figure out what I had to do minimally to make the grade I wanted so that at least I wouldn't see a drop in my GPA.
I regularly advise my students to look hard at their options for instructors; get recommendations on who to take and who to avoid. The value of each of your courses is so dependent on how well an instructor's style and teaching methods work for you. For
those in college, I highly recommend going to as many "first day of class" sessions as you possibly can during the first week of classes (even the full ones), looking for which instructors you think you could get the most value from. Almost all the college
instructors I know would do what they could to add students to their rosters if the students took initiative and showed up more than once.
I will also echo the statements others posted above - your education is what YOU make of it. The more clear you are about your own life goals, the more easily you can see whether what you are learning can be relevant or not. One key is to look at whether
skills you are learning will matter outside the class you are learning them in. Math is often the most tricky - but you should be able to look at your goals and figure out which types of math are going to be the most important to put real effort into mastering.
I took AP Calculus in HS and more advanced math in college but have almost never used them. What I wish I'd taken earlier was statistics - that is relevant every day.
Frank B.
When I was your age I thought exactly the same thing. I thought math was suppose to be numbers, why were there letters involved.
But when you grow up, get a job and need to support a family (which by the way is not as far off as you think) time passes quickly, it becomes very important to know math, all types of math, as you never know where your life may take you.
Chef's use math in recipe's cooking, construction workers use math when building, manufacturing employees use math when producing the products we purchase at the stores everyday, even clerks at the retail stores use math when exchanging money. We all use
math when we get our paychecks and figure out our bills.
Heck when I became a freight manager I needed to use math to figure out fuel surcharges on shipments. At 45 I decided to go back to school and get my college degrees - YES degrees - one in CIS (Computer Information Systems) Network Security and the other
in Accounting.
If you think that you are in hard math now -- wait until you try Binary or Hexadecimal math for computers, or when you start doing ratios in accounting.
You never know where life will take you, when I was your age I never thought that I would obtain degrees in Accounting or CIS but at the age of 45 I went back to college and at 51 walked down the isle to graduate.
Good luck and contact me if I can help you in anyway. I know how you feel sort of been there did that kind of thing and rose above it.
Kevin S.
Unfortunately, no one can see into the future. School is trying to prepare you for what you may encounter in later years. As Thomas indicated, there are some daily applications of basic math. Some other daily applications are balancing your checkbook, knowing
how much a reasonable tip is at a restaurant (especially if you're the server), completing your income taxes. Which is a better buy - four 20 ounce sodas at $1.59 each or a two liter bottle at $3.00 ?
Math is used to explain and predict in the sciences - physics, chemistry, etc. Statistics are used in Biology, Insurance, and many other fields. Calculus is used by police departments for accident reconstruction. If you're looking at a career in the medical
field, math is essential - if you can't convert from cubic centimeters to milliliters to ounces on the fly, you could administer an incorrect dose of a medication. Trigonometry is used in construction and engineering.
Math is one of those topics that you can't get away from - it's all around you. Distances, weights and measures, money, even time. And sometimes it just takes a little extra effort (and the right teacher or tutor) to help explain topics a little differently.
Thomas W. a new stylish car. Lesser or less complicated forms of math could help your figure out your MPG, to ensure your engine is running efficiently, or how much money you'll take home from a days worth of work.
Ultimately, it's not about what good is math? Math is used everywhere. The question you should ask yourself is: how will I use or incorporate math into my life? This question is entirely up to you and your potential.
Robert Z.
There are a lot of high-paying careers in the sciences that require you have a good knowledge of math. I work at a transportation agency and I can tell you, civil engineers use math all the time. In order to get your engineering license, you have to take
two exams which require you be proficient in higher-level math. |
Synopses & Reviews
Publisher Comments:
Fractals for the Classroom breaks new ground as it brings an exciting branch of mathematics into the classroom. The book is a collection of independent chapters on the major concepts related to the science and mathematics of fractals. Written at the mathematical level of an advanced secondary student, Fractals for the Classroom includes many fascinating insights for the classroom teacher and integrates illustrations from a wide variety of applications with an enjoyable text to help bring the concepts alive and make them understandable to the average reader. This book will have a tremendous impact upon teachers, students, and the mathematics education of the general public. With the forthcoming companion materials, including four books on strategic classroom activities and lessons with interactive computer software, this package will be unparalleled.
Book News Annotation:
Not a textbook, and not just for the classroom, this book which comes in two parts, is a collection of largely independent chapters on the major concepts related to the science and mathematics of fractals, chaos, and dynamics, presented at the mathematical level of an advanced secondary school student. A wide range of illustrations (including 14 color plates) are integrated with an enjoyable text, bringing the concepts alive and making them understandable to a large audience. Annotation c. Book News, Inc., Portland, OR (booknews.com)
Synopsis:
"Synopsis"
by Ingram, |
Prealgebra - 4th edition
Addressing individual learning styles,Tom Carsonpresents targeted learning strategies and a complete study system to guide students to success. Carson's Study System, presented in the ''To the Student'' section at the front of the text, adapts to the way each student learns, and targeted learning strategies are presented throughout the book to guide students to success. Tom speaks to students in everyday language and walks them through the concepts, ex...show moreplaining not only how to do the math, but also where the concepts come from and why they work98.99 +$3.99 s/h
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Overview
The Advantage Math Series helps prepare students to better understand mathematics and acquire and apply the skills they need to solve problems they encounter in their daily lives. Students will receive instruction and practice in key skills for all strands of the national ...
More About
This Book
Overview
The Advantage Math Series helps prepare students to better understand mathematics and acquire and apply the skills they need to solve problems they encounter in their daily lives. Students will receive instruction and practice in key skills for all strands of the national curriculum math standards, including:
numeration and number theory
operations
geometry
measurement, patterns, functions, and algebra
data analysis and probability
problem solving.
The series offers strong skill instruction along with motivational features in an easy-to-use format.
Skill Review and Reinforcement If you choose a grade-level book equal to the grade your child is currently in or has just finished, use the book to:
reinforce skills being taught in the classroom throughout the year
keep your child's skills fresh over the summer
Extra Challenge and Enrichment If you choose a grade level book beyond the grade your child is currently in or has just finished, use the book to:
provide enrichment and an extra challenge during the school year
give your child a summer preview of what's to come in school next year
Whether you use this resource for your child to practice familiar skills or to learn new ones, the extra advantage will make a difference |
Precalculus: Mathematics for Calculus
This best selling author team explains concepts simply and clearly, without glossing over difficult points. Problem solving and mathematical modeling ...Show synopsisThis best selling author team explains concepts simply and clearly, without glossing over difficult points. Problem solving and mathematical modeling are introduced early and reinforced throughout, providing students with a solid foundation in the principles of mathematical thinking. Comprehensive and evenly paced, this book provides complete coverage of the function concept, and integrates a significant amount of graphing calculator material to help students develop insight into mathematical ideas. The authors' attention to detail and clarity, the same as found in James Stewart's market-leading Calculus text, is what makes this text the market leader Package. New Condition. SKU: 9780495319733-1-0-3...Paperback. Package. New Condition. SKU: 978049531975319733.
Description:New. 0534434215 AtAGlance Books--Orders ship next business day,...New. 0534434215 AtAGlance Books--Orders ship next business day, with tracking numbers, from our warehouse in upstate NY. This book is in brand new condition.
Reviews of Precalculus: Mathematics for Calculus
Book is actually pretty good, I like the fact that its first chapter is a review of stuff from intermediate algebra...if you are using this book...and you have just finished intermediate/college algebra...get this book and do each section of ch. 1...this will set you up for a good foundation for the rest of the class. I like the visuals of the book...I have the most recent edition.
A beautiful presentation and treatment of all math required before studying calculus. Comprehensive, and a strong focus on theory. Lots of problems to test yourself. Get through this and then star in your calculus study, as you are now VERY well prepared.
This book is one of the best out there in the current markets. Dr. Stewart explains this subject with geometrical shapes to better understand the subject. For example he explains and proves the the phytogorean theorem, laws of sines and cosines, and alot more. Highly recommend this book as well as his other ones like college algebra and calculus. He is an expert in this subjects. The best way to master a theorem in math is to prove and applied its existence and this book aims at teaching this principles |
Complex Analysis
9780387950693
ISBN:
0387950699
Pub Date: 2001 Publisher: Springer Verlag
Summary: The book provides an introduction to complex analysis for students with some familiarity with complex numbers from high school. The first part comprises the basic core of a course in complex analysis for junior and senior undergraduates. The second part includes various more specialized topics as the argument principle the Poisson integral, and the Riemann mapping theorem. The third part consists of a selection of to...pics designed to complete the coverage of all background necessary for passing Ph.D. qualifying exams in complex analysis.
Gamelin, Theodore W. is the author of Complex Analysis, published 2001 under ISBN 9780387950693 and 0387950699. Six hundred sixty seven Complex Analysis textbooks are available for sale on ValoreBooks.com, one hundred fourteen used from the cheapest price of $26.16, or buy new starting at $55.23 An introduction to complex analysis for students with some knowledge of complex numbers from high school. It contains sixteen chapters, the fi [more]
This item is printed on demand. An introduction to complex analysis for students with some knowledge of complex numbers from high school. It contains sixteen chapters, the first eleven of which are aimed at an upper division undergraduate audience. The r |
97805217819Calculus, Second Edition
Gilbert Strang's clear, direct style and detailed, intensive explanations make this textbook ideal as both a course companion and for self-study. Single variable and multivariable calculus are covered in depth. Key examples of the application of calculus to areas such as physics, engineering and economics are included in order to enhance students' understanding. New to the second edition is a chapter on the 'Highlights of Calculus', which accompanies the popular video lectures by the author on MIT's OpenCourseWare. (These can be accessed from math.mit.edu/~gs |
Mathematics
Precise Calculator has arbitrary precision and can calculate with complex numbers, fractions, vectors and matrices. Has more than 150 mathematical functions and statistical functions and is programmable (if, goto, print, return, for).
KinetDS is a software for curve fitting particularly designed for kinetic (mechanistic and empirical) description of a substance dissolution from solid state. It was primarily designed for handling pharmaceutical dissolution tests
Expression Solver computes the value of a mathematical equation/expression. Works with operators, numbers, variables, and functions. Users can define their own variables and functions. hosted: |
-a-Graph - Brain Waves Software
A graphing program for use in the teaching and learning of high school mathematics and Calculus. Features include the ability to transform any
relation, take the nth derivative of any function, and compose any two functions in nine different ways. TheZebragraph - Jonathan Choate
Download Choate's conference speeches, lecture notes, and other resources such as his Exeter courses and talks. "Enriching the Traditional Geometry Curriculum Syllabus" features the Problem of Regiomontanus. Investigations include the tetrahedral geometry
...more>>
Zeno's Paradoxes - Franz Kiekeben
Statements of the following paradoxes: The Racetrack (or Dichotomy); Achilles and the Tortoise; The Arrow; A brief analysis of the motion paradoxes; The Thomson Lamp; Kiekeben's Odd/Even Paradox; Benardete's Paradox of the Gods;
The Paradox of the Spaceship;entralblatt MATH Database - Springer-Verlag
A database edited by the European Mathematical Society, the FIZ Karlsruhe, and the Heidelberg Academy of Sciences, and established in cooperation with Math Doc Cell (France). Search for abstracts of papers on any math subject by author, title, classification,
...more>>
Zero-Knowledge Proofs - John Scammell
"Random thoughts about teaching math" from a high school math consultant who taught math for 18 years prior. Scammell's blog posts, which date back to May, 2010, have included "Lecture Method vs. Peer Instruction," "Mozartís Dice Game," "Let the Kidsome System - Zometool, Inc.
Available in kits of various sizes, Zometool is a 31-zone design tool and educational toy for kids, hobbyists, students, teachers and researchers. Precision-molded in ABS plastic, Zometool models spatial structures representing hyperspaces of up to 31
...more>>
Zoom Math - I.Q. Joe
The makers of Zoom Algebra, an application with an intuitive visual interface and numerous typing shortcuts. Their apps for TI-83 Plus or TI-84 Plus graphing calculators let you "type problems in just the way they look in your textbook, and then they
...more>>
ZooWhiz - EdAlive
A curriculum-linked learning system for parents and teachers of children ages 5-15. Sample fun, colorful learning activities for math, word skills, and reading before registering a free "keeper" account or purchasing a premium "zoologist" subscription.
...more>>
Z's Roguelike Stuff - Zeno Rogue
Mathematical games and projects include HyperRogue, a treasure-collecting, goblin-killing adventure set in non-Euclidean space -- specifically, a hyperbolic plane of negative curvature filled with elixirs and slime beasts; and Hydra Slayer, which begins
...more>>
Zêtre - Arte Merritt
A card game for the whole family that combines math, an element of chance, and a time limit. Players can compete against each other, against the clock, or against themselves. Zetre helps enhance players' math skills, including their knowledge of the order
...more>> |
Overview
Problem solving practice, writing opportunities, test prep and reading strategies for analyzing word problems are all offered in this convenient, consumable math workbook that relates back to the lessons covered in the Student Edition.
More About
This Book
Overview
Problem solving practice, writing opportunities, test prep and reading strategies for analyzing word problems are all offered in this convenient, consumable math workbook that relates back to the lessons covered in the Student |
Matlab: An Introduction WithMATLAB: An Introduction with Applications 5th Edition walks readers through the ins and outs of this powerful software for technical computing. The text describes basic features of the program and shows how to use it in simple arithmetic operations with scalars. The topic of arrays (the basis of MATLAB) is examined, along with a wide range of other applications. MATLAB: An Introduction with Applications 5th Edition is presented gradually and in great detail, generously illustrated through computer screen shots and step-by-step tutorials, and applied in problems in mathematics, science, and engineering |
Lesson
Title: Math/Algebra/Test S2:C4:PO1;S3:C3:PO1,PO2
2.
Guide, when your student has completed the test, have
him/her go to and
go to the Vertex-Edge Graphs (Color Maps) lesson (d.)
in Section 3. Data Analysis, Probability & Discrete
Mathematics. Your student should read the lesson and
then print the worksheet (Session Type: Printable Worksheet)
and complete 15 problems. Go over your student's work
with the printable answer sheet and explanations.
3. Guide, when your student is ready, have him/her complete 15 questions
in test mode in for
the Vertex-Edge Graphs lesson.
4. Guide, have your student
play one of the games in from
the Number Sentences lesson (b.) of Section 4. Patterns, Algebra & Functions.
When your student has completed with the game practice, have him/her
go on to test mode and complete 15 questions in this section of
studyisland. |
Everything about linear functions is contained in the Special Topic titled
Linear Functions listed under Special Topics.
It is strongly recommented that you study that document.
A few other miscellaneous formulas related to lines are
presented in this section of the text and are repeated here
Recommended Exercises Page 14619-28, 29-40, 41-48, 53-70.
Recommended Exercises Page 167 3, 13-18, 19-30, 31-46, 47-56.
Section 2.3: Quadratic Functions (page 172)
Everything about quadratic functions is contained in the Special Topic titled
Quadratic Functions listed under Special Topics.
It is strongly recommented that you study that document.
Recommended Exercises Page 186 41-48, 49, 51, 53.
Section 2.4: omitted --- Complex Numbers (page 190)
Be sure to study and understand Figure 2 on Page 194.
Be sure to study and understand the "Caution" on Page 202.
Recommended Exercises Page 204 5, 6, 7-12, 13-46.
Section 2.5: Quadratic Functions and Models (Page 206)
Be sure you can solve quadratic equations using factoring and the Zero Factor Property or with the Quadratic Formula.
Recommended Exercises Page 221 1-6, 19-26, 89
1, 3, 4, 5, 6, 7-16, 29-38, 51-60, 65-80.
Recommended Exercises Page 239-8, 19, 20, 21, 22, 27, 32, 57, 58.
Section 2.7: Solving Inequalities (page 210)
Properties of Inequalities:
The remaining material is presented as a review of absolute value. It is more extensive that the text coverage of this topic and is presented here for you convenience.
Other Inequalities
Click on the following link for a discussion of other kinds of inequalities.
The basis for solving all equations involving absolute values is the definition of absolute value.
Notice the precise definition of absolute value has two cases:
Case 1: The expression inside the absolute value symbol is positive or zero.
Case 2: The expression inside the absolute value symbol is negative.
Procedure:
To
solve equations involving absolute values of variable expressions, it
is necessary to solve the two equations which naturally result from the definition
of absolute value. The solution set contains the solution set of the original
equation.
Therefore an integral part of the solution process is to test each of the possible
solutions in the original equation.
Every equation involving absolute value is solved by considering the two cases as in this general example: Generic Example: To solve an equation involving |something|, two cases must be considered. The two cases arise from the definition
of absolute value. Therefore to solve any equation involving |something|, we consider:
Case 1: The equation that results from relacing |something| with (something)
Case 2: The equation that results from relacing |something| with -(something).
The following two properties of equations are also important tools when solving equations involving
absolute value.
These two properties might be used to:
Simplify the equation before considering the two cases.
Solve the equations in Case 1 and Case 2.
In this section of the textbook the expresion inside the absolute value symbol is
always a linear expresion in one variable. All of these equations can
be solved using the two cases that come from the definition of absolute value together with
the two properties for generating equivalent equations.
Solving an equation of the type |ax + b| = c.
Consider the two cases.
Case 1: (when ax + b is positive or zero) Solve the equation ax + b = c.
Case 2: (when ax + b is negative) Solve the equation -(ax + b) = + c = d.
Add the expression -c to both sides to obtain the equivalent equation
|ax + b| = d - c.
Consider the two cases.
Case 1: (when ax + b is positive or zero) Solve the equation ax + b = d - c.
Case 2: (when ax + b is negative) Solve the equation -(ax + b) = d - = |cx + d|.
Consider the two cases.
Case 1: (when ax + b is positive or zero) Solve the equation ax + b = cx + d.
Case 2: (when ax + b is negative) Solve the equation -(ax + b) = cx + d.
In each case, the two properties for generating equivalent equations may be used to find a simplest
equation.
The union of the solution sets for the two cases is the solution set for the original equation.
It may seem that there should be four cases in this last type of equation, but there are duplications
as shown here.
Case 1: (when ax + b is positive or zero) Solve the equation ax + b = |cx + d|.
Case 1A: (when cx + d is positive or zero) Solve the equation ax + b = cx + d.
Case 1B: (when cx + d is negative) Solve the equation ax + b = -(cx + d).
Case 2: (when ax + b is negative) Solve the equation -(ax + b) = |cx + d|.
Case 2A: (when cx + d is positive or zero) Solve the equation -(ax + b) = cx + d.
Case 2B: (when cx + d is negative) Solve the equation -(ax + b) = -(cx + d).
If both sides of the equation in Case 2B is multiplied by -1, the equivalent equation obtained is identical to the
equation in Case 1A.
If both sides of the equation in Case 2A is multiplied by -1, the equivalent equation obtained is identical to the
equation in Case 1B.
For that reason only two cases are necessary.
Solving an equation of the type |ax + b| = c where c is a negative number.
Simply observe that absolute value is always non-negative (positive or zero)
and therefore this type of equation does not have a solution. It solution set is the empty set.
Solving an equation of the type |ax + b| = 0.
The only way an absolute value can be zero is for the expression inside the
absolute value symbol to be zero. So only Case 1 needs to be considered. No harm is done if Case 2 is considered,
because in this type of equation it will always give the same solution set as Case 1.
Definition: Two inequalities joined by the words AND or OR are called compound inequalities. FACT: The solution set
of a compound inequality formed by the word AND is the intersection of the solution sets of the two inequalities. FACT:The solution set
of a compound inequality formed by the word OR is the union of the solution sets of the two inequalities.
Warning: When using the compact form of a compound inequality such as a < x < b,
be careful to write only meaningful statements. The inequalities must both "point"
in the same direction. The end expressions must be related as indicated by the
inequality symbols. For example, in a < x < b, it must be true that a
< b.
FACT: If k is a positive number and X is either a single variable or a variable expression,
then the inequality |
X | < k is equivalent to - k < X < k
Note this is a compound inequality formed with the word AND. Therefore the solution
set is the intersection of the solution sets of the two individual inequalities.
FACT: If k is a positive number and X is either a single variable or a variable expression,
then the inequality | X | > k is equivalent to X < - k OR X > k
Note this is a compound inequality formed with the word OR. Therefore the solution
set is the union of the solution sets of the two individual inequalities.
Recommended Exercises Page 254 25-38, 39-64.
Copyright 2007 by
All Rights Reserved. Use of text, images and other content on this website are subject to the terms and conditions specified on our Copyright and Fair Use page. |
EnglishFirst published in 1976, this book has been widely acclaimed as a major and enlivening contribution to the history of mathematics. The updated and corrected paperback contains extracts from the original writings of mathematicians who contributed to the foundations of graph theory. The author's commentary links each piece historically and frames the whole with explanations of the relevant mathematical terminology and notation.
The only text available on graph theory at the freshman/sophomore level, it covers properties of graphs, presents numerous algorithms, and describes actual applications to chemistry, genetics, music, linguistics, control theory and the social sciences. Illustrated.
Get in on the next sudoku trend! Godoku—or sudoku word puzzles—are already beginning to appear in national newspapers. They offer an alphabetical variation on the conventional version, complete with a hidden word thats revealed only when the puzzles complete. Anyone who knows sudoku will catch on quickly, because the games almost identical: the only difference is that instead of nine numbers, godoku has nine letters. And the grid still can be solved by logic alone. Some are straightforward, some are super-challenging, and all the words are arranged by category. A comprehensive introduction explains how these devilish new puzzles work.
More editions of Hidden Word Sudoku: The Last Word in Sudoku Puzzles! (52 Brilliant Ideas):
In recent years graph theory has emerged as a subject in its own right, as well as being an important mathematical tool in such diverse subjects as operational research, chemistry, sociology and genetics. Robin Wilsons book has been widely used as a text for undergraduate courses in mathematics, computer science and economics, and as a readable introduction to the subject for non-mathematicians.
The opening chapters provide a basic foundation course, containing definitions and examples, connectedness, Eulerian and Hamiltonian paths and cycles, and trees, with a range of applications. This is followed by two chapters on planar graphs and colouring, with special reference to the four-colour theorem. The next chapter deals with transversal theory and connectivity, with applications to network flows. A final chapter on matroid theory ties together material from earlier chapters, and an appendix discusses algorithms and their efficiency.
Includes a large number of examples, problems and exercises.
More editions of Graphs and Applications: An Introductory Approach (with CD-ROM):
The astonishing variety and beauty of mathematical elements in stamp design is brought to life in this collection of more than 350 stamps, illustrated with mathematical figures, people, and content, each reproduced in enlarged format, in full color. It's a perfect gift book for anyone interested in stamps, or in the surprising use of mathematics in the real world. The author is widely known in the math community for his regular column on stamps in the magazine The Mathematical Intelligencer. |
lesson received an honorable mention for the 2011 SoftChalk Lesson Challenge.'Differential equations show up in many...
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This lesson received an honorable mention for the 2011 SoftChalk Lesson Challenge.'Differential equations show up in many areas of science and technology. In fact, they turn up any time there is a relation involving some continuously varying quantities and ther rates of change. We have actually dealt with differntial equations before. A common modeling problem involving differential equations is the determination of the velocity of a ball falling which has an acceleration which is the acceleration due to gravity minus the acceleration due to air resistance. This is a differential equation because the derivative of the velocity of the ball depends on the velocity, thus finding the velocity as a function of time involves.' solving a differential equation.'In this section we willExamine the basic form of differential equationsVerify solutions to differential equationsDetermine slope fields for differential equationsFind solutions to differential equations numerically using Euler's methodFind solutions to differential equations using seperation of variables.'
Complete searcher for scientific material. Many papers and reviews with purchase or full free. Simple search and options for...
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Complete searcher for scientific material. Many papers and reviews with purchase or full free. Simple search and options for complete and acqurate searching. Excellent performance and easy tool for use.
This course introduces the student to the study of linear algebra. Practically every modern technology relies on linear...
see more
This course introduces the student to the study of linear algebra. Practically every modern technology relies on linear algebra to simplify the computations required for internet searches, 3-D animation, coordination of safety systems, financial trading, air traffic control, and everything in between. This free course may be completed online at any time. See course site for detailed overview and learning outcomes. (Computer Science 105, Mathematics 211)
This website is a lesson plan to help teachers make computer science and engineering fun for students. By stating the...
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This website is a lesson plan to help teachers make computer science and engineering fun for students. By stating the problem and need for making CSE fun, letting teachers know how it relates to real-world situations, and giving 4 objectives (performing skills, recalling facts, identifing concepts, & applying principles)This lesson plan will also have content of basic ideas to make CSE more interesting for students. It will have evaluations of test and/or observations for skills learned and technology applied. Lastly, this website will give teachers/educators a method using motivation, socialization, audience this lesson is geared towards, and the technology needs to successfully complete this lesson |
Geometry : a high school course by Serge Lang(
Book
) 40
editions published
between
1983
and
2011
in
3
languages
and held by
601 WorldCat member
libraries
worldwide
From the reviews: "A prominent research mathematician and a high school teacher have combined their efforts in order to produce a high school geometry course. The result is a challenging, vividly written volume which offers a broader treatment than the traditional Euclidean one, but which preserves its pedagogical virtues. The material included has been judiciously selected: some traditional items have been omitted, while emphasis has been laid on topics which relate the geometry course to the mathematics that precedes and follows. The exposition is clear and precise, while avoiding pedantry. There are many exercises, quite a number of them not routine. The exposition falls into twelve chapters: 1. Distance and Angles.- 2. Coordinates.- 3. Area and the Pythagoras Theorem.- 4. The Distance Formula.- 5. Some Applications of Right Triangles.- 6. Polygons.- 7. Congruent Triangles.- 8. Dilatations and Similarities.- 9. Volumes.- 10. Vectors and Dot Product.- 11. Transformations.- 12. Isometries. This excellent text, presenting elementary geometry in a manner fully corresponding to the requirements of modern mathematics, will certainly obtain well-merited popularity. Publicationes Mathematicae Debrecen#1
Geometry by Serge Lang(
Book
) 3
editions published
between
1997
and
2010
in
English and German
and held by
10 WorldCat member
libraries
worldwide |
Algebra Antics
Presenting algebra exercises in which FUN is the "unknown quantity!" Each page features 12 to 24 skill-building algebra problems. After students have simplified expressions or solved equations, the answers provide clues for drawing lines to reveal a secret picture in the coordinate grid. Algebra Antics is a unique and fresh approach to algebra practice; great for students who enjoy visual challenges and direct feedback.
for Algebra Antics
Author: Barbara Smead Date: 01/18/2014 Demographic: Parent Rating: 4 out of 5 stars My daughter has had a great time working the book. She has almost finished. Not bad for a non math freak!
Author: Sheryl Reed Date: 01/15/2014 Demographic: Parent Rating: 5 out of 5 stars She has done several pages already and really loves it. Great car activist when there is no homework.
Author: LS Date: 11/07/2013 Demographic: Teacher Rating: 5 out of 5 stars My middle-school students love this book's way of combining hard problems with a picture. I love how the books make my students practice difficult pre-algebra problems and graphing coordinates, so their brains have to switch gears and thereby make stronger memories. |
This study probes the thinking of students at different stages of mathematical experience: college students who have taken calculus; college students who have not taken calculus; current high school mathematics teachers; graduate students in a discipline-based mathematics education program. The study asks:
what is the nature of student thinking when solving a distance/time graph problem?
do students with different levels of mathematical experience solve graph problems differently from each other? |
07682025Pre-Algebra Made Simple, Middle School (Teaching Resource)
Now it's simple to make Pre-Algebra fun, relevant, interesting, and exciting. This book is designed to help students develop a basic understanding of algebraic concepts using everyday applications. Includes activities on whole numbers and integers, solving equations, geometry, logic, problem solving and patterning, and statistics and probability. Background information, extension activities, group learning, and school-home connections are provided along with an answer key |
Mathematics A Discrete Introduction
9780534398989
ISBN:
0534398987
Edition: 2 Pub Date: 2005 Publisher: Thomson Learning
Summary: With a wealth of learning aids and a clear presentation, this book teaches students not only how to write proofs, but how to think clearly and present cases logically beyond this course. All the material is directly applicable to computer science and engineering, but it is presented from a mathematician's perspective.
Scheinerman, Edward R. is the author of Mathematics A Discrete Introduction, published 2005... under ISBN 9780534398989 and 0534398987. Three hundred seventy five Mathematics A Discrete Introduction textbooks are available for sale on ValoreBooks.com, one hundred twenty four used from the cheapest price of $14.85, or buy new starting at $55.19.[read more]
Ships From:Salem, ORShipping:Standard, ExpeditedComments:Almost new condition. SKU:9780534398989-2-0-3 Orders ship the same or next business day. Expedite... [more] |
authors have created a book to really help students visualize mathematics for better comprehension. By creating algebraic visual side-by-sides to solve various problems in the examples, the authors show students the relationship of the algebraic solution with the visual, often graphical, solution. In addition, the authors have added a variety of new tools to help students better use the book for maximum effectiveness to not only pass the course, but truly understand the material. |
Geometry Workbook For Dummies
9780471799405
ISBN:
0471799408
Pub Date: 2006 Publisher: Wiley & Sons, Incorporated, John
Summary: From proofs to polygons -- solve geometry problems with ease Got a grasp on the terms and concepts you need to know, but get lost halfway through a problem or worse yet, not know where to begin? No fear -- this hands-on guide focuses on helping you solve the many types of geometry problems you encounter in a focused, step-by-step manner. With just enough refresher explanations before each set of problems, you'll shar...pen your skills and improve your performance. You'll see how to work with proofs, theorems, triangles, circles, formulas, 3-D figures, and more! 100s of Problems! Step-by-step answer sets clearly identify where you went wrong (or right) with a problem Get the inside scoop on geometry shortcuts and strategies Know where to begin and how to solve the most common problems Use geometry in practical applications with confidence
Ryan, Mark is the author of Geometry Workbook For Dummies, published 2006 under ISBN 9780471799405 and 0471799408. Two hundred eighty three Geometry Workbook For Dummies textbooks are available for sale on ValoreBooks.com, one hundred twenty |
Learning--and teaching!--math does not have to be difficult! Give your students and yourself the tools to succeed with this Saxon Teacher and Saxon 8/7 kit combination! Introduce your middle-schoolers to the concepts they'll need for upper-level algebra and geometry, including functions and coordinate graphing; integers; multiplying decimals and fractions; radius, circumference, and pi; and more. This kit includes Saxon's 3rd Edition Math 8/7 textbook, solutions manual, and tests/worksheets book.
Saxon Teacher provides support and reinforcement are included.
Content of CD-Rom's
We have used the Teaching Tape CD's which are great, but much more expensive. What content is on the Saxon CD-Rom? Does it include instruction to all lessons, lesson practice and investigations? Will the CD-Rom's play on the DVD player or must it be the computer?
The Saxon CD-Rom includes comprehensive lesson instructions. It features complete solutions to every practice problem, problem set, and test problem with step-by-step explanations and helpful hints. It is to be used on a computer. For more details, see description at CBD Stock # WW773667. |
Computer Graphics - First Mathematical Steps will help students to master basic Computer Graphics and the mathematical concepts which underlie this subject. They will be led to develop their own skills, and appreciate Computer Graphics techniques in both two and three dimensions. The presentation of the text is methodical, systematic and gently paced - everything translates into numbers and simple ideas. Sometimes students experience difficulty in understanding some of the mathematics in standard Computer Graphics books; this book can serve as a good introduction to more advanced texts. It starts from first principles and is sympathetically written for those with a limited mathematical background. Computer Graphics - First Mathematical Steps is suitable for supporting undergraduate programmes in Computers and also the newer areas of Computer Graphics and Visualization. It is appropriate for post-graduate conversion courses which develop expertise in Computer Graphics and CAD. It can also be used for enrichment topics for high-flying pre-college students, and for refresher/enhancement courses for computer graphics technicians. |
Shipping prices may be approximate. Please verify cost before checkout.
About the book:
This book aims to teach the methods of numerical computing, and as such it is a practical reference and textbook for anyone using numerical analysis. The authors provide the techniques and computer programs needed for analysis and also advice on which techniques should be used for solving certain types of problems. The authors assume the reader is mathematically literate and is familiar with FORTRAN and PASCAL programming languages, but no prior experience with numerical analysis or numerical methods is assumed. The book includes all the standard topics of numerical analysis (linear equations, interpolation and extrapolation, integration, nonlinear rootfinding, eigensystems and ordinary differential equations). The programs in the book are in ANSI-standard FORTRAN-77 for the main text, and are repeated in UCSDPASCAL at the end. They are available on discs for use on IBM-PC microcomputers and their compatibles. A workbook providing sample programs that illustrate the use of each subroutine and procedure is available, as well as discs giving programs listed in the book in USCD-PASCAL and FORTRAN-77 for use on IBM-PC microcomputers and their compatibles.
Hardcover, ISBN 0521308119 Publisher: Cambridge University Press, 1986 Cambridge University Press. Used - Good. Shows some signs of wear, and may have some markings on the inside. 100% Money Back Guarantee.
Hardcover, ISBN 0521308119 Publisher: Cambridge University Press, 1986 Cambridge University Press. Used - Very Good. Former Library book. Great condition for a used book! Minimal wear. 100% Money Back Guarantee.
Hardcover, ISBN 0521308119 Publisher: Cambridge University Press21308119 Publisher: Cambridge University Press Acceptable, Usually ships in 1-2 business days, Reading copy. May have notes, underlining or highlighting. Dust jacket may be missing.
Hardcover, ISBN 0521308119 Publisher: Cambridge University Press, 1986 Used - Good, Usually ships in 1-2 business days, Good condition with some wear of the hardcover. But no missing pages and a little marking inside. 100% Guaranteed and fast shipping.
Hardcover, ISBN 0521308119 Publisher: Cambridge University Press, 1986 Used - Very Good. Appearance of only slight previous use. Cover and binding show a little wear. All pages are undamaged with potentially only a few, small markings.
Hardcover, ISBN 0521308119 Publisher: Cambridge University Press, 1986 Cambridge University Press. Used - Good. Shows some signs of wear, and may have some markings on the inside. 100% Money Back Guarantee. |
More About
This Textbook
Overview
Kaufmann and Schwitters have built this text's reputation on clear and concise exposition, numerous examples, and plentiful problem sets. This traditional text consistently reinforces the following common thread: learn a skill; practice the skill to help solve equations; and then apply what you have learned to solve application problems. This simple, straightforward approach has helped many students grasp and apply fundamental problem-solving skills necessary for future mathematics courses. Algebraic ideas are developed in a logical sequence, and in an easy-to-read manner, without excessive vocabulary and formalism. The open and uncluttered design helps keep students focused on the concepts while minimizing distractions. Problems and examples reference a broad range of topics, as well as career areas such as electronics, mechanics, and health, showing students that mathematics is part of everyday life. The text's resource package—anchored by Enhanced WebAssign, an online homework management tool—saves instructors time while providing additional help and skill-building practice for students outside of class.
Editorial Reviews
From the Publisher
"This text is a very straightforward approach to learning skills needed in college algebra. There are plenty of examples and problems to help students prepare for college algebra."- Joseph Eyles, Morehouse College
"The one thing about Kaufmann books is that they are very clear in their intent. This is one of the strongest assets of the book. It's not all fancy; it just says this is what we're going to do and then it does it . . ."- Patrick Webster, El Camino College
Booknews
A text for college students who need an algebra course that bridges the gap between elementary algebra and the more advanced courses in precalculus mathematics, covering intermediate algebra topics. Algebraic ideas are developed in logical sequence in an easy-to-read manner without excessive formalism. Concepts are developed through examples and problem solving. Includes chapter summaries, problems, and chapter and cumulative tests, with answers. This sixth edition contains word problems, coverage of all real numbers, incorporation of a graphing approach, and earlier introduction of the graphing calculator. The authors are affiliated with Seminole Community College. Annotation c. Book News, Inc., Portland, OR booknews.com
Product Details
Meet the Author
Jerome E. Kaufmann received his Ed.D. in Mathematics Education from the University of Virginia. Now a retired Professor of Mathematics from Western Illinois University, he has more than 30 years of teaching experience at the high school, two-year, and four-year college levels. He is the author of 45 college mathematics textbooks.
Karen L. Schwitters graduated from the University of Wisconsin with a B.S. in Mathematics. She earned an M.S. Ed. in Professional Secondary Education from Northern Illinois University. Schwitters is currently teaching at Seminole Community College in Sanford, Florida, where she is very active in multimedia instruction and is involved in planning distance learning courses with multimedia materials. She is an advocate for Enhanced WebAssign and uses it in her classroom. In 1998, she received the Innovative Excellence in Teaching, Learning, and Technology |
Precalculus
9780073312637
ISBN:
0073312630
Edition: 6 Pub Date: 2007 Publisher: McGraw-Hill College
Summary: The Barnett, Ziegler, Byleen College Algebra series is designed to be user friendly and to maximize student comprehension. The goal of this series is to emphasize computational skills, ideas, and problem solving rather than mathematical theory. Precalculus introduces a unit circle approach to trigonometry and can be used in one or two semester college algebra with trig or precalculus courses. The large number of peda...gogical devices employed in this text will guide a student through the course. Integrated throughout the text, students and instructors will find Explore-Discuss boxes which encourage students to think critically about mathematical concepts. In each section, the worked examples are followed by matched problems that reinforce the concept being taught. In addition, the text contains an abundance of exercises and applications that will convince students that math is useful. A Smart CD is packaged with the seventh edition of the book. This CD reinforces important concepts, and provides students with extra practice problems.
Barnett, Raymond A. is the author of Precalculus, published 2007 under ISBN 9780073312637 and 0073312630. Twenty one Precalculus textbooks are available for sale on ValoreBooks.com, eleven used from the cheapest price of $80.95, or buy new starting at $120.14.[read more]
Ships From:Multiple LocationsShipping:Standard, ExpeditedComments:RENTAL: Supplemental materials are not guaranteed (access codes, DVDs, workbooks). Precalculus. This book is in Good condition. Buy with confidence. We ship from multiple location. |
...Therefore, in Precalculus, students will be introduced to the important and basic mathematical concepts inquired before in algebra with deeper and higher details. They comprise, but not limited in, inequalities, equations, absolute values, and graphs of lines and circles. Students also focus on functions and their graphs. |
Basic Linear Algebra
Synopses & Reviews
Publisher Comments:
This book is an introductory linear algebra text. The topics are covered in the traditional order: linear systems, matrix algebra, vector spaces, linear transformations, orthogonality, and the eigenvalue/eigenvector problem.
The book seeks to explain and discover the important concepts of linear algebra using well-chosen examples. Proofs of many important results are done if the proof can be accomplished in a succinct, straightforward fashion. There are over 100 worked-out examples in the text as well as a complete outline of every chapter and extensive review problems which test the student's knowledge of the material covered in each chapter.
There are over 1200 exercises in the text, which were created by the author with care to include exercises which resemble problems done in the text. Definitions are well marked and theorems are stated as briefly as possible in order to enhance the ability of the student to remember the result. In a slightly novel way, the answers to the odd problems are given directly next to the problem. Applications such as heat transfer, spring/mass systems, population dynamics, and systems of differential equations are sprinkled throughout the |
This Saxon Math Homeschool 7/6 Solutions Manual provides answers for all problems in the textbook lesson (including warm-up, lesson practice, and mixed practice exercises), as well as solutions for the investigations and supplemental practice found in the back of the student text. It also includes answers for the facts practice tests, activity sheets, and tests in the tests & worksheets book. Answers are line-listed, and are organized by type (lessons & investigations, facts practice tests, tests, etc.).
333 perforated pages, softcover. 3rd Edition Math 76, Fourth Edition, Solutions Manual
Review 1 for Math 76, Fourth Edition, Solutions Manual
Overall Rating:
2out of5
Date:August 22, 2010
Sarah Wade
As with the textbook, the solutions manual is now being printed on a thinner, cheaper, newspaper-like paper. I would rather pay more for a better quality book. |
Optional Primer Sessions
An Overview of MATLAB and the MATLAB Product Family
What is MATLAB®? Why do so many companies use MATLAB as their primary tool for data analysis? Why do so many universities use MATLAB for both teaching and research?
MATLAB is an interactive development environment as well as a technical computing language. This means you can use a graphical user interface, or a programmatic textual interface to perform your complex data analysis and visualisation, as well as your algorithm and application development. During this demo, we show how MATLAB fits into a typical data analyst's workflow, from data access, to data analysis and exploration, through to application deployment and reporting.
If you've never been exposed to MATLAB, or you are still new to MATLAB, this session provides a high-level overview of the major capabilities and provides an ideal foundation for the presentations that follow.
Introduction to Simulink for Modelling and Simulation
Why are so many companies using Simulink® as an integral part of their development processes? Why do all the universities that use MATLAB for both teaching and research, also use Simulink as part of this process?
If you're new to Simulink, or just need a refresher, attend this session where we introduce you to Simulink, an environment for multi-domain simulation and Model-Based Design for dynamic and embedded systems. Product demos give you a high-level overview of the major capabilities and how you can use Simulink to design, simulate, implement, and test a variety of time-varying systems, including communications, controls, signal processing, video processing, and image processing.
Keynote Presentation
MathWorks – from Research to Production
Over the last quarter century, engineers, scientists, economists, financial quantitative analysts, and others have turned their workflows upside down as a result of vast shifts in the engineering, scientific, economic, financial, and computing landscapes. These shifts include:
The economic availability (relative to 1980) of vast computing power to suit all requirements, from the desktop to the high-performance grid or cluster, has allowed them to rethink how they do things.
The growing sophistication of technical computing tools has revolutionised the practice of research (whether academic or commercial), data analysis and manipulation, algorithm development, and collaboration and communication whilst completing these tasks.
More rapid and flexible methods for deploying or distributing the results of research and development enables collaboration and interaction with groups that may not previously have been part of a process, including different departments or groups in an organisation, customers, or suppliers.
The growing use of microprocessors, driven by embedded software, continues to increase explosively as our access to computing power enables us to design and build ever more complex systems and system of systems.
The growth of MathWorks, the MATLAB technical computing platform, and the Simulink simulation platform have been intertwined with these changes. Technical computing is now part of "how things are done." As a result MathWorks technology has evolved to provide a platform able to support a broad workflow – from research to production. This workflow often spans multiple individuals and in many cases, organisations.
In this keynote presentation, Andrew Clay will discuss how MathWorks technologies have evolved to support this workflow. He will show how they remain powerful 'point solution' enablers but have also evolved so that they can be considered comprehensive platforms enabling broad workflows, from research to production.
Technical Sessions
Streamline Your Data Analysis Workflow with MATLAB
This presentation reviews the data analysis workflow and how MATLAB technology can be leveraged to streamline and automate your analyses. This session features a walk-through explanation of the data analysis workflow with MATLAB featuring the access, visualisation and analysis, and reporting stages. Addition highlights include discussion of industry- and application-specific MATLAB add-ons to complement your data analysis tasks and reduce your development costs.
Data Analysis and Modelling with MATLAB: A Practical Approach
Scientists, engineers, and analysts are presented with increasing quantities of data, so being able to sort through the mass of information quickly and easily is critical. This demo takes you through a practical example of how to apply MATLAB to the data analysis workflow discussed in the Streamline Your Data Analysis Workflow with MATLAB presentation.
Model-Based Design Turns 10
Are you having trouble meeting your product or system development deadlines? Does finding issues late in the development process result in budget blowouts? Are you interested in streamlining your development process? Model-Based Design has been integrated into many companies' development processes both to streamline them and to address these critical issues.
NASA's definition of Model-Based Design is the following: "Model-Based Design is a mathematical and visual method of addressing problems associated with designing complex control systems. It is used in motion control, industrial equipment, aerospace, and automotive applications.
"[Model-Based Design] provides an efficient approach for establishing a common framework for communication throughout the design process while supporting the development cycle ("V" diagram). In Model-Based Design, development is manifested in these steps: modeling a system, analyzing and synthesizing a controller for the system, simulating the system, and integrating all these phases by implementing the system. The model-based design paradigm is significantly different from traditional design methodology. Rather than using complex structures and extensive software code, designers can use [Model-Based Design] to define models with advanced functional characteristics using continuous-time and discrete-time building blocks. These built models used with simulation tools can lead to rapid prototyping, software testing, and verification. Not only is the testing and verification process enhanced, but also, in some cases, hardware-in-the-loop simulation can be used with the new design paradigm to perform testing of dynamic effects on the system more quickly and much more efficiently than with traditional design methodology."
This session provides a high-level overview of Model-Based Design by introducing the workflow and its application using customer success stories. It highlights the benefits Model-Based Design has provided to customers from a variety of industries over the last 10 years.
Pathways to Production – Taking MATLAB and Simulink Algorithms from Research and Design to Production
MATLAB and Simulink are used extensively in the research, analysis, and design of algorithms and systems. As high-level language design environments, they provide advantages in the rapid development of ideas when compared to lower level-languages such as C, HDL, Structured Text, or Java.
So what happens when you want to run your algorithms away from the MATLAB or Simulink environments, targeting a prototype or production application? Attend this session to understand the options available to rapidly transition from design to prototype or production systems, without having to go through the time-consuming and error-prone process of manually converting your designs to other environments.
This session covers deployment options for both production and prototype systems. Some examples include:
Developing a MATLAB based application with a graphical user interface, Web application, or back-end production system
Using Simulink to develop algorithms for an embedded system, hardware design, or real-time testing environment
Converting a MATLAB algorithm to generic C code for integration into your existing production systems
Industry Sessions
Research to Reality
Linda Davis, Institute for Telecommunications Research, University of South Australia
The Institute for Telecommunications Research (ITR) is internationally recognised for its research and technology development for wireless communications. ITR conducts its research in three main areas: satellite communications, high-speed data communications, flexible networks encompassing fixed and mobile, satellite- and terrestrial-based applications. ITR has experts in information theory, security, networks, distributed systems, signal processing, optical, and wireless communication systems. ITR has a long history of collaborative projects with industry and transforming research to reality.
At the core of ITR's fundamental research and product development are tasks of analysis, design, optimisation, and performance simulation. In this talk Linda highlights how ITR uses MATLAB and Simulink to support both theoretical work and product prototyping. She also gives an example of integrated tools (MATLAB, Simulink, and third-party tools) used as the development environment for hardware including custom FPGAs, DSPs, or an off-the-shelf platform such as Lyrtech's small form factor software defined radio.
Industry Case Study
The Rooftop loading model is used throughout the world in the assessment and development of passenger rail timetables. This session describes how a team of mathematics graduates implemented this model at QueenslandRail using MATLAB to examine the loading dynamics on the network.
This tool can be used to assess the impacts of a minor train plan alteration or a major timetable revision.
Real-Time Research Platform Applied to Sound Processing Research in Cochlear™ Implants and Hearing Aids
John Heasman, Cochlear Ltd.
The multichannel cochlear implant is a unique technological achievement, representing the application of a novel combination of science, technology, and medicine. It brings functional hearing to severely or profoundly deaf individuals, transforming not only their lives but those of their families.
In 1985 Cochlear™ released the first commercial multichannel Nucleus® implant system. Four further generations have been released in the 25 years since. Innovations in mechanical design, electronics, and signal processing have brought successive improvements in device reliability and clinical outcomes. As of April 2011, over 144,000 registered Cochlear Nucleus® Implant Systems were in use globally.
Recent advances in acoustic signal processing for cochlear implants have produced incremental, but significant, improvements in cochlear implant recipients' speech understanding. In the past, these improvements were constrained by laborious, time consuming algorithmic implementation using proprietary digital signal processor (DSP) devices. This investment in process can limit the time available for creative work and hence restrict technological innovation.
Cochlear circumvented this innovation bottleneck with a rapid prototyping platform built with MathWorks products. The accelerated development process greatly reduced the time from conception to realization and increased the potential for future innovation.
This session describes how Simulink and xPC Target™ were integrated into a PC-based system with real-time capability. It will also provide examples of day-to-day contributions to people with Cochlear implants or hearing aids.
Dr. Peter W. Gibbens, Senior Lecturer in Aerospace Engineering, School of Aerospace, Mechanical and Mechatronic Engineering, University of Sydney
At the University of Sydney, faculty at the School of Aerospace, Mechanical and Mechatronic Engineering use a motion-based flight simulation facility to help train engineering students in aspects of aircraft design and systems development. It also offers students experiential learning about impacts on flight dynamics, handling qualities, and aircraft operation. The variable stability flight simulator enables students to study how variations in aerodynamic parameters affect flight stability and handling characteristics. Students can design and implement flight control systems and assess the behaviour of these systems in real-time operation by simulating aircraft flight and systems.
This presentation gives an overview of the simulator and discusses its educational purpose and uses. MATLAB and Simulink are key to the success of the programme. MATLAB is the students' main analytical and development tool. Simulink and xPC Target™ provide a real-time implementation framework. The presentation also describes the system hardware and software, and the interactions between them, with particular emphasis on modelling and communications. Dr. Gibbens will address educational outcomes and describe why these tools provide an effective solution to the problem.
Workshops
Automatically Converting MATLAB Code to C Code Using MATLAB Coder
Many engineers, scientists, and researchers rely on the flexibility of the MATLAB language to prototype their ideas and algorithms. As a high-level language, MATLAB provides an easy-to-use environment that facilitates rapid design iterations compared to lower-level languages like C. What happens when these algorithms need to run independently of MATLAB and any MATLAB runtime libraries? For example, what if these algorithms need to be integrated within some external C environment or run on an embedded system?
New to R2011a, MATLAB Coder™ allows you to close the gap between a MATLAB prototype and implementation by automatically generating C code. With this capability, MATLAB algorithms can be exported to C environments completely independent of MATLAB. You can also compile the generated C code into a MATLAB friendly executable called a MEX file. This is useful for verification purposes, and may also provide speed improvements on your existing MATLAB code (most likely in the case of fixed-point algorithms).
This presentation walks you through the process of converting MATLAB code to C code, with demos that show:
Floating-point C code generation
Fixed-point C code generation
MEX (MATLAB executable) file generation
Dealing with inputs of variable sizes
Integrating generated C code into an external C environment
Note: Some of you may know this technology as Embedded MATLAB®, and if so, you are partially correct. MATLAB Coder is the current and more mature technology for automatic C code generation. This presentation shows some advances in the technology that make the generated C code more suitable outside of embedded applications, and also a more user-friendly interface.
Data Analysis for Design with MATLAB
Data analysis for design and quality assurance is a critical task in many industries. Scientists, analysts, and engineers working in fields as diverse as automotive, industrial automation, finance, mining and exploration, and biopharmaceuticals will benefit from attending this overview of data analysis for design with MATLAB.
This presentation reviews the MATLAB data analysis workflow for accessing, analysing, and modelling data, and documenting results. The presenter uses a Six Sigma approach to both measure and ensure quality in product designs. Demos show several ways to use MATLAB, including:
Using MATLAB for Signal Analysis
In signal processing, whether in academia, research, or industry, MATLAB is the de facto standard analysis tool. Whether you are a data analyst exploring time-series measurements or a seasoned DSP engineer, MATLAB provides rich functionality and a friendly development environment to help you become successful in your work. This presentation covers some of the major capabilities useful in a signal processing workflow, from data access through data exploration.
Product demos show how MATLAB can help users tackle a range of signal processing problems and challenges. If you're still new to MATLAB, don't be afraid to attend this presentation. Several demos illustrate how MATLAB can be used as an interactive point-and-click environment to simplify the analysis process, and then generate the required MATLAB code. Topics include:
Spectral analysis
Efficient techniques to handle data streams
Filter design and implementation techniques
Interfaces to instruments and data acquisition hardware (we use an Agilent arbitrary waveform generator feeding into a PC sound card)
Model-Based Design for High-Integrity and Business-Critical System Development
Do you want to understand how to refine your system development practices to improve quality? Are you identifying critical design issues or bugs late in the design process? Does this create difficulty when meeting project deadlines or budgets? Are you required to meet system quality standards such as ISO 26262, IEC 61508, IEC 62304, DO-178B, or MISRA? Learn how many organisations are using a development process based upon an executable system-level model, otherwise known as Model-Based Design, to improve the quality of their designs and code through verification and validation using Simulink.
Model-Based Design is a powerful development method that allows you to reduce time to market and improve quality. This session builds on the modelling and simulation capabilities of the Simulink platform and introduces tools and techniques that enable you to verify and validate your models early in the development process, trace requirements to your models and code, check modelling standards automatically, manage regression and unit testing, and apply formal methods to prove the absence of run-time errors in both models and source code. The adoption of these formal verification and validation techniques ensures that your embedded systems meet your quality and safety goals. |
W. D. Wallis
I was very excited when I received W. D. Wallis's Mathematics in the Real World to review. The previous evening, I had attended a gathering with a friend that teaches high school social studies. At one point in the night, he declared, "No offense, Kara, but I don't use algebra in my daily life and I don't see the point of my students having to learn it." I argued that it was okay to study a subject that you don't use in your daily life. Studying algebra helped students look for patterns, think logically, and communicate in a clear, concise manner. I also agreed with my friend to some extent, however, and suggested that more students would benefit from studying mathematical topics that were directly applicable in the real world. This textbook contains a treasure trove of such topics, such as sampling, cryptography, voting, and probability.
While Mathematics in the Real World is intended for students with a minimal background in mathematics, the mathematics in the textbook is not dumbed down. The student will be performing calculations, using formulas, and drawing graphs.
In each chapter, Wallis includes a set of multiple choice questions, as well as a set of exercises. As a teacher, I really like the option of being able to choose from both types of questions. I do have two small complaints about the problem sets. Wallis has included answers to all of the exercises, even and odd. I would prefer that only the odd exercises had answers. There are times when I want to see what my students can do, without the aid of an answer key. My other complaint is that most of his problems are fairly straightforward. If the students diligently read the chapter, then they should be able to complete the exercises. I wish that the author had written a few challenge exercises (and marked them with an asterisk), so that teachers could assign problems where students have a chance to do some original critical thinking. One or two challenges in each chapter would suffice. Perhaps in the next edition?
All in all, I think Wallis has written a wonderful text. He has chosen exciting topics and explained them in a straightforward manner. I enjoyed reviewing it and I wonder if my social studies friend would enjoy it as well.
Kara Shane Colley teaches math at Portland Community College in Portland, Oregon. She also leads a math circle at the Oregon Museum of Science and Industry (OMSI). |
Product Description
In this program we are going to raise the hood and show you how algebra works! With some things in algebra, it's just a step-by-step process to get to where you want to go and that's exactly what we are going to work on today, the mechanics of algebra! Topics Covered: Rule of Equality Solving for Solving Equations with Variables on Both Sides Absolute Value Simplifying Expressions Solving InequalitiesGrade Level: 8 - 12. |
This course will provide an introduction to the use of computers to solve problems arising in the physical, biological, and engineering sciences. Various computational approaches commonly used to solve mathematical problems (including systems of linear equations, curve fitting, integration, and differential equations) will be presented. Both the theory and application of each numerical method will be demonstrated. The student will gain mathematical judgment in selecting tools to solve scientific problems through in-class examples and programming homework assignments. MATLAB will be used as the primary environment for numerical computation. An overview of MATLAB's syntax, code structure, and algorithms will be given. Although the subject matter of Scientific Computing has many aspects that can be made rather difficult, the material in this course is an introduction to the field and will be presented in a simple as possible way. Theoretical aspects will be mentioned throught the course, but more complicated issues such as proofs of relevant theorems/schemes will not be presented. Applications will be emphasized.
Objectives
To make MATLAB superstars!
Syllabus
Week 1 - Basics of MATLAB and Introduction
We will begin with a brief review of MATLAB and its basic functionality, including plotting, implementing IF and FOR logicals and the construction of matrices and vectors.
(a) constructing matrices and vectors in MATLAB (text 1.1)
(b) FOR and IF statements for program logic (text 1.2)
(c) inputing/exporting data and plotting (text 1.5)
Week 2
Linear Algebra and Ax=b
We discuss direct solutions to matrix systems of equations Ax=b. Computational complexity, i.e. how methods scale with size of system, is considered. |
More About
This Textbook
Overview
This text is designed for students in a broad range of career programs that require a solid understanding of basic math, elementary algebra, trigonometry, and geometry. The 2009 Update of this popular text continues a tradition of evolving to reflect today's work environment and deliver math instruction using the best print and electronic teaching methods and delivery systems |
often referred to as industrial mathematics is becoming a more important focus of applied mathematics. An increased interest in undergraduate control theory courses for mathematics students is part of this trend. This is due to the fact that control theory is both quite mathematical and very important in applications. Introduction to Feedback Control provides a rigorous introduction to input/output, controller design for linear systems to junior/senior level engineering and mathematics students. All explanations and most examples are single-input, single-output for ease of exposition. The student is assumed to have knowledge of linear ordinary differential equations and complex variables. Written specifically for the undergraduate mathematics student Concise and clear examples that illustrate theoryAuthor is faculty member at University of Waterloo; the largest mathematics department in the worldContains exercisesIncludes MATLAB |
Mathematician's Delight -
Summary: 'The main object of this book is to dispel the fear of mathematics. Many people regard mathematicians as a race apart, possessed of almost supernatural powers. While this is very flattering for successful mathematicians, it is very bad for those who, for one reason or another, are attempting to learn the subject.'W.W. Sawyer's deep understanding of how we learn and his lively, practical approach have made this an ideal introduction to mathematics for generations of readers. By starting at the l...show moreevel of simple arithmetic and algebra and then proceeding step by step through graphs, logarithms and trigonometry to calculus and the dizzying world of imaginary numbers, the book takes the mystery out of maths. Throughout, Sawyer reveals how theory is subordinate to the real-life applications of mathematics - the Pyramids were built on Euclidean principles three thousand years before Euclid formulated them - and celebrates the sheer intellectual stimulus of mathematics at its best99 +$3.99 s/h
Good
Dream Books Company, LLC Englewood, CO
1991 Paperback Good Used book with normal wear and tear and may contain writing. Stock photo may be different from actual book cover.
London, England 1991 Trade paperback Good. The book has been read but remains in clean condition. All pages are intact and the cover is intact. Some minor wear to the spine. Trade paperback (US). Gl...show moreued binding. 240 p. Penguin04/25/1991 Paperback New edition ***SIMPLY BRIT*** We have dispatched from our UK warehouse books of good condition to over 1 million satisfied customers worldwide. we1991 |
You are currently viewing the 2013 - 2014 Student Catalog. The current catalog is located here.
Operations With Integers
Course
Course Title
Class
Lab/ Shop
Clinical/ Co-op
Credit
DMA 010
Operations With Integers
.75
.50
0
1
Prerequisites: None
Corequisites: None
Effective Term: Spring 2012
This course provides a conceptual study of integers and integer operations. Topics include integers, absolute value, exponents, square roots, perimeter and area of basic geometric figures, Pythagorean theorem, and use of the correct order of operations. Upon completion, students should be able to demonstrate an understanding of pertinent concepts and principles and apply this knowledge in the evaluation of expressions |
proven system of teaching math for classroom teachers, and for parents who are homeschooling. The easy-to-follow program requires only 20 minutes a day. Short, concise, and self-contained lessons help students to master, maintain and reinforce math skil
Customer Reviews
Most Helpful Customer Reviews on Amazon.com (beta)
Amazon.com:
61 reviews
36 of 37 people found the following review helpful
Our girls no longer complained!Feb. 24 2009
By
K. Webster
- Published on Amazon.com
Format: Paperback
Last year in our homeschool, we switched to "Mastering Essential Math Skills: Middle Grade/High School, and it was the best switch we could have made. Both girls were to the point of dreading math, and after changing to this program, they now looked forward to math class and could make sense of it because of the step-by-step instructions. Great for 7th-8th grades. Our 2nd daughter actually moved from 5th grade to 7th grade because this program was available.
65 of 75 people found the following review helpful
Not as advertisedFeb. 17 2007
By
Parent
- Published on Amazon.com
Format: Paperback
Verified Purchase
I loved the grade 6-8 edition of this book for reviewing math with my daughter. It is an excellent step by step approach to middle school math. This edition is the same book with very little added. It is not "high school" math. I wouldn't call it "Book Two". I was expecting something very different based on the title. Highly recommended for 6-7 grade math practice and review.
17 of 17 people found the following review helpful
Good math book, but has typo's.July 17 2007
By
The Way
- Published on Amazon.com
Format: Paperback
This book is helping me getting freshed up in math but it contains a few typo's. For example, on page 7 it says there are two #5 problems. There should be obviously one problem #5 and the other is problem #6. Nevertheless, it is a good math book. I just wished it gave more examples on how to actually DO some of the work.
9 of 9 people found the following review helpful
Miscategorized.Aug. 20 2011
By
Cautious Shopper
- Published on Amazon.com
Format: Paperback
Verified Purchase
This book seemed more geared toward elementary/middle school than middle school/high school. Anyone going into 9th grade might find this way too simple.
10 of 11 people found the following review helpful
Exactly what I was looking for!June 16 2009
By
L. Halling
- Published on Amazon.com
Format: Paperback
There's nothing like having a personal tour guide to lead you through the jungle of math skills. Richard Fisher's 30 years of experience teaching math shines through in his Mastering Essential Math Skills Books 1 and 2, shedding light on the dark and daunting paths to the realm of math comprehension. Book One for Grades 4 and 5, takes us all the way from basic adding, subtracting, multiplying and dividing to number theory and Algebra: effectively, concisely, and simply. I'm not a math expert but I would say both books will not only bring your student up to grade level (if they weren't there already) but, I dare say, will accelerate them. In the companion DVDs included with both books, Mr. Fisher's easy-going, encouraging manner walks us, at a comfortable pace, through the maze of math. A feature I consider ingenious is the built in speed drills on each one page lesson. It is so cleverly designed, it leaves one thinking, "why didn't I think of that?" Book Two is designated for Middle Grades and High school. It contains the same sequence of development but goes a little further into each skill giving students a more comprehensive understanding of the necessary skills to tackle Algebra I (coming soon). I plan to use Book One with my 14 year old as a review to begin with this fall before taking on Algebra second semester. After going through Mr. Fisher's Pre-Algebra Concepts book with DVD this year, my son has actually requested, nay, begged, to use Mr. Fisher's math curriculum for the remainder of his schooling! Need I say more? |
Course Descriptions
This course, in the calculus of a single variable, concerns
recognizing, analyzing, and calculating problems in the
following topic areas: the calculus of inverse trigonometric
functions, integration techniques, application of
integration, L'Hopital's Rule, improper integrals, infinite
sequences and series, plane curves, parametric equations,
polar coordinates, and polar curves.
PR: MAT 180 or consent of the department |
The Number Base converter is a tool which converts numbers from one base to another such as binary,decimal,hex,octal. The program is quite simple so even if you are a beginner you can use it without any trouble.
This very user-friendly desktop tool simplifies common statistical calculations and also allows quick analysis of raw data whether you enter it yourself, use a spreadsheet, or maintain a database. Copy data from spreadsheets, database tables, import text files, or enter data directly into the spreadsheet to quickly produce basic statistical analyses.
This aims to supply a Borland Delphi translation of Alan Miller's Random Module for FORTRAN-90. This translation has been done with Dr Miller's approval and is being made FREELY available to all Delphi Developers, though we do ask the Alan Miller and ESB Consultancy be given due credit. Random Number Generators include Normal, Poisson, Gamma, Chi-Squared, Exponential, Beta, Weibull, Binomial and more. Includes full Delphi Source and Demo˘˘˘ only. Educational discount: 20%.
The free Basic Facts Worksheet Factory can generate worksheets that will provide students with the practice they need to become proficient in the basic math operations. You can easily create unlimited customized worksheets for the practice of arithmetic facts in addition, subtraction, multiplication, and division.
This program is an help to work with fom-digests groups of messages.
Fom is an automated e-mail list for discussing Foundations Of Mathematics.
Fom is also avalaible in digest format. This means that, instead of receiving each fom message as an individual e-mail, you will receive groups of messages clumpsed together.
It runs on Windows 95/98 or Windows NT. No database is required.
EconoModeler is an Intuitively-Easy-To-Use software dealing with Analytic Geometry for Mathematical Economics and Management applications.
EconoModeler can be used as a Virtual Laboratory that very quickly and very easily generates Plane Analytic Geometry objects like Points, straight lines, Circumferences, Parabolas, Ellipses, Hyperbolas, Circular Arcs, Lines between two points, etc, to be applied on Economic Modeling and in Management.
Curvilinear: Easy Learning Plane Analytic Geometry
An Intuitively-Easy-To-Use visual interactive software, oriented to overcome the abstraction that exists in the Plane Analytic Geometry (PAG), this is a tool that makes it easy to learn and master the PAG, the user is allowed to literally "see" the mathematics while carrying out operations and problems by hundreds in a very short time with the help of the mouse.
Panageos is oriented to the intensive solution of problems on Plane Analytic Geometry.
The main feature of Panageos is its power to read the user's equations and interpret them, for this reason the data input is exclusively through the keyboard (coefficients of several types of equations are entered via keyboard).
Visual Mathematics is a highly interactive visualization software (containing -at least- 67 modules) addressed to High school, College and University students. This is a very powerful tool that helps to learn and solve problems by the hundreds in a very short time.
Visual Mathematics, a member of the Virtual Dynamics Mathematics Virtual Laboratory, is an Intuitively-Easy-To-Use Mathematics visualization software. |
getting in groups and being able to hear others explain the problems in a different way
It involved real life situations
This course has prepared me for future math courses.
What I find most valuable and helpful about this
course is how to calculate interest rate,
compound interest, and more.
I found using math lab is really helpful because if I get stuck on the homework, theirs links I can get help (online tutor) or a link that says "show me how" and the computer will show me how to work the problem out. Also, theirs practice problems you could do before the tests so you could get better at it and it will show you what you need to work on. It's really great and the teacher is always there to help you also when you get stuck. He encourages you to ask questions.
The thing I found most valuable as well as helpful about the course was the MyMathLab. It helped me practice how to do the assignments that were to be introduced in class before we even start doing it in class. Since I haven't been much of critical thinker, and I wasn't very good at doing word problems before coming into the class, the MyMathLab really helped a lot with this problem, so that's one thing I found most helpful in the class.
I enjoyed his fun ways of teaching. I don't know what it is but he knows how to make it fun and
entertaining. Like his examples of each problem discussed in class. I love the probability
section we learned in class because of the visual aid provided (we got to play with dices).
The slideshows that he uses in class.
The slideshows and before exams, getting into groups and going over what we learned and then
switching groups to share with the other people in the class.
he is very easy to understand and is very nice and out going.
-the assignments in MyMathLab gave feedback for incorrect answers
-Mr.Girard actually uses the whole class period to teach, unlike some
teachers who finish a lot earlier.
-slide shows/power points were very helpful
-working in groups often helped
/becoming more logical.
The knowledge my teacher showed about the subject
N/A
the mymathlab. It was easier than having to use a computer that had to have the downloaded cd disk in order to submit homework88
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22. What did you find most valuable and helpful about the instructor?
he knows what hes teaching and knows how to explain everything in different ways
Mr. Girard was very helpful in explaining almost every thing. Although our class was quiet, he made
it fun to learn and involved real life situations in math.
Ryan really cares about his students. He has a way of explaining things that help to ease
understanding. He listens.
He is very helpful when you dont understand the
problem. He makes you think and challenge your
brain. He is nice and understandable when
student doesnt quite get the new chapters are.
Always encourage us to do our best.
I found that he is really helpful and he makes the work seem so easy by all the examples he shows us. I like how he uses real life situations in his examples. It makes learning easier knowing I might be able to use this in the real life. Thank You Mr.Girard.
The thing I found most valuable and helpful about Mr. Girard was that he was incredibly enthusiastic, which everyday is a great relief on the nerve when I'm nervous about something, and he always made sure that we all understood how to do an assignment step by step before continuing with his lecture.
Again, I just enjoyed the way he teaches. I don't know how he does it, but it's entertaining. The
discussions we have in class help me understand things better.
He put us into groups and help each other
He is very easy to get together with outside of class for extra help and clarification.
always available and willing to help with any questions from students about
the lessons
For a test review, you should make review sheets instead of going in groups and letting us teach one another.
I think that Mr. Girard is doing everything very well, I don't think that their are any improvements that need to be made to his teaching methods, because as far as I know it's been completely perfect and helpful.
Just keep up the lively conversations and the helpful visual aids.
I have nothing because I like his way of teaching.
it's pretty efficient the way it is now! but maybe not have all
assignments/quizzes on MyMathLab because sometimes internet access isn't
always available for a student28. The course is83
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10 (6331. Other comments:
I'd recommend Mr. Girard to anyone.
Ryan Girard is the best math 100 teacher. I love
math but I dont like critical thinking problems.
After taking this class, I understand more and I
love it. I strongly suggest to take ryan class.
:)
Thank you for a fun semester Mr.Girard :).
I really enjoyed this course with Kumu Ryan, because he has been nothing but nice and helpful as well as an incredible instructor. His enthusiastic personality in class would make anyone's nerve calm some. My classmates were also very helpful when I didn't understand completely how to do the homework so I'm very thankful about that. Overall I really enjoyed my times spent in this class, and would recommend this class to anyone who needs to take MATH100 next semester.
I love that all the homework is online! And the quizzes, too!
Mr. Girard, you are by far one of my favorite college teachers. :)
I think he is very nice to all students and his explanation were very good to
understand to me. When I asked some questions I didnt know to him, he
explained them very carefully until i understood. I was glad to take his
class in this semester.
I found it helpful that Mr. Girard had a variety of options to choose from for help. For example we could email him anytime, there were videos, he informed us about different sites to use
This course helped me not only to be more confident in my math skills, but also confident in being
able to teach math when I become an elementary teacher. I mostly enjoyed the flexibility of the class
as well as the support from the my teacher and classmates.
it was very convenient and the my tsmath lab questions got me prepared to take the tests
I loved that I was able to do the course online.
It was very organized and had a good time-work ratio.
The most valuable and helpful thing about this course is the group projects we have weekly. My partners are extremely helpful when I am concerned in certain areas. Also, the teacher responds to email promptly.
I found most helpful throughout this course was being being able to do the
MyMathLab homework assignments and watch the videos on how it is done.
Being able to be in the minds of the children and to remember that each
person has a different learning style.
Working together in groups and having blogs to help explain answers.
I really liked the organization of the course. You always knew what was due and you knew where you could find it. I learned a lot in this course from the MyMathLab website and from my group members56
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22. What did you find most valuable and helpful about the instructor?
This course was online and I could do it at my own time.
Mr. Girard is so open to helping every student. He makes it very comfortable to ask question. On time to answering emails and questions. Eager to be very helpful so we can understand the lessons.
He wasn't pushy, but was willing to help his students and help them adjust to the classwork.
although he seemed very busy, he was nice and helpful when needed.
That he always made himself available to help and answer questions.
His enthusiasm was inspirational. He was also willing to meet on campus.
He was helpful.
Because this is an online class the only help he could give us is through email and he was very helpful with responding to my questions.
He was consistently emailing and updating us
He was there to answer our questions.
He offers support if he sees you struggling in his
course and tries to reccomend solutions to solving
problem.
Always available for help thru email, and would provide helpful feedback.
He was always willing to answer questions and he always gave good feedback on our work. He was prompt and that made it easy for us to succeed.
Having a more clear calendar/announcement system on Laulima, do not rely so
heavily on MML. Be available more often. When a student asks a specific
question do not respond with a reference back to the homework or readings, we
need clarification obviously. It would be nice if we were allowed to practice
the assignments once they have been submitted and graded.. MML can be useful
is we are allowed to utilize it's services.
Being that this was my first online class let a lone in a subject I struggle with, Math, I found this class to be a challenge, however Mr. Girard did really well with teaching it. He assigned Modules every Wednesdays and I thought maybe assigning them on Mondays would give us a better schedule to work on them. or maybe even Tuesdays.
Keep it up
if he could grade work earlier and not all at the last minute, it would help in guaging how we are doing
Give All homework assignments early and send out email reminders for due dates
In my opinoion the required meetings should have been optional, they were not
worth missing work over when it was informations I can find out on my own. I
would have rather used the meeting times to review and talk about the
material. Also, having group assignments were almost a joke. No one really
worked together as far as my experience went; dissapointing.
Perhaps meetings more as a class online.
Removing the group activity altogether. Its good having a group to ask
questions but it makes it very difficult to rely on others that dont do the
work, because then the others in the group are just doing more work and it
gets very frustrating.
Being more organized about releasing the modules at the beginning of the
semesters. I like working ahead with math.
Less group work because it seems group projects affect
your course grade more than the actual math material
and sometimes you can't depend on your group members.
NA
Sometimes the group projects are hard. But other than that, I think it was a great course06
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32. What did you like most about your online course experience?
That I had an awesome group that worked really hard while going through this
course together. I felt like if it wasn't for my group I may not have been
able to understand all the material.
I did like how I could do a lesson at my own paste.
The flexibility as well as the convenience.
it was convenient in my schedule
That I could work at my own pace.
It was convenient
It was easy to find and do the work, the due dates were clearly stated,
selfperpetuated learning was encouraged.
Our teacher was VERY helpful with everything.
I liked how it was online and on laulima and also my math lab.
I liked being able to do the MyMathLab course work.
Learning Blackboard.
The ability to go at your own pace at the comfort of your own home to
complete your math assignments
my mathlab, modules and group work
I was so organized and routine, it made it easy to keep track of so I could get a good grade and learn all that I could.
33. What did you like least about your online course experience?
The heavy use of MML. I was also discouraged from asking questions early in
the semester because I felt like I was just getting redirected back to the
material on MML that I already did not understand. When we have our class
meetings I feel that it is more in regards to the different programs that UH
systems offer in education, which is nice to know.. I feel that we are being
pitched a sale and not being taught a class.
I'm very knowledgeable when it comes to computers and folders on computers etc. So I was learning not only math but how to do a online course with computers. In the begging I fell behind because it wasn't really clear to me that modules was in a totally different tab as MML. However when I scheduled a meeting with Mr. Girard he was very helpful and didn't make me feel intimidated with things I didn't know
It's a bit tough if you're not good with tech stuff like me... takes a bit of time to work things out.
sometimes hard to show work even though i know what i'm trying to say inserting equations and tables are harder than just drawing them
I had some trouble with the online groups. I took an online course because I didn't have time to meet every week so trying to meet and make time with a group every week was a little difficult since we all had separate schedules.
Lack of communication
I found it difficult to contact someone quickly to clarify questions. I could
not access the online meeting on my computer. The homework was Wed-Thurs, and
threw my schedule off. Sun-Sat is preffered.
Having to meet with group members and staying online for a great amount of time.
The group work was hard trying to get together with group members. I think maybe the group
assignments should be due on a different day?
I disliked the group assignments and I dont know where I stand in this course
because hardly any of my work has been graded and that also gets frustrating.
I did not like the group projects. It worked well but its hard to meet up.
The group projects
na
Group Projects.
34. Other comments:
Mr. Girard was very good at sending us informative email I just wish I knew more on how to use computers. Maybe send out an informative video at the begging of the semester with step by step things that we will be looking at for the semester. For example in the video show where our modules will be held, where mml is and guide us through mml on informative video. There for we can always look back at it if we forget how to get to a certain page. In others words, like an instructional DVD on how to navigate Mr. Girards class online
If it is possible to change from BlackboardCollaborate it would be good.
From what I understood, many people had problems with it. My computer never
got it working, I had to use the app on my phone which is not as good.
Thank you!
na
Ryan is a great teacher and I learned a lot in his course. I look forward to having him again next semester.
The most helpful was Ryan's patience and willingness to help all the students
and encouraged us to work together on many class assignments.
being confident in my math skills in order to teach in my own class
It teaches you the relevance of the course. Even though it is a math class,
it's the way that it teaches you how to teach the material.
It was nice to have such a small class because then we got all very familiar with each other. We
helped each other with homework and most of the time we all didn't understand it so it was nice to
be able to get students involved9. The instructor7
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5 (71622. What did you find most valuable and helpful about the instructor?
Was very helpful and patient with my many questions!
He gets excited about the subject matter and is very encouraging even when we
are having difficulty understanding a concept. He never makes you feel like
you are asking a dumb question.
He really helped me understand math. All our guest speakers helped me be more informed into my
career path.
He helps you understand the materials addressed and helps you to want to be a
better teacher.
He is a very smart teacher and knows a lot about his math. He is always around campus so it was
easy to find him when I had questions. Sometimes using the my math lab is fun and other times its
not.
Perhaps mini units and quizzes to review material as the semester goes on.
n/a
I think if we actually went over the chapters/sections first instead of getting the homework. A lot of
us did not do math 100, or even algebra in high school and so it was hard seeing new material only
in the homework. If we were able to learn how to actually do probability or other assignments first,
then do the homework I would have done better because I feel like i still don't get a lot of the new
material. Doing the student presentations were good, but I still think having the teacher explain it is
better because we actually grasp the idea.
Mr. Girard is always willing to help whenever a difficult question arises. Not only is he a very willing
teacher, he explains material that could potentially be confusing in a relatable and understandable
way.
It was very challenging, and I have learned a lot.
Applied math to real world applications.
Part B was valuable and helpful being able to work with all the problems I
did not do or missed.
Gaining my confidence in the beginning since it had been awhile since I had
taken a math class and then being able to work the problems mostly on my own
towards the end of the semester, the problems got easier even though the
class stayed pretty much at the same difficulty level.
What I found most valuable and helpful about this course is doing extra
problems than what was assigned.
Ryan explains the ideas and concepts clearly. He also provides examples and
practice problems to get you thinking.
Probably the same as last semester, whatever I wrote.
The encouragement and support of the teacher.
It taught me a lot of concepts that I was
not previously comfortable with.
It made me feel comfortable and ready for calculus III11
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19. The instructor uses class time well7
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22. What did you find most valuable and helpful about the instructor?
He presents the material in a very understandable way. I felt as though I have a solid understanding
of the material I have learned.
He is willing to help the students out, and answer any questions that we
have.
He was always around and available, when needed.
Being able to breakdown problems
His enthusiasm for the subject as well as his desire for his students to
understand the material and be able to advance further
He is willing to help you if you are struggling. All you need to do is ask.
The homework problems and group work during class helped grasp the course
material. He is also enthusiastic about math in general and is very helpful.
Probably the same as last semester, whatever I wrote.
He gave chances for redemption, such as in the bonus point sheet and the part B of exams.
To provide helpful suggestions when students are stuck rather than putting
students on the spot to find the answer on their own, I would have preferred
if we took problem step by step so I could learn the whole process instead
of just figuring out the answer to the problems
There should be study groups to help students in this class.
Part B of the exams seems to be more about following instructions to the letter, rather than actually
learning from your mistakes. Not that following instructions isn't important.
not much to improve on.
Adding more worksheets or written
assignments
A more consistent number of homework problems so we know what to expect each time.
It made me really see how it's done in teaching students at a younger age.
It made me think about things carefully to make sure that I understood it
carefully, then to project that to the students.
I feel this course really prepared me for becoming a teacher. It exposed me to different styles of teaching as well as learning. Ryan was always available to clear up any questions you may have.
That we have to go up in front of the class to speak to the students. It
helped me to build confidence.
Our professor masterfully engaged each of us into discussion groups and encouraged us to
ask questions. He used manipulatives, technology, interesting facts, internet stories, videos,
etc to make this otherwise very boring subject more interesting...
The fact that our class gelled really well. We were able to ask Mr. Girrard queastions while keeping eachother accountable for our actions.
I learned a lot more about what goes on behind the scenes in math, and I
learned about what can go wrong when preparing for a presentation.
Gave a glimpse of what my future might look like, having us present our lesson plans to our class gave me confidence, practice and understanding what it takes to give a effective lesson to a class and for them to understand it.
How we are able to learn concepts. How we have to understand simple math in order to teach it. And
the different ways we can teach,
we could learn from one another if we didn;t understand a problem
I found the interaction with students in class, and Ryan's enthusiasm for
the content very valuable and helpful. Guest speakers were also relevant
and willing to assist students in their area of expertise36
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7 (6422. What did you find most valuable and helpful about the instructor?
He was always there to answer your questions and give you help in doing
starting our own lesson plans that we did as part of our class. He was good
in describing things and explaining.
Made it clear that we could ask as many questions as necessary. Always took time to address any concerns or questions. Was very patient when we were learning difficult concepts. Actually made a math course enjoyable, and that is not easily done!
Ryan is very intelligent and he uses real life situations to explain math
problems, which is confusing at first to me but then I understand. I like
that everything is submitted online, its easier.
Oh I think I already answered this in the previous section. I think only our professor could have
made this class interesting. He has a very unique and engaging way of presenting the
material...he made the class very interesting and helped us not only with class materials but
assisting us in planning for a better future illustrating more opportunity and options that are out
there for us to take... Ryan Girard is a wonderful professor... He cares about each of his
students. The environment he provides is nurturing yet academic...I am so happy that He was
my teacher.
Mr. Girrard was very knowledgeable about the subject. Any question we threw at him he answered if he didn't know he had the answer the moment we walked into class the next day.
Girard just wanted the students to learn. He didn't mind if we messed up on
our presentations or tests as long as we were able to see what we did wrong
and keep it in mind.
The lesson plans, the chance to correct the exams we took and learn from our mistakes! Also all the guest speakers were so helpful to understand and begin our journey in the UH system for teachers
This Dude just knows his stuff. Im really glad that we have EXAM part B's its helpful to learn from
your mistakes.
he helps show there isn't just one way to do something
Ryan is professional, always willing to assist students, knowledgeable about
the content and school resources, and very encouraging. His enthusiastic
way of explaining content and accessibility both on and off campus is very
supportive also. He has been very encouraging and fair to all students and I
have enjoyed his class very much.
Perhaps going over more material before the homework instead of after.
Going through and talking about each chapter would help us more before taking
a test rather than just questions being answered.
The course was unimaginative dry and confusing, but our professor made it very interesting
and clear. Our professor could not have been more perfect!
Girard as a teacher was awesome, but I think he should use less "mymathlab"
if he can. The site is just really clunky, breaks hen you need it to work,
and sometimes doesn't take correct answers written in a different way.
nothing, i loved how we had time for projects, time for online homework and a good dynamic of collaboration in class and helped build a special bond with our classmates
It would be awesome if this class was in the day time.
No suggestions for improvement. Ryan's teaching style and use of technology
in the classroom to appeal to students and provide elaboration is awesome.
In regards to this course, in particular, Ryan is a very good model of how
potential elementary school teachers can approach students and a classroom
atmosphere29. The guest lecturers addressed issues relevant to the course303132. Other comments:
Thank you Ryan, first math class I actually didn't dread! You made it a fun learning experience which prepared me for becoming a teacher myself.
I am glad that Yale didn't think of stealing our Ryan Girard!
Thank you for the time and effort you put into our class, the guest speakers were so helpful and you were so good about letting us know what all is out there by giving us plenty of options with our education path!
Ryan set clear objectives for the course and offered several opportunities
for extra credit and outside application of coursework. Rubrics and work
samples also helped a lot. Mahalo.
Not do much homework, because being a full time student it is hard to complete the homework of
this class combined with the homework from other classes.
I wish he would have a little bit more control of the class when other students are being loud and
obnoxious about non-math related topics.
Helping the students to prepare for exams by doing examples of the harder questions would be
helpful.
Go through each step of the equation when solving. Sometimes it isn't clear and there are
some students who haven't taken math in over a semester or a year, it's sometimes difficult to
remember certain steps even if it should be a "review"
Nothing
none
Let students know what assignments are due online because it is very easy for students to forget
Before exams it would be helpful to have a study guide or some sort of
guide. Before exams its a bit tough to figure out exactly what to study.
What I found most valuable was the availability of instructors and tutors that could help me to understand the material more thoroughly.
Throughout the semester Prof. Girard gave us sites and programs that could help us with
homework and lessons outside of the classroom. He also made us aware of tutoring hours and
times in which he was available to help us with any questions that we had.
I felt that I learned the most from doing the homework assignments. They were
relevant to the lectures. In fact, I'd say that the lectures perfectly
complemented the homework. I liked that the students could bring up a
difficult homework question in class and the instructor would have the class
do it as a whole.
having an in class tutor to help out
It helped me think more carefully about word problems, and the homework
helped with practice.
Mr. Girard didn't just give us one way to solve things. He showed us a few methods and let us pick
which ones we were most comfortable with. He encouraged us to try different methods.
Creating Study groups and going to tutoring sessions.
It provided me a chance to experience the rigor associated with college courses; I now feel better prepared to handle college next year.
Using self analysis on exams to rethink problems on why you got it wrong and how to
correct it.
The in-class discussions of problems.
I found that the Part B of his testing was the most helpful towards my
learning.
The part where I learned calculus.
The "Help Me Solve It"'s on My Math Lab and working with others6
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22. What did you find most valuable and helpful about the instructor?
Ryan is always ready to help answer questions or to tutor students through problems with a cheerful attitude. He makes you feel like he really wants you to succeed.
He gave us opportunities to improve our grades and was always willing to stay after class to assist
us on any issues we were having with any particular lesson.
He allows many extra credit opportunities and gives a chance of redemption
on tests by adding a part B.
He never hesitated to spend class time answering student questions, even if
it meant slowing the class down.
He is easy to approach and talk to when I need help.
He is always willing to help the students.
His enthusiasm for the course material is quite catching and by the end of every lesson I always am
ecstatic to learn more. He also seeks out the student and inquires if they have any questions even
when not prompted to do so.
He is willing to provide help outside of class if you are willing to meet
with him.
He has a wonderful way of explaining thing that would have otherwise made absolutely no sense.
Getting our books replaced when it fell apart.
His enthusiasm and knowledge of the subject matter are unmatched.
I found that Girard's organized notes helped break down difficult sections to
understand.
I found that the most valuable thing about Ryan, within the context of our
relationship as teacher and student, is that he is the person who taught me
calculus, and he did so very well.
The hardest part about the course was that some of the homework problems took us too far out of spectrum of the topic we were learning, so that too much time was taken to work the problem that you get lost and lose sight on what you're supposed to be getting out of it.
None. Great method
better audiovisual equipment
The class schedule should follow the traditional MWF or TTR instead of TWTR.
Also some of the homework problems weren't things we learned in class.
It was difficult to finish weekend homework, having no WiFi at home. I would
try to finish it on Monday or literally search for someone else's internet
around the neighborhood, which barely had any signal.
Probably lower the homework load a bit.
I was very content with Mr. Girard's teaching methods.
To improve the course there should be study groups available for students to
attend that are hosted by tutors.
Make the homework less time-consuming
encourage more work done on paper, as opposed to online.
Creating Study Groups, and Review Sheets before tests/exams
Maybe try not having homework so I can have a better grade. I apparently
learned the material without consistently doing my homework, but we'll see
how I do on the final.
Less problems on My Math Lab. More problems with the "Help Me Solve It"
option. Study guides before tests or more specification on what will be on
tests63
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7 (4434. Other comments:
Great class, great instructor!
I am very thankful for the professor I got this semester. Despite outside factors that affected my
studies, when I was in class I felt engaged and time flew by in the classroom. The course is very
demanding but encourages you to raise your hand and ask questions and be willing to admit that
you don't understand something.
It was fun.
Overall i enjoyed the class and think you're doing a good job. I could
understand better this semester than my first time taking the course with
another instructor.
None
This was the best class I have taken in my brief time at KCC.
Ryan's class is easy if your willing to learn the course material.
Good Team work this semester on math faculty in providing math clubs and opening the door
to math with a friendly atmosphere.
It is an online course and also met at night. If not for the setup of this
class I don't know how I would have gotten these required credits.
It was a good refresher before going into the higher math classes.
Its online.
Being that this course was mainly completed online, the most valuable and helpful part of this course was meeting face to face with the instructor and other students. Learning the material from Ryan gave a better understanding of what was not retained or learned from the online assignments.
Working on my own and figuring out how to do course work from the computer
I thought that Ryan Girard was able to always help me when I had questions about a math problem. He also explained it very well so that I could understand it.
As always, I say I know more than I did last semester but I don't think that
way of it now. This hybrid course kicked my butt this semester. I felt like I
was taking a 100 level course. Its only Math 26! Why did it feel like I was
being treated as if it were a 100 level course? This didn't make any sense.
If I retake this course next semester and not a hybrid course and I pass,
then I definitely believe it was because it was a hybrid class. I would
definitely not advise anyone to take the hybrid course25
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22. What did you find most valuable and helpful about the instructor?
Mr. Girard is great, a real positive for KCC. Mr. Girard is understanding,
patient and compassionate. People like me (who find math challenging to say
the least) can quickly become intimidated or afraid of an instructor. Mr.
Girard never made me feel this way. Part of an instructors goal is to make a
student believe that they can do the work, Mr. Girard does this.
He was able to identify where problems occurred, and was able to address it
promptly.
When you need help and have questions he is willing to help you and makes
time.
The way he uses real life situations and objects as tools to better understand math. Also, he is willing to assist in any way possible. He teaches in a way to make it more interesting, especially for people who really don't get math and have difficulties.
I found that the in class participation was valuable because I was able to ask questions when I needed help on the homework questions.
I would say that Professor Girard is a wonderful math teacher. I would say
one of the best. But the hybrid course does not let you learn at your own
pace and definitely does not allow enough time for you to absorb it before
going into the next chapter. And when you are learning the next chapter
you're taking an exam about the last two chapters. The flow of this course is
not equivalent to a fair learning style that allows you to process the
information learned throughout the semester.
I would have preferred to have more class time to go over drills, working a
selection of problems over and over (different sets, but same concept). We
currently spend the entire class time working two or three problem sets. I
wish were could have came to class everyday and completely gone through the
homework, not just spending the entire class to go over one or two problems.
There are so many concepts in this course, working two or three a class was
not enough.
The only improvement I would like to see is during class time there should be more reviews, not just on the current chapter but previous chapters. Especially before the chapter exams.
More in class instruction, not enough time to cover all of the material learned in the homework.
I thought there was nothing to improve. I mean you really just get what you get when you take this class. The online version is really helpful as well as Ryan Girard in class.
This has been a difficult course. It is jam packed with a lot of information. I think Mr. Girard should
use a style of reinforcement and summerize the math concepts..even with usage of the videos, I had
difficulty with a lot of the math concepts, because there was so much information.Even with the
homework, there seemed to not be enough time to be able to study enough on each Math segment.
Professor Girard's teaching style is awesome. I have nothing against
Professor Girard, it's the hybrid course that is affecting my grade, not the
Professor. He is only following the syllabus of the two courses combined37
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34. Other comments:
I feel that this class could better serve students if held once a week on
Wednesdays for 2.5 hours instead of the 1 hours and 15 minutes on Mon. &
Tues.
It takes awhile for students to settle every time we arrive for class, a
good 5-7 minutes are lost just settling in for class (announcements,
information). It also takes awhile to settle into the "groove" of math,
once we are on a roll and are understanding and working the concepts, class
is over. Also, traveling to and from campus takes much time and effort for
working people like me. I spend an hour commuting to and from campus that
takes time from my studies. I really feel a 2.5 hour class once a week
would be more beneficial to students.
From my understanding, this was the first time for the hybrid course. Not only was the course mainly online, but meeting face to face during an evening course worked out perfectly. Ryan is very knowledgeable and it helps when a teacher is very excited to teach it. Makes a student who does not like math or has a hard time, somewhat changes the frame of mind about math. The only other comment is that I wish Ryan could be available to do the same type of teaching for Math 103. Math is very hard for some and he made it very worthwhile being in his class.
I thought that Ryan Girard was a great teacher and really smart.
This hybrid course needs to be reevaluated and not suggested for slower
learners like me who have to be coached through every class on the material
just covered.
I found the textbook and computer book very help full. I loved how they had videos explaining the most difficult problems in the chapters. And the textbook explained really well.
all online assignments, organizers and text book was very helpful through the chapters and problem.
I found the help from the teacher and aid and all the helpful tutor guides real good.
The online tools really helped when not89
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22. What did you find most valuable and helpful about the instructor?
Mr. Girard is really an awesome teacher. He respects everyones way of learning so therefore he would explain to where he or she could understand. He would really break down a problem and not move on to the next step unless he saw that you really understood it. He is vey patient meaning before he would explain a problem that an individual doesn't understand he would start from square one regardless if it meant to go back 5 chapters ago.
his examples by using relatively to my personal life at work for examples to my school work.
He was a fun and ethusiastic person as well as well knowledgable which made it easy to ask or talk to him when I needed help.
Mr. Girard is very smart in his field. I cant wait to take another one of his
math classes!
I probably wouldn't recommend this summer course to someone who just took a compass test and qualifies for these math classes.
nothing
Maybe a little more HS teacher type like taking a little time to go over problems students are learning and any given point to show easier more effective ways of doing the problem instead of having to repeat himself to each student so he doesn't have to waste his time repeating himself. But one on one is also good sometimes.
I wouldnt change anything about this summer program.
24. Any comments about the TA?
our TA was awesome! I really loved going to class having a TA and a teacher.
computer is a little slow but other than that environment is great and everyone was willing to learn.
It was a nice learning enviroment. Queit, cool, and had everything you would need for the course.
envirioment was perfect!
34. Other comments:
I really enjoyed Mr. Girards teaching and the way he set up the class. Hes a very positive patient math teacher. I would totally recommend this course to anyone who previously took a math class prior to summer school.
KEEP UP THE GOOD WORK AS A TEACHER!!!! MR GIRARD IS VERY HELPFUL AND VERY KNOWLEDGABLE.....
I had a great time and experience in this course. It was challenging, but also help at any time that you needed it. I also liked how the teacher made sure their was a way to get in touch with him if you needed to during non class hours. It was also a nice enviroment where you didn't feel over welmed or stressful while you study or did your work. I really liked the set up of the course because it was chellenging but not over bored to where you couldn't keep up with the course outline and also how you can work ahead so you were able to go at your own pace.
Mr Girard is the MAN!!!!!! thank you for all your hard work and dedication!
I loved the fact that it was all self-paced so that you could work ahead of the class. It was also kind
of nice to have a small amount of students so that we all got a lot of one on one time with the
teacher or tutor.
between the teacher, the tutor, and classmates there was always someone to help me9. The instructor clearly stated at the beginning of the semester the objectives of the course and requirements20. The instructor was consistently well-prepared and organized for class.
I can easily apply what I have learned from this class to the real world.
I find what was most valuable about this course was that Professor Girard created examples
that could relate to my daily life, which gave me a better understanding.
Didn't expect for the math in this class to be like this. It was based more on theories,
probabilities, and logic. It was tough but Mr Girard did his best at explaining the presentations
Everything was valuable and helpful.
Mymath lab and extra credit
Being able to do a part 2 on the test
Mr. Girard always made sure that each topic was explained clear before moving on to the next topic.
I ended up looking more into starting a ROTH IRA, and i begin this month.
It was great being able to use all the different math materials you'd need to
know for the real world instead of doing geometry and algebra.
he was able to explain things very clearly. I've learned many things that I
will use in the future and that is the first time I can actually say that.
My Math Lab is a really great program to use. It is a little awkward at first transitioning from
Hawkes to My Math Lab, but I find that My Math Lab is better program. Not only that, Ryan is a
really great teacher. He is friendly and really easy to talk to. If you have any questions or problems
you can ask him and he'll try his best to help you.
The powerpoints in class helped with proffesor Girard lectures and My math was very easy to use on heplful.
Ryan was very helpful when I didn't understand a problem and he made sure that I understood it17
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12 (7122. What did you find most valuable and helpful about the instructor?
Good explaining about everything
He is knowledgeable and passionate about the subject. He is also always prompt at returning exams.
I found what was most valuable about Mr.Girard was that he was willing to help a student
understand more if needed. Also he wouldn't give up on the student.
I liked that Mr Girard showed concern for his students to effectively learn what he was
teaching
He is happy to provide further explanation of procedures when requested
Ryan Girard's examples are very valuable. He tries to use real life situations and tries to instill them
in the math problems he is teaching.
the online homework is much better than using the Hawkes system. help me learn much better
He knows what he is talking about, and can explain in a understandable why.
He was able to break everything down into simple steps.
very very knowledgeable about each topic at hand and took the time to answer all questions in a way that each individual could understand it better.
I liked the finance chapter, it made me realize how to look out for interests, when either trying to get a loan, or looking to save for the future.
He was always very enthusiastic and passionate when teaching us, and was very
helpful when we'd be stumped on certain criteria.
Every now and then on his lecture he stops and ask us if we need help and
most of the time we have questions and he is immideatly teaches us an easier
way to learn the topic.
Ryan is overall a really great math professor. He is one of the only teachers I am able to follow and
understand when the material is presented to me. He also posts the lecture notes ahead of time
which is also really helpful to have in class during lecture time. When he presents the material, he
presents it clearly and easily for the class to understand.
He was knowledgable of the subject, and the matrieals were made easy to follow which allowed for his students to do well.
He really wants us to succeed in his class so he tries his best to help us out with understanding difficult problems.
All in all, math 100 was a little tough for me, I think chapter 11 really was the most difficult
chapter which could've been explained slower, and clearer. I was just confused on this
chapter
N/A
Have students do stuff on the boards.
keep up the good work!
I enjoyed your class.
Honestly, I liked it just the way it was.
During group work when we have a test or an exam to study for I had hope to
atleast pick our own group once because I was working with the same people
the entire time we did the group works.
The only problem I had sometimes in his class is that sometimes the lecture notes are blank on
certain slides. I suggest saving it without any transitions or added effects so that it shows up on the
pdf file for printing. Then adding the transitions and effects and saving it under a different file
name41
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31. Other comments:
Good teaching skills
Thank you!
Great semester!
i really enjoyed this class
Thanks for a great semester.
Honestly, I was afraid of taking this class. It may not be the highest Math course available on
campus but Math is not my strongest subject. But because Ryan is so easy to follow and work with I
was able to learn a lot of the material with ease. If you're considering taking a math class take it
with Ryan, you won't regret it. :)
This class was great and I learned quite a lot of info from this class. Thank you!
MyMathLab was very useful in understanding the subject. The videos were most helpful.
The videos!
discussions with the entire group
Since this is a course on how to teach math to elementary students, I found
very helpful and useful the class videos where I could see different
approaches to teach the same topic.
Group work helped me a lot. My group members were really great and helped me to learn subjects and pushed me to do better.
My math lab. The example tab was extremly helpful
The most valuable thing was my instructors flexibility and patience. There were many difficulties I
encountered but none he was not able to walk me through.
Just like Math 111 from last semester I found this course to be very well organized and taught. There weren't any surprises on tests that weren't covered in the modules, due dates were according to the syllabus, the instructor was easy to reach with questions, modules were available when they were expected to be available. I found the content itself to be challenging but the course was just really well taught.
What I found most valuable and helpful about this course was becoming more confident in my math skills. Especially after having to answer a question more than once if I got it wrong. It can sometimes be frustrating. But when you get the answer its very rewarding when you see the green button that says well done, or good job!
What I found most helpful about the course was how Mr. Girard set up each module so that we were able to learn the content in a variety of ways; vidoes the book, slides. It allowed for all learning types to learn.
All the different methods of learning and the resources that we had. Each week had a couple of tasks that needed to be done, wether it was working with our groups, discussions, or our math lab homework. The resources were always availible to us especially on my math lab and when we had special assignments like finding videos online that relates to what is being taught that week in class or finding online games that relates to what is being learned.
The support of the professor76
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22. What did you find most valuable and helpful about the instructor?
Preparing for exams with other students on the discussion boards were very helpful. I also like that he uses homework questions on exams.
He is quick to email back and is very knowledgeable.
always available to answer questions and challenged thinking
He's very helpful and selects very interesting material for the class. He's
always willing to help students and he answers emails with questions right
away.
He graded very appropriately and gave helpful feedback.
Ryan was very meticulous, and gave clear detailed instructions of what he expected from us as
students. It was nice he got back to emails almost instantly and was open to many suggestions. I
really appreciated Ryan taking the time to help me with specific problems I would email to him. It
was nice to know he cared about his students even from a different island. Ryan knew his content
and could give detailed descriptions of how to solve problems we got wrong.
The comments on the group projects about areas to improve. This shows that the coursework is reviewed and isn't just given to receive credit. Reviewing problems from the previous test also showed that the instructor was concerned with the students learning the content of the course and not just getting the right answer. I also found the videos of actual student exercising to be extremely helpful!
What I found most valuable and helpful about Ryan Girard was that when there were questions that I had about problems I was having he responded back to me in a timely fashion.
He always responded to my questions quickly and with a good answer.
The he is knowledgable about his subject. Some teachers don't bother with making online classes fun or even interesting, he is determined to watch you succeed and will go out of his way to help you or find a way to solve your problem, he never gives up on you.
make the modules available with a link on laulima like math 111. it was harder to go through gmail every time i needed to find assignment info
I think he teaches very well.
update the videos you use in mymathlab, khan acedemy is much better
i really disliked the group projects, it was just a bunch of busy work
My only suggestion would be notes for tests, we barely go over each homework assignment and its
near impossible for us to memorize everything. There are no "study sessions" and it is very hard to
make time to meet up with people we don't know.
...
Sometimes the over load of module tasks can be a little over whelming but overall the course is great and since I took his online math class last semester I am already used to it85
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32. What did you like most about your online course experience?
working with groups was helpful in learning the material.
We were able to do our assignments on our own time. We knew exactly what was due and when it was due.
online classes best fit my schedule
I could study late at night time, that's the only time available I have. I
can't go to any class during the day, this is the only way for me to continue
college.
Having a week to do assignments was very helpful time wise i could finish the work on my time when it was best for me. he was also consistent with when assignments were due which was helpful.
I liked the flexibility of time, I could choose when and where to do my homework.
This course was taught in a way that I feel I received the same instruction as a student taking the same course, but not online. I like the online alternative because I am a nontraditional student and have other life commitments that make traditional college schedules challenging. I am also on a different island than where this course is taught and, in fact, I am not near a campus. The nearest satellite facility is 45 minutes away.
What I liked most about my online class experience is that I was able to learn and not even step foot in a classroom. I prefer online classes because I get to be home with my family, even it is means I am working in a different room so I can have my piece and quiet.
The ability to learn on my time.
The class interactions that we had, for an online course it was fun being able to work with groups and have online discussions like a virtual class room.
I liked the freedom of creating my own schedule. Being a working mother of two it is not always
easy to get it all done. But having the ease to create my own schedule has really helped me with
this.
33. What did you like least about your online course experience?
collab sessions on sunday. yes I could make these but sundays are family time in my house and it was at a time i would be trying to put my son to bed.
NOthing
the group projects! i didnt sign up for an online class to work around other peoples schedules
I had a hard time getting group work done, it was difficult when you are not all on the same island.
This is the best means for my life at this time.
Group Projects.
The lack of socialization.
I know there is a ton of material to cover , but the weekly module over loads are sometimes hectic.
Group work is a little tricky with online course.
34. Other comments:
None
This was a surprisingly decent course, unlike the gruesome course you take at UH manoa, those
instructors only confuse us.
I really feel like I learned so much in this course and I am so grateful for the positive experience. As I mentioned before, I found the coursework to be challenging and frustrating at times but the actual course instruction was superb8
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31. Other comments:
Thank you I'm looking forward to taking the next class in the fall with him.
None
Good teacher!
Ryan is the teacher to have if you are needing help with your math skills. I
would recommend his classes to anyone. I am looking forward to taking his
class in the fall. I was disappointed that I missed the deadline for
nominating instructors for excellence reports. He deserves recognition for
his enthusiasm and his patient, kind way of helping his students.
The class was well organized and there were many built in systems to help me along the way while doing the homework. The tests were accurate reflections of what we did in class, so I felt confident that if I studied the materials, I would be prepared.
It gave me a fresh perspective on math.
I think that the most valuable part about this course was a combination of
everything. From the online homework, to the discussions, to the group work,
to the online classes.
*Relating each subject to the ways it will be taught in the classroom.
*The online aspect!!!!!
*The quizzes gave options to assist in finding the answer if I was having trouble
*The instruction videos weren't boring
*I appreciated the instruction videos that featured real students
*The analysis of students' mistakes
*Mixed media style of learning
*Personalization of the course - this is especially difficult because it is online
it was a refresher for math, even though it was a course to teach math it helped like a refresher with the basic algebra and math concepts.
The projects we did throughout the semester were very helpful and actually made me feel lie I was
on the road to becoming a teacher!
I liked the videos of the children because they show their thinking and their
challenges with math. Great tool for when we become elementary teachers.
I like the set up of the online course work in my math lab. It has organized notes and lots of videos to help students understand the material.
when doing homework online i really like how we got view an example of the problem and if we didnt know how to do the problem we could press on the bottom that said "help me solve this problem". I also like how if we had any questions at all we could post them in the discussion post were we could get answers right away.
I found the videos and pdfs to be helpful along with reading the textbook. Also being able to do the homework in My Math Lab is great because you can also click view example if you get stuck.
Having the opportunity to learn different strategies in teaching todays math in schools today67
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0 (0%)
4 (33%)
7 (58%)
22. What did you find most valuable and helpful about the instructor?
I felt that the Ryan Girard really cared and was there for us, which is rare for an online instructor. He answered all of my questions promptly, and always made me feel like it was easy to contact him with any concerns at any time.
He knows how to teach math!
Thanks for giving feed back on the exams. I also liked how on some questions
you asked for an explanation because it challenges us as students to make
sure that we have a complete understanding of the content.
*Responded to emails in a timely manner
*The class was well thought out, interesting, fair, and balanced
mr girard is very knowledgable about math and expecially with communication if you have any questions or problems he gets right back to you through email. He is helpful if you made a mistake or dont understand one of the problems he will work with you.
The lessons and assignments he had us do were very helpful and really got me thinking!
He responded immediately to questions and emails. He's always very helpful.
I like how he replies back to students quickly and he gives through explanations.
He really took the time to look over our problems and see if we were doing the problems right, using the right methods, etc. He didnt just base the grade off of the answer he based the grade off of how we worked the problem which i though was really good that way we could see what exactly we did wrong and how we could fix it.
I found it helpful when he would put instructor comments in My Math Lab to make sure we understood what was being asked.
His response to questions via email were done promptly which in turn gave the student the advantage to move on in their assignments.
I thought there were too many group projects and what happens is that one person does all of the work and everyone is getting credit for it. I also think that as a "Math for Educators" course, it would be beneficial to have more lessons on how to actually structure a lesson and teach the material we were learning.
he has a very effective style of teaching.
group projects were very inconvienant because i was not able to do my work on my own time like
you would like to do when taking an ONLINE class. i found them really annoying. there must be
another option
In the beggining of the course maybe state more clearly what each weeks assignments will look like
and include details about blogs and discussions
The course has too much homework for the 3 credits. There are too many
questions in MML for each of the modules, they are supposed to be a review of
the topic we already learned in school. It's time consuming!
As for Ryan Girard, I think he does a great job teaching this class!5
12
0.71
Freq(%)
0 (0%)
0 (0%)
1 (8%)
3 (25%)
6 (50%)
0 (0%)
32. What did you like most about your online course experience?
I learned a bunch of new computer skills.
It can be done on my time.
the different resources that was availible to us and the feedback that we got each week.
That I could work at my own pac
I can work on my homework whenever is convenient for me.
I liked how we got to interact with our classmates, especially through the use of google documents.
It was very time efficient even though I am a full time student and have a job I still had time to do homework for math online. Deadlines were great and provided me time to work the homework on my own schedule. Google docs allowed me and my partner as well as my teacher to interact in problems and working out the material.
I liked that the homework questions were in My Math Lab.
Working on my own pace, work schedule and my other classes throughout the week.
33. What did you like least about your online course experience?
having to input math things typed out on the computer during the exams . . . it is hard for me.
I have no time :-)
sometimes it was hard to communicate with other classmates for the group assignments.
Couldn't ask questions and get a response immediately
It was my first online course. I think it was good and I will continue taking
classes online.
n/a
Group projects
The group projects I found to be difficult b/c you are not physically together doing the work. It's hard to meet up online because people all have different schedules. I think it should be mandatory for each group to meet at least once a week on google chat or in the chat room.
None
34. Other comments:
This class was great! It wasn't easy for me by any means and I didn't get all the questions right or understand the concepts immediately. However, it taught me the concepts, I understood what I was doing wrong, taught me about children's learning styles, and My Math Lab was easy to use. I would highly recommend the course taught by this instructor.
I've never taken a math course online because i thought it would be too complicated but it wasn't that bad.
I Liked the way he taught us he engaged all his students and made us feel apart of the class.
multiplication
I liked hawks it made sure i did the home work right befor you were finised very useful tool
Hawks, notes of chapter, and warm ups that relate to hwaks problems.
What I found most valuable about this course is the fact that I can apply Math to everywhere and anywhere that I go. So it is very useful/ valuable.
The teacher's commitment and fun explanation of algebra.
Ryan would start out the class with warmups which are things people are having problems with.
He was there to answer any questions
the tutors coming in to the class for help44
Freq(%)
0 (0%)
0 (0%)
0 (0%)
7 (78%)
2 (2213
Freq(%)
0 (0%)
2 (22%)
2 (22%)
3 (33%)
2 (2263
9
0.74
Freq(%)
0 (0%)
1 (11%)
1 (11%)
6 (67%)
0 (067
9
1.22
Freq(%)
1 (11%)
0 (0%)
2 (22%)
4 (44%)
2 (2220. The instructor22. What did you find most valuable and helpful about the instructor?
teaching skills
he hade us think for ourselfs in stead of giving us the answers I felt he was a good teacher I would recomend him I hope to have his class for math 25 I learned a lot in his class
Class discussion.
What I found most valuable and helpful about Ryan Girard is he has a true passion for teaching Math so I know I would learn something from his course. And he made it helpful when he uses reality examples like population of the U.S.A. Things like that.
His love of the subject makes it fuun to learn. His availability and his eloquewnce helped me to proceed when I am stuck. Makes me want to join the math club
The course worked for me I learned math that I did not know in a resonable time frame
-Don't be so up tight sometimes and go with the flow
-Yes it's possible to be a FUN instructor as well as a FIRM instructor
-Make more group work assignments
-Do more class hands on activities (personally I think I learned and remembered a little more through this method)
-Just make everything more FUN and ENJOYABLE to Learn
dim the lights.
Making sure all materials are introduced in class before exams. Not make the question on the test the exception to the rule and we have not seen it before31. Other comments:
none
Keep teaching you have a real passion for it, but remember to try and incoorperate hands on activites like in group work (example: When you put us in groups and gave us a small white-board and a dry-erase marker to do our work on it. That was fun and allowed us to be more active)
We went over the basics of algebra problems to let him know what we know
already.
Hawkes is very helpful, and the teacher helps the students understand whats being taught, and
encourages participation.
What I find most valuable and helpful is the Hawks Learning System
The course really helped me learn what I could not learn in high school.
This specific Math 25 class has made me more confident in my math skills. I
admit to have taken math 25 two semesters ago, but dropped out due to issues
with the teacher as well as for some reason, not being able to retain
information. This math class, however, has made a lot better. Now that I
think back on it - I regret not finishing math 25 when I had a chance - at
the same time, not really, because I really enjoyed and learned a lot from
this course.
The grouping thing. so we could ask and learn from others.
I found Ryan Girard very intelligent and energetic. I liked the hands on
warm-ups we did every beginning of the class. It's a great way to re-freshen
what we learned about the previous classes.
nothing
What I found most valuable and helpful from this course is the fact that it was not too big of a class; so therefore, there was enough time for one on one with the teacher. Also the different methods shown through each chapter as we learned about new topics. The ooportunity to ave more than one option to simplify the work so we could choose an option that may be easier than another.
Cheerful attitude of the instructor made learning the subject more enjoyable. The instructor
clearly explained the methods and steps required
It helped me understand the concepts of mathematics that i missed in highschool
This course refreshed me in algebra.
This course taught me how to do math that I will need in the next math class
that I have to take.
I found that doing HAWKS was most helpful for me because I got to
do my work learning as an individual and independently on my own time. It also helped a lot with the tutor section on it and providing examples, which showed a way of explaining and gave me a better understanding in what topic I was required to learn.
The course was able to refresh my mind on some math concepts that I had
forgotten.
it helped me feel confident about math and that I could actually do the math once I learned the steps to the various problems.
I find his class work and notes very useful when I can't comprehend Hawk's
instructions. I like the flexibility of doing my homework offline if I'm not
able to get connected to the internet. Also the online tutoring is very
useful.
Really helped
Hawkes was a really nice way to practice. I felt like having physical homework would make learning
far more long-lasting.
This course helped me brush up on and strengthen my algebra 2 skills, as I did not learn them very well in high school.
I liked his method of teaching and the Hawkes online math problems really prepared me for the exams.
I liked how he has you interact with the people around you to figure out the answers77
13
0.44
Freq(%)
0 (0%)
0 (0%)
0 (0%)
3 (23%)
10 (7728
Freq(%)
0 (0%)
0 (0%)
0 (0%)
1 (8%)
12 (9229
Freq(%)
0 (0%)
0 (0%)
0 (0%)
1 (8%)
11 (858 (6225
16
0.93
Freq(%)
0 (0%)
1 (6%)
2 (13%)
5 (31%)
894
16
1.24
Freq(%)
1 (6%)
1 (6%)
3 (19%)
4 (25%)
7 (442
16
1.15
Freq(%)
0 (0%)
2 (13%)
2 (13%)
2 (13%)
9 (56%)
16. The instructor suggests specific ways students can improve.
Mean
N-Size
Std Dev
Strongly Disagree
Disagree
Neutral
Agree
Strongly Agree
4.0
16
1.25
Freq(%)
1 (6%)
1 (6%)
2 (13%)
4 (2533
16
0.98
Freq(%)
0 (0%)
1 (6%)
2 (13%)
3 (19%)
9 (56%)
18. The instructor was available and willing to help with individual problems outside of class. |
Functions and Change: Model Approach to...
9780618858040
ISBN:
0618858040
Edition: 1 Publisher: Houghton Mifflin Company
Summary: Intended for precalculus courses requiring a graphing calculator, Functions and Change emphasizes the application of mathematics to real problems students encounter each day. Applications from a variety of disciplines, including Astronomy, Biology, and the Social Sciences, make concepts interesting for students who have difficulty with more theoretical coverage of mathematics. In addition to these meaningful applicat...ions, the authors' easy-to-read writing style allows students to see mathematics as a descriptive problem-solving tool. An extended version of the successful Functions and Change: A Modeling Approach to College Algebra, this text includes three chapters of trigonometry.
Crauder, Bruce is the author of Functions and Change: Model Approach to..., published under ISBN 9780618858040 and 0618858040. Five hundred eighteen Functions and Change: Model Approach to... textbooks are available for sale on ValoreBooks.com, one hundred forty one used from the cheapest price of $17.80, or buy new starting at $121.34.[read more]
Ships From:Multiple LocationsShipping:Standard, ExpeditedComments:RENTAL: Supplemental materials are not guaranteed (access codes, DVDs, workbooks). 0618858040 This book may have some creased pages or bent cover corners. There may be highlightin... [more]RENTAL: Supplemental materials are not guaranteed (access codes, DVDs, workbooks).Seller Rating:(0)
Ships From:Rushville, ILShipping:StandardComments: 0618858040 This book may have some creased pages or bent cover corners. There may be highlightin... [more].[The course was great at giving concrete, real world examples. Actual data was used so you could understand why you were learning the material, and that makes a big difference in your understanding. The many examples help students learn for themselves the material.
The section on logs and exponentials really needs some beefing up with more basic abstract work. Too little time was spend on those subjects. |
Product Description
The Algebra 1: The Complete Course DVD Series will help students build confidence in their ability to understand and solve algebraic problems.
In this episode, students will learn to use concrete examples and practical applications to understand algebraic concepts. Students will learn about sets of natural and whole numbers, sets of integers, sets of rational numbers and sets of real numbers. Grades 5-9. 30 minutes on DVD.
DVD Playable in Bermuda, Canada, United States and U.S. territories. Please check if your equipment can play DVDs coded for this region. Learn more about DVDs and Videos |
More About
This Textbook
Overview
Lucid examples throughout draw students into the text and clarify the key material at hand.
End-of-chapter exercises allow students to test their understanding of the concepts discussed in that chapter.
Answers to selected exercises appear as an appendix and allow students to check their work
Zill's "no nonsense" approach provides a straightforward, concise, and modern introduction to trigonometric topics at a level appropriate for today's undergraduate student.
Notes from the Classroom are remarks aimed at students and address a variety of student issues, such as alternative terminology, reinforcement of important concepts, misinterpretations, common errors |
18,742Linear equations, slope, direct variation, and functions are major concepts in Algebra 1. Students may encounter simultaneous equations, quadratic equations, parabolas, and inequalities for the first time.
For these and other reasons, students often feel uneasy...
read more
Offering
10+ subjects
including precalculus precalculus |
This applet encourages students to use the mathematics of the motion of a projectile to choose angle and velocity necessary to hit a roving target. The source code is available from the link within the Sharing Area of the Flash Forum.
article explores the symmetry method in elementary differential equations, which uses the invariance of the equation under certain transformations to create a coordinate system in which the equation greatly simplifies.
This Java applet plots up to 10 functions simultaneously with zoom and trace features. It can be used to study properties of functions or approximate solutions to equations. The syntax for entering the functions is given on the accompanying web page. |
I am a teacher assistant for a teacher that plans her lessons around this workbook. Instead of taking notes she goes through this workbook to teach the lesson. This is a good idea because the students are taking more time learning instead of worrying about writing everything down. However, this is NOT the workbook to do this with. I will agree that it covers a lot of good material and has some good examples, but the parts meant to "teach" are terrible. All of the problems are fill in the blank. Any math educator should know that there are MANY different ways to solve one algebra problem. Well, when solving for an unknown angle in a triangle, this workbook has fill in the blank spaces for solving the equation. Something like: __x + 2 = 3__ + x. I even had to step back and try and figure out what the workbook wanted for each blank because it sometimes would solve equations in a completely different way than I would think to do. Therefore, the students only see one way and we miss out on class discussions of how their classmates solved the same equation. The worst part is, when they get to a problem on their homework where they have to do everything themselves (no fill in the blanks) they are lost. This is especially prevalent in the sections on proofs. The workbook presents 3 different ways to write proofs, 2-column, flowchart, and paragraph (though 2-column is the mostly used) Anyway, the workbook also does fill in the blank proofs. Even MORE so than with algebra equations, proofs use reasoning and creativity. The fill-in-the-blank method takes both of these aspects out of the proof writing process. When my students took their first quiz where the proof was NOT fill in the blank, they were stuck, didn't know what to do and I had to make the proofs extra credit. After that I got rid of this workbook and started teaching them how to do work on their own.
I understand that this is just a 2 dollar workbook, but if you plan on getting a resource for your students to use, do NOT order this workbook! The only thing this workbook teaches the students to do is be dependent on being given TOO much information. Get something else that teaches students to reason on their own and make connections between relationships in ways that makes sense to them... not just the one way that this workbook forces the students to do.Read more › |
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Product Description
By Matthew LarsonFocusing on similarities between these new standards and those outlined in NCTM's influential Principles and Standards for School Mathematics, this handbook efficiently highlights reasoning processes that are essential in any high-quality mathematics program. Students who know these processes are set to "do math" in any context, real or abstract.
Research indicates that many individuals within a school can and do contribute to the work of leading and managing a school. This guide can support anyone who is working to improve mathematics education—either alone or with others.
Make the most of the rare opportunity that the Common Core State Standards offer for rethinking school mathematics and creating exciting new pathways from high school to college and beyond.
Includes answers to "Frequently Asked Questions" and detailed lists of resources to support school mathematics programsCustomers Who Bought This Also Bought... proof and the process of proving. It is organized around five big ideas, supported by multiple smaller, interconnected ideas—essential understandings"Powerfully affirmative to teachers who may feel less comfortable teaching math than reading . . . equally appealing to teachers who are math gurus and those who are less confident in their math skills." —Sally Moomaw, Ed.D., Assistant Professor of Early Childhood Education, University of Cincinnati
The National Council of Teachers of Mathematics is the public voice of mathematics education, supporting teachers to ensure equitable mathematics learning of the highest quality for all students through vision, leadership, professional development, and research. |
Designed to engage the student as an active participant in the classroom, the Mathematics in Action series presents concepts using real-life activities and reinforces those concepts with practice exercises in the text. The second book of a three-part series, An Introduction to Algebraic, Graphical, and Numerical Problem Solving, Third Edition, illustrates how mathematics arises naturally from everyday situations through updated and revised real-life activities and the accompanying practice exercises. This unique approach helps students increase their knowledge of mathematics, sharpen their problem-solving skills, and raise their overall confidence in their ability to learn. Technology integrated throughout the text helps students interpret real-life data algebraically, numerically, symbolically, and graphically. The active style of this book develops students' mathematical literacy and builds a solid foundation for future study in mathematics and other disciplines. CourseSmart textbooks do not include any media or print supplements that come packaged with the bound book. |
Students then move into measurement of plane and solid figures through use of formulas as they
find area and volume. Right triangles are explored through a study of the Pythagorean Theorem as it relates to square roots, irrational numbers and special right triangles. An overview of data analysis and probability teaches students how to represent data and determine probabilities. |
Students will practice the new skills using cooperative learning games and real–world applications. ... Reteaching Properties of Logarithms Name: ... Logarithms 32 Student answers should include comparing the original population and the population in.
large for these calculators to handle. Using the properties of logarithms and algebra, we can solve for the correct answer. On the TMSCA state calculator contest in ... Practice problems with answers ... {.3,.6,1.5,1.8,2.5} Calculate the x value given y = 5; x2 = 71.192… x = 8 ....
software provides students hands-on practice with these concepts. Additionally, self-check activities are ... Reading and writing skills are incorporated throughout the course as students participate in discussion ... Properties of Logarithms and Logarithmic Equations.
In this unit the student will need to learn the following skills ... Use a slide rule Information from web site Internet site Properties of logarithms Practice problems using Windows Journal ... 8 Exponential and logarithmic equations Practice Problems Text ....
Basic skills Quiz 2 is over exponential and logarithmic functions. You need to know the ... We can use these properties to solve equations involving exponents or logarithms: • If e3t+1 = 4, ... Practice Problems.
MATH 70 TEACHING IDEAS FOR CHAPTERS 2, 3, 4, 5, 6, and 8. Topics Sections Number of Weeks ... Student Skills for Chapter 5: ... This section in the book is really more practice with laws of logarithms, ....
Schools that utilize these standards "enroll" students in a mathematical apprenticeship in which they practice skills, solve ... 2.1 Use concrete objects to determine the answers to addition and ... 14.0 Students understand and use the properties of logarithms to simplify logarithmic numeric ....
asked to use algebra and arithmetic skills from previous math courses. To prepare for this exam refer to the Texas Essential Knowledge and Skill for Algebra 1, ... • use properties of logarithms to solve problem situations (for example, population growth problems) ALG 2B.
... 8-2, 8-5 (A) develop the definition of logarithms by exploring and describing the relationship between exponential functions ... Texas Assessment of Knowledge and Skills Where You Practice TAKS Tune-Up features provide ongoing TAKS ... use properties of transformations and their compositions to.
Practice Workbook (receive with your text book) ... Answers only or questions only will not be counted as a completed homework assignment. ... 1 day Sections 8.5/8.6 Properties of Logarithms/ Solving Exponential and Logarithm Equations.
• Prerequisite Skills Practice • Previews of Math 3 topics ... Answers for all of the Prerequisite Practice problems are provided at the end of each unit. 10. ... 12 8 5? The elements of a matrix are given by row number, .... |
This site has has interactive explanations and simulations of math from alegrbra to trigonometry. Just click the...
see more
This site has has interactive explanations and simulations of math from alegrbra to trigonometry. Just click the "interactive" tab on the top left menu and you can choose different simulations. It includes, the complete definition of parabolas, reaching beyond the ability to graph into the realm of why the graph appears as it does. It also has vivid descriptions of angles including circle angles for geometry. It also has calculators for principal nth roots, gdc, matrices, and prime factorization. It's definitely worth checking out. Quote from site: "A parabola is actually a locus of a point and a line. The point is called the focus and the line the directrix. That means that all points on a parabola are equidistant from the focus and the directrix. To change the equation and the graph of the interactive parabola below just click and drag either the point A, which is the focus, or point B, which controls the directrix." This is an interactive site that allows people to change the graph to understand why directrix and focus dictate parabolic graphs.
Math and science search engine, particularly focusing on clarification of lessons and providing lessons for teachers, called...
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Math and science search engine, particularly focusing on clarification of lessons and providing lessons for teachers, called TEKS toolkit. It also provides worksheets and games for classroom enhancement. The parent site is a research project for learning improvement, so papers and news concerning learning improvement can also be found on this site. (The TEKS toolkit can also be purchased for more information than just online pdf's. Just go to the parent site click "products" on the toolbar. |
COURSE DESCRIPTION
Calculus II is the payoff for mastering Calculus I. This second course in the calculus sequence introduces you to exciting new techniques and applications of one of the most powerful mathematical tools ever invented. Equipped with the skills of Calculus II, you
can solve a wide array of problems in the physical, biological, and social sciences, engineering, economics, and other areas. Success at Calculus II also gives you a solid foundation for the further study of mathematics, and it meets the math requirement for many undergraduate majors.
But beyond these advantages, you will find that the methods you learn in Calculus II are practical, interesting, and elegant, involving ideas that are beautifully simple. Because it can model real-life situations, calculus has an amazing range of uses, and these applications come into full flower in Calculus II.
Understanding Calculus II: Problems, Solutions, and Tips takes you on this exhilarating journey in 36 intensively illustrated half-hour lectures that cover all the major topics of the second full-year calculus course in high school at the College Board Advanced Placement BC level or a second-semester course in college. Drawing on decades of teaching experience, Professor Bruce H. Edwards of the University of Florida enriches his lectures with crystal-clear explanations, frequent study tips, pitfalls to avoid, and—best of all—hundreds of examples and practice problems that are specifically designed to explain and reinforce key concepts.
Few calculus teachers are as qualified, accessible, or entertaining as Professor Edwards, who has won multiple teaching awards and coauthored a best-selling series of calculus textbooks. Many calculus students give up trying to understand why a particular procedure works and resort to memorizing the steps to a solution. With Professor Edwards, the underlying concepts are always clear and constantly reinforced, which greatly eases the path to learning the material.
Get behind the Wheel of the "Limit Machine"
Professor Edwards begins with a three-lecture review of the fundamental ideas of calculus. He also includes brief reviews of major concepts throughout the course, which makes Understanding Calculus II a self-contained lecture series for anyone who is already familiar with the two main operations of calculus, differentiation and integration. Professor Edwards takes these ideas beyond the definitions, rules, and formulas that are the focus of first-semester calculus and applies them in intriguing ways. For example:
Differential equations: This far-reaching field puts derivatives to work—modeling population growth, nuclear decay, falling objects, and countless other processes involving change. Professor Edwards recalls that as a young mathematician, he spent summers working for NASA, solving differential equations for aircraft in flight.
Infinite series: Does adding an infinite sequence of numbers give an infinite result? Not necessarily. The series may converge on a specific value, or it may diverge to infinity. Calculus can provide the answer for different types of infinite series and represent familiar functions from algebra or trigonometry in surprising ways.
Vectors: Among the geometric applications of calculus is the analysis of vectors. These are quantities, such as velocity, that have both magnitude and direction. In Calculus II, you learn techniques for evaluating vectors in the plane, allowing you to solve problems involving moving and accelerating objects, whether they are on a straight or curved path.
Understanding Calculus II covers the above subjects in considerable depth, particularly infinite series, which you explore in 11 lectures. You also study other standard topics in second-semester calculus, including
integration formulas and techniques,
integrating areas and volumes,
Taylor and Maclaurin polynomials,
L'Hôpital's rule for evaluating limits,
evaluating improper integrals,
calculus applied to parametric equations, and
calculus applied to polar coordinates.
These very different applications of calculus each involve the essential idea of the limit. Professor Edwards notes that calculus can be thought of as a "limit machine"—a set of procedures for approaching infinitely close to a value. One of the interesting features of calculus is its logical rigor combined with its creative use of the mysterious entity of the infinite. From this unusual marriage emerge astonishingly precise solutions to otherwise inaccessible problems.
Explore the Immense Riches of Calculus
Calculus is full of fascinating properties, baffling paradoxes, and entertaining problems. Among the many you investigate in Understanding Calculus II are these:
Gabriel's Horn: Rotate a simple curve around its axis and you get a three-dimensional shape that looks like an infinitely long trumpet. Called Gabriel's Horn, this geometric figure has an unusual property: It has infinite surface area but finite volume. See calculus prove that this must be so.
Baseball thriller: A baseball 3 feet above home plate is hit at 100 feet per second and at an angle of 45 degrees. Employ Newton's second law of motion and the derivative of the position function to determine if the ball will be a home run, clearing a 10-foot-high fence 300 feet away.
Cantor Set: Remove the middle third of a line segment. Repeat with the two pieces that remain. Repeat again ad infinitum. The end points of all the pieces will form an infinite set. But what about the total length of all the line segments? Summing this infinite series reveals the surprising answer.
Master Calculus on Your Own Schedule
Understanding CalculusII is an immensely rewarding experience that you can study at your own pace. Professor Edwards often encourages you to pause the video and test yourself by solving a problem before he reveals the answer. Those who will benefit from this engaging and flexible presentation include
high school or college students currently, or about to be, enrolled in Calculus II who want personal coaching from an outstanding teacher;
high school students preparing for the College Board Advanced Placement test in Calculus at the BC level;
students in higher-level math courses or professionals who want a rigorous review of calculus; and
anyone interested in pursuing one of life's greatest intellectual adventures, which has been solving difficult problems for over 300 years.
A three-time Teacher of the Year at the University of Florida, Professor Edwards knows how to help students surmount the stumbling blocks on their path to mastering calculus. In this course, he uses a steady stream of on-screen equations, graphs, and other visual aids to document the key steps in solving sample problems. The accompanying workbook is designed to reinforce each lecture with more practice problems and worked-out solutions, as well as lecture summaries, tips, and pitfalls; and formulas for derivatives, integration, and power series.
Professor Edwards's lectures also include a feature he calls "You Be the Teacher," in which he reverses roles, challenging you to answer a typical question posed in the classroom, design a suitable problem to illustrate a principle, or otherwise put yourself in the instructor's shoes—an invaluable exercise in learning to think for yourself in the language of calculus.
Open Doors with Your New Fluency
The place of calculus at the end of the high school math curriculum makes it seem like a final destination. But it is also only a beginning. Calculus is a world unto itself, an ever-expanding collection of tools that can solve the most intractable problems in ingenious and often surprising ways. The deeper you go into calculus, the richer it gets and the better you are prepared for even more advanced math courses that open doors of their own.
In his last lecture, Professor Edwards looks ahead to where your math studies may take you after this course. It's exciting terrain. Imagine arriving in a foreign country equipped with the ability to speak the nation's language. Your opportunities for exploration, interaction, and further learning are almost limitless. That's what Understanding Calculus II does for your fluency in one of the greatest achievements of the human mind.
LECTURES
36Lectures
Learn what distinguishes Calculus II from Calculus I. Then embark on a three-lecture review, beginning with the top 10 student pitfalls from precalculus. Next, Professor Edwards gives a refresher on basic functions and their graphs, which are essential tools for solving calculus problems.
In your second warm-up lecture, review the concept of derivatives, recalling the derivatives of trigonometric, logarithmic, and exponential functions. Apply your knowledge of derivatives to the analysis of graphs. Close by reversing the problem: Given the derivative of a function, what is the original function?
Complete your review by going over the basic facts of integration. After a simple example of integration by substitution, turn to definite integrals and the area problem. Reacquaint yourself with the fundamental theorem of calculus and the second fundamental theorem of calculus. End the lecture by solving a simple differential equation.
In the first of three lectures on differential equations, learn various techniques for solving these very useful equations, including separation of variables and Euler's method, which is the simplest numerical technique for finding approximate solutions. Then look at growth and decay models, with two intriguing applications.
Continue your study of differential equations by examining orthogonal trajectories, curves that intersect a given family of curves at right angles. These occur in thermodynamics and other fields. Then develop the famous logistic differential equation, which is widely used in mathematical biology.
Investigate linear differential equations, which typically cannot be solved by separation of variables. The key to their solution is what Professor Edwards calls the "magic integrating factor." Try several examples and applications. Then return to an equation involving Euler's method, which was originally considered in Lecture 4.
Use integration to find areas and volumes. Begin by trying your hand at planar regions bounded by two curves. Then review the disk method for calculating volumes. Next, focus on ellipses as well as solids obtained by rotating ellipses about an axis. Finally, see how your knowledge of ellipsoids applies to the planet Saturn.
Continue your exploration of the power of integral calculus. First, review arc length computations. Then, calculate the areas of surfaces of revolution. Close by surveying the concept of work, answering questions such as, how much work does it take to lift an object from Earth's surface to 800 miles in space?
Study moments and centers of mass, developing formulas for finding the balancing point of a planar area, or lamina. Progress from one-dimensional examples to arbitrary planar regions. Close with the famous theorem of Pappus, using it to calculate the volume of a torus.
Begin a series of lectures on techniques of integration, also known as finding anti-derivatives. After reviewing some basic formulas from Calculus I, learn to develop the method called integration by parts, which is based on the product rule for derivatives. Explore applications involving centers of mass and area.
Explore integrals of trigonometric functions, finding that they are often easy to evaluate if either sine or cosine occurs to an odd power. If both are raised to an even power, you must resort to half-angle trigonometric formulas. Then look at products of tangents and secants, which also divide into easy and hard cases.
Trigonometric substitution is a technique for converting integrands to trigonometric integrals. Evaluate several cases, discovering that you can conveniently represent these substitutions by right triangles. Also, what do you do if the solution you get by hand doesn't match the calculator's answer?
Put your precalculus skills to use by splitting up complicated algebraic expressions to make them easier to integrate. Learn how to deal with linear factors, repeated linear factors, and irreducible quadratic factors. Finally, apply these techniques to the solution of the logistic differential equation.
Revisit the concept of limits from elementary calculus, focusing on expressions that are indeterminate because the limit of the function may not exist. Learn how to use L'Hôpital's famous rule for evaluating indeterminate forms, applying this valuable theorem to a variety of examples.
So far, you have been evaluating definite integrals using the fundamental theorem of calculus. Study integrals that appear to be outside this procedure. Such "improper integrals" usually involve infinity as an end point and may appear to be unsolvable—until you split the integral into two parts.
Start the first of 11 lectures on one of the most important topics in Calculus II: infinite series. The concept of an infinite series is based on sequences, which can be thought of as an infinite list of real numbers. Explore the characteristics of different sequences, including the celebrated Fibonacci sequence.
Look at an example of a telescoping series. Then study geometric series, in which each term in the summation is a fixed multiple of the previous term. Next, prove an important convergence theorem. Finally, apply your knowledge of geometric series to repeating decimals.
Explore an important test for divergence of an infinite series: If the terms of a series do not tend to zero, then the series diverges. Solve a bouncing ball problem. Then investigate a paradoxical property of the famous Cantor set.
Does the celebrated harmonic series diverge or converge? Discover a proof using the integral test. Then generalize to define an entire class of series called p-series, and prove a theorem showing when they converge. Close with the sum of the harmonic series, the fascinating Euler-Mascheroni constant, which is not known to be rational or irrational.
Develop more convergence tests, learning how the direct comparison test for positive-term series compares a given series with a known series. The limit comparison test is similar but more powerful, since it allows analysis of a series without having a term-by-term comparison with a known series.
Having developed tests for positive-term series, turn to series having terms that alternate between positive and negative. See how to apply the alternating series test. Then use absolute value to look at the concepts of conditional and absolute convergence for series with positive and negative terms.
Finish your exploration of convergence tests with the ratio and root tests. The ratio test is particularly useful for series having factorials, whereas the root test is useful for series involving roots to a given power. Close by asking if these tests work on the p-series, introduced in Lecture 19.
Try out techniques for approximating a function with a polynomial. The first example shows how to construct the first-degree Maclaurin polynomial for the exponential function. These polynomials are a special case of Taylor polynomials, which you investigate along with Taylor's theorem.
Discover that a power series can be thought of as an infinite polynomial. The key question with a power series is to find its interval of convergence. In general, this will be a point, an interval, or perhaps the entire real line. Also examine differentiation and integration of power series.
Learn the steps for expressing a function as a power series. Experiment with differentiation and integration of known series. At the end of the lecture, investigate some beautiful series formulas for pi, including one by the brilliant Indian mathematician Ramanujan.
Finish your study of infinite series by exploring in greater depth the Taylor and Maclaurin series, introduced in Lecture 23. Discover that you can calculate series representations in many ways. Close by using an infinite series to derive one of the most famous formulas in mathematics, which connects the numbers e, pi, and i.
Review parabolas, ellipses, and hyperbolas, focusing on how calculus deepens our understanding of these shapes. First, look at parabolas and arc length computation. Then turn to ellipses, their formulas, and the concept of eccentricity. Next, examine hyperbolas. End by looking ahead to parametric equations.
Parametric equations consider variables such as x and y in terms of one or more additional variables, known as parameters. This adds more levels of information, especially orientation, to the graph of a parametric curve. Examine the calculus concept of slope in parametric equations, and look closely at the equation of the cycloid.
In the first of two lectures on polar coordinates, review the main properties and graphs of this specialized coordinate system. Consider the cardioids, which have a heart shape. Then look at the derivative of a function in polar coordinates, and study where the graph has horizontal and vertical tangents.
Continue your study of polar coordinates by focusing on applications involving integration. First, develop the polar equation for the area bounded by a polar curve. Then turn to arc lengths in polar coordinates, discovering that the formula is similar to that for parametric equations.
Begin a series of lectures on vectors in the plane by defining vectors and their properties, and reviewing vector notation. Then learn how to express an arbitrary vector in terms of the standard unit vectors. Finally, apply what you've learned to an application involving force.
Deepen your skill with vectors by exploring the dot product method for determining the angle between two nonzero vectors. Then turn to projections of one vector onto another. Close with some typical applications of dot product and projection that involve force and work.
Use your knowledge of vectors to explore vector-valued functions, which are functions whose values are vectors. The derivative of such a function is a vector tangent to the graph that points in the direction of motion. An important application is describing the motion of a particle.
Combine parametric equations, curves, vectors, and vector-valued functions to form a model for motion in the plane. In the process, derive equations for the motion of a projectile subject to gravity. Solve several projectile problems, including whether a baseball hit at a certain velocity will be a home run.
Use the unit tangent vector and normal vector to analyze acceleration. The unit tangent vector points in the direction of motion. The unit normal vector points in the direction an object is turning. Learn how to decompose acceleration into these two components.
See how the concept of curvature helps with analysis of the acceleration vector. Come full circle by using ideas from elementary calculus to determine the point of maximum curvature. Then close by looking ahead at the riches offered by the continued study of calculus.
Dr. Bruce H. Edwards is Professor of Mathematics at the University of Florida. Professor Edwards received his B.S. in Mathematics from Stanford University and his Ph.D. in Mathematics from Dartmouth College. After his years at Stanford, he taught mathematics at a university near Bogotá, Colombia, as a Peace Corps volunteer.
Professor Edwards has won many teaching awards at the University of Florida, including Teacher of the Year in the College of Liberal Arts and Sciences, Liberal Arts and Sciences Student Council Teacher of the Year, and the University of Florida Honors Program Teacher of the Year. He was selected by the Office of Alumni Affairs to be the Distinguished Alumni Professor for 1991–1993.
Professor Edwards has taught a variety of mathematics courses at the University of Florida, from first-year calculus to graduate-level classes in algebra and numerical analysis. He has been a frequent speaker at research conferences and meetings of the National Council of Teachers of Mathematics. He has also coauthored a wide range of mathematics textbooks with Professor Ron Larson. Their textbooks have been honored with various awards from the Text and Academic Authors Association. |
9780871508676
ISBN:
0871508672
Edition: 2 Publisher: Brooks/Cole
Summary: This outstanding text starts off using vectors and the geometric approach, featuring a computational emphasis. The authors provide students with easy-to-read explanations, examples, proofs, and procedures. Elementary Linear Algebra can be used in both a matrix-oriented course, or a more traditionally structured course.
ISBN-13:9780871508676
ISBN:0871508672
Edition:2nd
Publisher:Brooks/Cole
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Starting at $87 new edition of BEGINNING & INTERMEDIATE ALGEBRA is an exciting and innovative revision that takes an already successful book and makes it more compelling for today's users. The new edition has been thoroughly updated with a new interior design and other pedagogical features that make the book both easier to read and easier to use. Known for its clear writing and an engaging, accessible approach that makes algebra relevant, BEGINNING & INTERMEDIATE ALGEBRA helps users to develop problem-solving skills and strategies that they can use in their everyday lives. The new edition welcomes two new co-authors Rosemary Karr and Marilyn Massey who along with David Gustafson have developed a learning plan to help users succeed in Beginning Algebra and transition to the next level in their coursework.
Table of Contents
Real Numbers and Their Basic Properties
Real numbers and their graphs
Fractions
Exponents and order of operations
Adding and subtracting real numbers
Multiplying and dividing real numbers
Algebraic expressions
Properties of real numbers
Projects
Chapter summary
Test
Equations and Inequalities
Solving basic linear equations in one variable
Solving more linear equations in one variable
Simplifying expressions to solve linear equations in one variable
Formulas
Introduction to problem solving
Motion and mixed problems
Solving linear inequalities in one variable
Projects
Chapter summary
Chapter test Cumulative review exercises
Graphing and Solving Systems of Linear Equations and Linear Inequalities |
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Math 911
Professor Martin Weissman has been teaching Mathematics for 50 years. Passionate about the subject, he desires to help students overcome their fear of Mathematics and to master it. In his years of teaching, he felt that students needed fewer explanatory textbooks and more illustrative examples to help them understand how to do higher math problems. To that end, he developed the Math911 tutorial software.
The Math911 tutorial software download generates problems for students to solve in Introductory Algebra, Intermediate Algebra, College Algebra, Statistics, Trigonometry, and Pre-Calculus. No answers are marked "wrong." The student must get a certain number of problems "correct" to prove mastery and move to the next level. If the student does not understand how to do the problem, he may click on a button for step-by-step instructions to see the problem being solved. He may follow these step-by-step instructions for as many problems as necessary to understand the process. Professor Weissman wanted to give students practice, practice, practice in all types and variations of problems.
Schoolhouse Crew members were given the complete download of this product, and some received codes to review it with multiple students. We then each reviewed the level of math at which our students were working, from Introductory Algebra through Pre-Calculus. For Introductory Algebra only, there are also free PDF downloads with some explanations of how to do the problems.
The complete Math911 program is available at for the discounted price of $49.95. That includes all the subjects listed above. As a special back-to-school offer, you may try a free download of the Introductory Algebra course from the website. Tech support is always free. The age range is 12 and up, or any student who is ready for Introductory Algebra.
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This text is appropriate for any one-semester junior/senior level course in Modern Algebra, Abstract Algebra, Algebraic Structures, or Groups, Rings and Fields. Durbin has two main goals: to introduce the most important kinds of algebraic structures, and to help students improve their ability to understand and work with abstract ideas. The first six chapters present the core of the subject; the remainder are designed to be as flexible as possible. Durbin covers groups before rings, which is a matter of personal preference for instructors. The course is mostly comprised of mathematics majors, but you will find engineering and computer science majors as wellBooks WorldWide Express via United States
Hardcover, ISBN 0471433357 Publisher: Wiley, John, & Sons, Inc33357 Publisher: Wiley, John, & Sons, Inc33357 Publisher: Wiley, John, & Sons, Inc, 2004 533357 Publisher: wiley, 2004 wiley. Paperback. New. International Edition Softcover Brand New.ISBN may be Different but Same Content As US Edition Great Packaging and Great Customer Service please contact us for any Query.
Hardcover, ISBN 0471433357 Publisher: Wiley, 2004 Wiley. Paperback. New. International Edition Softcover Brand New.ISBN may be Different but Same Content As US Edition Great Packaging and Great Customer Service please contact us for any Query.
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Numerical Analysis, 8th Edition
Learn numerical analysis the easy way! NUMERICAL METHODS provides an introduction to the modern approximation techniques and explains how, why, and when the techniques can be expected to work. With a wealth of examples and exercises, this text demonstrates the relevance of numerical analysis to a variety of disciplines and provides ample practice. The applications chosen demonstrate concisely how numerical methods can be, and often must be, applied in real-life situations. Use this text, and you'll gain a firm basis for the future study of numerical analysis and scientific computing335.95
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Elementary and Intermediate Algebra - 4th edition
Summary: Ideal for lecture-format courses taught at the post-secondary level, ELEMENTARY AND INTERMEDIATE ALGEBRA, Fourth Edition, makes algebra accessible and engaging. Author Charles ''Pat'' McKeague's passion for teaching mathematics is apparent on every page, and with many years of experience teaching mathematics, he knows how to write in a way that you will understand and appreciate.
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This is a free course offered by the Saylor Foundation.'Math for Economists will help you assemble a toolkit of skills and...
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This is a free course offered by the Saylor Foundation.' For example, an economist may be called upon to determine the right mix allocation of capital to a production process. The tools in this course will help you evaluate the options and select from the best alternatives. Advanced courses in economics typically utilize mathematical techniques beyond basic calculus; so, gaining practice in fundamental skills can serve as a good basis for further study. Of note, this course applies precalculus and calculus; this is different from "applied math," which economists typically use to refer to probability and statistics.This course begins with a survey of basic optimization tools and then applies them to solve problems over several periods in time. These optimization tools describe feasible choices and then direct you to the best possible solution. In essence, they will help you evaluate an economic environment and determine the best course of action. The role of risk in financial decisions is explored in relation to individual choices and macroeconomic processes. The equitable distribution of resources is then considered. In other words, this class will explore whether an optimal solution is indeed also fair to the participants and society. A specific application, game theory, is presented as one of the major recent advances in economic theory. The final topic returns to microeconomic problems such as taxes, elasticity, and specific types of supply and demand curves.The materials included in the course offer a framework that you can use to apply quantitative skills. You should liberally use Saylor course materials (MA001, MA003, MA005, ECON103/MA101, and MA102) to refresh their general techniques. This course fits into the major as a bridge between quantitative theory and specific applications to problems in economics. Completing this course can greatly help students successfully build a toolkit for the intermediate courses, ECON201 and ECON202.'
This online textbook "covers the basic mathematical tools used in economic theory. Knowledge of elementary calculus is...
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This online textbook "covers the basic mathematical tools used in economic theory. Knowledge of elementary calculus is assumed; some of the prerequisite material is reviewed in the first section. The main topics are multivariate calculus, concavity and convexity, optimization theory, differential equations, and difference equations. The emphasis throughout is on techniques rather than abstract theory.״
Three multiple-response tests which are marked using JavaScript. The number of correct answers in each test is given, but the...
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Three multiple-response tests which are marked using JavaScript. The number of correct answers in each test is given, but the student has to work out for each question whether one or multiple answers are correct.
It is a JavaScript that performs matrix multiplication with up to 10 rows and up to 10 columns. Moreover, it computes the...
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It is a JavaScript that performs matrix multiplication with up to 10 rows and up to 10 columns. Moreover, it computes the power of a square matrix, with applications to the large Markov chains computations.
By means of three spin buttons, the user changes the values of the parameters that affect the orientation of the isoquants,...
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By means of three spin buttons, the user changes the values of the parameters that affect the orientation of the isoquants, the degree of substitutability and the degree of homogeneity; the accompanying chart automatically updates. Click on the image, above left; "how to use" this worksheet should be intuitive.
By means of six spin buttons, the user sets the parameters for the Cobb-Douglas preferences for each trader, the total...
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By means of six spin buttons, the user sets the parameters for the Cobb-Douglas preferences for each trader, the total endowment shared by the two traders and the initial endowment for Trader A. The initial endowment for Trader B and the accompanying graph automatically update, showing the contract curve, the initial endowment point and the Walrasian equilibrium. Click the image above left to see a screen capture of the interface. |
Emphasizing the finite difference approach for solving differential equations, the second edition of Numerical Methods for Engineers and Scientists presents a methodology for systematically constructing individual computer programs. Providing easy access to accurate solutions to complex scientific and engineering problems, each chapter begins with objectives, a discussion of a representative application, and an outline of special features, summing up with a list of tasks students should be able to complete after reading the chapter- perfect for use as a study guide or for review. The AIAA Journal calls the book "…a good, solid instructional text on the basic tools of numerical analysis."
Most Helpful Customer Reviews
A great textbook for a first course in Numerical Methods as it gives an extensive yet detailed coverage of numerical techniques that form the base for more advanced work in CFD. This book is based almost entirely on the finite difference method for solving differential equations. A chapter addresses finite element techniques but it only a primer; you will need a textbook solely devoted to FEM. This book is written for a mechanical engineer, as most of the governing equations are invariably from the areas of heat transfer, fluid flow, gas dynamics and solid mechanics.
The good points are 1. Each method described comes with a index notated formula that takes the head ache out of programming. Plus there are plenty of FORTRAN subroutines to look at.
2. Not only does Hoffman give you the finite difference equation he also throws in a solved example with one or two iterations worked out in full detail; the benefit of this cannot be overstated.
3. Plenty of practice problems with results at the back of the book.
4. Enough math to give the reader an insight into how the method works. If you care for rigor this is not the book.
The drawbacks are 1. Hoffman has condensed the portions dealing with PDE's from previous editions cutting out some theoretical development. Since most wouldn't have had a course in PDE's (like me) a few more pages might have better squared away a few difficult concepts (eg. characteristic lines of PDE's).
2. Could use another round of proof-reading. This book is littered with typos; which one runs into even in key formulas. This is unacceptable in what is otherwise a pedagogically sound book.
I had Dr. Hoffman as the instructor. He is an excellent teacher and writer. Many math books are horrible for self-study and first timers. But this book sets up a standard. A lot of details are given with plenty of detailed examples. I am sure first timers will appreciate his huge effort put into the book.
But keep in mind one feature (not drawback). This book uses heavily finite difference method (400 pages in the 1st edition and 200 pages in the 2nd). This method is good for only 1d problems. Coordinate transformation needed to extend this to 2d or 3d (even just non-uniform 1d) is not easy especially for 3d. I wish there was an equally good book on finite volume method, which is popular for 3d CFD. Anyways, this book is intended for beginners and thus the choice of finite difference method is an appropriate one.
I spend couple of months on searching a good book on Numerical method specifically a introductory text on Finite Element Methods and finally found this one. This is a good book for any one who do not have any experience of Finite Element Methods before.
To understand Finite Element Methods, it would be help full if one flows along the progression of the book rather then reading specific chapters.
Overall, This is a good book for Numerical Method and best Introduction to Finite element Methods I have read so far.
I am so glad that I had the opportunity to take the course offered by Prof. Joe D. Hoffman where he used this textbook at Purdue University. Here agin a fabulous teacher and an expert has written a fabulous book, what more you can expect! It was a phenomenal class where I learned (from my non-engineering background) the basics and advance of numerical methods that I am able to still apply today. The algorithms are fluently explained in very simple language with great examples. I would recommend this book to anyone who is looking to pursue a career in science and engineering. |
Discrete Mathemetics
9780618415380
0618415386
Summary: Discrete Mathematics combines a balance of theory and applications with mathematical rigor and an accessible writing style. The author uses a range of examples to teach core concepts, while corresponding exercises allow students to apply what they learn. Throughout the text, engaging anecdotes and topics of interest inform as well as motivate learners. The text is ideal for one- or two-semester courses and for studen...ts who are typically mathematics, mathematics education, or computer science majors. Part I teaches student how to write proofs; Part II focuses on computation and problem solving. The second half of the book may also be suitable for introductory courses in combinatorics and graph theory.
Ferland is the author of Discrete Mathemetics, published 2008 under ISBN 9780618415380 and 0618415386. Six hundred fifty seven Discrete Mathemetics textbooks are available for sale on ValoreBooks.com, one hundred fifty seven used from the cheapest price of $29.98, or buy new starting at $91Pages are clean. The spine of this book has been repaired and there is water damage on the botto... [more]) |
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four-function calculator. To this end calculators will not be allowed in the following:
Math 27 (self-paced basic skills mathematics)
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Following this, students will be allowed to use a four-function calculator during class and during tests in
any math course offered by the college, provided they supply their instructors with a letter explaining their
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The square root of 2 is a fascinating number, if a little less famous than such mathematical stars as pi, the number e, the golden ratio, or the square root of 1. This book shows why v2 is an important number, and how, in puzzling out its special qualities, mathematicians gained insights into the illusive nature of irrational numbers. more...
This monograph provides a thorough treatment of module theory, a subfield of algebra. The authors develop an approximation theory as well as realization theorems and present some of its recent applications, notably to infinite-dimensional combinatorics and model theory. The book is devoted to graduate students interested in algebra as well as to experts... more...
Many basic ideas of algebra and number theory intertwine, making it ideal to explore both at the same time. Certain Number-Theoretic Episodes in Algebra focuses on some important aspects of interconnections between number theory and commutative algebra. Using a pedagogical approach, the author presents the conceptual foundations of commutative algebra... more...
Absolute values and their completions - such as the p-adic number fields - play an important role in number theory. In valuation theory, the notion of a completion has to be replaced by that of "Henselization". This book develops the theory of valuations as well as of Henselizations. more...
This book provides a comprehensive exposition of the use of set-theoretic methods in abelian group theory, module theory, and homological algebra, including applications to Whitehead's Problem, the structure of Ext and the existence of almost-free modules over non-perfect rings. This second edition is completely revised and udated to include major... more...
Arranged to present the evolution of concepts and ideas of the subject, this is a collection of about 500 problems in algebraic number theory. Containing solved problems, it is designed for students with the usual background of undergraduate algebra. more...
Provides a study of supplements and projectivity conditions needed to investigate classes of modules related to lifting modules. This book introduces small submodules, the radical, variations on projectivity, and hollow dimension. It considers preradicals and torsion theories, decompositions of modules, supplements in modules, and lifting modules. more...
Ever since the analogy between number fields and function fields was discovered, it has been a source of inspiration for new ideas. This title contains articles which explore various aspects of the parallel worlds of function fields and number fields, ranging from Arakelov geometry to Drinfeld modules, and t-motives. more... |
PREFACE
This book is a revision of the sixth edition, published in 1996. That edition has served,
just as the earlier ones did, as a textbook for a one-term introductory course in the
theory and application of functions of a complex variable. This edition preserves the
basic content and style of the earlier editions, the first two of which were written by
the late Ruel V. Churchill alone.
In this edition, the main changes appear in the first nine chapters, which make up
the core of a one-term course. The remaining three chapters are devoted to physical
applications, from which a selection can be made, and are intended mainly for self-
study or reference.
Among major improvements, there are thirty new figures; and many of the old
ones have been redrawn. Certain sections have been divided up in order to emphasize
specific topics, and a number of new sections have been devoted exclusively to exam-
ples. Sections that can be skipped or postponed without disruption are more clearly
identified in order to make more time for material that is absolutely essential in a first
course, or for selected applications later on. Throughout the book, exercise sets occur
more often than in earlier editions. As a result, the number of exercises in any given
set is generally smaller, thus making it more convenient for an instructor in assigning
homework.
As for other improvements in this edition, we mention that the introductory
material on mappings in Chap. 2 has been simplified and now includes mapping
properties ofthe exponential function. There has been some rearrangement of material
in Chap. 3 on elementary functions, in order to make the flow of topics more natural.
Specifically, the sections on logarithms now directly fo11ow the one on the exponential
XV
•
XVI PREFACE
function; and the sections on trigonometric and hyberbolic functions are now closer
to the ones on their inverses. Encouraged by comments from users of the book in the
past several years, we have brought some important material out of the exercises and
into the text. Examples of this are the treatment of isolated zeros of analytic functions
in Chap. 6 and the discussion of integration along indented paths in Chap. 7.
The first objective of the book is to develop those parts of the theory which
are prominent in applications of the subject. The second objective is to furnish an
introduction to applications of residues and conformal mapping. Special emphasis
is given to the use of conformal mapping in solving boundary value problems that
arise in studies of heat conduction, electrostatic potential, and fluid flow. Hence the
book may be considered as a companion volume to the authors' "Fourier Series and
Boundary Value Problems" and Ruel V. Churchill's "Operational Mathematics," where
other classical methods for solving boundary value problems in partial differential
equations are developed. The latter book also contains further applications ofresidues
in connection with Laplace transforms.
This book has been used for many years in a three-hour course given each term at
The University ofMichigan. The classes have consisted mainly ofseniors and graduate
students majoring in mathematics, engineering, or one ofthe physical sciences. Before
taking the course, the students have completed at least a three-term calculus sequence,
a first course in ordinary differential equations, and sometimes a term of advanced
calculus. In order to accommodate as wide a range of readers as possible, there are
footnotes referring to texts that give proofs and discussions of the more delicate results
from calculus that are occasionally needed. Some of the material in the book need not
be covered in lectures and can be left for students to read on their own. If mapping
by elementary functions and applications of conformal mapping are desired earlier
in the course, one can skip to Chapters 8, 9, and 10 immediately after Chapter 3 on
elementary functions.
Most of the basic results are stated as theorems or corollaries, followed by
examples and exercises illustrating those results. A bibliography of other books,
many of which are more advanced, is provided in Appendix I. A table of conformal
transformations useful in applications appears in Appendix 2.
In the preparation of this edition, continual interest and support has been provided
by a number of people, many of whom are family, colleagues, and students. They
include Jacqueline R. Brown, Ronald P. Morash, Margret H. Hoft, Sandra M. Weber,
Joyce A. Moss, as well as Robert E. Ross and Michelle D. Munn of the editorial staff
at McGraw-Hill Higher Education.
James Ward Brown
COMPLEX VARIABLES AND APPLICATIONS
CHAPTER
1
COMPLEX NUMBERS
In this chapter, we survey the algebraic and geometric structure ofthe complex number
system. We assume various corresponding properties of real numbers to be known.
1. SUMS AND PRODUCTS
Complex numbers can be defined as ordered pairs (x, y) of real numbers that are to
be interpreted as points in the complex plane, with rectangular coordinates x and y,
just as real numbers x are thought of as points on the real line. When real numbers
x are displayed as points (x, 0) on the real axis, it is clear that the set of complex
numbers includes the real numbers as a subset. Complex numbers of the form (0, y)
correspond to points on the y axis and are called pure imaginary numbers. The y axis
is, then, referred to as the imaginary axis.
It is customary to denote a complex number (x, y) by z, so that
(1) z=(x,y).
The real numbers x and y are, moreover, known as the real and imaginary parts of z,
respectively; and we write
(2) Re z = x, Im z = y.
Two complex numbers z1 =(x1, y1) and z2 =(x2, y2) are equal whenever they have
the same real parts and the same imaginary parts. Thus the statement z1 = z2 means
that z1 and z2 correspond to the same point in the complex, or z, plane.
1
SEC. 2 BASIC ALGEBRAIC PROPERTIES 3
Observe that the right-hand sides of these equations can be obtained by formally
manipulating the terms on the left as if they involved only real numbers and by
replacing i 2
by -1 when it occurs.
2. BASIC ALGEBRAIC PROPERTIES
Various properties of addition and multiplication of complex numbers are the same as
for real numbers. We list here the more basic of these algebraic properties and verify
some of them. Most of the others are verified in the exercises.
The commutative laws
(1)
and the associative laws
(2)
follow easily from the definitions in Sec. 1 of addition and multiplication of complex
numbers and the fact that real numbers obey these laws. For example, if z1 =(xi> y1)
and z 2 = (x2, y2), then
Zt + Z2 = (xl +x2, Y1 + Y2) = (x2 + Xt. Y2 + Yt) = Z2 + Zt·
Verification of the rest of the above laws, as well as the distributive law
(3)
is similar.
According to the commutative law for multiplication, iy = yi. Hence one can
write z =x + yi instead of z =x + iy. Also, because of the associative laws, a sum
z1+ z2 + z3 or a product z1z2z3 is well defined without parentheses, as is the case with
real numbers.
The additive identity 0 = (0, 0) and the multiplicative identity 1 = (1, 0) for real
numbers carry over to the entire complex number system. That is,
(4) z +0 = z and z · 1= z
for every complex number z. Furthermore, 0 and 1 are the only complex numbers with
such properties (see Exercise 9).
There is associated with each complex number z = (x, y) an additive inverse
(5) -z = (-x, -y),
satisfying the equation z + (-z) = 0. Moreover, there is only one additive inverse
for any given z, since the equation (x, y) + (u, v) = (0, 0) implies that u =-x and
v =-y. Expression (5) can also be written -z = -x - iy without ambiguity since
SEC.3 FURTHER PROPERTIES 5
7. Use the associative law for addition and the distributive law to show that
8. By writing i =(0, 1) andy= (y, 0), show that -(iy) = (--i)y =i (-y).
9. (a) Write (x, y} + (u, v) =(x, y) and point out how it follows that the complex number
0 = (0, 0) is unique as an additive identity.
(b) Likewise, write (x, y)(u, v) =(x, y) and showthatthenumber 1=(1, 0) is a unique
multiplicative identity.
10. Solve the equation z2 +z + l =0 for z = (x, y) by writing
(x, y)(x, y) + (x, y) + (1, 0) =(0, 0)
and then solving a pair of simultaneous equations in x and y.
Suggestion: Use the fact that no real number x satisfies the given equation to show
that y ::/= 0.
Ansz = (-~, ±-;)
3. FURTHER PROPERTIES
In this section, we mention a number of other algebraic properties of addition and
multiplication of complex numbers that follow from the ones already described in
Sec. 2. Inasmuch as such properties continue to be anticipated because they also apply
to real numbers, the reader can easily pass to Sec. 4 without serious disruption.
We begin with the observation that the existence ofmultiplicative inverses enables
us to show that if a product z1z2 is zero, then so is at least one of the factors z1 and
z2. For suppose that z1z2 = 0 and z1 f= 0. The inverse z11
exists; and, according to the
definition of multiplication, any complex number times zero is zero. Hence
That is, if z1z2 = 0, either z1 =0 or z2 =0; or possibly both z1 and z2 equal zero.
Another way to state this result is that iftwo complex numbers z1 and z2 are nonzero,
then so is their product z1z2.
Division by a nonzero complex number is defined as follows:
"'q -1
- =ZtZ,
zz -
(1) (zz I= 0).
If z1=(xb y1) and z2 = (x2, y2), equation (1) here and expression (8) in Sec. 2 tell us
that
6 COMPLEX NUMBERS CHAP. I
That is,
(2) (zz :/= 0).
Although expression (2) is not easy to remember, it can be obtained by writing (see
Exercise 7)
(3)
Z1 (XI+ iy1)(x2- iy2)
Z2 - (xz + iy2)(x2 - iyz)'
multiplying out the products in the numerator and denominator on the right, and then
using the property
(4)
The motivation for starting with equation (3) appears in Sec. 5.
There are some expected identities, involving quotients, that follow from the
relation
(5)
1 -1
--"- .(.2
Z2
(Z2 f= 0),
which is equation (1) when z1 = 1. Relation (5) enables us, for example, to write
equation (1) in the form
(6) (Z2 f= 0).
Also, by observing that (see Exercise 3)
(Zl f= 0, Z2 f= 0),
and hence that (z1z2)-1
= zi1
z21
, one can use relation (5) to show that
(7) (Zl f= 0, Z2 f= 0).
Another useful identity, to be derived in the exercises, is
(8) (z3 f= 0, Z4 f= 0).
SEC.3 EXERCISES 7
EXAMPLE. Computations such as the following are now justified:
1 1 5+i 5+i
, (2- 3i)(l +i) 5- i 5 + i (5- i)(5 +i)
5+i 5 i 5 1.
26 = 26 + 26 = 26 + 26l •
Finally, we note that the binomialformula involving real numbers remains valid
with complex numbers. That is, if z1 and z2 are any two complex numbers,
(9) (n=1,2, ...)
where
(~) - k!(nn~k)!
(k = 0, 1, 2, ... , n)
and where it is agreed that 0! = 1. The proof, by mathematical induction, is left as an
exerctse.
EXERCISES
1. Reduce each of these quantities to a real number:
( )
1+2i 2 - i (b) 5i
a 3- 4i + 5i ; (1 - i)(2- i)(3- i);
(c)(l-i)4.
Ans. (a) -2/5; (b) -1/2; (c) -4.
2. Show that
(a) (-l)z = -z;
1
(b) - =z (z :f: 0).
Ijz
3. Use the associative and commutative laws for multiplication to show that
4. Prove that if z1z2z3 =0, then at least one of the three factors is zero.
Suggestion: Write (z1z2)z3 =0 and use a similar result (Sec. 3) involving two
factors.
5. Derive expression (2), Sec. 3, for the quotient zIIz2 by the method described just after
it.
6. With the aid of relations (6) and (7) in Sec. 3, derive identity (8) there.
7. Use identity (8) in Sec. 3 to derive the cancellation law:
(Z2 :f: 0, Z :f: 0).
8 COMPLEX NUMBERS CHAP. I
8. Use mathematical induction to verify the binomial formula (9) in Sec. 3. More precisely,
note first that the formula is true when n = 1. Then, assuming that it is valid when n = m
where m denotes any positive integer, show that it must hold when n =m + l.
4. MODULI
It is natural to associate any nonzero complex number z = x +iy with the directed line
segment, or vector, from the origin to the point (x, y) that represents z (Sec. 1) in the
complex plane. In fact, we often refer to z as the point z or the vector z. In Fig. 2 the
numbers z =x + iy and -2 +i are displayed graphically as both points and radius
vectors.
(-2. 1)
•
VI
J (x, y)
•
-2 0 X
FIGURE2
According to the definition of the sum of two complex numbers z1 = x1 +(v1
and z2 = x2 + iy2, the number z1 + z2 corresponds to the point (x 1 +x2, y1 +y2). It
also corresponds to a vector with those coordinates as its components. Hence z1+z2
may be obtained vectorially as shown in Fig. 3. The difference z1 - z2 =z1+ (-z2)
corresponds to the sum of the vectors for z1 and -z2 (Fig. 4).
y
0 X
FIGURE3
Although the product of two complex numbers z1 and z2 is itself a complex
number represented by a vector, that vector lies in the same plane as the vectors for z1
and z2. Evidently, then, this product is neither the scalar nor the vector product used
in ordinary vector analysis.
The vector interpretation of complex numbers is especially helpful in extending
the concept of absolute values of real numbers to the complex plane. The modulus,
or absolute value. of a complex number z =x + iy is defined as the nonnegative real
SEC. 4 MODULI 9
y
0 X
FIGURE4
number )x2 + y2 and is denoted by lzl; that is,
(l) lzl = )x2 + y2•
Geometrically, the number lzl is the distance between the point (x, y) and the origin,
or the length of the vector representing z. It reduces to the usual absolute value in the
real number system when y = 0. Note that, while the inequality z1 < z2 is meaningless
unless both z1and z2 are real, the statement lz11< lz21means that the point z1 is closer
to the origin than the point z2 is.
EXAMPLE 1. Since 1- 3 +2iI=J13 and 11 +4i I=v17, the point -3 +2i is
closer to the origin than l +4i is.
The distance between two points z1 = x1+iYI and z2 = x2 +iY2 is Iz1 - z2l· This
is clear from Fig. 4, since lz1 - z21is the length of the vector representing z1 - z2; and,
by translating the radius vector z1 - z2, one can interpret z1 - z2 as the directed line
segment from the point (x2, y2) to the point (xl> y1). Alternatively, it follows from the
expression
and definition (1) that
The complex numbers zcorresponding to the points lying on the circle with center
zo and radius R thus satisfy the equation lz - zoI= R, and conversely. We refer to this
set of points simply as the circle lz - zol =R.
EXAMPLE 2. The equation lz- 1+3il = 2 represents the circle whose center is
zo = (1, -3) and whose radius is R =2.
It alsofollowsfromdefinition (1) that the real numbers lzl, Re z=x, andlm z= y
are related by the equation
(2)
10 COMPLEX NUMBERS CHAP. I
Thus
(3) Re z <IRe zl < lzl and Im z <lim zl < lzl.
We turn now to the triangle inequality, which provides an upper bound for the
modulus of the sum of two complex numbers z1 and z2:
(4)
This important inequality is geometrically evident in Fig. 3, since it is merely a
statement that the length of one side of a triangle is less than or equal to the sum
of the lengths of the other two sides. We can also see from Fig. 3 that inequality (4)
is actually an equality when 0, Zt> and z2 are collinear. Another, strictly algebraic,
derivation is given in Exercise 16, Sec. 5.
An immediate consequence of the triangle inequality is the fact that
(5)
To derive inequality (5), we write
which means that
(6)
This is inequality (5) when lz11> lz2j. If lz11< iz21, we need only interchange z1 and
z2 in inequality (6) to get
which is the desired result. Inequality (5) tells us, of course, that the length of one side
of a triangle is greater than or equal to the difference of the lengths of the other two
sides.
Because 1- z21= lz2 1, one can replace z2 by -z2 in inequalities (4) and (5) to
summarize these results in a particularly useful form:
(7)
(8)
lz1 ± zzl < lz11 + lzzl,
lz1 ± zzl >lizII- lzzll·
EXAMPLE 3. If a point z lies on the unit circle lzl = 1about the origin, then
lz-21<izl+2=3
and
lz- 21 >liz!- 21 = 1.
12 COMPLEX NUMBERS
y
(x, y)
0 X
FIGURES
So the conjugate of the sum is the sum of the conjugates:
(2)
In like manner, it is easy to show that
(3)
(4)
and
(5)
(;~) = ;~ (zz f. 0).
CHAP. I
The sum z + z of a complex number z =x +iy and its conjugate z = x - iy is
the real number 2x, and the difference z- z is the pure imaginary number 2iy. Hence
(6)
z+z
Rez = ,
2
-
z-z
lm z=- .
2i
An important identity relating the conjugate of a complex number z = x + iy to
its modulus is
(7) -.7- 17'12
---- - - '
where each side is equal to x 2 + y2. It suggests the method for determining a quotient
zJiz2 that begins with expression (3), Sec. 3. That method is, of course, based on
multiplying both the numerator and the denominator of zdz2 by z2, so that the
denominator becomes the real number lz2 i2
.
EXAMPLE 1. As an illustration,
-1 + 3i = (-1 + 3i)(2 +i) = -5 +5i = -5 +5i = -1 +i.
2- i (2 - i)(2 + i) 12- i12 5
See also the example near the end of Sec. 3.
SEC. 5 ExERCISES 13
Identity (7) is especially useful in obtaining properties ofmoduli from properties
of conjugates noted above. We mention that
(8)
and
Z1 lz1l=-
Zz lzzl
(1}
Property (8) can be established by writing
and recalling that a modulus is never negative. Property (9) can be verified in a similar
way.
EXAMPLE 2. Property (8) tells us that lz2
1= lzl2
and lz3
1= lz!3• Hence if z is a
point inside the circle centered at the origin with radius 2, so that lzl < 2, it follows
from the generalized form (9) of the triangle inequality in Sec. 4 that
!z3
+ 3z2
- 2z + 11 ~ lzl3
+ 31zl2
+ 21zl + 1 < 25.
EXERCISES
1. Use properties of conjugates and moduli established in Sec. 5 to show that
(a) z+3i =z - 3i;
(c) (2 +i)2 =3- 4i;
(b) iz = -iz;
(d) 1(2z + 5)(J2- i)l =.J312z +51.
2. Sketch the set of points determined by the condition
(a) Re(z- i) =2; (b) i2z- il =4.
3. Verify properties (3) and (4) of conjugates in Sec. 5.
4. Use property (4) of conjugates in Sec. 5 to show that
(a) Z1Z2Z3 =Zl Zz Z3; (b) z
4
=z4
•
5. Verify property (9) of moduli in Sec. 5.
6. Use results in Sec. 5 to show that when z2 and z3 are nonzero,
7. Use established properties of moduli to show that when lz31::/= lz41,
14 COMPLEX NUMBERS CHAP. I
8. Show that
when lzl < 1.
9. It is shown in Sec. 3 that if z1z2 =0, then at least one of the numbers z1 and z2 must be
zero. Give an alternative proof based on the corresponding result for real numbers and
using identity (8), Sec. 5.
10. By factoring z4
- 4z2
+3 into two quadratic factors and then using inequality (8), Sec. 4,
show that if z lies on the circle lzl =2, then
11. Prove that
1 1
<-
z4- 4z2 + 3 - 3
(a) z is real if and only if z=z;
(b) z is either real or pure imaginary if and only if z2 =z2
.
12. Use mathematical induction to show that when n = 2, 3, ... ,
(a) Z1 +Z2 + ···+Zn =Z1 +Z2 + · · ·+Zn; (b) ZJZ2 · · · Zn =ZJ Z2 · · · Zw
13. Let a0 , a!> a2, ... , an (n > 1) denote real numbers, and let z be any complex number.
With the aid of the results in Exercise 12, show that
2 n - -2 -n
ao +a1z + a2z + ···+ anz =ao +a1z +azz + ···+anz .
14. Show that the equation jz - zol = R of a circle, centered at z0 with radius R, can be
written
lzl2
- 2 Re(zzo) + iz012
= R
2
.
15. Using expressions (6), Sec. 5, for Re z and Im z, show that the hyperbola x2
- y 2 = I
can be written
16. Follow the steps below to give an algebraic derivation of the triangle inequality (Sec. 4)
(a) Show that
'l - -
lz, + zzl"" = (z1 + zz)(Zi +z2) = z,Z] + (z,zz + z,zz) + zzzz.
(b) Point out why
Z1Z2 +Z1Z2 = 2 Re(z,zz) < 21zdlz21·
(c) Use the results in parts (a) and (b) to obtain the inequality
lzt +z2i
2
< Clztl + lz2l)
2
,
and note hQw the triangle inequality follows.
SEC.6 EXPONENTIAL FoRM 15
6. EXPONENTIAL FORM
Let rand B be polar coordinates of the point (x, y) that corresponds to a nonzero
complex number z =x +iy. Since x =r cos () and y = r sin (), the number z can be
written in polarform as
(1) z = r(cos () + i sin()).
If z = 0, the coordinate () is undefined; and so it is always understood that z =f:. 0
whenever arg z is discussed.
In complex analysis, the real number r is not allowed to be negative and is the
length ofthe radius vector for z; that is, r = Izj. The real number() represents the angle,
measured in radians, that z makes with the positive real axis when z is interpreted as
a radius vector (Fig. 6). As in calculus, () has an infinite number of possible values,
including negative ones, that differ by integral multiples of 2n. Those values can be
determined from the equation tan () = yIx, where the quadrant containing the point
corresponding to z must be specified. Each value of(} is called an argument of z, and
the set of all such values is denoted by arg z. The principal value of arg z, denoted by
Arg z. is that unique value E> such that -n < E> < n. Note that
(2) arg z = Arg z +2mr (n = 0, ±I. ±2, ...).
Also, when z is a negative real number, Arg z has value 7T, not -n.
y
z=x + iy
X
FIGURE6
EXAMPLE 1. The complex number -1 - i, which lies in the third quadrant, has
principal argument -3nj4. That is,
Arg(-1- i) =-
3
n.
4
It must be emphasized that, because of the restriction -n < e < n of the principal
argument e, it is not true that Arg(-1- i) = 5nj4.
According to equation (2),
arg(-1- i) =-
3
7T +2nn
4
(n =0, ±1, ±2, ...).
16 COMPLEX NUMBERS CHAP. I
Note that the term Arg z on the right-hand side of equation (2) can be replaced by any
particular value of arg z and that one can write, for instance,
(3)
arg(-1- i) = S;r +2nn
4
(n = 0, ±1, ±2, ...).
The symbol ei8 , or exp(iO), is defined by means of Euler's formula as
i 8
=cos 0 +i sin 0,
where 0 is to be measured in radians. It enables us to write the polar form (1) more
compactly in exponential form as
(4)
The choice of the symbol ei8 will be fully motivated later on in Sec. 28. Its use in Sec.
7 will, however, suggest that it is a natural choice.
EXAMPLE 2. The number -1 - i in Example 1has exponential form
(5)
With the agreement that e-ie = ei(-B), this can also be written -1- i = J2e-i3Ir/4.
Expression (5) is, of course, only one of an infinite number of possibilities for the
exponential form of -1 - i:
(6) -1 - i = h exp [i(-3
; + 2mr)J (n. = 0, ± 1, J:2, ...).
Note how expression (4) with r = 1tells us that the numbers eie lie on the circle
centered at the origin with radius unity, as shown in Fig. 7. Values of ei8 are, then,
immediate from that figure, without reference to Euler's formula. It is, for instance,
v•
X
FIGURE7
SEC. 7 PRODUCTS AND QUOTIENTS IN EXPONENTIAL FORM 17
geometrically obvious that
ei"=-1, e-irr/2=-i, and e-i4rr=l.
Note, too, that the equation
(7)
is a parametric representation of the circle lzl = R, centered at the origin with radius
R. As the parameter(} increases from (} = 0 to () =2n, the point z starts from the
positive real axis and traverses the circle once in the counterclockwise direction. More
generally, the circle lz- zol = R, whose center is zo and whose radius is R, has the
parametric representation
(8) z = zo + Reifi (0 < (} < 2n).
This can be seen vectorially (Fig. 8) by noting that a point z traversing the circle
lz - zol =R once in the counterclockwise direction corresponds to the sum of the
fixed vector zo and a vector of length R whose angle of inclination (} varies from 8 = 0
toe= 2n.
y
0 X
FIGURES
7. PRODUCTS AND QUOTIENTS IN EXPONENTIAL FORM
Simple trigonometry tells us that ew has the familiar additive property of the exponen-
tial function in calculus:
ei81ei82
=(cos 81+ i sin 01)(cos 02 + i sin 82)
=(cos 01cos 8z - sin 81 sin 8z) + i (sin 01 cos Oz +cos 81 sin 02)
= cos(01 +82) + i sin(81+ 02) = ei(B,+Bz).
Thus, if z1 = r 1ei8t and z2 =r2eifh, the product z1z2 has exponential form
(1)
18 COMPLEX Nt:MBERS
Moreover,
(2)
zI rl eifhe-ith
Zz = rz . eiR2e-ifl2
CHAP. I
Because 1 = leiO, it follows from expression (2) that the inverse of any nonzero
complex number z =rei() is
(3)
-1 1 1 -ie
z =- = -e
z r
Expressions (1), (2), and (3) are, of course, easily remembered by applying the usual
algebraic rules for real numbers and ex.
Expression (1) yields an important identity involving arguments:
(4)
It is to be interpreted as saying that if values of two of these three (multiple-valued)
arguments are specified, then there is a value of the third such that the equation holds.
We start the verification of statement (4) by letting 81 and 82 denote any values
of arg z1 and arg z2, respectively. Expression (1) then tells us that e1+fh is a value of
arg(z1z2). (See Fig. 9.) If, on the other hand, values ofarg(z1z2) andarg z1are specified,
those values correspond to particular choices of n and n1 in the expressions
(n = 0, ±1, ±2, ...)
and
(n 1 = 0, ±1, ±2, ...).
Since
FIGURE9
SEC. 7 PRoDUCTS AND QuoTIENTS IN ExPONENTIAL FoRM 19
equation (4) is evidently satisfied when the value
is chosen. Verification when values of arg(z1z2) and arg z2 are specified follows by
symmetry.
Statement (4) is sometimes valid when arg is replaced everywhere by Arg (see
Exercise 7). But, as the following example illustrates, that is not always the case.
EXAMPLE 1. When z1 = -1 and z2 = i,
j( 3Jr
Arg z1 + Arg zz = rr +- =-.
2 2
If, however, we take the values of arg z1 and arg z2 just used and select the value
j( 3Jr
Arg(z1zz) +2n = -- + 2rr =-
2 2
of arg(z1z2), we find that equation (4) is satisfied.
Statement (4) tells us that
(
z1) -1) ( -1)arg Zz = arg(ztz2 =arg z1 +arg z2 ,
and we can see from expression (3) that
(5)
Hence
(6) arg(;~) = arg z1 - arg z2.
Statement (5) is, of course, to be interpreted as saying that the set of all values on the
left-hand side is the same as the set of all values on the right-hand side. Statement (6)
is, then, to be interpreted in the same way that statement (4) is.
EXAMPLE 2. In order to find the principal argument Arg zwhen
-2
z= '
1+ J3i
observe that
arg z = arg(-2) - arg(l +.J3i).
20 CoMPLEX NUMBERS
Since
Arg(-2) = rr and Arg(l + .J3i) =::,
3
CHAP. I
one value of arg z is 2n/3; and, because 2rr/3 is between -n and n, we find that
Arg z =2rr j3.
Another important result that can be obtained formally by applying rules for real
numbers to z = reie is
(7) (n = 0, ±1, ±2, ...).
It is easily verified for positive values of n by mathematical induction. To be specific,
we first note that it becomes z = reie when n = 1. Next, we assume that it is valid
when n = m, where m is any positive integer. In view ofexpression (1) for the product
of two nonzero complex numbers in exponential form, it is then valid for n = m + 1:
Expression (7) is thus verified when n is a positive integer. It also holds when n = 0,
with the convention that z0 = 1. If n = -1, -2, ... , on the other hand, we define zn
in terms of the multiplicative inverse of z by writing
~n _ (.,-l)m
<(. - ,"'- where m = -n = 1, 2, ....
Then, since expression (7) is valid for positive integral powers, it follows from the
exponential form (3) of z-1 that
zn = [: ei(-t'))Jm = c~)m eim(-tJ) = (:)-n ei(-n)(-fJ) = rneine
(n = -1, -2, ...).
Expression (7) is now established for all integral powers.
Observe that if r = 1, expression (7) becomes
(8) (n = 0, ±1. ±2, ...).
When written in the form
(9) (cos e+i sin e)n =cos ne + i sin ne (n = 0, ±1, ±2, ...),
this is known as de Moivre'sfomzula.
Expression (7) can be useful in finding powers of complex numbers even when
they are given in rectangular form and the result is desired in that form.
22 COMPLEX NUMBERS
and [see Exercise 2(b)J
10. Establish the identity
') 1- zn+l
1+ z + c + ... + zn = ---
1- z
and then use it to derive Lagrange's trigonometric identity:
CHAP. I
(z i= 1)
1+cos() +cos 2() + · · ·+cos n() = ~ + _si_n_[(:...Z_n_+--.:..I)_e.:....;2...::J
2 2 sin(B/2)
(0 < () < 2rr).
Suggestion: As for the first identity, write S = 1+ z + z2 + · · · + zn and consider
the differenceS - zS. To derive the second identity,. write z = eiB in the first one.
11. (a) Use the binomial formula (Sec. 3) and de Moivre's formula (Sec. 7) to write
cos n(J + i sin n(J =t (n) cosn-k O(i sin (J)k
k=O k
Then define the integer m by means of the equations
{
n/2 if n is even,
m = (n- 1)/2 if n is odd
(n = I, 2, ...).
and use the above sum to obtain the expression [compare Exercise 5(a)]
cos nO =t (n)(-l)k cosn-Zk () sin2k()
k=O 2k
(n = l, 2, ...).
(b) Write x =cos() and suppose that 0 < e < rr, in which case -1 < x < I. Point out
how it follows from the final result in part (a) that each of the functions
(n = 0, 1, 2, ...)
is a polynomial of degree n in the variable x .*
8. ROOTS OF COMPLEX NUMBERS
Consider now a point z = rei0 , lying on a circle centered at the origin with radius r (Fig.
10). As() is increased, z moves around the circle in the counterclockwise direction. In
particular, when () is increased by 2rr, we arrive at the original point; and the same is
*These polynomials are called Chebyshev polynomials and are prominent in approximation theory.
SEC. 8 ROOTS OF COMPLEX NUMBERS 23
Y'
X
FIGURE 10
true when() is decreased by 2rr. It is, therefore, evident from Fig. 10 that two nonzero
complex numbers
are equal ifand only if
r 1 =r2 and 81 =82 +2k:rr,
where k is some integer (k = 0, ±1, ±2, ...).
This observation, together with the expression zn = rneine in Sec. 7 for integral
powers of complex numbers z = reiB, is useful in finding the nth roots of any nonzero
complex number zo =r0eiBo, where nhas one of the values n =2, 3, .... The method
starts with the fact that an nth root of zo is a nonzero number z = reie such that zn = z0,
or
According to the statement in italics just above, then,
rn = r0 and nO = 00 +2k:rr,
where k is any integer (k = 0, ±1, ±2, ...). So r = .vrr<J, where this radical denotes
the unique positive nth root of the positive real number r0, and
8 = 00 +2k:rr = 80 + 2k:rr (k = 0, ±1, ±2, ...).
n n n
Consequently, the complex numbers
[·(e0 2krr)Jz = .zy'rQ exp z --;; + --;;- (k = 0, ±1, ±2, ...)
are the nth roots of z0. We are able to see immediately from this exponential form of
the roots that they all lie on the circle lzI= .vtri) about the origin and are equally spaced
every 2rrIn radians, starting with argument 80/ n. Evidently, then, all of the distinct
24 COMPLEX NUMBERS CHAP. I
roots are obtained when k = 0, 1, 2, ... , n - 1, and no further roots arise with other
values of k. We let ck (k = 0, 1, 2, ... , n - 1) denote these distinct roots and write
(1) ck =Faexp[i ( ~; +
2
~n)J (k =0, 1, 2, ... , n - 1).
(See Fig. 11.)
X
FIGURE 11
The number ~ is the length of each of the radius vectors representing the n
roots. The first root c0 has argument 00 /n; and the two roots when n =2 lie at the
opposite ends of a diameter of the circle IzI= yrro, the second root being -c0. When
n > 3, the roots lie at the vertices ofa regular polygon ofn sides inscribed in that circle.
We shall let z~n denotethe set of nth roots of z0 . If, in particular, zo is a positive
real number r0 , the symbol r~/n denotes the entire set of roots; and the symbol yrrQ in
expression (1) is reserved for the one positive root. When the value of80 that is used in
expression (1) is the principal value of arg zo (-n < 80 < n ), the number c0 is referred
to 'as the principal root. Thus when zo is a positive real number r0, its principal root is
V'Yo·
Finally, a convenient way to remember expression (1) is to write z0 in its most
general exponential form (compare Example 2 in Sec. 6)
(2) (k = 0, ±1, ±2, ...)
and to formally apply laws offractional exponents involving real numbers, keeping in
mind that there are precisely n roots:
ljn _ [ i(11o+2br)JI/n _ n!v': [i(8o+2kn)]z0 - r0 e - ...; r0 exp
n
= Fo exp[i ( ~; +
2
~n)J (k =0, 1, 2, ... , n- 1).
The examples in the next section serve to illustrate this method for finding roots of
complex numbers.
26 COMPLEX NUMBERS CHAP. I
it is worthwhile observing that if c is any particular nth root of a nonzero complex
number z0, the set of nth roots can be put in the form
This is because multiplication of any nonzero complex number by wn increases the
argument of that number by 2rrjn, while leaving its modulus unchanged.
EXAMPLE 2. Let us find all values of (-8i)113, or the three cube roots of -8i. One
need only write
(k = 0, ±1, ±2, ...)
to see that the desired roots are
(3) ck = 2 exp[i (- ~ +
2
~Jr)J (k = 0, 1, 2).
They lie at the vertices of an equilateral triangle, inscribed in the circle IzI =2, and
are equally spaced around that circle every 2rr/3 radians, starting with the principal
root (Fig. 13)
c0 =2 exp[i (- ~)] =2(cos ~ - i sin~)= v'3- i.
Without any further calculations, it is then evident that c1 = 2i; and, since c2 is
symmetric to c0 with respect to the imaginary axis, we know that c2 = -v'3 - i.
These roots can, of course, be written
co, cow3, c0wi where w3=exp(i
2
;).
(See the remarks at the end of Example 1.)
)'
2 X
FIGURE 13
28 COMPLEX NUMBERS CHAP. I
Consequently,
Since c1 = -c0, the two square roots of v'3 +i are, then,
EXERCISES
1. Find the square roots of (a) 2i; (b) 1 - ../3i and express them in rectangular coordinates.
Ans. (a) ±(1 +i); (b)±~~ i.
2. In each case, find all of the roots in rectangular coordinates, exhibit them as vertices of
certain squares, and point out which is the principal root:
(a) (-16)114; (b) (-8- 8../3i)1/4.
Ans. (a) ±VIO + i), ±VI(l- i); (b) ±(../3- i), ±(1 + ../3i).
3. In each case, find all of the roots in rectangular coordinates, exhibit them as vertices of
certain regular polygons, and identify the principal root:
(a) (-l)lf3; (b) gl/6.
r;:; 1+ J3i 1- J3i
Ans. (b) ±vL., ± VI , ± VI .
4. According to Example 1 in Sec. 9, the three cube roots of a nonzero complex number zo
can be written c0, cow3, c0w~, where c0 is the principal cube root of zo and
_ (.2rr) _-1+../3iw3 - exp 1- - .
3 2
Show that if z0 =-4J2 +4VIi, then c0 =VIO +i) and the other two cube roots are,
in rectangular form, the numbers
5. (a) Let a denote any fixed real number and show that the two square roots of a + i are
where A= Ja2 + 1and a= Arg(a + i).
SEC. IO REGIONS IN THE COMPLEX PLANE 29
(b) With the aid of the trigonometric identities (5) in Example 3 of Sec. 9, show that the
square roots obtained in part (a) can be written
[Note that this becomes the final result in Example 3, Sec. 9, when a= ./3.]
6. Find the four roots of the equation z4 +4 =0 and use them to factor z4
+4 into quadratic
factors with real coefficients.
Ans. (z2
+2z + 2)(z2 - 2z +2).
7. Show that if cis any nth root of unity other than unity itself, then
1+c + c2 + ···+cn-l = 0.
Suggestion: Use the first identity in Exercise 10, Sec. 7.
8. (a) Prove that the usual formula solves the quadratic equation
")
az~ +bz +c =0 (a f. 0)
when the coefficients a, b, and care complex numbers. Specifically, by completing
the square on the left-hand side, derive the quadratic formula
-b + (b2 - 4ac)lf2
Z=
2a
where both square roots are to be considered when b2 - 4ac =I= 0,
(b) Use the result in part (a) to find the roots of the equation z2 +2z + (1- i) =0.
Ans.(b)(-1+ ~)+ ~· (-1- ~)- ~·
9. Let z = rei0 be any nonzero complex number and n a negative integer (n = -1, -2, ...).
Then define zl!n by means of the equation zlln =(z-1) 1/m, where m = -n. By showing
that them values of (zlfm)-1 and (z-1) 1/m are the same, verify that zl!n = (z11m)-1.
(Compare Exercise 8, Sec. 7.)
10. REGIONS IN THE COMPLEX PLANE
In this section, we are concerned with sets ofcomplex numbers, or points in the zplane,
and their closeness to one another. Our basic tool is the concept of an s neighborhood
(1) lz- zol < s
of a given point z0. It consists of all points zlying inside but not on a circle centered at
30 COMPLEX NUMBERS CHAP. 1
y
lz - Zol0,....--....,,I
I !'-l-.-£_.1
l z I
I Zo I
I
' /
.... "'---
0 X
FIGURE 15
zo and with a specified positive radius e(Fig. 15). When the value ofeis understood or
is immaterial in the discussion, the set (1) is often referred to as just a neighborhood.
Occasionally, it is convenient to speak of a deleted neighborhood
(2) 0 < lz- zol < s,
consisting of all points z in an e neighborhood of zo except for the point z0 itself.
A point zo is said to be an interior point of a set S whenever there is some
neighborhood of zo that contains only points of S; it is called an exterior point of
S when there exists a neighborhood of it containing no points of S. If z0 is neither of
these, it is a boundary point of S. A boundary point is, therefore, a point all of whose
neighborhoods contain points in S and points not in S. The totality of all boundary
points is called the boundary of S. The circle lzl = 1, for instance, is the boundary of
each of the sets
(3) lzl < 1 and lzl < 1.
A set is open if it contains none of its boundary points. It is left as an exercise
to show that a set is open if and only if each of its points is an interior point. A set is
closed if it contains all of its boundary points; and the closure of a set S is the closed
set consisting of all points in S together with the boundary of S. Note that the first of
the sets (3) is open and that the second is its closure.
Some sets are, of course, neither open nor closed. For a set to be not open,
there must be a boundary point that is contained in the set; and if a set is not closed,
there exists a boundary point not contained in the set. Observe that the punctured disk
0 < lzl < 1is neither open nor closed. The set of all complex numbers is, on the other
hand, both open and closed since it has no boundary points.
An open set S is connected if each pair of points z1 and z2 in it can be joined
by a polygonal line, consisting of a finite number of line segments joined end to end,
that lies entirely in S. The open set lzl < 1 is connected. The annulus 1 < lzl < 2 is,
of course, open and it is also connected (see Fig. 16). An open set that is connected
is called a domain. Note that any neighborhood is a domain. A domain together with
some, none, or all of its boundary points is referred to as a region.
SEC. IO
y
_,.
I
/
il
I
/
/
,.-/
'2I
I
X
EXERCISES 31
FIGURE 16
A setS is bounded if every point of S lies inside some circle lzl = R; otherwise,
it is unbounded. Both of the sets (3) are bounded regions, and the half plane Re z > 0
is unbounded.
A point zo is said to be an accumulation point of a set S if each deleted neigh-
borhood of z0 contains at least one point of S. It follows that if a set S is closed, then
it contains each of its accumulation points. For if an accumulation point z0 were not
inS, it would be a boundary point of S; but this contradicts the fact that a closed set
contains all of its boundary points. It is left as an exercise to show that the converse
is, in fact, true. Thus, a set is closed if and only if it contains all of its accumulation
points.
Evidently, a point z0 is not an accumulation point of a set S whenever there exists
some deleted neighborhood of zo that does not contain points of S. Note that the origin
is the only accumulation point of the set Zn = i In (n = 1, 2, ...).
EXERCISES
1. Sketch the following sets and determine which are domains:
(a) lz- 2 +i I < 1; (b) !2z +31 > 4;
(c) Im z > 1;
(e) 0 < arg z -:;: Tl /4 (z f 0);
Ans. (b), (c) are domains.
(d) Imz = 1;
if) !z- 41 > lzl.
2. Which sets in Exercise 1 are neither open nor closed?
Ans. (e).
3. Which sets in Exercise 1 are bounded?
Ans. (a).
4. In each case, sketch the closure of the set:
(a) -n < arg z < Tl (z f 0); (b) IRezl < lzl;
(c) Re(;) -:;: ~; (d) Re(z2
) > 0.
32 COMPLEX NTJMBERS CHAP. I
5. LetS be the open set consisting of all points z such that Jzl < 1or Jz- 21 < 1. State why
S is not connected.
6. Show that a set S is open if and only if each point in S is an interior point.
7. Determine the accumulation points of each of the following sets:
(a) Zn =in (n =1, 2, ...); (b) Zn =in jn (n =1, 2, ...);
n-1
(c) 0 < arg z < rr/2 (z f. 0); (d) Zn =(-ItO+ i) (n = 1, 2, ...).
n
Ans. (a) None; (b) 0; (d) ±(1 +i).
8. Prove that if a set contains each of its accumulation points, then it must be a closed set.
9. Show that any point zo of a domain is an accumulation point of that domain.
10. Prove that a finite set of points ZJ. z2, ..• , Zn cannot have any accumulation points.
CHAPTER
2
ANALYTIC FUNCTIONS
We now consider functions of a complex variable and develop a theory of differenti-
ation for them. The main goal of the chapter is to introduce analytic functions, which
play a central role in complex analysis.
11. FUNCTIONS OF A COMPLEX VARIABLE
Let S be a set of complex numbers. A function f defined on S is a rule that assigns to
each z in S a complex number w. The number w is called the value of f at z and is
denoted by j(z); that is, w = j(z). The setS is called the domain ofdefinition of f.*
It must be emphasized that both a domain of definition and a rule are needed in
order for a function to be well defined. When the domain ofdefinition is not mentioned,
we agree that the largest possible set is to be taken. Also, it is not always convenient
to use notation that distinguishes between a given function and its values.
EXAMPLE 1. Iff is defined on the set z f::. 0 by means of the equation w = 1/z, it
may be referred to only as the function w =1/z, or simply the function 1/z.
Suppose that w = u + i vis the value of a function f at z = x + iy, so that
u+iv=f(x+iy).
*Although the domain of definition is often a domain as defined in Sec. I0, it need not be.
33
34 ANALYTIC FUNCTIONS CHAP. 2
Each of the real numbers u and v depends on the real variables x and y, and it follows
that f(z) can be expressed in terms of a pair of real-valued functions of the real
variables x and y:
(1) f(z) = u(x, y) +iv(x, y).
If the polar coordinates r and e, instead of x and y, are used, then
u +iv = f(rei8
),
where w = u + iv and z = rei8. In that case, we may write
(2) /(z) = u(r, 8) + iv(r, e).
EXAMPLE 2. If /(z) = z2
, then
f(x +iy) = (x +iy)
2
=x
2
- i + i2xy.
Hence
u(x, y) = x2
- y2
and v(x, y) = 2xy.
When polar coordinates are used,
f(rei8
) = (rei8) 2 = r2eiZe = r 2
cos 28 +ir2 sin 28.
Consequently,
u(r,8)=r2
cos2e and v(r,8)=r2
sin28.
If, in either of equations (1) and (2), the function v always has value zero, then
the value off is always real. That is, f is a real-valuedfunction of a complex variable.
EXAMPLE 3. A real-valued function that is used to illustrate some important
concepts later in this chapter is
f(z) = lzl2
= x2
+ l + iO.
If n is zero or a positive integer and if a0, ab a2, .•. , an are complex constants,
where an i= 0, the function
P(z) = ao a1z +a2z2
+ ···+ anzn
is apolynomial ofdegree n. Note that the sum here has a finite number ofterms and that
the domain of definition is the entire zplane. Quotients P (z) j Q(z) ofpolynomials are
called rational/unctions and are defined at each point z where Q(z) i= 0. Polynomials
and rational functions constitute elementary, but important, classes of functions of a
complex variable.
SEC. I I EXERCISES 35
A generalization of the concept of function is a rule that assigns more than one
value to a point z in the domain of definition. These multiple-valuedfunctions occur
in the theory of functions of a complex variable, just as they do in the case of real
variables. When multiple-valued functions are studied, usually just one of the possible
values assigned to each point is taken, in a systematic manner, and a (single-valued)
function is constructed from the multiple-valued function.
EXAMPLE 4. Let z denote any nonzero complex number. We know from Sec. 8
that z1
12
has the two values
(e)z1/Z = ±JTexp i "2 ,
where r =!zi and 8(-rr < 8 < rr) is the principal value of arg z. But, if we choose
only the positive value of ±JTand write
(3) (r > 0, -j( < e < rr),
the (single-valued) function (3) is well defined on the set of nonzero numbers in the z
plane. Since zero is the only square root ofzero, we also write f (0) =0. The function
f is then well defined on the entire plane.
EXERCISES
1. For each of the functions below, describe the domain of definition that is understood:
(a) f(z) = 2
1
; (b) /(z) =Arg(~);
z + 1 z
z 1
(c)f(z)= _; (d)f(z)=
2
.
z +z 1-lzl
Ans. (a) z :f: ±i; (c) Re z ;f 0.
2. Write the function f(z) = z3
+z + 1in the form f(z) =u(x, y) + iv(x, y).
Ans. (x3 - 3xy2 +x + 1) + i(3x2y- y3 + y).
3. Suppose that f(z) =x 2
- y2
- 2y + i(2x- 2xy), where z = x +iy. Use the expres-
sions (see Sec. 5)
z+z
x=--
2
and
z-z
y ::::: 2i
to write f(z) in terms of z. and simplify the result.
A -2 2'ns.z + tz.
4. Write the function
f
' ·. 1
Z) =Z +-
z
(z ;f 0)
36 ANALYTIC FUNCTIONS CHAP. 2
in the form f(z) =u(r' (J) +iv(r' e).
Ans. (r + :) cos e+ i (r - ;)sin e.
12. MAPPINGS
Properties of a real-valued function of a real variable are often exhibited by the graph
of the function. But when w = f(z), where z and ware complex, no such convenient
graphical representation of the function f is available because each of the numbers
z and w is located in a plane rather than on a line. One can, however, display some
information about the function by indicating pairs of corresponding points z =(x, y)
and w =(u, v). To do this, it is generally simpler to draw the z and w planes separately.
When a function f is thought of in this way, it is often referred to as a mapping,
or transformation. The image of a point z in the domain of definition S is the point
w = f (z), and the set of images of all points in a set T that is contained in S is called
the image of T. The image of the entire domain of definition S is called the range of
f. The inverse image of a point w is the set of all points z in the domain of definition
of f that have w as their image. The inverse image of a point may contain just one
point, many points, or none at all. The last case occurs, of course, when w is not in the
range of f.
Terms such as translation, rotation, and reflection are used to convey dominant
geometric characteristics ofcertain mappings. In such cases, it is sometimes convenient
to consider the z and w planes to be the same. For example, the mapping
w = z + 1= (x + 1) +iy,
where z = x + iy, can be thought of as a translation of each point z one unit to the
right. Since i = eirr:/2, the mapping
where z = reie, rotates the radius vector for each nonzero point z through a right angle
about the origin in the counterclockwise direction; and the mapping
- .W = Z =X -ly
transforms each point z = x + iy into its reflection in the real axis.
More information is usually exhibited by sketching images of curves and regions
than by simply indicating images of individual points. In the following examples, we
illustrate this with the transformation w = z2•
We begin by finding the images of some curves in the z plane.
SEC. 12 MAPPINGS 37
EXAMPLE 1. According to Example 2 in Sec. 11, the mapping w = z2
can be
thought of as the transformation
(1)
from the xy plane to the uv plane. This form of the mapping is especially useful in
finding the images of certain hyperbolas.
It is easy to show, for instance, that each branch of a hyperbola
(2) (c1 > 0)
is mapped in a one to one manner onto the vertical line u = c1. We start by noting
from the first of equations (1) that u = c1 when (x, y) is a point lying on either branch.
When, in particular, it lies on the right-hand branch, the second of equations (1) tells
us that v =2yJy2 +c1• Thus the image of the right-hand branch can be expressed
parametrically as
U =CJ, V =2yJy2
+Ct (-oo < y < oo);
and it is evident that the image of a point (x, y) on that branch moves upward along the
entire line as (x, y) traces out the branch in the upward direction (Fig. 17). Likewise,
since the pair of equations
(-oo < y < oo)
furnishes a parametric representation for the image of the left-hand branch of the
hyperbola, the image of a point going downward along the entire left-hand branch
is seen to move up the entire line u = c1.
On the other hand, each branch of a hyperbola
(3)
is transformed into the line v =c2, as indicated in Fig. 17. To verify this, we note from
the second of equations (1) that v =c2 when (x, y) is a point on either branch. Suppose
y
I
I
'
..., 0
I
I
I
' ...
X
v
u = c1 >0
------- -- --v;:;: c2 > 0
0 u
FIGURE 17
w =z2
.
38 ANALYTIC FUNCTIONS CHAP. 2
that it lies on the branch lying in the first quadrant. Then, since y =c2j(2x), the first
of equations (1) reveals that the branch's image has parametric representation
(0 < x < oo).
Observe that
lim u =-oo and lim u =oo.
X-+00
Since u depends continuously onx, then, it is clear that as (x, y) travels down the entire
upper branch of hyperbola (3), its image moves to the right along the entire horizontal
line v = c2. Inasmuch as the image of the lower branch has parametric representation
(-00 < y < 0)
and since
lim u = -oo and lim u = oo,
y-+-oo y~O
y<O
it follows that the image of a point moving upward along the entire lower branch also
travels to the right along the entire line v = c2 (see Fig. 17).
We shall now use Example 1 to find the image of a certain region.
EXAMPLE 2. The domain x > 0, y > 0, xy < 1consists of all points lying on the
upper branches of hyperbolas from the family 2xy = c, where 0 < c < 2 (Fig. 18). We
know from Example 1 that as a point travels downward along the entirety of one of
these branches, its image under the transformation w = z2 moves to the right along
the entire line v = c. Since, for all values of c between 0 and 2, the branches fill out
y
A
B
D
D'
E
c X A'
v
2i E'
FIGURE 18
B' C' u
w=z2
.
SEC. 12 MAPPINGS 39
the domain x > 0, y > 0, xy < 1, that domain is mapped onto the horizontal strip
0 < v < 2.
In view of equations (1), the image of a point (0, y) in the z plane is (-y2, 0).
Hence as (0, y) travels downward to the origin along the y axis, its image moves to the
right along the negative u axis and reaches the origin in the w plane. Then, since the
image of a point (x, 0) is (x2
, 0), that image moves to the right from the origin along
the u axis as (x, 0) moves to the right from the origin along the x axis. The image
of the upper branch of the hyperbola xy = 1 is, of course, the horizontal line v =2.
Evidently, then, the closed region x > 0, y > 0, xy < 1is mapped onto the closed strip
0 < v < 2, as indicated in Fig. 18.
Our last example here illustrates how polar coordinates can be useful in analyzing
certain mappings.
EXAMPLE 3. The mapping w = z2 becomes
when z =rew. Hence if w = peit/1, we have pei¢1 =r2
ei2e; and the statement in italics
near the beginning of Sec. 8 tells us that
p = r2
and ¢ = 2(} +2kn,
where k has one of the values k = 0, ± 1, ±2, .... Evidently, then, the image of any
nonzero point z is found by squaring the modulus of z and doubling a value of arg z.
Observe that points z = r0ew on a circle r = r0 are transformed into points
w = rJei2
e on the circle p = rJ. As a point on the first circle moves counterclockwise
from the positive real axis to the positive imaginary axis, its image on the second
circle moves counterclockwise from the positive real axis to the negative real axis (see
Fig. 19). So, as all possible positive values of r0 are chosen, the corresponding arcs
in the z and w planes fill out the first quadrant and the upper half plane, respectively.
The transformation w =z2 is, then, a one to one mapping of the first quadrant r > 0,
0 < e< j(!2 in the z plane onto the upper half p > 0, 0 < ¢ < j( of the w plane, as
indicated in Fig. 19. The point z = 0 is, of course, mapped onto the point w = 0.
The transformation w = z2
also maps the upper half planer > 0, 0 < e< n onto
the entire w plane. However, in this case, the transformation is not one to one since
v
FIGURE 19
w =z2
•
40 ANALYTIC FuNCTIONS CHAP. 2
both the positive and negative real axes in the z plane are mapped onto the positive
real axis in the w plane.
When n is a positive integer greater than 2, various mapping properties of the
transformation w = zn, or pei¢> = rneine, are similar to those of w = z2• Such a
transformation maps the entire z plane onto the entire w plane, where each nonzero
point in the w plane is the image of n distinct points in the z plane. The circle r =r0
is mapped onto the circle p = r0;and the sector r < r0, 0 < e< 2n/n is mapped onto
the disk p < r0,but not in a one to one manner.
13. MAPPINGS BY THE EXPONENTIAL FUNCTION
In Chap. 3 we shall introduce and develop properties of a number ofelementary func-
tions which do not involve polynomials. That chapter will start with the exponential
function
(1) (z=x+iy),
the two factors ex and eiY being well defined at this time (see Sec. 6). Note that
definition (1), which can also be written
is suggested by the familiar property
of the exponential function in calculus.
The object of this section is to use the function ez to provide the reader with
additional examples of mappings that continue to be reasonably simple. We begin by
examining the images of vertical and horizontal lines.
EXAMPLE 1. The transformation
(2)
can be written pei¢> = exeiY, where z = x + iy and w = peiif>. Thus p =ex and
<P = y +2mr, where n is some integer (see Sec. 8); and transformation (2) can be
expressed in the form
(3)
The image of a typical point z = (ci> y) on a vertical line x = c1 has polar
coordinates p = exp c1and <P =yin thew plane. That image moves counterclockwise
around the circle shown in Fig. 20 as z moves up the line. The image of the line is
evidently the entire circle; and each point on the circle is the image of an infinite
number of points, spaced 2n units apart, along the line.
SEC. 13
y
---- - -- y = c2
0
v
X
MAPPINGS BY THE EXPONENTIAL FuNCTION 41
FIGURE20
w = expz.
A horizontal line y = c2 is mapped in a one to one manner onto the ray cp = c2. To
see that this is so, we note that the image of a point z = (x, c2) has polar coordinates
p =ex and¢= c2. Evidently, then, as that point z moves along the entire line from
left to right, its image moves outward along the entire ray ¢ = c2, as indicated in
Fig. 20.
Vertical andhorizontal line segments are mapped ontoportions ofcircles and rays,
respectively, and images of various regions are readily obtained from observations
made in Example 1. This is illustrated in the following example.
EXAMPLE 2. Let us show that the transformation w = ez maps the rectangular
region a< x < b, c < y < d onto the region ea < p < eb, c <¢<d. The two regions
and corresponding parts of their boundaries are indicated in Fig. 21. The vertical line
segment AD is mapped onto the arc p = ea, c < ¢ < d, which is labeled A'D'. The
images of vertical line segments to the right of AD and joining the horizontal parts
of the boundary are larger arcs; eventually, the image of the line segment BC is the
arc p = eb, c < ¢ < d, labeled B'C'. The mapping is one to one if d- c < 2rr. In
particular, if c = 0 and d = rr, then 0 < ¢ < rr; and the rectangular region is mapped
onto half of a circular ring, as shown in Fig. 8, Appendix 2.
y
d
D
c
0 a
FIGURE21
w=expz.
I
I
t
:
c
B
b
v
X u
42 ANALYTIC FUNCTIONS CHAP. 2
Our final example here uses the images of horizontal lines to find the image of a
horizontal strip.
EXAMPLE 3. When w = e2
, the image of the infinite strip 0 < y < 7r is the upper
half v > 0 of the w plane (Fig. 22). This is seen by recalling from Example 1 how
a horizontal line y = c is transformed into a ray <P = c from the origin. As the real
number c increases from c =0 to c = 7r, the y intercepts of the lines increase from
0 to n and the angles of inclination of the rays increase from <P = 0 to <P = n. This
mapping is also shown in Fig. 6 of Appendix 2, where corresponding points on the
boundaries of the two regions are indicated.
y
Cl
----------- ----.------
FIGURE22
w=expz.
EXERCISES
0 X
v
I
I
I
I
0
I
I
I
I
II
I
I
I
¢=c
u
1. By referring to Example 1 in Sec. 12, find a domain in the z plane whose image under
the transformation w =z2 is the square domain in the w plane bounded by the lines
u = 1, u =2, v = 1, and v =2. (See Fig. 2, Appendix 2.)
2. Find and sketch, showing corresponding orientations, the images of the hyperbolas
x2
- i = c1 (c1 < 0) and 2xy =c2 (cz < 0)
under the transformation w =z2.
3. Sketch the region onto which the sector r < 1, 0 < () < n /4 is mapped by the transfor-
mation (a) w =z2
; (b) w =z3; (c) w = z4.
4. Show that the lines ay = x (a f 0) are mapped onto the spirals p =exp(a¢) under the
transformation w =exp z, where w = p exp(i¢).
5. By considering the images of horizontal line segments, verify that the image of the
rectangular region a < x < b, c < y < d under the transformation w =exp z is the region
ea < p < eb, c < ¢ < d, as shown in Fig. 21 (Sec. 13).
6. Verify the mapping of the region and boundary shown in Fig. 7 of Appendix 2, where
the transformation is w =exp z.
7. Find the image of the semi-infinite strip x > 0, 0 < y < n under the transformation
w =exp z, and label corresponding portions of the boundaries.
SEC. 14 LtMrrs 43
8. One interpretation ofa function w =f (z) =u(x, y) +i v(x, y) is that ofa vectorfield in
the domain ofdefinition off. The function assigns a vector w, with components u(x, y)
and v(x, y), to each point z at which it is defined. Indicate graphically the vector fields
represented by (a) w =iz; (b) w =z/lzl.
14. LIMITS
Let a function f be defined at all points z in some deleted neighborhood (Sec. 10) of
z0. The statement that the limit off(z) as z approaches z0 is a number w0, or that
(1) lim f(z) = wo.
z-+zo
means that the point w = f{z) can be made arbitrarily close to w0 if we choose the
point z close enough to z0 but distinct from: it. We now express the definition of limit
in a precise and usable form.
Statement (1) means that, for each positive number£, there is a positive number
8 such that
(2) lf(z)- w0 1< £ whenever 0 < lz- z01< 8.
Geometrically, this definition says that, for each £ neighborhood Iw - w0 ! < £ of w0,
there is a deleted 8 neighborhood 0 < lz - z01 < 8 of z0 such that every point z in it
has an image w lying in the£ neighborhood (Fig. 23). Note that even though all points
in the deleted neighborhood 0 < lz- zol < 8 are to be considered, their images need
not fill up the entire neighborhood lw- w0 1<£.Iff has the constant value w0, for
instance, the image of z is always the center ofthat neighborhood. Note, too, that once
a 8 has been found, it can be replaced by any smaller positive number, such as 8J2.
y v
..... --/ 'I '
/"-8'
;•w~,
Wo /
I Zoe?: '.... /
I .~,
__....
0
, __ X 0 u FIGURE23
It is easy to show that when a limit of a function f (z) exists at a point z0, it is
unique.. To do this, we suppose that
lim f(z) = w0 and lim f(z) = w1•
Z~Zo Z-+Zo
Then, for any positive number£, there are positive numbers 80 and 81 such that
!f(z)- wo! < £ whenever 0 < z- z01 < 80
44 ANALYTIC FUNCTIONS CHAP. 2
and
1/(z)- wtl < 8 whenever 0 < lz- zol < 81.
So if 0 < lz - zol < 8, where 8 denotes the smaller of the two numbers 8o and 81o we
find that
lwi- wol =1[/(z)- wo]- [f(z)- Wt]l < 1/(z)- wol +1/(z)- wtl < 8 +8 =28.
But lw1 - w01is a nonnegative constant, and 8 can be chosen arbitrarily small. Hence
Definition (2) requires that f be defined at all points in some deleted neighbor-
hood of z0. Such a deleted neighborhood, ofcourse, always exists when z0 is an interior
point ofa region on which f is defined. We can extend the definition oflimit to the case
in which zo is a boundary point of the region by agreeing that the first of inequalities
(2) need be satisfied by only those points z that lie in both the region and the deleted
neighborhood.
EXAMPLE 1. Let us show that if f(z) = iz/2 in the open disk lzl < 1, then
(3) lim f(z) =!..,
z-,rl 2
the point 1being on the boundary of the domain of definition off. Observe that when
z is in the region lzl < 1,
•
l
f(z)--
2
iz i
2 2
lz - 11
2
Hence, for any such z and any positive number 8 (see Fig. 24),
y
~- -..../
/
/
I
I
0
'' ....
' - --....
'
l
f(z)-- < 8 whenever 0 < lz- 11 < 28.
2
v
...
e'I' •
I L I
2 I
' /
-X 0 u
FIGURE24
SEC. 14 LIMITS 45
Thus condition (2) is satisfied by points in the region lzl < 1when 8 is equal to 2£ or
any smaller positive number.
If z0 is an interior point of the domain of definition of f, and limit (1) is to
exist, the first of inequalities (2) must hold for all points in the deleted neighborhood
0 < lz- zo! < 8. Thus the symbol z ~ z0 implies that z is allowed to approach z0
in an arbitrary manner, not just from some particular direction. The next example
emphasizes this.
EXAMPLE 2. If
(4)
the limit
(5)
f(z) = ~,
z
lim f(z)
z~o
does not exist. For, if it did exist, it could be found by letting the point z = (x, y)
approach the origin in any manner. But when z = (x, 0) is a nonzero point on the real
axis (Fig. 25),
f(z)=x+iO=l;
X- iO
and when z= (0, y) is a nonzero point on the imaginary axis,
f(z)= O+iy =-1.
0- iy
Thus, by letting zapproach the origin along the real axis, we would find that the desired
limit is l. An approach along the imaginary axis would, on the other hand, yield the
limit -1. Since a limit is unique, we must conclude that limit (5) does not exist.
y
z=(0, y)
(0, 0) z = (x, 0) X
FIGURE25
48 ANALYTIC fUNCTIONS CHAP. 2
This important theorem can be proved directly by using the definition ofthe limit
of a function of a complex variable. But, with the aid of Theorem 1, it follows almost
immediately from theorems on limits of real-valued functions of two real variables.
To verify property (9), for example, we write
f(z) =u(x, y) +iv(x, y), F(z) =U(x, y) +iV(x, y),
z0 =x0 +iy0, w0 =u0 +iv0• W0 =U0 +iV0•
Then, according to hypotheses (7) and Theorem 1. the limits a..<.; (x, y) approaches
(x0 , y0) of the functions u, v, U, and V exist and have the values u0, v0, U0, and V0,
respectively. So the real and imaginary components of the product
f(z)F(z) = (uU- vV) + i(vU + uV)
have the limits u0U0 - v0V0 and v0U0 + u0V0, respectively, as (x, y) approaches
(x0 , y0). Hence, by Theorem 1 again, f(z)F(z) has the limit
(u0U0 - v0 V0) +i(v0U0 + u0 Vo)
as z approaches z0: and this is equal to w0 W0. Property (9) is thus established.
Corresponding verifications of properties (8) and (10) can be given.
It is easy to see from definition (2), Sec.14, of limit that
lim c = c and lim z=z0 ,
z-+zo z-+zo
where zo and c are any complex numbers; and, by property (9) and mathematical
induction, it follows that
I
. n n
1m z = z0Z~Zu
(n=l,2, ...).
So, in view of properties (8) and (9), the limit of a polynomial
P(z) = ao +a1z +a2z
2
+ ···+anZ
11
as z approaches a point z0 is the value of the polynomial at that point:
(11) lim P(z) =P(zo).
z-+zo
16. LIMITS INVOLVING THE POINT AT INFINITY
It is sometimes convenient to include with the complex plane the point at infinity,
denoted by oo, and to use limits involving it. The complex plane together with this
point is called the extended complex plane. To visualize the point at infinity, one can
think of the complex plane as passing through the equator of a unit sphere centered at
the point z = 0 (Fig. 26). To each point z in the plane there corresponds exactly one
point P on the surface of the sphere. The point P is determined by the intersection of
the line through the point z and the north pole N of the sphere with that surface. In
SEC. 16 LIMITS INVOLVING THE POINT AT INFINITY 49
r------------------------7I I
I I
I I
I I
I I
I I
I I
I ------- II I
I • z I
I
0
I
I 1 ...... _ 1
I 1 ·, I
I I ' I
I " I I
I '- / I
I '-----~ /I I
~------------------------~ FIGURE26
like manner, to each point P on the surface of the sphere, other than the north pole N,
there corresponds exactly one point z in the plane. By letting the point N ofthe sphere
correspond to the point at infinity, we obtain a one to one correspondence between the
points ofthe sphere and the points ofthe extended complex plane. The sphere is known
as the Riemann sphere, and the correspondence is called a stereographic projection.
Observe that the exterior of the unit circle centered at the origin in the complex
plane corresponds to the upper hemisphere with the equator and the point N deleted.
Moreover, for each small positive numbers, those points in the complex plane exterior
to the circle lzl = 1/e correspond to points on the sphere close toN. We thus call the
set lzl > 1/s an s neighborhood, or neighborhood, of oo.
Let us agree that, in referring to a point z, we mean a point in the finite plane.
Hereafter, when the point at infinity is to be considered, it will be specifically men-
tioned.
A meaning is now readily given to the statement
lim /(z) = wo
z~zo
when either zo or w0, or possibly each of these numbers, is replaced by the point
at infinity. In the definition of limit in Sec. 14, we simply replace the appropriate
neighborhoods of z0 and w0 by neighborhoods of oo. The proof of the following
theorem illustrates how this is done.
Theorem. Ifzo and w0 are points in the z and w planes, respectively, then
(1) lim f (z) =oo if and only if
Z~Zo
1
lim =0
z~zo f(z)
and
(2) lim f(z) = w0 if and only if lim f (~) =w0 .
z~oo z-+0 z
Moreover,
(3) lim /(z) =oo if and only ifz~oo
1
lim = 0.
z~o /(1/z)
SEC. 17 CONTINUITY 51
Furthermore,
]. 2z3 - 1
1m = oo smce
z-+oo z2 + 1
lim (1/z2) + 1 =lim z +z3 = 0.
z-+0 (2/z3) - 1 z-+O 2 - z3
17. CONTINUITY
A function f is continuous at a point z0 if all three of the following conditions are
satisfied:
(1)
(2)
(3)
lim f(z) exists,
z-+zo
f (z0) exists,
lim f(z) = f(zo).
Z-+Zo
Observe that statement (3) actually contains statements (1) and (2), since the existence
of the quantity on each side of the equation there is implicit. Statement (3) says that,
for each positive number £, there is a positive number osuch that
(4) lf(z)- f(zo)l < c: whenever lz- zol < 8.
A function of a complex variable is said to be continuous in a region R if it is
continuous at each point in R.
Iftwo functions are continuous at a point, their sum and product are also continu-
ous at that point; their quotient is continuous at any such point where the denominator
is not zero. These observations are direct consequences of Theorem 2, Sec. 15. Note,
too, that a polynomial is continuous in the entire plane because of limit (11), Sec. 15.
We tum now to two expected properties of continuous functions whose verifica-
tions are not so immediate. Our proofs depend on definition (4), and we present the
results as theorems.
Theorem 1. A composition ofcontinuous functions is itselfcontinuous.
A precise statement of this theorem is contained in the proof to follow. We let
w = f (z) be a function that is defined for all z in a neighborhood lz - zol < 8 of a
point zo, and we let W = g(w) be a function whose domain of definition contains the
image (Sec. 12) ofthat neighborhood under f. The composition W = g[f(z)] is, then,
defined for all z in the neighborhood Iz - zo I < 8. Suppose now that f is continuous at
zo and that g is continuous at the point f (z0) in the w plane. In view of the continuity
of gat f(z0), there is, for each positive number c:, a positive number y such that
ig[f(z)]- g[f(zo)]l < s whenever lf(z) f(zo)l < y.
52 ANALYTIC FUNCTIONS CHAP. 2
y v v
-- -----/
"/
'--- --- /
'/ ' / ..- -.....
/
'/ •z ' '/
'
I
/
I I I
• g[f(z)]
~
I
I~ I
' zo I
: /(zo)
I I
'
I
I I g[f(zo)] I
'
___.,.,.
II I
0 X 0 I u 0 I u
' •/(z) I
I
' /
I.... _ / I
- ' ,f
----- ' /
' /
' ..... /
- ------
FIGURE27
(See Fig. 27.) But the continuity off at zo ensures that the neighborhood lz- zol < 8
can be made small enough that the second of these inequalities holds. The continuity
of the composition g[f(z)] is, therefore, established.
Theorem 2. Ifafunction f (z) is continuous and nonzero at a point z0 , then f (z) =!= 0
throughout some neighborhood ofthat point.
Assuming that f (z) is, in fact, continuous and nonzero at z0, we can prove
Theorem 2 by assigning the positive value lf(zo)l/2 to the numbers in statement
(4). This tells us that there is a positive number 8 such that
lf(z)- f(z0)1 < lf(zo)l whenever lz- zol < 6.
2
So if there is a point z in the neighborhood lz - zol < 8 at which f(z) = 0, we have
the contradiction
and the theorem is proved.
lf(zo)l < lf(zo)l;
2
The continuity of a function
(5) f(z) = u(x, y) + iv(x, y)
is closely related to the continuity of its component functions u(x, y) and v(x, y).
We note, for instance, how it follows from Theorem 1 in Sec. 15 that the function
(5) is continuous at a point z0 = (x0, Yo) if and only if its component functions are
56 ANALYTIC FUNCTIONS CHAP. 2
!l.y
(0, !l.y)
(0, 0) (!l.x, 0) !l.x
FIGURE29
If the limit of b. w1b.z exists, it may be found by letting the point b.z = (b.x, b.y)
approach the origin in the b.z plane in any manner. In particular, when b.z approaches
the origin horizontally through the points (b.x, 0) on the real axis (Fig. 29),
In that case,
b.z = b.x + iO =b.x- iO =b.x + iO =b.z.
b.w -
- = z + b.z +z.
D.z
Hence, if the limit of b.wlb.z exists, its value must be z + z. However, when b.z
approaches the origin vertically through the points (0, b.y) on the imaginary axis, so
that
we find that
b.z =0 + i b.y = -(0 + i b.y) = -b.z,
b.w -
- =z+ b.z -z.
b.z
Hence the limit must be z- z if it exists. Since limits are unique (Sec. 14), it follows
that
Z+ Z = Z- Z,
or z = 0, if dwIdz is to exist.
To show that dwldz does, in fact, exist at z =0, we need only observe that our
expression for b.wIAz reduces to b.z when z =0. We conclude, therefore, thatdwldz
exists only at z = 0, its value there being 0.
Example 2 shows that a function can be differentiable at a certain point but
nowhere else in any neighborhood of that point. Since the real and imaginary parts
of f(z) = lzl2
are
(4) U (X , y) =X
2
+ i and V (X , y) = 0,
SEC. 19 DIFFERENTIATION FoRMULAS 57
respectively, it also shows that the real and imaginary components of a function of a
complex variable can have continuous partial derivatives of all orders at a point and
yet the function may not be differentiable there.
The function f(z) = lzi2 is continuous at each point in the plane since its com-
ponents (4) are continuous at each point. So the continuity of a function at a point
does not imply the existence of a derivative there. It is, however, true that the existence
ofthe derivative ofa function at a point implies the continuity ofthe function at that
point. To see this, we assume that f'(zo) exists and write
lim [f(z)- f(zo)] = lim f(z)- f(zo) lim (z- z0) =!'(z0) · 0 =0,
z___,.zo z-+zo z - zo z-+zo
from which it follows that
lim f(z) = f(zo).
7.-i-'Zo
This is the statement of continuity of f at zo (Sec. 17).
Geometric interpretations of derivatives of functions of a complex variable are
not as immediate as they are for derivatives of functions of a real variable. We defer
the development of such interpretations until Chap. 9.
19. DIFFERENTIATION FORMULAS
The definition of derivative in Sec. 18 is identical in form to that of the derivative of a
real-valued function of a real variable. In fact, the basic differentiation formulas given
below can be derived from that definition by essentially the same steps as the ones used
in calculus. In these formulas, the derivative of a function f at a point z is denoted by
either
d
-f(z) or f'(z),
dz
depending on which notation is more convenient.
Let c be a complex constant and let f be a function whose derivative exists at a
point z. It is easy to show that
(1)
d
-c=O.
dz ·
Also, if n is a positive integer,
(2)
d
- .. - 1.
dz - :}cf(z)] = cf'(z).
d -_.,n =nzn 1<. • •
dz
This formula remains valid when n is a negative integer, provided that z # 0.
58 ANALYTIC FUNCTIONS CHAP. 2
If the derivatives of two functions f and F exist at a point z, then
(3) :
2
[J(z) + F(z)]= j'(z) + F'(z),
(4)
d I /
dz [f(z)F(z)] = j(z)F (z) + f (z)F(z);
and, when F(z) # 0,
(5)
!!_ [ f(z) J= F(z)j'(z)- j(z)F'(z).
dz F(z) [F(z)l2
Let us derive formula (4). To do this, we write the following expression for the
change in the product w = j(z)F(z):
~w = f(z + ~z)F(z + ~z)- j(z)F(z)
= J(z)[F(z + ~z)- F(z)l + [j(z + ~z)- f(z)]F(z + ~z).
Thus
~w = f(z) F(z + ~z)- F(z) + j(z + ~z)- f(z) F(z + ~z);
~z Az ~z
and, letting ~z tend to zero, we arrive at the desired formula for the derivative of
f(z)F(z). Here we have used the fact that F is continuous at the point z, since F'(z)
exists; thus F(z + ~z) tends to F(z) as ~z tends to zero (see Exercise 8, Sec. 17).
There is also a chain rule for differentiating composite functions. Suppose that f
has a derivative at zo and that g has a derivative at the point f(z0). Then the function
F(z) = g[f(z)] has a derivative at z0, and
(6) F'(zo) =g'[f (zo)]f'(zo) ·
If we write w = f(z) and W = g(w), so that W = F(z), the chain rule becomes
dW dW dw
- = - -
dz dw dz
EXAMPLE. To find the derivative of (2z2
+ i)5, write w = 2z2
+ i and W = w5
.
Then
To start the proof of formula (6), choose a specific point zo at which f'(z0)
exists. Write w0 = f(zo) and also assume that g'(w0) exists. There is, then, some
e neighborhood !w- w0 i < e of w0 such that, for all points win that neighborhood,
60 ANALYTIC FUNCTIONS CHAP. 2
(b) the coefficients in the polynomial P(z) in part (a) can be written
a0 = P(O),
P'(O)
al = 1!
P''(O)
az= 2! '
. . . '
3. Apply definition (3), Sec. 18, of derivative to give a direct proof that
I 1
f (z) =--z2
1
when /(z) =- (z i= 0).
z
4. Suppose that /(z0) =g(z0) =0 and that f'(z0) and g'(z0) exist, where g'(z0) i= 0. Use
definition (1), Sec. 18, of derivative to show that
lim /(z) = /'(zo).
z-..zn g(z) g'(zo}
5. Derive formula (3), Sec. 19, for the derivative ofthe sum of two functions.
6. Derive expression (2), Sec. 19, for the derivative of zn when n is a positive integer by
using
(a) mathematical induction and formula (4), Sec. 19, for the derivative of the product of
two functions;
(b) definition (3), Sec. 18, of derivative and the "binomial formula (Sec .3).
7. Prove that expression (2), Sec. 19, for the derivative of zn remains valid when n is a
negative integer (n =-1, -2, ...), provided that z i= 0.
Suggestion: Write m = -n and use the formula for the derivative of a quotient of
two functions.
8. Use the method in Example 2, Sec. 18, to show that f'(z) does not exist at any point z
when
(a) f(z) =z; (b) f(z) =Re z; (c) /(z) =Im z.
9. Let f denote the function whose values are
{
-2
f(z) = ; when z i= 0,
when ~-o<.- .
Show that if z =0, then 6.wj 6.z =1 at each nonzero point on the real and imaginary
axes in the 6.z, or 6.x 6.y, plane. Then show that 6.wj6.z =-1 at each nonzero point
(6.x, 6.x) on the line 6.y = Ll.x in that plane. Conclude from these observations that
f'(0) does not exist. (Note that, to obtain this result, it is not sufficient to consider only
horizontal and vertical approaches to the origin in the 6.z plane.)
20. CAUCHY-RIEMANN EQUATIONS
In this section, we obtain a pair of equations that the first-order partial derivatives of
the component functions u and v of a function
(1) /(z) = u(x, y) + iv(x, y)
62 ANALYTIC FUNCTIONS CHAP. 2
Evidently, then,
R
6.w
1
. v(xo, Yo+ 6.y)- v(xo, Yo) ( . )
lim e - = 1m =Vv xo. Yo
(t...x, t..y)--+(O,O) 6.z t..y...... o 6.y ~
and
I
6.w -- 1' u(xo, Yo+ 6.y) - u(xo, Yo) - ( )
lim m Az - 1m - -uy x 0, Yo .
(t..x,t..y)-+(0,0) u t..y--+0 6.y
Hence it follows from expression (3) that
(5)
where the partial derivatives of u and v are, this time, with respect to y. Note that
equation (5) can also be written in the form
Equations (4) and (5) not only give !'(z0) in terms of partial derivatives of the
component functions u and v, but they also provide necessary conditions for the
existence of f'(z0). For, on equating the real and imaginary parts on the right-hand
sides of these equations, we see that the existence of f'(zo) requires that
.
Equations (6) are the Cauchy-Riemann equations, so named in honor of the French
mathematician A. L. Cauchy (1789-1857), who discovered and used them, and in
honor of the German mathematician G. F. B. Riemann (1826-1866), who made them
fundamental in his development of the theory of functions of a complex variable.
We summarize the above results as follows.
Theorem. Suppose that
f(z) = u(x, y) + iv(x, y)
and that !'(z) exists at a point z0 =x0 + iy0. Then the first-order partial derivatives
ofu and v must exist at (x0, y0), and they must satisfy the Cauchy-Riemann equations
(7)
there. Also, J'(zo) can be written
(8)
where these partial derivatives are to be evaluated at (x0, y0).
SEC. 21 SUFFICIENT CONDITIONS FOR DIFFERENTIABILITY 63
EXAMPLE 1. In Example 1, Sec. 18, we showed that the function
f(z) = z2
= x2
- i + i2xy
is differentiable everywhere and that f'(z) = 2z. To verify that the Cauchy-Riemann
equations are satisfied everywhere, we note that
u(x, y) =x2
- i and v(x, y) = 2xy.
Thus
Moreover, according to equation (8),
J'(z) =2x +i2y = 2(x +iy) =2z.
Since the Cauchy-Riemann equations are necessary conditions for the existence
of the derivative of. a function f at a point z0, they can often be used to locate points
at which f does not have a derivative.
EXAMPLE 2. When f(z) = lzl2
, we have
u(x, y) =x2
+ i and v(x, y) = 0.
If the Cauchy-Riemann equations are to hold at a point (x,-y), it follows that 2x = 0
and 2y = 0, or that x =y =0. Consequently, f' (z) does not exist at any nonzero point,
as we already know from Example 2 in Sec. 18. Note that the above theorem does not
ensure the existence of f'(0). The theorem in the next section will, however, do this.
21. SUFFICIENT CONDITIONS FOR DIFFERENTIABILITY
Satisfaction ofthe Cauchy-Riemann equations at a point zo =(x0, y0) is not sufficient
to ensure the existence ofthe derivative ofa function f (z) at that point. (See Exercise 6,
Sec. 22.) But, with certain continuity conditions, we have the following useful theorem.
Theorem. Let the function
f(z) = u(x, y) + il!(x, y)
be defined throughout some s neighborhood of a point zo = x0 + iy0, and suppose
that the first-order partial derivatives ofthe functions u and v with respect to x andy
exist everywhere in that neighborhood. If those partial derivatives are continuous at
(xo, Yo) and satisfy the Cauchy-Riemann equations
at (x0, Yo). then f'(z0) exists.
SEC.22 POLAR COORDINATES 65
But J(.6.x)2 + (.6.y)2 = [8z[, and so
JC.6.x)2 + (.6.y)2
=1.
Also, e1 + ie2 tends to 0 as (.6.x, .6.y) approaches (0, 0). So the last term on the right
in equation (5) tends to 0 as the variable .6.z = .6.x + i .6.y tends to 0. This means that
the limit of the left-hand side of equation (5) exists and that
(6)
where the right-hand side is to be evaluated at (x0, y0).
EXAMPLE 1. Consider the exponential function
(z = x +iy),
some of whose mapping properties were discussed in Sec. 13. In view of Euler's
formula (Sec. 6), this function can, of course, be written
J(z) =ex cosy+ iex sin y,
where y is to be taken in radians when cos y and sin y are evaluated. Then
u(x,y)=excosy and v(x,y)=exsiny.
Since ux = vy and uy = -vx everywhere and since these derivatives are everywhere
continuous, the conditions in the theorem are satisfied at all points in the complex
plane. Thus f'(z) exists everywhere, and
f'(z) = ux + ivx =ex cosy+ iex sin y.
Note that f'(z) = j(z).
EXAMPLE 2. It also follows from the theorem in this section that the function
f(z) = lzl2
, whose components are
u(x, y) = x 2
+ i and v(x, y) = 0,
has a derivative at z= 0. In fact, f'(O) =0 + iO =0 (compare Example 2, Sec. 18). We
saw in Example 2, Sec. 20, that this function cannot have a derivative at any nonzero
point since the Cauchy-Riemann equations are not satisfied at such points.
22. POLAR COORDINATES
Assuming that zo ::f:. 0, we shall in this section use the coordinate transformation
(1) x =r cos () , y = r sin ()
66 ANALYTIC FUNCTIONS CHAP. 2
to restate the theorem in Sec. 21 in polar coordinates.
Depending on whether we write
z=x+iy or z=reie (z f. 0)
when w = f(z), the real and imaginary parts of w =u + iv are expressed in terms of
either the variables x and y orr and(). Suppose that the first-order partial derivatives
of u and v with respect to x and y exist everywhere in some neighborhood of a given
nonzero point zo and are continuous at that point. The first-order partial derivatives
with respect to r and () also have these properties, and the chain rule for differentiating
real-valued functions of two real variables can be used to write them in terms of the
ones with respect to x and y. More precisely, since
au au ox au oy
-=--+--,
ar ox or oy or
one can write
au au ox au oy
-=--+--
8() ax 8() 8y ae'
(2) Ur = ux cos() + uy sin(), Ue = -ux r sin() + uy r cos().
Likewise,
(3) Vr = Vx cos() + Vy sin(), Ve = -vx r sin() + Vy r cos().
Ifthe partial derivatives with respect to x andy also satisfy the Cauchy-Riemann
equations
(4)
at z0, equations (3) become
(5) vr = -uy cos()+ ux sin(), Ve = uy r sin()+ ux r cos()
at that point. It is then clear from equations (2) and (5) that
(6)
at the point zo.
If, on the other hand, equations (6) are known to hold at z0, it is straightforward
to show (Exercise 7) that equations (4) must hold there. Equations (6) are, therefore,
an alternative form of the Cauchy-Riemann equations (4).
We can now restate the theorem in Sec. 21 using polar coordinates.
Theorem. Let the function
f(z) = u(r, ()) +iv(r, ())
be defined throughout some e. neighborhood ofa nonzero point zo =r0 exp(i00), and
suppose that thefirst-orderpartial derivatives ofthefunctions u and v with respect tor
SEC.22
5. Show that when f (z) =x3 +i (1 - y)3, it is legitimate to write
f'(z) =ux +ivx =3x
2
only when z =i.
EXERCISES 69
6. Let u and v denote the real and imaginary components of the function f defined by the
equations
{
-2
f(z) ~ ~ when z :j::. 0,
when z= 0.
Verify that the Cauchy-Riemann equations ux = vy and uy = -vx are satisfied at the
origin z = (0, 0). [Compare Exercise 9, Sec. 19, where it is shown that f'(O) nevertheless
fails to exist.]
7. Solve equations (2), Sec. 22, for ux and uy to show that
sin 0
Ux =Ur COS 0 - Ue ,
r
. cos 0
uy =ur sm 0 +u0 - -
r
Then use these equations and similar ones for Vx and vy to show that, in Sec. 22, equations
(4) are satisfied at a point zo if equations (6) are satisfied there. Thus complete the
verification that equations (6), Sec. 22, are the Cauchy-Riemann equations in polar form.
8. Let a function f(z) =u +iv be differentiable at a nonzero point z0 = r0 exp(i!10). Use
the expressions for ux and vx found in Exercise 7, together with the polar form (6), Sec.
22, of the Cauchy-Riemann equations, to rewrite the expression
in Sec. 21 as
! '( ) -ie( . )zo =e Ur + 1Vr ,
where u, and vr are to be evaluated at (r0 , 80).
9. (a) With the aid of the polar form (6), Sec. 22, ofthe Cauchy-Riemann equations, derive
the alternative form
f'(zo) = -i (uo +ivo)
zo
of the expression for f'(z0) found in Exercise 8.
(b) Use the expression for f'(z0) in part (a) to show that the derivative of the function
f(z) =ljz (z :j::. 0) in Example 1, Sec. 22, is f'(z) =-ljz2
.
10. (a) Recall (Sec. 5) that if z =x +iy, then
z+zX=--
2
-z z
and y = 2
i .
70 ANALYTIC FUNCTIONS CHAP. 2
By formally applying the chain rule in calculus to a function F(x, y) of two real
variables, derive the expression
oF oF ax oF oy 1 (oF .aF)
az: = ax az: + ay ai = 2 ax + 1
ay :
(b) Define the operator
a 1(a .a)
az: = 2 ax + 1
ay '
suggested by part (a), to show that if the first-order partial derivatives of the real
and imaginary parts of a function f (z) =u(x, y) + i v(x, y) satisfy the Cauchy-
Riemann equations, then
~i = ~[(ux- Vy) +i(vx +Uy)] =0.
Thus derive the complexform ajjaz = 0 ofthe Cauchy-Riemann equations.
23. ANALYTIC FUNCTIONS
We are now ready to introduce the concept of an analytic function. A function f of the
complex variable z is analytic in an open set if it has a derivative at each point in that
set.* If we should speak of a function f that is analytic in a set S which is not open,
it is to be understood that f is analytic in an open set containing S. In particular, f is
analytic at a point zo if it is analytic throughout some neighborhood of z0.
We note, for instance, that the function f(z) = 1/z is analytic at each nonzero
point in the finite plane. But the function f(z) = lzl2 is not analytic at any point since
its derivative exists only at z=0 and not throughout any neighborhood. (See Example
2, Sec. 18.)
An entire function is a function that is analytic at each point in the entire finite
plane. Since the derivative of a polynomial exists everywhere, it follows that every
polynomial is an entire function.
If a function f fails to be analytic at a point zo but is analytic at some point
in every neighborhood of z0, then z0 is called a singular point, or singularity, of f.
The point z =0 is evidently a singular point of the function f (z) =1/z. The function
f(z) =lzl2, on the other hand, has no singular points since it is nowhere analytic.
A necessary, but by no means sufficient, condition for a function f to be analytic
in a domain Dis clearly the continuity off throughout D. Satisfaction ofthe Cauchy-
Riemann equations is also necessary, but not sufficient. Sufficient conditions for
analyticity in D are provided by the theorems in Sees. 21 and 22.
Other useful sufficient conditions are obtained from the differentiation formulas
in Sec. 19. The derivatives of the sum and product of two functions exist wherever the
*The terms regular and holomorphic are also used in the literature to denote analyticity.
SEC. 23 ANALYTIC FUNCTIONS 71
functions themselves have derivatives. Thus, iftwo functions are analytic in a domain
D, their sum and their product are both analytic in D. Similarly, their quotient is
analytic in D provided thefunction in the denominator does not vanish at any point in
D. In particular, the quotient P(z)/ Q(z) of two polynomials is analytic in any domain
throughout which Q(z) ::j=. 0.
From the chain rule for the derivative of a composite function, we find that
a composition of two analytic functions is analytic. More precisely, suppose that a
function f(z) is analytic in a domain D and that the image (Sec. 12) of D under the
transformation w = f(z) is contained in the domain of definition of a function g(w).
Then the composition g[f(z)] is analytic in D, with derivative
d
dzg[f(z)] = g'[f(z)]f(z).
The following theorem is especially useful, in addition to being expected.
Theorem. If f' (z) = 0 everywhere in a domain D, then f (z) must be constant
throughout D.
We start the proofby writing f(z) = u(x, y) + iv(x, y). Assuming that f'(z) = 0
in D, we note that ux +ivx =0; and, in view of the Cauchy-Riemann equations,
vy - iuy =0. Consequently,
at each point in D.
Next, we show that u(x, y) is constant along any line segment L extending from
a point P to a point P' and lying entirely in D. We lets denote the distance along L
from the point P and let U denote the unit vector along L in the direction ofincreasing
s (see Fig. 30). We know from calculus that the directional derivative dujds can be
written as the dot product
(1)
y
0
du
- = (grad u) · U,
ds
---------
X
FIGURE30
72 ANALYTIC FUNCTIONS CHAP. 2
where grad u is the gradient vector
(2) grad u = uxi +uy j.
Because ux and uy are zero everywhere in D, then, grad u is the zero vector at all
points on L. Hence it fo1lows from equation (1) that the derivative dujds is zero along
L; and this means that u is constant on L.
Finally, since there is always a finite number of such line segments, joined end
to end, connecting any two points P and Q in D (Sec. 10), the values of u at P and
Q must be the same. We may conclude, then, that there is a real constant a such that
u(x, y) = a throughout D. Similarly, v(x, y) = b; and we find that f (z) = a + bi at
each point in D.
24. EXAMPLES
As pointed out in Sec. 23, it is often possible to determine where a given function is
analytic by simply recalling various differentiation formulas in Sec. 19.
EXAMPLE 1. The quotient
f(z) = z3 +4
. (z2 - 3)(z2 + 1)
is evidently analytic throughout the z plane except for the singular points z = ±.J3
and z= ± i. The analyticity is due to the existence of familiar differentiation formulas,
which need be applied only if the expression for j'(z) is wanted.
When a function is given in terms ofits component functions u(x, y) and v(x, y),
its analyticity can be demonstrated by direct application of the Cauchy-Riemann
equations.
EXAMPLE 2. When
f (z) = cosh x cosy +i sinh x sin y,
the component functions are
u(x, y) =cosh x cosy and v(x, y) =sinh x sin y.
Because
ux =sinh x cosy= Vy and uy =-cosh x sin y = -vx
everywhere, it is clear from the theorem in Sec. 21 that f is entire.
SEC.24 EXERCISES 73
Finally, we illustrate how the theorems in the last four sections, in particular the
one in Sec. 23, can be used to obtain some important properties of analytic functions.
EXAMPLE 3. Suppose that a function
f(z) =u(x, y) +iv(x, y)
and its conjugate
f(z) = u(x, y)- iv(x, y)
are both analytic in a given domain D. It is easy to show that f (z) must be constant
throughout D.
To do this, we write f(z) as
f(z) = U(x, y) +iV(x, y),
where
(1) U(x, y) = u(x, y) and V(x, y) = -v(x, y).
Because of the analyticity of J(z), the Cauchy-Riemann equations
(2)
hold in D, according to the theorem in Sec. 20. Also, the analyticity off(z) in D tells
us that
Ux =Vy, Uy =-Vx-
In view of relations (1), these last two equations can be written
(3)
By adding corresponding sides of the first of equations (2) and (3), we find that
ux = 0 in D. Similarly, subtraction involving corresponding sides of the second of
equations (2) and (3) reveals that vx =0. According to expression (8) in Sec. 20, then,
J'(z) =ux + ivx =0 + iO =0;
and it follows from the theorem in Sec. 23 that f (z) is constant throughout D.
EXERCISES
1. Apply the theorem in Sec. 21 to verify that each of these functions is entire;
(a) f(z) =3x +y +i(3y - x); (b) f(z) =sin x cosh y +i cos x sinh y;
(c) f(z) =e-Y sin x - ie-Y cos x; (d) f(z) = (z2 - 2)e-xe-iY.
SEC. 25 HARMONIC FUNCTIONS 75
25. HARMONIC FUNCTIONS
A real-valued function H of two real variables x andy is said to be harmonic in a given
domain ofthe xy plane if, throughout that domain, it has continuous partial derivatives
of the first and second order and satisfies the partial differential equation
(1) Hxx(x, y) + Hyy(x, y) = 0,
known as Laplace's equation.
Harmonic functions play an important role in applied mathematics. For example,
the temperatures T(x, y) in thin plates lying in the xy plane are often harmonic. A
function V (x, y) is harmonic when it denotes an electrostatic potential that varies
only with x and y in the interior of a region of three-dimensional space that is free of
charges.
EXAMPLE 1. It is easy to verify that the function T (x, y) = e-Y sin x is harmonic
in any domain of the xy plane and, in particular, in the semi-infinite vertical strip
0 < x < n, y > 0. It also assumes the values on the edges of the strip that are indicated
in Fig. 31. More precisely, it satisfies all of the conditions
Txx(X, y) + Tyy(x, y) = 0,
T(O, y) = 0, T(n, y) = 0,
T(x,O)=sinx, lim T(x,y)=O,
y-+oo
which describe steady temperatures T (x, y) in a thin homogeneous plate in the xy
plane that has no heat sources or sinks and is insulated except for the stated conditions
along the edges.
y
0 T=sinx n x
FIGURE31
The use of the theory offunctions of a complex variable in discovering solutions,
such as the one in Example 1, of temperature and other problems is described in
76 ANALYTIC FuNCTIONS CHAP. 2
considerable detail later on in Chap. 10 and in parts of chapters following it.* That
theory is based on the theorem below, which provides a source of harmonic functions.
Theorem 1. lfafunction f(z) = u(x, y) +iv(x, y) is analytic in a domain D, then
its component functions u and v are harmonic in D.
To show this, we need a result that is to be proved in Chap. 4 (Sec. 48). Namely,
if a function of a complex variable is analytic at a point, then its real and imaginary
components have continuous partial derivatives of all orders at that point.
Assuming that f is analytic in D, we start with the observation that the first-
order partial derivatives ofits component functions must satisfy the Cauchy-Riemann
equations throughout D:
(2)
Differentiating both sides of these equations with respect to x, we have
(3)
Likewise, differentiation with respect to y yields
(4)
Now, by a theorem in advanced calculus,t the continuity of the partial derivatives of
u and v ensures that uyx = Uxy and vyx = Vxy· It then follows from equations (3) and
(4) that
Uxx + Uyy = 0 and Vxx + Vyy = 0.
That is, u and v are harmonic in D.
EXAMPLE 2. The function J(z) =e-y sin x- ie-Y cos x is entire, as is shown
in Exercise 1(c), Sec. 24. Hence its real part, which is the temperature function
T (x, y) = e-Y sin x in Example 1, must be harmonic in every domain of the xy plane.
EXAMPLE 3. Since the function f (z) =i Iz2 is analytic whenever z ::j:. 0 and since
l i z2
i"z2
rz? 2xy +i(x2 - y2
)
z2 = z2 . zz = (zz)2 - lzl4 - (x2 + y2)2
*Another important method is developed in the authors' "Fourier Series and Boundary Value Problems,"
6th ed., 2001.
t See, for instance, A. E. Taylor and W. R. Mann, "Advanced Calculus," 3d ed., pp. 199-201, 1983.
SEC. 25 HARMONIC FUNCTIONS 77
the two functions
2xy
u(x, y) = 2 2 2
(x + y)
and
x2 _ y2
v(x' y) = 2 2 2
(x + y)
are harmonic throughout any domain in the xy plane that does not contain the origin.
If two given functions u and v are harmonic in a domain D and their first-order
partial derivatives satisfy the Cauchy-Riemann equations (2) throughout D, vis said
to be a harmonic conjugate of u. The meaning of the word conjugate here is, of course,
different from that in Sec. 5, where z is defined.
Theorem 2. Afunction f(z) =u(x, y) + iv(x, y) is analytic in a domain D ifand
only ifv is a harmonic conjugate ofu.
The proof is easy. If v is a harmonic conjugate of u in D, the theorem in Sec.
21 tells us that f is analytic in D. Conversely, iff is analytic in D, we know from
Theorem 1 above that u and v are harmonic in D; and, in view of the theorem in Sec.
20, the Cauchy-Riemann equations are satisfied.
The following example shows that if v is a harmonic conjugate of u in some
domain, it is not, in general, true that u is a harmonic conjugate of v there. (See also
Exercises 3 and 4.)
EXAMPLE 4. Suppose that
u(x, y) = x2
-l and v(x, y) = 2xy.
Since these are the real and imaginary components, respectively, of the entire function
f(z) = z2
, we know that vis a harmonic conjugate of u throughout the plane. But u
cannot be a harmonic conjugate of v since, as verified in Exercise 2(b), Sec. 24, the
function 2xy + i (x2 - y2) is not analytic anywhere.
In Chap. 9 (Sec. 97) we shall show that a function u which is harmonic in a
domain of a certain type always has a harmonic conjugate. Thus, in such domains,
every harmonic function is the real part of an analytic function. It is also true that a
harmonic conjugate, when it exists, is unique except for an additive constant.
EXAMPLE 5. We now illustrate one method of obtaining a harmonic conjugate of
a given harmonic function. The function
(5)
78 ANALYTIC FUNCTIONS CHAP. 2
is readily seen to be harmonic throughout the entire xy plane. Since a harmonic
conjugate v(x, y) is related to u (x, y) by means of the Cauchy-Riemann equations
(6)
the first of these equations tells us that
vy(x, y) = -6xy.
Holding x fixed and integrating each side here with respect toy, we find that
(7) v(x, y) = -3xl +¢(x),
where ¢ is, at present, an arbitrary function of x. Using the second of equations (6),
we have
3l- 3x
2
= 3l- ¢'(x),
or¢'(x) =3x2. Thus ¢ (x) =x3 + C, where C is an arbitrary real number. According
to equation (7), then, the function
(8) v(x, y) = -3xl +x
3
+C
is a harmonic conjugate of u(x, y).
The corresponding analytic function is
(9)
The form f(z) =i(z3 +C) ofthis function is easily verified and is suggested by noting
that when y = 0, expression (9) becomes f (x) = i (x 3 +C).
EXERCISES
1. Show that u(x, y) is harmonic in some domain and find a harmonic conjugate v(x, y)
when
(a) u(x, y) =2x(l- y); (b) u(x, y) =2x- x 3 +3xy2;
(c) u(x, y) =sinh x sin y; (d) u(x, y) = yj(x 2 +y2
).
Ans. (a) v(x, y) = x2
- y2
+2y; (b) v(x, y) = 2y- 3x2y +y3;
(c) v(x, y) =-cosh x cosy; (d) v(x, y) =xj(x2
+y2).
2. Show that if v and V are harmonic conjugates of u in a domain D, then v(x, y) and
V (x, y) can differ at most by an additive constant.
3. Suppose that, in a domain D, a function vis a harmonic conjugate of u and also that u
is a harmonic conjugate of v. Show how it follows that both u(x, y) and v(x, y) must be
constant throughout D.
4. Use Theorem 2 in Sec. 25 to show that, in a domain D, v is a harmonic conjugate of u
if and only if -u is a harmonic conjugate of v. (Compare the result obtained in Exer-
cise 3.)
SEC.25 EXERCISES 79
Suggestion: Observe that the function f(z) = u(x, y) + iv(x, y) is analytic in D
if and only if -ij(z) is analytic there.
5. Let the function .f(z) =u(r, 0) +iv(r, 0) be analytic in a domain D that does not
include the origin. Using the Cauchy-Riemann equations in polar coordinates (Sec. 22)
and assuming continuity of partial derivatives, show that, throughout D, the function
u(r, 0) satisfies the partial differential equation
r
2
urr(r, 0) +rur(r, 0) +uee(r, 0) =0,
which is the polarform ofLaplace's equation. Show that the same is true of the function
v(r, 0).
6. Verify that the function u(r, 0) =ln r is harmonic in the domain r > 0, 0 < 0 < 27( by
showing that it satisfies the polarform ofLaplace's equation, obtained in Exercise 5. Then
use the technique in Example 5, Sec. 25, but involving the Cauchy-Riemann equations
in polarform (Sec. 22), to derive the harmonic conjugate v(r, 0) =fJ. (Compare Exercise
6, Sec. 24.)
7. Let the function .f(z) = u(x, y) +iv(x, y) be analytic in a domain D, and consider the
families of level curves u(x, y) =c1 and v(x, y) =c2, where c1 and c2 are arbitrary
real constants. Prove that these families are orthogonal. More precisely, show that if
z0 = (x0, y0) is a point in D which is common to two particular curves u(x, y) = c1
and v (x, y) = c2 and if f 1
(zo) -I 0, then the lines tangent to those curves at (x0, y0) are
perpendicular.
Suggestion: Note how it follows from the equations u(x, y) =c1 and v(x, y) =c2
that
au + au dy = 0 and
ax ay dx
8. Show that when f(z) =z2
, the level curves u(x, y) =c1 and v(x, y) =c2 of the compo-
nent functions are the hyperbolas indicated in Fig. 32. Note the orthogonality of the two
y
---- X
FIGURE32
80 ANALYTIC FUNCTIONS CHAP. 2
families, described in Exercise 7. Observe that the curves u(x, y) = 0 and v(x, y) =0
intersect at the origin but are not, however, orthogonal to each other. Why is this fact in
agreement with the result in Exercise 7?
9. Sketch the families oflevel curves of the component functions u and v when f(z) = 1/z.
and note the orthogonality described in Exercise 7.
10. Do Exercise 9 using polar coordinates.
11. Sketch the families oflevel curves of the component functions u and v when
z-l
f(z) = ,
z+l
and note how the result in Exercise 7 is illustrated here.
26. UNIQUELY DETERMINED ANALYTIC FUNCTIONS
We conclude this chapter with two sections dealing with how the values of an analytic
function in a domain D are affected by its values in a subdomain or on a line segment
lying in D. While these sections are of considerable theoretical interest, they are not
central to our development of analytic functions in later chapters. The reader may pass
directly to Chap. 3 at this time and refer back when necessary.
Lemma. Suppose that
(i) a function f is analytic throughout a domain D;
(ii) f(z) = 0 at each point z ofa domain or line segment contained in D.
Then f(z) =0 in D; that is, f(z) is identically equal to zero throughout D.
To prove this lemma, we let f be as stated in its hypothesis and let zo be any
point of the subdomain or line segment at each point of which f (z) = 0. Since D is a
connected open set (Sec. 10), there is a polygonal line L, consisting of a finite number
of line segments joined end to end and lying entirely in D, that extends from zo to any
other point P in D. We let d be the shortest distance from points on L to the boundary
of D, unless D is the entire plane; in that case, d may be any positive number. We then
form a finite sequence of points
along L, where the point Zn coincides with P (Fig. 33) and where each point is
sufficiently close to the adjacent ones that
(k=l,2, ... ,n).
SEC. 26 UNIQUELY DETERMINED ANALYTic FuNCTIONS 81
-------- / ..,__
----<....---.::: ......., / --~ ....
----- .... ' ' ,Nn -I I
/
'/
N ',N,' N2 ' Nn 'I I
I 0 I
I I I
I
p
I I I
I Zn-l I
I zo z, I 221 I
I
Zn I
I I I II
'
I
/
I
' ' I
'
/ / '- '- /
/ / /
--- /
'-
/ ...c. __ .....
---- FIGURE33'-
-------
Finally, we construct a finite sequence of neighborhoods
where each neighborhood Nk is centered at zk and has radius d. Note that these
neighborhoods are all contained in D and that the center Zk of any neighborhood Nk
(k = 1, 2, ... , n) lies in the preceding neighborhood Nk-l·
At this point, we need to use a result that is proved later on in Chap. 6. Namely,
Theorem 3 in Sec. 68 tells us that since f is analytic in the domain No and since
f(z) = 0 in a domain or on a line segment containing z0, then f(z) =0 in N0. But
the point z1 lies in the domain N0. Hence a second application of the same theorem
reveals that f(z) =0 in N1; and, by continuing in this manner, we arrive at the fact
that f(z) = 0 in Nn- Since Nn is centered at the point P and since P was arbitrarily
selected in D, we may conclude that f(z) = 0 in D. This completes the proof of the
lemma.
Suppose now that two functions f and g are analytic in the same domain D and
that f(z) =g(z) at each point z of some domain or line segment contained in D. The
difference
h(z) = f(z) - g(z)
is also analytic in D, andh(z) = 0 throughout the subdomain or along the line segment.
According to the above lemma, then, h(z) = 0 throughout D; that is, f(z) = g(z) at
each point z in D. We thus arrive at the following important theorem.
Theorem. A function that is analytic in a domain D is uniquely determined over D
by its values in a domain, or along a line segment, contained in D.
This theorem is useful in studying the question of extending the domain of
definition of an analytic function. More precisely, given two domains D1 and D2,
consider the intersection D1 n D2, consisting of all points that lie in both D1 and D2.
If D1 and D2 have points in common (see Fig. 34) and a function ft is analytic in D1,
there may exist a function fz, which is analytic in D2, such that fz(z) = j 1(z) for each
z in the intersection D1 n D2. If so, we call fz an analytic continuation of !J into the
second domain D2.
Whenever that analytic continuation exists, it is unique, according to the theorem
just proved. That is, not more than one function can be analytic in D2 and assume the
82 ANALYTIC FuNCTIONS CHAP. 2
FIGURE34
value ft (z) at each point z of the domain D1n D2 interior to D2. However, if there is
an analytic continuation h of h from D2 into a domain D3 which intersects Db as
indicated in Fig. 34, it is not necessarily true that f3(z) = /1(z) for each z in D1n D3•
Exercise 2, Sec. 27, illustrates this.
If fz is the analytic continuation of / 1 from a domain D1 into a domain D2, then
the function F defined by the equations
F(z) = { ft(Z) when z ~s ~n Dt,
h(z) when z ISm D2
is analytic in the union D1 U D2, which is the domain consisting of all points that lie
in either D1 or D2• The function F is the analytic continuation into D1 U D2 of either
ft or fz; and f 1 and h are called elements of F.
27. REFLECTION PRINCIPLE
The theorem in this section concerns the fact that some analytic functions possess the
property that f(z) = fCi) for all points z in certain domains, while others do not. We
note, for example, that z+ 1and z2 have that property when D is the entire finite plane;
but the same is not true of z +i and iz2
• The theorem, which is known as the reflection
principle, provides a way of predicting when /(z) = f[i).
Theorem. Suppose that a function f is analytic in some domain D which contains
a segment of the x axis and whose lower half is the reflection of the upper halfwith
respect to that axis. Then
(1) f(z) = f(J)
for eachpoint zin the domain ifand only iff(x) is realfor eachpoint x on the segment.
We start the proof by assuming that f(x) is real at each point x on the segment.
Once we show that the function
(2) F(z) = f(J)
SEC.27 REFLECTION PRINCIPLE 83
is analytic in D, we shaH use it to obtain equation (1). To establish the analyticity of
F(z), we write
j(z) = u(x, y) iv(x, y), F(z) = U(x, y) +iV(x, y)
and observe how it follows from equation (2) that, since
(3) f(Z) = u(x, -y)- iv(x, -y),
the components of F(z) and f(z) are related by the equations
(4) U(x, y) = u(x, t) and V(x, y) = -v(x, t),
where t = -y. Now, because j(x +it) is an analytic function of x +it, the first-
order partial derivatives of the functions u(x, t) and v(x, t) are continuous throughout
D and satisfy the Cauchy-Riemann equations*
(5)
Furthermore, in view of equations (4),
and it follows from these and the first of equations (5) that Ux = VY" Similarly,
and the second of equations (5) tells us that Uy = - Vx- Inasmuch as the first-order
partial derivatives of U(x, y) and V(x, y) are now shown to satisfy the Cauchy-
Riemannequations and since those derivatives are continuous, we find that the function
F(z) is analytic in D. Moreover, since f (x) is real on the segment ofthe real axis lying
in D, v(x, 0) = 0 on that segment; and, in view of equations (4), this means that
F(x) = U(x, 0) iV(x, 0) = u(x, 0)- iv(x, 0) = u(x, 0).
That is,
(6) F(z) = f(z)
at each point on the segment. We now refer to the theorem in Sec. 26, which tells us
that an analytic function defined on a domain Dis uniquely determined by its values
along any line segment lying in D. Thus equation (6) actually holds throughout D.
*See the paragraph immediately following Theorem 1 in Sec. 25.
· 84 ANALYTIC FUNCTIONS CHAP. 2
Because of definition (2) of the function F(z), then,
(7) f(Z) = f(z);
and this is the same as equation (1).
To prove the converse ofthe theorem, we assume that equation (1) holds and note
that, in view of expression (3), the form (7) of equation (1) can be written
u(x, -y)- iv(x, -y) = u(x, y) + iv(x, y).
In particular, if (x, 0) is a point on the segment of the real axis that lies in D,
u(x, 0)- iv(x, 0) = u(x, 0) +iv(x, 0);
and, by equating imaginary parts here, we see that v(x, 0) =0. Hence f(x) is real on
the segment of the real axis lying in D.
EXAMPLES. Just prior to the statement of the theorem, we noted that
z + 1= z + 1 and z2 = z
2
for all zin the finite plane. The theorem tells us, of course, that this is true, since x + 1
and x2 are real when x is real. We also noted that z+i and iz2 do not have the reflection
property throughout the plane, and we now know that this is because x +i and ix2 are
not real when x is real.
EXERCISES
1. Use the theorem in Sec. 26 to show that if f(z) is analytic and not constant throughout
a domain D, then it cannot be constant throughout any neighborhood lying in D.
Suggestion: Suppose that f (z) does have a constant value w0 throughout some
neighborhood in D.
2. Starting with the function
and referring to Exercise 4(b), Sec. 22, point out why
(r > 0, ~ < e< 2rr)
is an analytic continuation of fi across the negative real axis into the lower half plane.
Then show that the function
(r > 0, rr < e<
5
;)
is an analytic continuation of fz across the positive real axis into the first quadrant but
that hCz) = - !1(z) there.
SEC. 27 EXERCISES 85
3. State why the function
f4(Z) = ..;rei0/2 (r > 0, -j( < e< j()
is the analytic continuation of the function f 1(z) in Exercise 2 across the positive real
axis into the lower half plane.
4. We know from Example 1, Sec. 21, that the function
J(z) = exeiy
has a derivative everywhere in the finite plane. Point out how it follows from the reflection
principle (Sec. 27) that
f(z) = !Ci)
for each z. Then verify this directly.
5. Show that if the condition that f (x) is real in the reflection principle (Sec. 27) is replaced
by the condition that f(x) is pure imaginary, then equation (1) in the statement of the
principle is changed to
--f(z) =-!Ci).
CHAPTER
3
ELEMENTARY FUNCTIONS
We consider here various elementary functions studied in calculus and define corre-
sponding functions of a complex variable. To be specific, we define analytic functions
of a complex variable z that reduce to the elementary functions in calculus when
z =x + iO. We start by defining the complex exponential function and then use it
to develop the others.
28. THE EXPONENTIAL FUNCTION
As anticipated earlier (Sec. 13), we define here the exponential function ez by writing
(1) (z=x+iy),
where Euler's formula (see Sec. 6)
(2) eiy =cosy+ i sin y
is used and y is to be taken in radians. We see from this definition that ez reduces to
the usual exponential function in calculus when y = 0; and, following the convention
used in calculus, we often write exp z for e'-.
Note that since the positive nth root ::(e of e is assigned to ex when x = 1/n
(n = 2, 3, ...), expression (1) tells us that the complex exponential function ez is also
::(e when z = ljn (n = 2, 3, ...). This is an exception to the convention (Sec. 8) that
would ordinarily require us to interpret e1fn as the set of nth roots of e.
87
88 ELEMENTARY FUNCTIONS CHAP. 3
According to definition (1), exeiy = ex+iy; and, as already pointed out in Sec. 13,
the definition is suggested by the additive property
of e-'' in calculus. That property's extension,
(3)
to complex analysis is easy to prove. To do this, we write
Then
But x 1 and x2 are both real, and we know from Sec. 7 that
Hence
and, since
(xl +x2) + i(Yl + Y2) = (xl + iyt) + (x2 + iy2) = Zt + Z2,
the right-hand side ofthis last equation becomes ez1+z2. Property (3) is now established.
Observe how property (3) enables us to write ez1-z2ez2 = ez', or
(4)
From this and the fact that e0 = 1, it follows that ljeZ = e-z.
There are a number of other important properties of ez that are expected. Accord-
ing to Example 1 in Sec. 21, for instance,
d • z
-e'"=e
dz
(5)
everywhere in the z plane. Note that the differentiability of ez for all z tells us that
ez is entire (Sec. 23). It is also true that
(6) ez ::j:. 0 for any complex number z.
This is evident upon writing definition (1) in the form
ez = pei<P where p = ex and </J =y,
SEC. 28 EXERCISES 89
which tells us that
(7) lezl =ex and arg(e2
) = y + 2mr (n = 0, ±1, ±2, ...).
Statement (6) then follows from the observation that 1ez1 is always positive.
Some properties of ez are, however, not expected. For example, since
we find that ez is periodic, with a pure imaginary period 2rri:
(8)
The following example illustrates another property of ez that ex does not have.
Namely, while ex is never negative, there are values of ez that are.
EXAMPLE. There are values of z, for instance, such that
(9) ez = -1.
To find them, we write equation (9) as exeiY = leirr. Then, in view of the statement
in italics at the beginning of Sec. 8 regarding the equality of two nonzero complex
numbers in exponential form,
ex=l and y=rr+2nrr(n=0,±1,±2, ...).
Thus x = 0, and we find that
(10) z = (2n + l)ni (n = 0, ±1, ±2, ...).
EXERCISES
1. Show that
(a) exp(2 ± 3ni) = -e2;
(
'2+ni) 1re(b) exp
4
= y2
(1 + i);
(c) exp(z + ni) =- exp z.
2. State why the function 2z2 - 3 - zez +e-z is entire.
3. Use the Cauchy-Riemann equations and the theorem in Sec. 20 to show that the function
f(z) =exp zis not analytic anywhere.
4. Show in two ways that the function exp(z2) is entire. What is its derivative?
.4ns. 2z exp(z2
).
5. Write lexp(2z + i)l and lexp(iz2
)1 in terms of x andy. Then show that
lexp(2z + i) + exp(iz2)1 < e2
x + e-2xY.
6. Show that lexp(z2
)1 < exp(lzl2
).
SEC.30
y
I
I
I
I
I
I
0
I
I
I
I
I
I
I
I
X
BRANCHES AND DERIVATIVES OF LOGARITHMS 93
FIGURE35
is single-valued and continuous in the stated domain (Fig. 35). Note that ifthe function
(2) were to be defined on the ray 8 =a, it would not be continuous there. For, if z is
a point on that ray, there are points arbitrarily close to z at which the values of v are
near a and also points such that the values of v are near a + 2rr.
The function (2) is not only continuous but also analytic in the domain r > 0,
a < () < a + 2rr since the first-order partial derivatives of u and v are continuous
there and satisfy the polar form (Sec. 22)
of the Cauchy-Riemann equations. Furthermore, according to Sec. 22,
that is,
(4)
In particular,
(5)
d 1 -iec . ) -ie ( 1 ·o) t- og z =e llr + l Vr =e - + l = -.-fl ;
dz r r~
d 1
-logz =-
dz z
d 1
-Logz=
dz z
(IzI> 0, a < arg z < a + 2rr).
(lzl > 0, -rr < Arg z < rr).
A branch of a multiple-valued function f is any single-valued function F that is
analytic in some domain at each point z of which the value F (z) is one of the values
f(z). The requirement of analyticity, of course, prevents F from taking on a random
selection of the values of f. Observe that, for each fixed a, the single-valued function
(2) is a branch of the multiple-valued function ( l). The function
(6) Log z=In r +i8 (r > 0, -rr < 8 < rr)
is called the principal branch.
A branch cut is a portion of a line or curve that is introduced in order to define a
branch F of a multiple-valued function f. Points on the branch cut for Fare singular
points (Sec. 23) ofF, and any point that is common to all branch cuts off is called a
SEC.3I SOME IDENTITIES INVOLVING LoGARITHMS 95
10. Show in two ways that the function ln(x2 + y2) is harmonic in every domain that does
not contain the origin.
11. Show that
1
Re[log(z- 1)] =- ln[(x- 1)2
+ y2
]
2
(Z f- 1).
Why must this function satisfy Laplace's equation when z i- 1?
31. SOME IDENTITIES INVOLVING LOGARITHMS
As suggested by relations (3) and (4) in.Sec. 29, as well as Exercises 3, 4, and 5 with
Sec. 30, some identities involving logarithms in calculus carry over to complex analysis
and others do not. In this section, we derive a few that do carry over, sometimes with
qualifications as to how they are to be interpreted. A reader who wishes to pass to Sec.
32 can simply refer to results here when needed.
If z1 and z2 denote any two nonzero complex numbers, it is straightforward to
show that
(l)
This statement, involving a multiple-valued function, is to be interpreted in the same
way that the statement
(2) arg(ztzz) = arg Zt + arg Zz
was in Sec. 7. That is, if values of two of the three logarithms are specified, then there
is a value of the third logarithm such that equation (1) holds.
The proof of statement (1) can be based on statement (2) in the following way.
Since lz1z21 = lztllz21 and since these moduli are all positive real numbers, we know
from experience with logarithms of such numbers in calculus that
So it follows from this and equation (2) that
(3) In lztzzl + i arg(ztZz) =(In lztl + i arg Zt) + (In lzzl + i arg zz).
Finally, because of the way in which equations (1) and (2) are to be interpreted,
equation (3) is the same as equation (1).
EXAMPLE. To illustrate statement (1), write z1 = z2 = -1 and note that z1z2 = 1.
If the values log z1 = 1ii and log z2 =-ni are specified, equation (1) is evidently
satisfied when the value log(z1z2) = 0 is chosen.
Observe that, for the same numbers z1 and z2,
Log(ztzz) =0 and Log z1 +Log z2 = 2ni.
Thus statement (1) is not, in general, valid when log is replaced everywhere by Log.
96 ELEMENTARY FUNCTIONS
(4)
Verification of the statement
log(~1
) =log z1 -log z2,
~2
CHAP. 3
which is to be interpreted in the same way as statement (1), is left to the exercises.
We include here two other properties of log z that will be of special interest in
Sec. 32. If z is a nonzero complex number, then
(5) (n = 0 ± 1, ±2, ...)
for any value of log z that is taken. When n = 1, this reduces, of course, to relation (3),
Sec. 29. Equation (5) is readily verified by writing z =re18
and noting that each side
becomes rneine.
It is also true that when z =I= 0,
(6) z1
1n =exp(~ log z) (n =1, 2, ...).
That is, the term on the right here has n distinct values, and those values are the nth
roots of z. To prove this, we write z = r exp(iE>), where E> is the principal value of
arg z. Then, in view of definition (2), Sec. 29, of log z,
(
1 ) [ 1 i (E> + 2kJT)]exp - log z = exp - In r + ,
n n 11
where k = 0, ±1. ±2, .... Thus
(7) exp(~ log z) = ~exp[i ( ~ +
2
:rr)J (k = 0, ±1, ±2, ...).
Because exp(i2krrj11) has distinct values only when k = 0, 1, ... , 11- 1, the right-
hand side of equation (7) has only n values. That right-hand side is, in fact, an
expression for the nth roots of z (Sec. 8), and so it can be written zl!n. This establishes
property (6), which is actually valid when n is a negative integer too (see Exercise 5).
EXERCISES
1. Show that if Re z1 > 0 andRe z2 > 0, then
Log(ZJZ2) =Log Zt +Log Z2·
2. Show that, for any two nonzero complex numbers z1 and z2,
Log(ZtZ2) =Log z1 +Log z2 + 2Nrri
where N has one of the values 0, ±1. (Compare Exercise 1.)
3. Verify expression (4), Sec. 31, for log(zifz2) by
(a) using the fact that arg(ztfz2) =arg z1 - arg z2 (Sec. 7);
SEC.32 CoMPLEX EXPONENTS 97
(b) showing that log(l/z) =-log z (z f. 0), in the sense that log(1/z) and -log zhave
the same set of values, and then referring to expression (1), Sec. 31, for log(z1z2).
4. By choosing specific nonzero values of z1 and z2, show that expression (4). Sec. 31, for
log(z 1jz2) is not always valid when log is replaced by Log.
5. Show that property (6), Sec. 31, also holds when n is a negative integer. Do this by writing
z1/n =(z lfm) -I (m = -n), where n has any one of the negative values n = -1, -2, ...
(see Exercise 9, Sec. 9), and using the fact that the property is already known to be valid
for positive integers.
6. Let z denote any nonzero complex number, written z = reiE> (-n < e < n), and let n
denote any fixed positive integer (n = 1, 2, ...). Show that all of the values of log(zlfn)
are given by the equation
I ( lfn) 1 1 . e + 2(pn + k)n
og z =- n r +l ,
n n
where p = 0, ±1, ±2, ... and k = 0, 1, 2, ... , n- 1. Then, after writing
1 I . 8 + 2qn- log z = - In r +1 _ _..::._
n n n
where q = 0, ± 1, ±2, ... , show that the set of values oflog(z 1ln) is the same as the set
of values of (1/n) log z. Thus show that log(z11n) = (ljn) log z, where, corresponding
to a value oflog(z 11n) taken on the left, the appropriate value oflog z is to be selected on
the right, and conversely. fThe result in Exercise 5(a), Sec. 30, is a special case of this
one.]
Suggestion: Use the fact that the remainder upon dividing an integer by a positive
integer n is always an integer between 0 and n - 1, inclusive; that is, when a positive
integer n is specified, any integer q can be written q = pn + k, where pis an integer and
k has one of the values k =0, 1, 2, ... , n- 1.
32. COMPLEX EXPONENTS
When z =!= 0 and the exponent c is any complex number, the function zc is defined by
means of the equation
(1)
where Jog z denotes the multiple-valued logarithmic function. Equation (1) provides
a consistent definition of zc in the sense that it is already known to be valid (see Sec.
31) when c =n (n =0, ±1, ±2, ...) and c = 1/n (n = ±1, ±2, ...). Definition (1)
is, in fact, suggested by those particular choices of c.
EXAMPLE 1. Powers of z are. in general. multiple-valued, as illustrated by writing
i-2
i = exp(-2i log i)
CHAPTER
4
INTEGRALS
Integrals are extremely important in the study of functions of a complex variable. The
theory of integration, to be developed in this chapter, is noted for its mathematical
elegance. The theorems are generally concise and powerful, and most of the proofs
are simple.
36. DERIVATIVES OF FUNCTIONS w(t)
In order to introduce integrals off(z) in a fairly simple way, we need to first consider
derivatives of complex-valued functions w of a real variable t. We write
(1) w(t) =u(t) +iv(t),
where the functions u and v are real-valued functions oft. The derivative w'(t), or
d[w(t)]/dt, of the function (1) at a point tis defined as
(2) w'(t) = u'(t) +iv'(t),
provided each of the derivatives u' and v' exists at t.
From definition (2), it follows that, for every complex constant z0 =x0 +iy0,
~[z0w(t)] = [(x0 +iy0)(u +iv)]' = [(x0u- y0v) +i(y0u +x0v)]'
dt
= (xou- Yov)' +i(you +xov)' =(xou'- YoV') +i(you' +xov').
111
112 INTEGRALS
But
( I I) •( I I·) ( • ) ( I • ') I(. )x0u -y0v +t y0u +x0v = x0 +ty0 u +lv =z0w t,
and so
(3)
d I
-[zow(t)] = z0w (t).
dt
Another expected rule that we shall often use is
(4) d ezot - ""oezot
dt - "' '
where z0 =x0 + iy0. To verify this, we write
and refer to definition (2) to see that
!!:_ezot = (exot cos Yot>' + i (exot sin Yot)1
•
dt
CHAP. 4
Familiar rules from calculus and some simple algebra then lead us to the expression
:tezot = (xo +iy0)(exot cos Yot +iexor sin Yot),
or
d ezot = (x + iv ).exotei.vot.
dt . 0 -0
This is, of course, the same as equation (4).
Various other rules learned in calculus, such as the ones for differentiating sums
and products, apply just as they do for real-valued functions of t. As was the case
with property (3) and formula (4), verifications may be based on corresponding rules
in calculus. It should be pointed out, however, that not every rule for derivatives in
calculus carries over to functions of type (1). The following example illustrates this.
EXAMPLE. Suppose that w(t) is continuous on an interval a < t ~ b; that is, its
component functions u(t) and v(t) are continuous there. Even if w'(t) exists when
a < t < b, the mean value theorem for derivatives no longer applies. To be precise, it
is not necessarily true that there is a number c in the interval a < t < b such that
'( .) w(b)- w(a)
w c = .
b-a
To see this, consider the function w(t) = eit on the interval 0 < t < 2n. When that
function is used, lw'(t)l = !ieitl = 1; and this means that the derivative w1
(t) is never
zero, while w(2n) - w(O) = 0.
SEC.37 DEFINITE INTEGRALS OF FUNCTIONS w(t) 113
37. DEFINITE INTEGRALS OF FUNCTIONS w(t)
When w(t) is a complex-valued function of a real variable t and is written
(1) w(t) = u(t) + iv(t),
where u and v are real-valued, the definite integral of w(t) over an interval a < t < b
is defined as
(2) 1bw(t) dt = 1bu(t) dt +i 1bv(t) dt
when the individual integrals on the right exist. Thus
(3) Re lbw(t) dt = 1bRe[w(t)] dt and Im lbw(t) dt = lbIm[w(t)] dt.
EXAMPLE 1. For an illustration of definition (2),
(1 + it)2
dt = (1- t
2
) dt + i 2t dt = - + i.
lol lol ~o· 2
0 0 0 3
Improper integrals of w(t) over unbounded intervals are defined in a similar way.
The existence of the integrals of u and v in definition (2) is ensured if those
functions are piecewise continuous on the interval a < t < b. Such a function is
continuous everywhere in the stated interval except possibly for a finite number of
points where, although discontinuous, it has one-sided limits. Ofcourse, only the right-
hand limit is required at a; and only the left-hand limit is required at b. When both u
and v are piecewise continuous, the function w is said to have that property.
Anticipated rules for integrating a complex constant times a function w(t), for
integrating sums of such functions, and for interchanging limits of integration are all
valid. Those rules, as well as the property
lbw(t) dt =1cw(t) dt + [bw(t) dt,
are easy to verify by recalling corresponding results in calculus.
The fundamental theorem of calculus, involving antiderivatives, can, moreover,
be extended so as to apply to integrals of the type (2). To be specific, suppose that the
functions
w(t) =u(t) + iv(t) and W(t) = U(t) + iV(t)
114 INTEGRALS CHAP. 4
are continuous on the interval a < t <b. If W'(t) = w(t) when a < t < b, then
U'(t) =u(t) and V'(t) = v(t). Hence, in view of definition (2),
f w(t) dt =U(t{ +iV(t)J:
= [U(b) + iV(b)]- [U(a) + iV(a)].
That is,
(4) 1bw(t) dt = W(b) W(a) = W(t)J:.
EXAMPLE 2. Since (eit)' = ieit (see Sec. 36),
r:n:/4 ]:rr/4Jo eit dt = -ieit o = -iei:n:/4 +i
We finish here with an important property of moduli of integrals. Namely,
(5) 1bw(t) dt < 1blw(t)i dt (a< b).
This inequality clearly holds when the value of the integral on the left is zero, in
particular when a= b. Thus, in the verification, we may assume that its value is a
nonzero complex number. If r0 is the modulus and 80 is an argument of that constant,
then
Solving for r0, we write
(6) ro = ibe-it1ow dt.
Now the left-hand side of this equation is a real number, and so the right-hand side is
too. Thus, using the fact that the real part of a real number is the number itself and
referring to the first of properties (3), we see that the right-hand side of equation (6)
can be rewritten in the following way:
116 INTEGRALS CHAP. 4
Evaluate the two integrals on the right here by evaluating the single integral on the left
and then using the real and imaginary parts of the value found.
Ans. -(1 +err)/2, (1 +err)/2.
5. Let w(t) be a continuous complex-valued function oft defined on an interval a < t <b.
By considering the special case w(t) = eit on the interval 0 < t < 2rr, show that it is not
always true that there is a number c in the interval a < t < b such that
1bw(t) dt = w(c)(b- a).
Thus show that the mean value theorem for definite integrals in calculus does not apply
to such functions. (Compare the example in Sec. 36.)
6. Let w(t) =u(t) + iv(t) denote a continuous complex-valued function defined on an
interval -a < t < a.
(a) Suppose that w(t) is even; that is, w(-t) =w(t) foreachpointt in the given interval.
Show that
faw(t)dt=2 ra w(t)dt.
-a Jo
(b) Show that if w(t) is an odd function, one where w(-t) = -w(t) for each point tin
the interval, then
i:w(t) dt =0.
Suggestion: In each part of this exercise, use the corresponding property of
integrals of real-valued functions oft, which is graphically evident.
7. Apply inequality (5), Sec. 37, to show that for all values of x in the interval -1 < x < 1,
the functions*
(n =0, 1, 2, ...)
satisfy the inequality IPn(x)l < 1.
38. CONTOURS
Integrals of complex-valued functions of a complex variable are defined on curves in
the complex plane, rather than on just intervals of the real line. Classes of curves that
are adequate for the study of such integrals are introduced in this section.
*These functions are actually polynomials in x. They are known as Legendre polynomials and are
important in applied mathematics. See, for example, Chap. 4 of the book by Lebedev that is listed
in Appendix 1.
SEC.38 CONTOURS 117
A set of points z =(x, y) in the complex plane is said to be an arc if
(1) x = x(t), y = y(t) (a<t<b),
where x(t) and y(t) are continuous functions of the real parameter t. This definition
establishes a continuous mapping ofthe interval a < t < b into the xy, or z, plane; and
the image points are ordered according to increasing values of t. It is convenient to
describe the points of C by means of the equation
(2) z =z(t) (a<t<b),
where
(3) z(t) = x(t) + iy(t).
The arc Cis a simple arc, or a Jordan arc,* if it does not cross itself; that is, Cis
simple if z(t1) f. z(t2) when t1 f. t2• When the arc Cis simple except for the fact that
z(b) =z(a), we say that Cis a simple closed curve, or a Jordan curve.
The geometric nature of a particular arc often suggests different notation for the
parameter t in equation (2). This is, in fact, the case in the examples below.
EXAMPLE 1. The polygonal line (Sec. 10) defined by means of the equations
(4) z _ { x +ix when 0 < x < 1,
- x +i when 1< x < 2
and consisting of a line segment from 0 to 1+ i followed by one from 1+ i to 2 +i
(Fig. 36) is a simple arc.
y
1+i 2+i
1
0 1 2 X
FIGURE36
EXAMPLE 2. The unit circle
(5)
*Named for C. Jordan (1838-1922), pronouncedjor-don'.
118 INTEGRALS CHAP. 4
about the origin is a simple closed curve, oriented in the counterclockwise direction.
So is the circle
(6) (O<e<2n),
centered at the point zo and with radius R (see Sec. 6).
The same set of points can make up different arcs.
EXAMPLE 3. The arc
(7)
is not the same as the arc described by equation (5). The set of points is the same, but
now the circle is traversed in the clockwise direction.
EXAMPLE 4. The points on the arc
(8)
are the same as those making up the arcs (5) and (7). The arc here differs, however, from
each of those arcs since the circle is traversed twice in the counterclockwise direction.
The parametric representation used for any given arc C is, of course, not unique.
It is, in fact, possible to change the interval over which the parameter ranges to any
other interval. To be specific, suppose that
(9) t = ¢(r) (a< r < {3),
where ¢ is a real-valued function mapping an interval a: < r < f3 onto the interval
a< t <bin representation (2). (See Fig. 37.) We assume that¢ is continuous with a
continuous derivative. We also assume that ¢'(r) > 0 for each r; this ensures that t
increases with r. Representation (2) is then transformed by equation (9) into
(10)
t
b
a --
0
z = Z(r) (a:<r<{J),
FIGURE37
t = ¢(r)
SEC.38 CONTOURS 119
where
(11) Z(r) = z[¢(r)].
This is illustrated in Exercise 3, where a specific function 4> (r) is found.
Suppose now that the components x'(t) and y'(t) of the derivative (Sec. 36)
(12) z'(t) = x'(t) +iy'(t)
ofthe function (3), used to represent C, are continuous on the entire interval a < t < b.
The arc is then called a differentiable arc, and the real-valued function
lz'(t)l = J[x'(t)]2 + [y'(t)]2
is integrable over the interval a < t <b. In fact, according to the definition ofarc length
in calculus, the length of C is the number
(13) L = 1blz'(t)l dt.
The value of L is invariant under certain changes in the representation for C that
is used, as one would expect. More precisely, with the change of variable indicated in
equation (9), expression (13) takes the form [see Exercise l(b)]
L = £13
lz'[¢(r)]l¢'(r) dr.
So, if representation (10) is used for C, the derivative (Exercise 4)
(14) Z'(r) = z'[¢(r)]¢'(r)
enables us to write expression (13) as
Thus the same length of C would be obtained if representation (10) were to be used.
If equation (2) represents a differentiable arc and if z'(t) f. 0 anywhere in the
interval a < t < b, then the unit tangent vector
T = z'(t)
lz'(t) I
is well defined for all t in that open interval, with angle of inclination arg z'(t).
Also, when T turns, it does so continuously as the parameter t varies over the entire
interval a < t <b. This expression forT is the one learned in calculus when z(t) is
120 INTEGRALS CHAP. 4
interpreted as a radius vector. Such an arc is said to be smooth. In referring to a smooth
arc z = z(t)(a < t <b), then, we agree that the derivative z'(t) is continuous on the
closed interval a < t <band nonzero on the open interval a < t <b.
A contour, or piecewise smooth arc, is an arc consisting of a finite number of
smooth arcs joined end to end. Hence if equation (2) represents a contour, z(t) is
continuous, whereas its derivative z'(t) is piecewise continuous. The polygonal line
(4) is, for example, a contour. When only the initial and final values of z(t) are the
same, a contour Cis called a simple closed contour. Examples are the circles (5) and
(6), as well as the boundary of a triangle or a rectangle taken in a specific direction.
The length of a contour or a simple closed contour is the sum of the lengths of the
smooth arcs that make up the contour.
The points on any simple closed curve or simple closed contour Care boundary
points of two distinct domains, one of which is the interior of C and is bounded. The
other, which is the exterior of C, is unbounded. It will be convenient to accept this
statement, known as the Jordan curve theorem, as geometrically evident; the proof is
not easy.*
EXERCISES
1. Show that if w(t) = u(t) +iv(t) is continuous on an interval a< t < b, then
1-a lb(a) w(-t) dt = w(r) dr;
-b a
l
b 1fJ .(b) a w(t) dt = a w[</>(r)]</>1
(r) dr, where ¢('r) is the function in equation (9),
Sec. 38.
Suggestion: These identities can be obtained by noting that they are valid for
real-valued functions oft.
2. Let C denote the right-hand half of the circle lzI= 2, in the counterclockwise direction,
and note that two parametric representations for C are
z =z(O) =2ei0
(- rt: < e< rt:)
2- - 2
and
z= z (y) = J4 - y2 + iy (-2<y<2).
*See pp. 115-116 of the book by Newman or Sec. 13 of the one by Thron, both of which are cited in
Appendix 1. The special case in which C is a simple closed polygon is proved on pp. 281-285 of Vol.
1 of the work by Hille, also cited in Appendix 1.
SEC.38 EXERCISES 121
Verify that Z(y) = z[I/J(y)], where
4> (y) =arctan y
)4- y2
(-~ <arctant<~).
Also, show that this function cp has a positive derivative, as required in the conditions
following equation (9), Sec. 38.
3. Derive the equation ofthe line through the points (a, a) and ({3, b) in the r t plane, shown
in Fig. 37. Then use it to find the linear function 4>(r) which can be used in equation (9),
Sec. 38, to transform representation (2) in that section into representation (10) there.
b- a af3- ba
Ans. ¢(-r) = r + .
f3-a {3-a
4. Verify expression (14), Sec. 38, for the derivative of Z(r) =z[¢(-r)].
Suggestion: Write Z(r) =x[¢(-r)] +iy[¢(-r)] and apply the chain rule for real-
valued functions of a real variable.
5. Suppose that a function j(z) is analytic at a point z0 =z(t0) lying on a smooth arc
z =z(t) (a< t <b). Show that if w(t) = f[z(t)], then
w'(t) = f'[z(t)]z'(t)
whent =t0.
Suggestion: Write j(z) = u(x, y) + iv(x, y) and z(t) = x(t) + iy(t), so that
w(t) =u[x(t), y(t)] +iv[x(t), y(t)].
Then apply the chain rule in calculus for functions of two real variables to write
w' = (uxx' +Uyy') +i(vxx' + vyy
1
),
and use the Cauchy-Riemann equations.
6. Let y(x) be a real-valued function defined on the interval 0 < x < l by means of the
equations
(a) Show that the equation
z = x + iy(x)
when 0 < x < 1,
whenx =0.
(0 <X< 1)
represents an arc C that intersects the real axis at the points z =1/n (n =1, 2, ...)
and z =0, as shown in Fig. 38.
(b) Verify that the arc C in part (a) is, in fact, a smooth arc.
Suggestion: To establish the continuity of y(x) at x = 0, observe that
122 INTEGRALS CHAP. 4
y
0
when x > 0. A similar remark applies in finding y'(O) and showing that y'(x) is
continuous at x =0.
1 X
FIGURE38
39. CONTOUR INTEGRALS
We tum now to integrals of complex~valued functions f of the complex variable z.
Such an integral is defined in terms of the values f(z) along a given contour C,
extending from a point z =z1 to a point z =z2 in the complex plane. It is, therefore,
a line integral; and its value depends, in general, on the contour C as well as on the
function f. It is written
fc f(z) dz or
1z
2
f(z) dz,
"I
the latter notation often being used when the value of the integral is independent of
the choice of the contour taken between two fixed end points. While the integral may
be defined directly as the limit of a sum, we choose to define it in terms of a definite
integral of the type introduced in Sec. 37.
Suppose that the equation
(1) z = z(t) (a< t <b)
represents a contour C, extending from a point z1 = z(a) to a point z2 = z(b). Let the
function f(z) be piecewise continuous on C; that is, f[z(t)] is piecewise continuous
on the interval a < t < b. We define the line integral, or contour integral, off along
C as follows:
(2) fc f(z) dz = lbf[z(t)]z'(t) dt.
Note that, since C is a contour, z'(t) is also piecewise continuous on the interval
a < t < b; and so the existence of integral (2) is ensured.
The value of a contour integral is invariant under a change in the representation of
its contour when the change is of the type (11), Sec. 38. This can be seen by following
the same general procedure that was used in Sec. 38 to show the invariance of arc
length.
SEC.39 CONTOUR INTEGRALS 123
It follows immediately from definition (2) and properties ofintegrals ofcomplex-
valued functions w(t) mentioned in Sec. 37 that
(3) Lzo/(z) dz =zo Lf(z) dz,
for any complex constant z0, and
(4) fc [f(z) +g(z)] dz = fc f(z) dz + Lg(z) dz.
Associated with the contour C used in integral (2) is the contour - C, consisting of
the same set of points but with the order reversed so that the new contour extends from
the point z2 to the point z1 (Fig. 39). The contour -C has parametric representation
z=z(-t) (-b < t <-a);
and so, in view of Exercise 1(a), Sec. 37,
f(z)dz= f[z(-t)]-z(-t)dt=- f[z(-t)]z'(-t)dt,
1 1-a d 1-a-C -b dt -b
where z'(-t) denotes the derivative of z(t) with respect tot, evaluated at -t. Making
the substitution r =-tin this last integral and referring to Exercise l(a), Sec. 38, we
obtain the expression
1f(z)dz=- fb f[z(r)z'(r)dr,
-C Ja
which is the same as
(5)
1f(z) dz =- f f(z) dz.
-c lc
y
0 x FIGURE39
Consider now a path C, with representation (1), that consists of a contour C1from
z1 to z2 followed by a c~mtour C2 from z2 to z3, the initial point of C2 being the final
point of C1 (Fig. 40). There is a value c oft, where a < c < b, such that z(c) =z2•
124 INTEGRALS
y
0 X
FIGURE40
C=C1+Cz
Consequently, C1 is represented by
z =z(t) (a< t <c)
and C2 is represented by
z = z(t) (c<t<b).
Also, by a rule for integrals of functions w(t) that was noted in Sec. 37,
1bf[z(t)]z'(t) dt = 1cf[z(t)]z'(t) dt + lbf(z(t)]z'(t) dt.
a . a c
Evidently, then,
(6) rf(z) dz = rf(z) dz + r f(z) dz.
lc lc1 lc2
CHAP. 4
Sometimes the contour C is called the sum of its legs C1 and C2 and is denoted by
C1 + C2. The sum of two contours C1 and -C2 is well defined when C1 and C2 have
the same final points, and it is written C1 - C2.
Definite integrals in calculus can be interpreted as areas, and they have other in-
terpretations as well. Except in special cases, no corresponding helpful interpretation,
geometric or physical, is available for integrals in the complex plane.
40. EXAMPLES
The purpose of this section is to provide examples of the definition in Sec. 39 of
contour integrals and to illustrate various properties that were mentioned there. We
defer development of the concept of antiderivatives of the integrands f (z) in contour
integrals until Sec. 42.
EXAMPLE 1. Let us find the value of the integral
(I) I= fc zdz
SEC.40
y
0 X
FIGURE43
we note that, according to Exercise l(b), Sec. 37,
Thus
!!:_ [z(t)]z = z(t)z'(t).
dt 2
I= [z(t)]z]b
2 a
[z(b)] 2 - [z(a)] 2
2
EXAMPLES 127
But z(b) = z2 and z(a) = z1; and so I = (z~ - zi)/2. Inasmuch as the value of I
depends only on the end points of C, and is otherwise independent of the arc that
is taken, we may write
(6)
1z2 zi- z?z dz = .
Zt 2
(Compare Example 2, where the value of an integral from one fixed point to another
depended on the path that was taken.)
Expression (6) is also valid when C is a contour that is not necessarily smooth
since a contour consists of a finite number of smooth arcs Ck (k = I, 2, ... , n), joined
end to end. More precisely, suppose that each Ck extends from zk to Zk+1. Then
(7)
n n 2 2
1 ""1 ""zk+l - zk
z dz = L...t z dz = L...t
c k=l c* k=I 2
z1 being the initial point of C and Zn+I its final point.
It follows from expression (7) that the integral of the function f(z) =z around
each closed contour in the plane has value zero. (Once again, compare Example 2,
where the value of the integral of a given function around a certain closed path was not
zero.) The question of predicting when an integral around a closed contour has value
zero will be discussed in Sees. 42, 44, and 46.
EXAMPLE 4. Let C denote the semicircular path
128 INTEGRALS CHAP. 4
y
-3 0 3 X FIGURE44
from the point z = 3 to the point z = -3 (Fig. 44). Although the branch (Sec. 30)
(8) (r > 0, 0 < (:) < 2n)
ofthe multiple-valued function z1
12
is not defined at the initial point z= 3ofthe contour
C, the integral
(9) I= Lzl12 dz
of that branch nevertheless exists. For the integrand is piecewise continuous on C. To
see that this is so, we observe that when z(B) = 3ei8, the right-hand limits of the real
and imaginary components of the function
/[z(B)] = J3ei012 = J3 cos e +i J3 sin e
2 2
at(:)= 0 are J3and 0, respectively. Hence f[z(e)] is continuous on the closed interval
0 < (} < n when its value at (:) = 0 is defined as J3. Consequently,
I= fon J3ei8123iei8 de= 3J3i fo1T ei3fJ12 dB;
and
Finally, then,
I= -2J3(1 +i).
EXERCISES
For the functions f and contours C in Exercises I through 6, use parametric
representations for C, or legs of C, to evaluate
Lf(z) dz.
130 INTEGRALS CHAP. 4
respectively. Use these parametric representations to show that
[ f(z) dz = [ f(z- zo) dz
lc leo
when f is piecewise continuous on C.
10. Let C0 denote the circle lz - zol = R, taken counterclockwise. Use the parametric
representation z =z0 + Rew(-rr < (J < rr) for C0 to derive the following integration
formulas:
(a) i dz =2Jri;
C 7- "'oo- "'
(b) [ (z- z0)n-1 dz =0 (n =±1, ±2, ...).
leo
11. Use the parametric representation in Exercise 10 for the oriented circle C0 there to show
that
[
( )
a-1 d .2Ra . ( .)z - zo z = 1 ··- sm an ,
c0 a
where ais any real numberotherthan zero and where the principal branch ofthe integrand
and the principal value of Ra are taken. [Note how this generalizes Exercise lO(b).]
12. (a) Suppose that a function f(z) is continuous on a smooth arc C, which has a parametric
representation z = z(t) (a < t < b); that is, f[z(t)] is continuous on the interval
a < t <b. Show that if </>(r)(a < r < {J) is the function described in Sec. 38, then
1bf[z(t)]z'(t) dt = i/3f[Z('r)]Z'(r) dr,
where Z(r) =z[¢(r)].
(b) Point out how it follows that the identity obtained in part (a) remains valid when C
is any contour, not necessarily a smooth one, and f(z) is piecewise continuous on
C. Thus show that the value of the integral of f(z) along Cis the same when the
representation z=Z(r) (a< r < {J) is used, instead of the original one.
Suggestion: In part (a), use the result in Exercise l(b), Sec. 38, and then refer
to expression (14) in that section.
41. UPPER BOUNDS FOR MODULI
OF CONTOUR INTEGRALS
When C denotes a contour z = z(t)(a < t <b), we know from definition (2), Sec. 39,
and inequality (5) in Sec. 37 that
[ J(z) dz = 1bf[z(t)]z'(t) dt < 1bjf[z(t)]IJz'(t)j dt.
So, for any nonnegative constant M such that the values off on C satisfy the inequality
//(z)/ < M,
[ f(z) dz < M 1b/z'(t)/ dt.
SEC. 41 UPPER BOUNDS FOR MODUU OF CONTOUR INTEGRALS 131
Since the integral on the right here represents the length L of the contour (see Sec. 38),
it follows that the modulus of the value of the integral of f along C does not exceed
ML:
(1) fc f(z) dz < ML.
This is, of course, a strict inequality when the values of f on C are such that
1/(z)l < M.
Note that since all of the paths of integration to be considered here are contours
and the integrands are piecewise continuous functions defined on those contours, a
number M such as the one appearing in inequality (1) will always exist. This is
because the real-valued function If[z(t)11 is continuous on the closed bounded interval
a < t < b when f is continuous on C; and such a function always reaches a maximum
value M on that interval.* Hence 1/(z)l has a maximum value on C when f is
continuous on it. It now follows immediately that the same is true when f is piecewise
continuous on C.
EXAMPLE 1. Let C be the arc of the circle IzI= 2 from z = 2 to z = 2i that lies in
the first quadrant (Fig. 45). Inequality (1) can be used to show that
(2)
1z +4 dz < 6n.
c z3 - 1 - 7
This is done by noting first that if z is a point on C, so that IzI= 2, then
lz+41 < lzl +4=6
and
lz3
- 11 > llzl3
- 11 = 7.
y
2i
0 2 X
FIGURE45
*See, for instance A. E. Taylor and W. R. Mann, "Advanced Calculus," 3d ed., pp. 86-90, 1983.
SEC.41 EXERCISES 133
Since the length of CR is the number L = n R, it follows from inequality (1) that
But
M L= nR~. 1jR
2
= n/~
R R2 - 1 1/R2 1- (1/R2)'
and it is clear that the term on the far right here tends to zero as R tends to infinity.
Limit (3) is, therefore, established.
EXERCISES
1. Without evaluating the integral, show that
{ dz < n
lc z2 - 1 - 3
when C is the same arc as the one in Example 1, Sec. 41.
2. Let C denote the line segment from z = i to z = 1. By observing that, of all the points
on that line segment, the midpoint is the closest to the origin, show that
{ dz <4h
lc z4 -
without evaluating the integral.
3. Show that if C is the boundary of the triangle with vertices at the points 0, 3i, and -4,
oriented in the counterclockwise direction (see Fig. 47), then
L(eZ - z) dz < 60.
y
3i
X
FIGURE47
134 INTEGRALS CHAP. 4
4. Let CR denote the upper half of the circle lzl = R (R > 2), taken in the counterclockwise
direction. Show that
1 2z2
- 1 d rrR(2R2
+ 1)
z < .
cR z4 + 5z2 + 4 - (R2 - 1)(R2- 4)
Then, by dividing the numerator and denominator on the right here by R4, show that the
value of the integral tends to zero as R tends to infinity.
5. Let CR be the circle lzl =R (R > 1), described in the counterclockwise direction. Show
that
and then use I'Hospital's rule to show that the value of this integral tends to zero as R
tends to infinity.
6. Let CP denote the circle lzl =p (0 < p < 1), oriented in the counterclockwise direction,
and suppose that f(z) is analytic in the disk lzl < 1. Show that if z-1/ 2 represents any
particular branch of that power ofz, then there is a nonnegative constant M, independent
of p, such that
Thus show that the value of the integral here approaches 0 as p tends to 0.
Suggestion: Note that since f(z) is analytic, and therefore continuous, throughout
the disk lzl < 1, it is bounded there (Sec. 17).
7. Let CN denote the boundary of the square formed by the lines
where N is a positive integer, and let the orientation of CN be counterclockwise.
(a) With the aid of the inequalities
Isin zl > Isin xi and lsin zl > Isinh yl,
obtained in Exercises lO(a) and ll(a) ofSec. 33, show that Isin zl > lon the vertical
sides of the square and that Isin zl > sinh(rr/2) on the horizontal sides. Thus show
that there is a positive constant A, independent ofN, such that Isin zl >A for all
points z lying on the contour CN.
(b) Using the final result in part (a), show that
{ dz 16
leN z2 sin z < (2N + l)rr A
and hence that the value of this integral tends to zero as N tends to infinity.
SEC.42 ANTIDERIVATIVES 135
42. ANTIDERIVATIVES
Although the value of a contour integral of a function f(z) from a fixed point z1 to a
fixed point z2 depends, in general, on the path that is taken, there are certain functions
whose integrals from z1 to z2 have values that are independent of path. (Compare
Examples 2 and 3 in Sec. 40.) The examples just cited also illustrate the fact that
the values of integrals around closed paths are sometimes, but not always, zero. The
theorem below is useful in determining when integration is independent of path and,
moreover, when an integral around a closed path has value zero.
In proving the theorem, we shall discover an extension of the fundamental theo-
rem of calculus that simplifies the evaluation ofmany contour integrals. That extension
involves the concept of an antiderivative of a continuous function f in a domain D,
or a function F such that F'(z) = f(z) for all z in D. Note that an antiderivative is, of
necessity, an analytic function. Note, too, that an antiderivative of a given function f
is unique except for an additive complex constant. This is because the derivative of the
difference F(z) - G(z) of any two such antiderivatives F(z) and G(z) is zero~ and,
according to the theorem in Sec. 23, an analytic function is constant in a domain D
when its derivative is zero throughout D.
Theorem. Suppose that a function f(z) is continuous on a domain D. If any one of
the following statements is true, then so are the others:
(i) f(z) has an antiderivative F(z) in D;
(ii) the integrals of f(z) along contours lying entirely in D and extending from any
fixed point z1 to any fixed point z2 all have the same value;
(iii) the integrals of f(z) around closed contours lying entirely in D all have value
zero.
It should be emphasized that the theorem does not claim that any of these
statements is true for a given function f and a given domain D. It says only that
all of them are true or that none of them is true. To prove the theorem, it is sufficient
to show that statement (i) implies statement (ii), that statement (ii) implies statement
(iii), and finally that statement (iii) implies statement (i).
Let us assume that statement (i) is true. If a contour C from z1 to z2, lying in D, is
just a smooth arc, with parametric representation z = z(t)(a < t <b), we know from
Exercise 5, Sec. 38, that
d
- F[z(t)] = F'[z(t)]z'(t) = f[z(t)]z'(t)
dt
(a < t <b).
Because the fundamental theorem of calculus can be extended so as to apply to
complex-valued functions of a real variable (Sec. 37), it follows that.
b ]bif(z) dz = 1f[z(t)]z'(t) dt = F[z(t)] a= F[z(b)]- F[z(a)].
136 INTEGRALS CHAP. 4
Since z(b) = z2 and z(a) = z1, the value of this contour integral is, then,
F(zz)- F(zt);
and that value is evidently independent of the contour C as long as C extends from z1
to z2 and lies entirely in D. That is,
(1) z ]~11
2
/(z) dz =F(z2)- F(z1) = F(z) .
ZJ
when Cis smooth. Expression (1) is also valid when Cis any contour, not necessarily
a smooth one, that lies in D. For, if C consists of a finite number of smooth arcs
Ck (k =1, 2, ... , n), each Ck extending from a point Zk to a point Zk+h then
n n
rJ(z) dz =I: [ /(z) dz =L (F(Zk+I)- F(zk)] =F(Zn+l)- F(zl).
lc k=l lck k=l
(Compare Example 3, Sec. 40.) The fact that statement (ii) follows from statement (i)
is now established.
To see that statement (ii) implies statement (iii), we let z1 and z2 denote any two
points on a closed contour C lying in D and form two paths, each with initial point
z1 and final point z2, such that C = C1 - C2 (Fig. 48). Assuming that statement (ii) is
true, one can write
(2)
1f(z) dz = [ J(z) dz,
c, Jc2
or
(3) [ f(z)dz+j f(z)dz=O.
Jc1 -C2
That is, the integral off(z) around the closed contour C = C1 - C2 has value zero.
y
0 ' /
..... ___ X
FIGURE48
It remains to show that statement (iii) implies statement (i). We do this by
assuming that statement (iii) is true, establishing the validity of statement (ii), and
then arriving at statement (i). To see that statement (ii) is true, we let C1and C2 denote
any two contours, lying in D, from a point z1 to a point z2 and observe that, in view of
SEC.42 ANTIDERIVATIVES 137
statement (iii), equation (3) holds (see Fig. 48). Thus equation (2) holds. Integration
is, therefore, independent of path in D; and we can define the function
F(z)- (" f(s) ds
lzo
on D. The proof of the theorem is complete once we show that F'(z) = f (z) every-
where in D. We do this by letting z + !lz be any point, distinct from z, lying in some
neighborhood of z that is small enough to be contained in D. Then
[
z+~z [z [z+~z
F(z + L'.z)- F(z) = f(s) ds- f(s) ds = f(s) ds,
~o zo z
where the path of integration from z to z + L'.z may be selected as a line segment (Fig.
49). Since
r+~z
lz ds = Llz
(see Exercise 5, Sec. 40), we can write
1 [z+~z
f(z) =- f(z) ds;
Llz z
and it follows that
F(z +Llz)- F(z) - f(z) = _1 f"+~z[j(s)- f(z)] ds.
!lz Llzh
y ----------/ .../
'/
'/
'/
-/
s z + ilz 1/
/ I
I
I
I
/
/
I D /
I /
/
I /
----~~---~-
X
FIGURE49
But f is continuous at the point z. Hence, for each positive number 8, a positive number
8 exists such that
lf(s)- f(z)l < 8 whenever Is- zl < 8.
Consequently, if the point z + Llz is close enough to z so that ILlzI< 8, then
138 INTEGRALS CHAP. 4
that is,
lim F(z + .:lz)- F(z) = f(z),
~z-o LlZ
or F'(z) = /(z).
43. EXAMPLES
The following examples illustrate the theorem in Sec. 42 and, in particular, the use of
the extension (1) of the fundamental theorem of calculus in that section.
EXAMPLE 1. The continuous function f (z) = z2
has an antiderivative F (z) =z3
/3
throughout the plane. Hence
f z2 dz=~ =!(l+i)3
=
2
(-l+i)
l+i 3]l+i
lo 3 0
3 3
for every contour from z =0 to z = 1+i.
EXAMPLE 2. The function f(z) = 1/z2
, which is continuous everywhere except
at the origin, has an antiderivative F (z) = -1/z in the domain lzI> 0, consisting of
the entire plane with the origin deleted. Consequently,
1dz =O
c z2
when Cis the positively oriented circle (Fig. 50)
(1) (-n<O<JT)
about the origin.
Note that the integral of the function f(z) = 1/z around the same circle cannot
be evaluated in a similar way. For, although the derivative of any branch F(z) of log z
y
2i
---- Ol
X
-2i
FIGURE 50
SEC. 43 EXAMPLES 139
is ljz (Sec. 30), F(z) is not differentiable, or even defined, along its branch cut In
particular, if a ray f) =a from the origin is used to form the branch cut, F'(z) fails to
exist at the point where that ray intersects the circle C (see Fig. 50). So C does not
lie in a domain throughout which F'(z) = 1/z, and we cannot make direct use of an
antiderivative. Example 3, just below, illustrates how a combination of two different
antiderivatives can be used to evaluate f(z) = 1/z around C.
EXAMPLE 3. Let C1 denote the right half
(2) (_n < e< n)2- - 2
of the circle C in Example 2. The principal branch
Log z = In r +i8 (r > 0, -n < e < n)
of the logarithmic function serves as an antiderivative of the function 1/z in the
evaluation of the integral of 1/z along C1 (Fig. 51):
1d !2i d ]2i~ = . ~ = Log z = Log(2i) - Log(-2i)
c, z -21 z -2i
= (In 2 +i ~) - (tn 2 - i ~) =ni.
This integral was evaluated in another way in Example 1, Sec. 40, where representation
(2) for the semicircle was used.
y
2i
X
-2i
FIGURE 51
Next, let C2 denote the left half
(3) (
TC 3n)-<{}<-
2- - 2
of the same circle C and consider the branch
log z = In r + if) (r > 0, 0 < f) < 2n)
SEC.43 EXERCISES 141
not defined on the ray e=0, in particular at the point z = 3. But another branch,
r > 0 -- <0 <-(
1l' 3Jr), 2 2 ,
is defined and continuous everywhere on C1. The values of j 1(z) at all points on c1
except z = 3 coincide with those of our integrand (5); so the integrand can be replaced
by ! 1(z). Since an antiderivative of ! 1(z) is the function
we can now write
(Compare Example 4 in Sec. 40.)
The integral
(6)
r > 0 -- < () <-
(
1r 3rr), 2 2 '
of the function (5) over any contour C2 that extends from z = -3 to z = 3 below the
real axis can be evaluated in a similar way. In this case, we can replace the integrand
by the branch
(
1l' 5Jr)r > 0, 2 < () < 2 '
whose values coincide with those of the integrand at z = -3 and at all points on C2
below the real axis. This enables us to use an antiderivative of h(z) to evaluate integral
(6). Details are left to the exercises.
EXERCISES
1. Use an antiderivative to show that, for every contour C extending from a point z1 to a
point Z2,
{ zn dz = 1 (zn+l - zn+l)
Jc n + 1
2 1 (n =0, 1, 2, ...).
2. By finding an antiderivative, evaluate each of these integrals, where the path is any
contour between the indicated limits of integration:
(a) [i/Z errz dz; (b) forr+Zi cos(~) dz; (c) 13
(z- 2)3 dz.
Ans. (a) (1 +i)j:rr; (b) e +(Ife); (c) 0.
142 INTEGRALS
3. Use the theorem in Sec. 42 to show that
{ (z - zot-1
dz =0
lc0
CHAP. 4
(n = ±1, ±2, ...)
when C0 is any closed contour which does not pass through the point z0. lCompare
Exercise lO(b), Sec. 40.]
4. Find an antiderivative F2(z) of the branch h(z) of z112 in Example 4, Sec. 43, to show
that integral (6) there has value 2J3(-1 +i). Note that the value of the integral of the
function (5) around the closed contour C2 - C1 in that example is, therefore, -4J3.
5. Show that
!1 i d 1+e-rr (1 ")
z z= -z,
-1 2
where zi denotes the principal branch
zi = exp(i Log z) (lzl > 0, -n < Arg z < n)
and where the path of integration is any contour from z = -1 to z= 1that, except for its
end points, lies above the real axis.
Suggestion: Use an antiderivative of the branch
zi = exp(i log z) (
n 3n)lzl > 0,-2 <argz < 2
of the same power function.
44. CAUCHY-GOURSAT THEOREM
In Sec. 42, we saw that when a continuous function f has an antiderivative in a domain
D, the integral off(z) around any given closed contour C lying entirely in D has value
zero. In tl1is section, we present a theorem giving other conditions on a function f,
which ensure that the value of the integral of f(z) around a simple closed contour
(Sec. 38) is zero. The theorem is central to the theory offunctions ofa complex variable;
and some extensions of it, involving certain special types of domains, will be given in
Sec. 46.
We let C denote a simple closed contour z = z(t) (a < t <b), described in the
positive sense (counterclockwise), and we assume that f is analytic at each point
interior to and on C. According to Sec. 39,
(1) Lf(z) dz =1bf[z(t)]z'(t) dt;
and if
f(z) = u(x, y) +iv(x, y) and z(t) = x(t) +iy(t),
SEC.44 CAUCHY-GOURSAT THEOREM 143
the integrand f[z(t)]z'(t) in expression (1) is the product ofthe functions
u[x(t), y(t)] +iv[x(t), y(t)], x'(t) + iy'(t)
of the real variable t. Thus
(2) [ f(z) dz = 1ux'- vy') dt + i 1b(vx' + uy') dt.
In terms of line integrals of real-valued functions of two real variables, then,
(3) [ f(z) dz = [ u dx- v dy + i [ v dx + u dy.
Observe that expression (3) can be obtained formally by replacing f (z) and dz on the
left with the binomials
u+iv and dx+idy,
respectively, and expanding their product. Expression (3) is, of course, also valid when
C is any contour, not necessarily a simple closed one, and f[z(t)] is only piecewise
continuous on it.
We next recall a result from calculus that enables us to express the line integrals
on the right in equation (3) as double integrals. Suppose that two real-valued functions
P (x, y) and Q (x, y), together with their first-order partial derivatives, are continuous
throughout the closed region R consisting of all points interior to and on the simple
closed contour C. According to Green's theorem,
Now f is continuous in R, since it is analytic there. Hence the functions u and
v are also continuous in R. Likewise, if the derivative f' off is continuous in R, so
are the first-order partial derivatives of u and v. Green's theorem then enables us to
rewrite equation (3) as
(4) [ f(z) dz = Jl(-vx- uy) dA i Jl (ux- vy) dA.
But, in view of the Cauchy-Riemann equations
the integrands of these two double integrals are zero throughout R. So, when f is
analytic in R and f' is continuous there,
(5) [ f(z) dz =0.
This result was obtained by Cauchy in the early part of the nineteenth century.
144 INTEGRALS CHAP. 4
Note that, once it has been established that the value of this integral is zero, the
orientation of C is immaterial. That is, statement (5) is also true if C is taken in the
clockwise direction, since then
1f(z) dz =- J f(z) dz = 0.
C -C
EXAMPLE. If C is any simple closed contour, in either direction, then
Lexp(z
3
) dz =0.
This is because the function f (z) =exp(z3
) is analytic everywhere and its derivative
f'(z) = 3z2 exp(z3) is continuous everywhere.
Goursat* was the first to prove that the condition of continuity on f' can be
omitted. Its removal is important and will allow us to show, for example, that the
derivative f' of an analytic function f is analytic without having to assume the
continuity off', which fo1lows as a consequence. We now state the revised form of
Cauchy's result, known as the Cauchy-Goursat theorem.
Theorem. Ifa function f is analytic at all points interior to and on a simple closed
contour C, then
l f(z) dz = 0.
The proof is presented in the next section, where, to be specific, we assume that
C is positively oriented. The reader who wishes to accept this theorem without proof
may pass directly to Sec. 46.
45. PROOF OF THE THEOREM
We preface the proof of the Cauchy-Goursat theorem with a lemma. We start by
forming subsets of the region R which consists of the points on a positively oriented
simple closed contour C together with the points interior to C. To do this, we draw
equally spaced lines parallel to the real and imaginary axes such that the distance
between adjacent vertical lines is the same as that between adjacent horizontal lines.
We thus form a finite number of closed square subregions, where each point of R lies
in at least one such subregion and each subregion contains points of R. We refer to
these square subregions simply as squares, always keeping in mind that by a square we
*E. Goursat (1858-1936), pronounced gour-sah'.
SEC.45 PROOF OF THE THEOREM 145
mean a boundary together with the points interior to it. If a particular square contains
points that are not in R, we remove those points and call what remains a panial square.
We thus cover the region R with a finite number of squares and partial squares (Fig.
54), and our proof of the folJowing lemma starts with this covering.
y
0 X
F'IGURE54
Lemma. Let f be analytic throughout a closed region R consisting of the points
interior to a positively oriented simple closed contour C together with the points on C
itself For any positive numbers, the region R can be covered with a finite number of
squares and partial squares, indexed by j = 1, 2, ... , n, such that in each one there
is a fixed point zj for which the inequality
(1) <e
is satisfied by all other points in that square or panial square.
To start the proof, we consider the possibility that, in the covering constructed
just prior to the statement of the lemma, there is some square or partial square in
which no point zj exists such that inequality (1) holds for all other points z in it. If
that subregion is a square, we construct four smaller squares by drawing line segments
joining the midpoints of its opposite sides (Fig. 54). If the subregion is a partial square,
we treat the whole square in the same manner and then let the portions that lie outside
R be discarded. If, in any one of these smaller subregions, no point zj exists such that
inequality (1) holds for all other point-; z in it, we construct still smaller squares and
partial squares, etc. When this is done to each of the original subregions that requires
it, it turns out that, after a finite number ofsteps, the region R can be covered with a
finite number of squares and partial squares such that the lemma is true.
146 INTEGRALS CHAP. 4
To verify this, we suppose that the needed points zj do not exist after subdividing
one of the original subregions a finite number of times and reach a contradiction. We
let a0 denote that subregion if it is a square; if it is a partial square, we let a0 denote
the entire square of which it is a part. After we subdivide a0, at least one of the four
smaller squares, denoted by a 1, must contain points of R but no appropriate point
Zj· We then subdivide a 1 and continue in this manner. It may be that after a square
ak-l (k = 1, 2, ...) has been subdivided, more than one of the four smaller squares
constructed from t can be chosen. To make a specific choice, we take ak to be the one
lowest and then furthest to the left.
In view of the manner in which the nested infinite sequence
(2)
of squares is constructed, it is easily shown (Exercise 9, Sec. 46) that there is a point zo
common to each ak; also, each of these squares contains points Rother than possibly
z0. Recall how the sizes of the squares in the sequence are decreasing, and note that
any 8 neighborhood lz - zol < 8 of zo contains such squares when their diagonals have
lengths less than o. Every 8 neighborhood lz- zol < 8 therefore contains points of R
distinct from z0, and this means that zo is an accumulation point of R. Since the region
R is a closed set, it follows that zo is a point in R. (See Sec. 10.)
Now the function f is analytic throughout R and, in particular, at z0. Conse-
quently, f'(zo) exists, According to the definition of derivative (Sec. 18), there is, for
each positive numbers, a 8 neighborhood lz - zol < 8 such that the inequality
J(z)- f(zo) j'( .)
- zo < s
'z- zo
is satisfied by all points distinct from zo in that neighborhood. But the neighborhood
lz- zol < ocontains a square aK when the integer K is large enough that the length of
a diagonal ofthat square is less than 8 (Fig. 55). Consequently, zo serves as the point zj
in inequality (1) for the subregion consisting of the square aK or a part of aK. Contrary
to the way in which the sequence (2) was formed, then, it is not necessary to subdivide
aK. We thus arrive at a contradiction, and the proof of the lemma is complete.
}'!
0 X
FIGURE 55
SEC.45 PROOF OF THE THEOREM 147
Continuing with a function f which is analytic throughout a region R consisting
of a positively oriented simple closed contour C and points interior to it, we are now
ready to prove the Cauchy-Goursat theorem, namely that
(3) fc f(z) dz = 0.
Given an arbitrary positive number c:, we consider the covering of R in the
statement of the lemma. Let us define on the j th square or partial square the following
function, where Zj is the fixed point in that subregion for which inequa1ity (1) holds:
(4)
/(z)- f(z1) f'(
-------'- - zj)
Z- Zj
when z =!= Zj,
0 when z = z1.
According to inequality (l),
(5)
at all points z in the subregion on which 8j(z) is defined. Also, the function 8/z) is
continuous throughout the subregion since f (z) is continuous there and
Next, let c1 (j = 1, 2, ... , n) denote the positively oriented boundaries of the
above squares or partial squares covering R. In view of definition (4), the value off
at a point z on any particular CJ can be written
and this means that
(6) r f(z) dz
lcj
= [/(Zj)- Zjf'(zj)] 1dz + J'(z;) ( Z dz +1(z- Zj)oj(Z) dz.
c lc. c1 J 1
But
fc. dz = 0 and
j
1zdz =0
cj
since the functions 1 and z possess antiderivatives everywhere in the finite plane. So
equation (6) reduces to
(7) r f(z)dz= r(z-zj)8/z)dz
lcj lcj
(J = 1, 2, ... , n).
148 INTEGRALS CHAP. 4
The sum of all n integrals on the left in equations (7) can be written
n
L 1f(z)dz= 1f(z)dz
J=l cj c
since the two integrals along the common boundary ofevery pair ofadjacent subregions
cancel each other, the integral being taken in one sense along that line segment in one
subregion and in the opposite sense in the other (Fig. 56). Only the integrals along the
arcs that are parts of C remain. Thus, in view of equations (7),
and so
(8)
n
[ f(z)dz <L [(z-zJ)DJ(z)dz.
lc ._1
lc,.j- .
y
----- s
0 X
FIGURE 56
Let us now use property (1), Sec. 41 to find an upper bound for each absolute
value on the right in inequality (8). To do this, we first recall that each Cj coincides
either entirely or partially with the boundary ofa square. In either case, we letsJ denote
the length of a side of the square. Since, in the jth integral, both the variable zand the
point zj lie in that square,
lz- ZJI < J2sj·
In view of inequality (5), then, we know that each integrand on the right in inequality
(8) satisfies the condition
(9)
SEC.46 SIMPLY AND MULTIPLY CONNECTED DOMAINS 149
As for the length ofthe path Ci, it is 4si if Cj is the boundary of a square. In that case,
we let Ai denote the area of the square and observe that
(10)
If Cj is the boundary of a partial square, its length does not exceed 4si + L i• where
Li is the length of that part of Ci which is also a part of C. Again letting Aj denote
the area of the full square, we find that
where S is the length of a side of some square that encloses the entire contour C as
well as all of the squares originally used in covering R (Fig. 56). Note that the sum of
all the Ai 's does not exceed S2
•
If L denotes the length of C, it now follows from inequalities (8), (10), and (11)
that
Lf(z) dz < (4J2S2
+ J2SL)s.
Since the value of the positive numbers is arbitrary, we can choose it so that the right-
hand side of this last inequality is as small as we please. The left-hand side, which
is independent of s, must therefore be equal to zero; and statement (3) fol1ows. This
completes the proof of the Cauchy-Goursat theorem.
46. SIMPLY AND MULTIPLY CONNECTED DOMAINS
A simply connected domain Dis a domain such that every simple closed contour within
it encloses only points of D. The set of points interior to a simple closed contour is an
example. The annular domain between two concentric circles is, however, not simply
connected. A domain that is not simply connected is said to be multiply connected.
The Cauchy-Goursat theorem can be extended in the following way, involving a
simply connected domain.
Theorem 1. ff a function f is analytic throughout a simply connected domain D,
then
(1) Lf(z) dz = 0
for every closed contour C lying in D.
150 INTEGRALS CHAP. 4
y
0 X
FIGURE 57
The proof is easy if C is a simple closed contour or if it is a closed contour that
intersects itselfafinite number oftimes. For, if Cis simple and lies in D, the function f
is analytic at each point interior to and on C; and the Cauchy-Goursat theorem ensures
that equation (1) holds. Furthermore, if C is closed but intersects itself a finite number
of times, it consists of a finite number of simple closed contours. This is illustrated
in Fig. 57, where the simple closed contours Ck (k = 1, 2, 3, 4) make up C. Since the
value of the integral around each Ck is zero, according to the Cauchy-Goursat theorem,
it follows that
4
[ f(z) dz = L [ f(z) dz =0.
lc k=l lck
Subtleties arise if the closed contour has an infinite number of self-intersection
points. One method that can sometimes be used to show that the theorem still applies
is illustrated in Exercise 5 below.*
Corollary 1. A function f that is analytic throughout a simply connected domain D
must have an antiderivative everywhere in D.
This corollary follows immediately from Theorem l because of the theorem in
Sec. 42, which tells us that a continuous function f always has an antiderivative in a
given domain when equation (l) holds for each closed contour C in that domain. Note
that, since the finite plane is simply connected, Corollary 1tells us that entirefunctions
always possess antiderivatives.
The Cauchy-Goursat theorem can also be extended in a way that involves inte-
grals along the boundary of a multiply connected domain. The following theorem is
such an extension.
*For a proofof the theorem involving more general paths of finite length, see, for example, Sees. 63-65
in Vol. I of the book by Markushevich, cited in Appendix I.
SEC. 46 SIMPLY AND MULTIPLY CONNECTED DOMAINS 151
Theorem 2. Suppose that
(i) Cis a simple closed contour, described in the counterclockwise direction;
(ii) Ck (k = 1, 2, ... , n) are simple closed contours interior to C, all described in
the clockwise direction, that are disjoint and whose interiors have no points in
common (Fig. 58).
If a function f is analytic on all of these contours and throughout the multiply
connected domain consisting ofall points inside C and exterior to each Ck, then
(2)
n
{ f(z) dz +L { f(z) dz = 0.
lc k=l lck
y
0 X
FIGURE 58
Note that, in equation (2), the direction of each path of integration is such that
the multiply connected domain lies to the left of that path.
To prove the theorem, we introduce a polygonal path L 1> consisting of a finite
number of line segments joined end to end, to connect the outer contour C to the inner
contour C1. We introduce another polygonal path L2 which connects C1 to C2; and we
continue in this manner, with Ln+l connecting Cn to C. As indicated by the single-
barbed arrows in Fig. 58, two simple closed contours r 1 and r2 can be formed, each
consisting of polygonal paths L k or - L k and pieces of C and Ck and each described
in such a direction that the points enclosed by them lie to the left. The Cauchy-
Goursat theorem can now be applied to f on r 1 and r 2, and the sum of the values
of the integrals over those contours is found to be zero. Since the integrals in opposite
directions along each path Lk cancel, only the integrals along C and Ck remain; and
we arrive at statement (2).
The following corollary is an especially important consequence of Theorem 2.
Corollary 2. Let C1and C2 denote positively oriented simple closed contours, where
C2 is interior to C1 (Fig. 59). Ifafunction f is analytic in the closed region consisting
ofthose contours and all points between them, then
(3) { f(z) dz = { f(z) dz.
lc, lc2
152 INTEGRALS CHAP. 4
y
c,2..1-------
0 X
FIGURE 59
For a verification, we use Theorem 2 to write
[ f(z)dz+j f(z)dz=O;
lc1 -c2
and we note that this is just a different form of equation (3).
Corollary 2 is known as the principle ofdeformation ofpaths since it tells us that
if C1 is continuously deformed into C2, always passing through points at which f is
analytic, then the value of the integral off over C1 never changes.
EXAMPLE. When C is any positively oriented simple closed contour surrounding
the origin, Corollary 2 can be used to show that
f dz =2ni.
lc z
To accomplish this, we need only construct a positively oriented circle C0 with center
at the origin and radius so small that C0 lies entirely inside C (Fig. 60). Since [Exercise
lO(a), Sec. 40]
y
X
FIGURE60
SEC.46 EXERCISES 155
(b) By accepting the fact that*
and observing that
roo 2 .jii
Jo e-x dx = 2
[b 2 [b 2
Jo eY sin 2ay dy < lo eY dy,
obtain the desired integration formula by letting a tend to infinity in the equation at
the end of part (a).
5. According to Exercise 6, Sec. 38, the path C1 from the origin to the point z = 1 along
the graph of the function defined by means of the equations
y
when 0 < x < 1,
whenx = 0
is a smooth arc that intersects the real axis an infinite number of times. Let C2 denote
the line segment along the real axis from z = 1 back to the origin, and let C3 denote
any smooth arc from the origin to z =1that does not intersect itself and has only its end
points in common with the arcs C1 and C2 (Fig. 63). Apply the Cauchy-Goursat theorem
to show that if a function f is entire, then
1f(z) dz = 1f(z) dz and
Ct c3 1f(z) dz = -1 f(z) dz.
Cz c3
1 X
FIGURE63
*The usual way to evaluate this integral is by writing its square as
roo 2 roo 2 roo roo 2 2
Jo e-x dx Jo e-Y dy = Jo Jo e-(x +y ) dx dy
and then evaluating the iterated integral by changing to polar coordinates. Details are given in, for
example, A. E. Taylor and W. R. Mann, "Advanced Calculus," 3d ed., pp. 680-681, 1983.
156 INTEGRALS CHAP. 4
Conclude that, even though the closed contour C = C1 + C2 intersects itself an infinite
number of times,
[ f(z)dz=O.
Jc
6. Let C denote the positively oriented boundary of the half disk 0 < r < 1, 0 < (} .::; rr, and
let f (z) be a continuous function defined on that half disk by writing f (0) = 0 and using
the branch
r > 0 -- <8 <-( rr 3rr)
' 2 2
of the multiple-valued function z1
12
• Show that
fc /(z) dz =0
by evaluating separately the integrals of f(z) over the semicircle and the two radii which
make up C. Why does the Cauchy-Goursat theorem not apply here?
7. Show that if Cis a positively oriented simple closed contour, then the area of the region
enclosed by C can be written
__!_ fzdz.
2i lc
Suggestion: Note that expression (4), Sec. 44, can be used here even though the
function /(z) =zis not analytic anywhere (see Exercise l(a), Sec. 22).
8. Nested Intervals. An infinite sequence of closed intervals an < x < bn (n =0, 1, 2, ...}
is formed in the following way. The interval a 1 < x < b1 is either the left-hand or right-
hand half of the first interval ao ::: x < b0, and the interval a2 < x < b2 is then one of the
two halves of a 1 < x < b1, etc. Prove that there is a point x0 which belongs to every one
of the closed intervals an < x ::; bn.
Suggestion: Note that the left-hand end points an represent a bounded nondecreas-
ing sequence of numbers, since ao <an <an+I < bo; hence they have a limit A as n
tends to infinity. Show that the end points bn also have a limit B. Then show that A= B,
and write x0 =A= B.
9. Nested Squares. A square u0 : a0 < x < b0 , c0 < y < d0 is divided into four equal squares
by line segments parallel to the coordinate axes. One of those four smaller squares
u 1 : a1 < x < b1, c1 < y < d1 is selected according to some rule. It, in tum, is divided
into four equal squares one of which, called u2, is selected, etc. (see Sec. 45). Prove
that there is a point (x0, Yo) which belongs to each of the closed regions of the infinite
sequence u0, u 1, o-2, •..•
Suggestion: Apply the result in Exercise 8 to each of the sequences of closed
intervals an< X< bn and en < y < dn (n =0, 1, 2, ...}. .
SEC.47 CAUCHY INTEGRAL FORMULA 157
47. CAUCHY INTEGRAL FORMULA
Another fundamental result will now be established.
Theorem. Let f be analytic everywhere inside and on a simple closed contour C,
taken in the positive sense. If z0 is any point interior to C, then
(1) f(zo) = _1_. rf(z) dz
2m lc z -zo
Formula (1) is called the Cauchy integral formula. It tells us that if a function f
is to be analytic within and on a simple closed contour C, then the values off interior
to C are completely determined by the values of f on C.
When the Cauchy integral formula is written
i f(z) dz _ 2 'f( )- m z0 ,
c z -zo
(2)
it can be used to evaluate certain integrals along simple closed contours.
EXAMPLE. Let C be the positively oriented circle lzl = 2. Since the function
z
f(z) = 9 2
-z
is analytic within and on C and since the point zo = -i is interior to C, formula (2)
tells us that
( Z dz ( zj (9 - z
2
) d
2
. ( -i) TC
lc (9-z2)(z+i) = lc z....,(-i) z= m 10 =5·
We begin the proofofthe theorem by letting CP denote a positively oriented circle
lz zol =p, where p is small enough that CPis interior to C (see Fig. 64). Since the
function f(z)/(z- z0) is analytic between and on the contours C and CP, it follows
y
Cp
@
0 X
FIGURE64
158 INTEGRALS CHAP. 4
from the principle of deformation of paths (Corollary 2, Sec. 46) that
{ f(z) dz = { f(z) dz.
Jc z - zo Jcp z - zo
This enables us to wtite
(3) { f(z) dz _ f(zo) 1 dz =1f(z)- f(z0) dz.
Jc z - zo cp z - z0 cp z - z0
But [see Exercise lO(a), Sec. 40]
1 dz
2
.
- - = 7l'l;
CP Z- Zo
and so equation (3) becomes
(4) { f(z) dz - 2nif(zo) = 1f(z)- f(zo) dz.
Jc z - zo cp z - zo
Now the fact that f is analytic, and therefore continuous, at zo ensures that,
corresponding to each positive number E, however small, there is a positive number 8
such that
(5) 1/(z)- f(zo)l < e whenever lz- zol < 8.
Let the radius p of the circle CP be smaller than the number 8 in the second of these
inequalities. Since lz - zoI= p when z is on CP' it follows that thefirst of inequalities
(5) holds when z is such a point; and inequality (1), Sec. 41, giving upper bounds for
the moduli of contour integrals, tells us that
1J(z)- f(zo) d
z
CP Z- Zo
In view of equation (4), then,
E
<- 2np = 2ne.
p
f J(z) dz - 2nif(z
0
) < 2ne.
lc z- zo
Since the left-hand side of this inequality is a nonnegative constant that is less than an
arbitrarily small positive number, it must equal to zero. Hence equation (2) is valid,
and the theorem is proved.
48. DERIVATIVES OF ANALYTIC FUNCTIONS
It follows from the Cauchy integral formula (Sec. 47) that if a function is analytic at a
point, then its derivatives of all orders exist at that point and are themselves analytic
160 INTEGRALS CHAP. 4
Next, we let M denote the maximum value of lf(s)l on C and observe that, since
Is- zl > d and l~zl < d,
Is- z- ~zl = l(s- z)- ~zl >lis- zl -l~zll > d -l~zl > 0.
Thus
1 ~zf(s) ds < l~ziM L,
c (s- z- ~z)(s- z)2 - (d- l~zl)d2
where L is the length of C. Upon letting ~z tend to zero, we find from this inequality
that the right-hand side of equation (3) also tends to zero. Consequently,
lim f(z + ~z)- f(z) _ _ 1_1 f(s) ds = O·
tlz-+0 ~z 2ni c (s - z)2 '
and the desired expression for f'(z) is established.
The same technique can be used to verify the expression for f"(z) in the statement
of the lemma. The details, which are outlined in Exercise 9, are left to the reader.
Theorem 1. Ifa function is analytic at a point, then its derivatives ofall orders exist
at that point. Those derivatives are, moreover, all analytic there.
To prove this remarkable theorem, we assume that a function f is analytic at a
point z0 . There must, then, be a neighborhood lz- zol < 8 ofzo throughout which f is
analytic (see Sec. 23). Consequently, there is a positively oriented circle C0, centered
at zo and with radius 8/2, such that f is analytic inside and on C0 (Fig. 66). According
to the above lemma,
f"(z) = ~ 1f(s) ds
Jrl c0 (s - z)3
at each point z interior to C0, and the existence of .f"(z) throughout the neighborhood
lz :.._ zol < e/2 means that .f' is analytic at z0. One can apply the same argument to the
y
0 X
FIGURE66
SEC. 48 DERIVATIVES OF ANALYTIC FUNCTIONS 161
analytic function f' to conclude that its derivative f" is analytic, etc. Theorem 1 is
now established.
As a consequence, when a function
f(z) =u(x, y) +iv(x, y)
is analytic at a point z = (x, y), the differentiability off' ensures the continuity of f'
there (Sec. 18). Then, since
f'(z) = Ux +ivx = Vy- iuy,
we may conclude that the first-order partial derivatives of u and v are continuous at
that point. Furthermore, since f" is analytic and continuous at z and since
f "( ) . .z =Uxx + lVxx =Vyx- IUyx•
etc., we arrive at a corollary that was anticipated in Sec. 25, where harmonic functions
were introduced.
Corollary. Ifa function f(z) = u(x, y) +iv(x, y) is defined and analytic at a point
z = (x, y) then the componentfunctions u and v have continuous partial derivatives
ofall orders at that point.
(4)
One can use mathematical induction to generalize formulas (1) to
f(n)(z) = ~ r f(s) ds
2:rri Jc (s- z)n+I
(n=1,2, ...).
The verification is considerably more involved than for just n = 1 and n = 2, and we
refer the interested reader to other texts for it.* Note that, with the agreement that
f(O)(z) = f(z) and 0! = 1,
expression (4) is also valid when n = 0, in which case it becomes the Cauchy integral
formula (2).
(5)
When written in the form
1 f(z) dz = 2:rri f(n)(zo)
c (z- zo)n+l n!
(n = 0, 1, 2, ...),
expression (4) can be useful in evaluating certain integrals when f is analytic inside
and on a simple closed contour C, taken in the positive sense, and zo is any point
interior to C. It has already been illustrated in Sec. 47 when n = 0.
*See, for example, pp. 299-301 in Vol. I of the book by Markushevich, cited in Appendix 1.
162 INTEGRALS CHAP. 4
EXAMPLE 1. If Cis the positively oriented unit circle lzl = 1and
/(z) =exp(2z),
then
rexp(2z) dz = r f(z) dz = 2ni f/1!(0) = 8:rri.
lc z4 lc (z- 0)3+1 3! 3
EXAMPLE 2. Let z0 be any point interior to a positively oriented simple closed
contour C. When /(z) = 1, expression (5) shows that
and
1 dz =2:rri
c z zo
r dz -0
lc (z- zo)n+l -
(n=l,2, ...).
(Compare Exercise 10, Sec. 40.)
We conclude this section with a theorem due to E. Morera (1856-1909). The proof
here depends on the fact that the derivative of an analytic function is itself analytic, as
stated in Theorem 1.
Theorem 2. Let f be continuous on a domain D. If
(6) £f(z) dz =0
for every closed contour C lying in D, then f is analytic throughout D.
In particular, when D is simply connected, we have for the class of continuous
functions on D a converse of Theorem 1 in Sec. 46, which is the extension of the
Cauchy-Goursat theorem involving such domains.
To prove the theorem here, we observe that when its hypothesis is satisfied, the
theorem in Sec. 42 ensures that f has an antiderivative in D; that is, there exists an
analytic function F such that F'(z) = f (z) at each point in D. Since f is the derivative
ofF, it then follows from Theorem 1above that f is analytic in D.
EXERCISES
1. Let C denote the positively oriented boundary of the square whose sides lie along the
lines x = ± 2 and y =± 2. Evaluate each of these integrals:
(a) { e-z dz . (b) { cos z dz· (c) { zdz .
lcz-(rt:i/2)' lcz(z2 +8) ' lc2z+1'
164 INTEGRALS CHAP. 4
8. (a) With the aid of the binomial formula (Sec. 3), show that, for each value of n, the
function
(n =0, I, 2, ...)
is a polynomial of degree n.*
(b) Let C denote any positively oriented simple closed contour surrounding a fixed
point z. With the aid of the integral representation (4), Sec. 48, for the nth derivative
of an analytic function, show that the polynomials in part (a) can be expressed in the
form
(ll = 0, 1, 2, ...).
(c) Point out how the integrand in the representation for Pn(Z) in part (b) can be written
(s + l)n j(s - 1) if z =I. Then apply the Cauchy integral formula to show that
(n =0, 1, 2, ...).
Similarly, show that
(n =0, I, 2, ...).
9. Follow the steps below to verify the expression
in the lemma in Sec. 48.
f"(z) = _1 1f(s) ds
n:i c(s-z)3
(a) Use the expression for f'(z) in the lemma to show that
f'(z +Llz)- f'(z) __l 1f(s) ds =_I_ [ 3(s- z)Llz- 2(Llz)
2
f(s) ds.
Llz rri c (s - z)3 2rri Jc (s- z- Llz)2(s- z)3
(b) Let D and d denote the largest and smallest distances, respectively, from z to points
on C. Also, let M be the maximum value of If(s) I on C and L the length of C. With
the aid of the triangle inequality and by referring to the derivation of the expression
for f'(z) in the lemma, show that when 0 < ILlzl < d, the value of the integral on
the right-hand side in part (a) is bounded from above by
(3DILlzl +2JLlzi2
)M L
(d-1Llzl)2d3 ·
(c) Use the results in parts (a) and (b) to obtain the desired expression for f"(z).
*These are the Legendre polynomials which appear in Exercise 7. Sec. 37. when z = x. See the footnote
to that exercise.
SEC.49 LIOUVILLE'S THEOREM AND THE FUNDAMENTAL THEOREM OF ALGEBRA 165
49. LIOUVILLE'S THEOREM AND THE FUNDAMENTAL
THEOREM OF ALGEBRA
This section is devoted to two important theorems that follow from the extension of
the Cauchy integral formula in Sec. 48.
Lemma. Suppose that a function f is analytic inside and on a positively oriented
circle CR• centered at z0 and with radius R (Fig. 67). If MR denotes the maximum
value oflf(z)l on CR. then
(l) f(n)(Zo) < n';R (n=l,2, ...).
y
0 X
FIGURE67
Inequality (1) is called Cauchy's inequality and is an immediate consequence of
the expression
f(n)(zo) = ~ 1 j(z) dz
2rd cR (z - z0)n+l
(n = 1, 2, ...),
which is a slightly different form of equation (5), Sec. 48. We need only apply
inequality (1), Sec. 41, which gives upper bounds for the moduli of the values of
contour integrals, to see that
f (n)( ) n! MR 2 Rzo <-. :rr
- 2n Rn+!
(n=1,2, ...),
where MR is as in the statement of the lemma. This inequality is, of course, the same
as inequality (1) in the lemma.
The lemma can be used to show that no entire function except a constant is
bounded in the complex plane. Our first theorem here, which is known as Liouville's
theorem, states this result in a somewhat different way.
Theorem 1. Iff is entire and bounded in the complex plane, then f (z) is constant
throughout the plane.
166 INTEGRALS CHAP. 4
To start the proof, we assume that f is as stated in the theorem and note that,
since f is entire, Cauchy's inequality (1) with n = 1holds for any choices of zo and
R:
(2)
Moreover, the boundedness condition in the statement of the theorem tells us that a
nonnegative constant M exists such that If(z)I< M for all z; and, because the constant
MR in inequality (2) is always less than or equal toM, it follows that
(3) lf'(zo)l < ~•
where zo is any fixed point in the plane and R is arbitrarily large. Now the number M
in inequality (3) is independent of the value of R that is taken. Hence that inequality
can hold for arbitrarily large values of R only iff'(zo) = 0. Since the choice of zo was
arbitrary, this means that f'(z) = 0 everywhere in the complex plane. Consequently,
f is a constant function, according to the theorem in Sec. 23.
The following theorem, known as the fundamental theorem ofalgebra, follows
readily from Liouville's theorem.
Theorem 2. Any polynomial
P(z) = ao +a1z +azz2
+ ···+anzn
ofdegree n (n > 1) has at least one zero. That is, there exists at least one point zo such
that P(z0) = 0.
The proof here is by contradiction. Suppose that P(z) is not zero for any value
of z. Then the reciprocal
1
f(z) = P(z)
is clearly entire, and it is also bounded in the complex plane.
To show that it is bounded, we first write
(4) w = ao + a1 + az + ... + an-1,
zn zn-1 zn-2 z
so that P(z) =(an+ w)zn. We then observe that a sufficiently large positive number
R can be found such that the modulus of each of the quotients in expression (4) is less
than the number lanl/(2n) when lzl > R. The generalized triangle inequality, applied
ton complex numbers, thus shows that lwl < lanl/2 for such values ofz. Consequently,
when lzl > R,
SEC.jO MAXIMUM MODULUS PRINCIPLE 167
and this enables us to write
(5) IP(z)l =ian+ wllznl > la;llzln > la;l Rn whenever lzl > R.
Evidently, then,
1 2
1/(z)l- < whenever lzl > R.
- IP(z)l laniRn
So f is bounded in the region exterior to the disk lzl < R. But f is continuous in that
closed disk, and this means that f is bounded there too. Hence f is bounded in the
entire plane.
It now follows from Liouville's theorem that f(z), and consequently P(z), is
constant. But P(z) is not constant, and we have reached a contradiction.*
The fundamental theorem tells us that any polynomial P(z) of degree n (n > 1)
can be expressed as a product of linear factors:
(6) P(z) =c(z- z1)(z- zz) · · · (z- zn),
where c and Zk (k = 1, 2, ... , n) are complex constants. More precisely, the theorem
ensures that P(z) has a zero z1• Then, according to Exercise 10, Sec. 50,
where Q1(z) is a polynomial of degree n- 1. The same argument, applied to Q1(z),
reveals that there is a number z2 such that
where Q2(z) is a polynomial of degree n - 2. Continuing in this way, we arrive at
expression (6). Some of the constants Zk in expression (6) may, of course, appear more
than once, and it is clear that P(z) can have no more than n distinct zeros.
50. MAXIMUM MODULUS PRINCIPLE
In this section, we derive an important result involving maximum values ofthe moduli
of analytic functions. We begin with a needed lemma.
Lemma. Suppose that lf(z)l < lf(z0)1 at each point z in some neighborhood
Iz - z0I < s in which f is analytic. Then f (z) has the constant value f (z0) throughout
that neighborhood.
"'For an interesting proof of the fundamental theorem using the Cauchy-Goursat theorem, see R. P.
Boas, Jr., Amer. Math. Monthly, Vol. 71, No.2, p. 180, 1964.
168 INTEGRALS CHAP. 4
y
0 X
FIGURE68
To prove this, we assume that f satisfies the stated conditions and let z1 be any
point other than zo in the given neighborhood. We then let p be the distance between
z1 and z0• If CP denotes the positively oriented circle lz- zol = p, centered at zo and
passing through z1 (Fig. 68), the Cauchy integral formula tells us that
(1) f(zo) = _1_.J f(z) dz;
2ro cP z- zo
and the parametric representation
z = z0 + peie
for CP enables us to write equation (1) as
(2) f(zo) = -
1
f
2
1f f(zo + pei9) dB.
2n lo
We note from expression (2) that when a function is analytic within and on a given
circle, its value at the center is the arithmetic mean of its values on the circle. This
result is called Gauss's mean value theorem.
From equation (2), we obtain the inequality
(3) 1 121f1/(zo)l <- lf(zo + pei8
)1 dB.
2n o
On the other hand, since
(4)
we find that
f21f f21f
lo lf(zo +pew)Id() < lo lf(zo)l d() = 2nlf(z0)1.
Thus
(5) 1 121flf(zo)l > - lf(zo + peiB)I d().
2n o
SEC. 50 MAXIMUM MODULUS PRINCIPLE 169
It is now evident from inequalities (3) and (5) that
1 1217"lf(zo)l =- lf(zo + pei8
)1 d(),
2Jr 0
or
{21i
lo [lf(zo)l - lf(zo + pei
9
)1] d() = 0.
The integrand in this last integral is continuous in the variable 0; and, in view of
condition (4), it is greater than or equal to zero on the entire interval 0 < () < 2n.
Because the value of the integra] is zero, then, the integrand must be identically equal
to zero. That is,
(6) lf(zo + pei9
)1 = lf(zo)l (0 < () < 2n).
This shows that lf(z)l = lf(z0)1 for all points z on the circle lz- zol = p.
Finally, since z1 is any point in the deleted neighborhood 0 < lz- zol < s, we
see that the equation lf(z)l = lf(zo)l is, in fact, satisfied by all points z lying on any
circle lz- zol = p, where 0 < p < s. Consequently, lf(z)l = lf(zo)l everywhere in
the neighborhood lz- zul <e. But we know from Exercise 7(b), Sec. 24, that when the
modulus of an analytic function is constant in a domain, the function itself is constant
there. Thus f(z) = f(z0) for each point z in the neighborhood, and the proof of the
lemma is complete.
This lemma can be used to prove the following theorem, which is known as the
maximum modulus principle.
Theorem. If a function f is analytic and not constant in a given domain D, then
lf(z)l has no maximum value in D. That is, there is no point zo in the domain such
that If(z)I < If(zo)Ifor all points z in it.
Given that f is analytic in D, we shall prove the theorem by assuming that lf(z)l
does have a maximum value at some point zo in D and then showing that f (z) must
be constant throughout D.
The general approach here is similar to that taken in the proof of the lemma in
Sec. 26. We draw a polygonal line L lying in D and extending from zo to any other
point P in D. Also, d represents the shortest distance from points on L to the boundary
of D. When Dis the entire plane, d may have any positive value. Next, we observe
that there is a finite sequence of points
along L such that Zn coincides with the point P and
(k= 1, 2, ... ,n).
170 INTEGRALS CHAP. 4
---......
----
/' _....... __--<..--~ ' / / "',.,.- ...... ' ' L N ,. '" ' ' N ' n-1/ N '/ N. 'N1 ' 2 ' ' I n I 0
I I p
I I I I
Zo Z1 I Z2l I Zn I
I I I II
'
I
I
I
' ' I
, I /
.... ' /
/ / /
--...... /
' / ...e. __ -
---""" FIGURE69' ------
On forming a finite sequence of neighborhoods (Fig. 69)
where each Nk has center Zk and radius d, we see that f is analytic in each of these
neighborhoods, which are all contained in D, and that the center of each neighborhood
Nk (k = 1, 2, ... , n) lies in the neighborhood Nk-I·
Since If(z) I was assumed to have a maximum value in D at z0, it also has a
maximum value in N0 at that point. Hence, according to the preceding lemma, j(z)
has the constant value f(zo) throughout N0. In particular, f(z1) = j(z0). This means
that 1/(z)l < 1/(z1)1 for each point z in N1; and the lemma can be applied again, this
time telling us that
f(z) = f(zl) = f(zo)
when z is in N1. Since z2 is in Nh then, j(z2) = j(zo). Hence 1/(z)l < l/(z2)1 when
z is in N2; and the lemma is once again applicable, showing that
when zis in N2. Continuing in this manner, we eventually reach the neighborhood Nn
and arrive at the fact that f(zn) = /(zo).
Recalling that Zn coincides with the point P, which is any point other than z0 in
D, we may conclude that j(z) = f(zo) for every point z in D. Inasmuch as j(z) has
now been shown to be constant throughout D, the theorem is proved.
If a function f that is analytic at each point in the interior of a closed bounded
region R is also continuous throughout R, then the modulus 1/(z)l has a maximum
value somewhere in R (Sec. 17). That is, there exists a nonnegative constant M such
that 1/(z)l < M for all points z in R, and equality holds for at least one such point.
Iff is a constant function, then 1/(z)l = M for all z in R. If, however, /(z) is not
constant, then, according to the maximum modulus principle, 1/(z)l ::j:. M for any
point z in the interior of R. We thus arrive at an important corollary of the maximum
modulus principle.
SEC. 50 EXERCISES 171
Corollary. Suppose that a function f is continuous on a closed bounded region R
and that it is analytic and not constant in the interior of R. Then the maximum value
oflf(z)l in R, which is always reached, occurs somewhere on the boundary ofRand
never in the interior.
EXAMPLE. Let R denote the rectangular region 0 < x < rr, 0 < y < 1. The corol-
lary tells us that the modulus of the entire function f (z) = sin z has a maximum value
in R that occurs somewhere on the boundary, and not in the interior, of R. This can be
verified directly by writing (see Sec. 33)
lf(z)l =Jsin2
x +sinh2
y
and noting that, in R, the term sin2
xis greatest when x = rr/2 and that the increasing
function sinh2
y is greatest when y = 1. Thus the maximum value of lf(z)l in R occurs
at the boundary point z= (rr/2, 1) and at no other point in R (Fig. 70).
y
~~------~------~
(n'/2,1)
0 X
FIGURE70
When the function fin the corollary is written f(z) = u(x, y) + iv(x, y), the
component function u(x, y) also has a maximum value in R which is assumed on
the boundary of R and never in the interior, where it is harmonic (Sec. 25). For the
composite function g(z) = exp[f(z)l is continuous in Rand analytic and not constant
in the interior. Consequently, its modulus lg(z)l = exp[u(x, y)], which is continuous
in R, must assume its maximum value in Ron the boundary. Because of the increasing
nature of the exponential function, it follows that the maximum value of u (x, y) also
occurs on the boundary.
Properties of minimum values of If (z)Iand u(x, y) are treated in the exercises.
EXERCISES
1. Let f be an entire function such that 1/(z)l < Aizl for all z, where A is a fixed positive
number. Show that f(z) =a1z, where a1 is a complex constant.
Suggestion: Use Cauchy's inequality (Sec. 49) to show that the second derivative
f''(z) is zero everywhere in the plane. Note that the constant M R in Cauchy's inequality
is less than or equal to A(izol + R).
172 INTEGRALS CHAP. 4
2. Suppose that f (Z) is entire and that the harmonic function u(x, y) = Re[f (z)] has an
upper bound u0; that is, u(x, y) < u0 for all points (x, y) in the xy plane. Show that
u (x, y) must be constant throughout the plane.
Suggestion: Apply Liouville's theorem (Sec. 49) to the function g(z) =exp[j(z)].
3. Show that, for R sufficiently large, the polynomial P(z) in Theorem 2, Sec. 49, satisfies
the inequality
IP(z)l < 21anllzln whenever lzl > R.
[Compare the first of inequalities (5), Sec. 49.]
Suggestion: Observe that there is a positive number R such that the modulus of
each quotient in expression (4), Sec. 49, is less than lanlfn when lzl > R.
4. Let a function f be continuous in a closed bounded region R, and let it be analytic and
not constant throughout the interior of R. Assuming that f(z) :f:. 0 anywhere in R, prove
that lf(z)l has a minimum value min R which occurs on the boundary of Rand never
in the interior. Do this by applying the corresponding result for maximum values (Sec.
50) to the function g(z) = 1/f(z).
5. Use the function f(z) =z to show that in Exercise 4 the condition f(z) :f:. 0 anywhere
in R is necessary in order to obtain the result of that exercise. That is, show that 1f (Z) I
can reach its minimum value at an interior point when that minimum value is zero.
6. Consider the function f(z) =(z + 1)2 and the closed triangular region R with vertices
at the points z =0, z= 2, and z = i. Find points in R where lf(z)l has its maximum and
minimum values, thus illustrating results in Sec. 50 and Exercise 4.
Suggestion: Interpret If(z) Ias the square of the distance between z and -1.
Ans. z =2, z = 0.
7. Let f(z) = u(x, y) +i v(x, y) be a function that is continuous on a closed bounded region
R and analytic and not constant throughout the interior of R. Prove that the component
function u (x, y) has a minimum value in R which occurs on the boundary of R and never
in the interior. (See Exercise 4.)
8. Let f be the function f(z) = ez and R the rectangular region 0 < x < 1, 0 < y < 1r.
Illustrate results in Sec. 50 and Exercise 7 by finding points in R where the component
function u(x, y) = Re[f(z)] reaches its maximum and minimum values.
Ans. z = I, z = 1 + rri.
9. Let the function f(z) = u(x, y) + iv(x, y) be continuous on a closed bounded region
R, and suppose that it is analytic and not constant in the interior of R. Show that the
component function v(x, y) has maximum and minimum values in R which are reached
on the boundary of R and never in the interior, where it is harmonic.
Suggestion: Apply results in Sec. 50 and Exercise 7 to the function g(z) = -if(z).
10. Let z0 be a zero of the polynomial
SEC. 50
of degree n (n > 1). Show in the following way that
P(z) =(z- z0)Q(z},
where Q(z) is a polynomial of degree n- 1.
(a) Verify that
i- z~ =(z- zo)(zk-
1
+ zk-
2
z0 + · · ·+ zz~-2
+ z~- 1
)
(b) Use the factorization in part (a) to show that
P(z)- P(z0) =(z- z0)Q(z),
EXERCISES 173
(k=2,3, ...).
where Q(z) is a polynomial of degree n - 1, and deduce the desired result from this.
CHAPTER
5
SERIES
This chapter is devoted mainly to series representations of analytic functions. We
present theorems that guarantee the existence of such representations, and we develop
some facility in manipulating series.
51. CONVERGENCE OF SEQUENCES
An infinite sequence
(1)
of complex numbers has a limit zif, for each positive number 8, there exists a positive
integer n0 such that
(2) lzn- zl < 8 whenever n > n0 .
Geometrically, this means that, for sufficiently large values of n, the points Zn lie in
any givens neighborhood of z (Fig. 71). Since we can chooses as small as we please,
it follows that the points Zn become arbitrarily close to z as their subscripts increase.
Note that the value of n0 that is needed will, in general, depend on the value of s.
The sequence (1) can have at most one limit. That is, a limit zis unique if it exists
(Exercise 5, Sec. 52). When that limit exists, the sequence is said to converge to z; and
we write
(3) lim Zn =z.n-+oo
Ifthe sequence has no limit, it diverges.
175
178 SERIES
52. CONVERGENCE OF SERIES
An infinite series
00
(1) LZn = Z1 + Z2 + · · ·+Zn +···
n=l
of complex numbers converges to the sum S if the sequence
N
(2) SN =L Zn =Zt +Z2 + ···+ ZN
n=l
ofpartial sums converges to S; we then write
00
LZn = S.
n=l
(N = 1, 2, ...)
CHAP. 5
Note that, since a sequence can have at most one limit, a series can have at most one
sum. When a series does not converge, we say that it diverges.
Theorem. Suppose that Zn = Xn +iyn (n = 1, 2, ...) and S = X +i Y. Then
(3)
ifand only if
00 00
(4) LXn=X and LYn = Y.
n=l n=l
This theorem tells us, of course, that one can write
00 00 00
L)xn +iyn) = L Xn +i L Yn
n=l n=I n=l
whenever it is known that the two series on the right converge or that the one on the
left does.
To prove the theorem, we first write the partial sums (2) as
(5)
where
N
XN= Lxn and
n=l
N
YN=LYn·
n=l
SEC. 52 CONVERGENCE OF SERIES 179
Now statement (3) is true if and only if
(6)
and, in view of relation (5) and the theorem on sequences in Sec. 51, limit (6) holds if
and only if
(7) lim XN = X and lim YN = Y.
N~oo N~oo
Limits (7) therefore imply statement (3), and conversely. Since XN and YN are the
partial sums ofthe series (4), the theorem here is proved.
By recalling from calculus that the nth term of a convergent series ofreal numbers
approaches zero as n tends to infinity, we can see immediately from the theorems in
this and the previous section that the same is true of a convergent series of complex
numbers. That is, a necessary conditionfor the convergence ofseries (1) is that
(8) lim Zn = 0.
n~oo
The terms of a convergent series of complex numbers are, therefore, bounded. To be
specific, there exists a positive constant M such that IZn I < M for each positive integer
n. (See Exercise 9.)
For another important property of series of complex numbers, we assume that
series (1) is absolutely convergent. That is, when Zn = Xn +iyn, the series
00 00
L lznl = L Jx; + Y;
n=l n=l
of real numbers Jx;+ y; converges. Since
lxnl < Jx~ +Y~ and IYnl < jx?;+ y'j;,
we know from the comparison test in calculus that the two series
00
LIYnl
n=l
must converge. Moreover, since the absolute convergence of a series of real numbers
implies the convergence ofthe series itself, it follows that there are real numbers X and
Y to which series (4) converge. According to the theorem in this section, then, series
(1) converges. Consequently, absolute convergence of a series of complex numbers
implies convergence ofthat series.
In establishing the fact that the sum of a series is a given number S, it is often
convenient to define the remainder PN after N terms:
(9)
180 SERIES CHAP. 5
Thus S = SN + PN; and, since ISN- Sl = lPN- 01, we see that a series converges
to a number S ifand only ifthe sequence ofremainders tends to zero. We shall make
considerable use of this observation in our treatment of power series. They are series
of the form
00
L an(Z- zo)n =ao +a1(z- Zo) +az(z- Zo)
2
+ ···+an(Z- zo)n + ···,
n=O
where zo and the coefficients an are complex constants and z may be any point in a
stated region containing z0. In such series, involving avariable z, we shall denote sums,
partial sums, and remainders by S(z), SN(Z), and PN(z), respectively.
EXAMPLE. With the aid of remainders, it is easy to verify that
00
(10) "zn -- 1 h~ w enever lzl < 1.
1-zn=O
We need only recall the identity (Exercise 10, Sec. 7)
? 1- zn+1
1+ z + z- + ···+zn = - - -
1-z
to write the partial sums
N-1
(z =I= 1)
SN(Z) = L zn = 1+ z +z2
+ ···+zN-l (z =1= 1)
n=O
as
If
1
S(z) = ,
1-z
then,
(z =1= 1).
Thus
and it is clear from this that the remainders PN(z) tend to zero when lzl < 1but not
when lzl > 1. Summation formula (10) is, therefore, established.
SEC. 52 EXERCISES 181
EXERCISES
1. Show in two ways that the sequence
. (-l)n
Zn = -2 +I
2n
(n = 1, 2, ...)
converges to -2.
2. Let rn denote the moduli and en the principal values of the arguments of the complex
numbers Zn in Exercise 1. Show that the sequence rn (n = 1, 2, ...) converges but that
the sequence en (n = 1, 2, ...) does not.
3. Show that
if lim Zn =z, then lim lznl = lzl.
n--+oo n--+oo
4. Write z =rei9
, where 0 < r < 1, in the summation formula that was derived in the
example in Sec. 52. Then, with the aid of the theorem in Sec. 52, show that
oo e 2
"'"" n () r cos - r
L..t r cos n = -------=-
n=l 1- 2r cos e+ r2
and
~ n • e r sin()
L..t r sm n = --------::-
n=l 1 - 2r cos(} + r2
when 0 < r < 1. (Note that these formulas are also valid when r =0.)
5. Show that a limit of a convergent sequence of complex numbers is unique by appealing
to the corresponding result for a sequence of real numbers.
6. Show that
00 00
if L Zn = S, then L Zn =S.
n=l n=l
7. Let c denote any complex number and show that
00 00
if LZn=S, then LCZn=cS.
n=l n=l
8. By recalling the corresponding result for series of real numbers and referring to the
theorem in Sec. 52, show that
if
00
LZn =Sand
n=l
00
then L(Zn + Wn) = S + T.
n=l
9. Let a sequence Zn (n =1, 2, ...) converge to a number z. Show that there exists a positive
number M such that the inequality lznl < M holds for all n. Do this in each of the ways
indicated below.
(a) Note that there is a positive integer n0 such that
lznl = lz +(Zn- z)l < lzl + 1
whenever n > no.
182 SERIES CHAP. 5
(b) Write z, = Xn + iyn and recall from the theory of sequences of real numbers that
the convergence of Xn and Yn (n = 1, 2, ...) implies that lxnl < M1 and IYnl < M2
(n = 1, 2, ...) for some positive numbers M1 and M2.
53. TAYLOR SERIES
We turn now to Taylor's theorem, which is one of the most important results of the
chapter.
Theorem. Suppose that a function f is analytic throughout a disk lz- zol < R0 ,
centered at zo and with radius R0 (Fig. 72). Then f (z) has the power series represen-
tation
00
(1) /(z) = L an(Z- zot (lz - zol < Ro),
n=O
where
(2)
f(n)(zo)
an =.;;_______...;;;_
n!
(n = 0, 1, 2, ...).
That is, series (1) converges to f(z) when z lies in the stated open disk.
y·
....------;' '
/ '/1' • '
/ z / ,
Zo ! I
I
I
0 ' /
' /..... /
"""----~
X
FIGURE72
This is the expansion of f (z) into a Taylor series about the point z0. It is the
familiar Taylor series from calculus, adapted to functions of a complex variable. With
the agreement that
t<O)(zo) = f(zo) and 0! =1,
series (1) can, of course, be written
!'(zo) !"(zo) "
(3) f(z) = J(zo) + 11
(z- zo) + 2
, (z- zo)- + · · · (lz- zol < Ro).
SEC. 53 TAYLOR SERIES 183
Any function which is analytic at a point zo must have a Taylor series about z0•
For, iff is analytic at z0, it is analytic throughout some neighborhood lz- zol < s of
that point (Sec. 23); and s may serve as the value of R0 in the statement of Taylor's
theorem. Also, iff is entire, R0 can be chosen arbitrarily large; and the condition of
validity becomes lz- zol < oo. The series then converges to j(z) at each point z in
the finite plane.
(4)
We first prove the theorem when zo = 0, in which case series (l) becomes
oo J(n)(Q)
f(z) = L 1
zn
0
n.
n=
(lzl < Ro)
and is called a Maclaurin series. The proof when z0 is arbitrary will follow as an
immediate consequence.
To begin the derivation of representation (4), we write lzl =rand let C0 denote
any positively oriented circle lzl = r0, where r < r0 < R0 (see Fig. 73). Since f is
analytic inside and on the circle C0 and since the point z is interior to C0, the Cauchy
integral formula applies:
(5)
y
j(z) =_1__ 1f(s) ds
2m c0 s- z
FIGURE73
Now the factor I/(s - z) in the integrand here can be put in the form
(6)
1 1 1
--=-· '
1- (z/s)s-z s
and we know from the example in Sec. 52 that
N-1 N
1 ='Lzn+_z_
1- Z n=O 1- Z
(7)
SEC. 54 EXAMPLES 185
the disk lzl < R0 ensures the existence of a Maclaurin series representation:
oo (n)(O)
() ''g n
g z = ~ z
n=O n!
(lzl < Ro).
That is,
oo /(n)( )
/( + ) '' Zo n
z zo =~
1
z (lzl < Ro).
n=0
n.
After replacing z by z - zo in this equation and its condition of validity, we have the
desired Taylor series expansion (1).
54. EXAMPLES
When it is known that f is analytic everywhere inside a circle centered at z0, conver-
gence of its Taylor series about zo to f (z) for each point z within that circle is ensured;
no test for the convergence of the series is required. In fact, according to Taylor's theo-
rem, the series converges to f (z) within the circle about z0 whose radius is the distance
from zo to the nearest point z1 where f fails to be analytic. In Sec. 59, we shall find
that this is actually the largest circle centered at z0 such that the series converges to
f (z) for all z interior to it.
Also, in Sec. 60, we shall see that if there are constants an (n = 0, 1, 2 ...) such
that
00
f(z) = L an(Z- Zo)n
n=O
for all points z interior to some circle centered at z0, then the power series here must
be the Taylor series for f about z0, regardless of how those constants arise. This
observation often allows us to find the coefficients an in Taylor series in more efficient
ways than by appealing directly to the formula an= f(n)(zo)/nl in Taylor's theorem.
In the following examples, we use the formula in Taylor's theorem to find the
Maclaurin series expansions of some fairly simple functions, and we emphasize the
use of those expansions in finding other representations. In our examples, we shall
freely use expected properties of convergent series, such as those verified in Exercises
7 and 8, Sec. 52.
EXAMPLE 1. Since the function f(z) = ez is entire, it has a Maclaurin series
representation which is valid for all z. Here f(n)(z) = ez~ and, because f(n)(O) = 1, it
follows that
(1) (lzl < oo).
188 SERIES CHAP. 5
This is, ofcourse, the summation formula that was found in another way in the example
in Sec. 52.
If we substitute -z for z in equation (6) and its condition of validity, and note
that lzl < 1when I- zl < 1. we see that
oc
1 =L:<-l)nzn
1+ z n=O
(lzl < 1).
If, on the other hand, we replace the variable z in equation (6) by 1- z, we have
the Taylor series representation
(z- 11 < 1).
This condition of validity follows from the one associated with expansion (6) since
11- zl < 1is the same as lz - 11 < 1.
EXAMPLE 5. For our final example, let us expand the function
f(z) = 1+2z
2
= ..!_ . 2(1 +z
2
) - 1 =..!_ ( 2 _ 1 )
z3 +z5 z3 1+z2 z3 1+z2
into a series involving powers of z. We cannot find a Maclaurin series for f(z) since
it is not analytic at z = 0. But we do know from expansion (6) that
1 1 2 4 6 8
-----:-= -z +z -z +z -···
1+z2
Clz < 1).
Hence, when 0 < lzl < 1,
f(z) =..!_(2- 1+ z2
- z4
+ z6
- z8
+ ···) =..!_ + l_ z + z3
- z5
+ ···.
z3 z3 z
We call such terms as 1/z3
and 1/z negative powers of z since they can be written z-3
and z-1, respectively. The theory of expansions involving negative powers of z - zo
will be discussed in the next section.
EXERCISES*
1. Obtain the Maclaurin series representation
00 4n+l
zcosh(z2
) :t:. L _z-
n=O (2n)!
(lzl < oo).
*In these and subsequent exercises on series expansions, it is recommended that the reader use, when
possible, representations (1) through (6) in Sec. 54.
SEC. 55
the domain, f(z) has the series representation
(1)
where
(2)
and
(3)
y
I
I
I
I
I
I
I
I
1 1 f(z) dz
an= 2rri c (z- z0
)n+1
bn = _1_·1 f(z) dz
21Tl c (z- Zo)-n+l
I
/
I
I
I
I
I
I
I
I
X
LAURENT SERIES 191
(n = 0, 1, 2, ...)
(n = 0, 1, 2, ...).
FIGURE 74
Expansion (1) is often written
(4)
where
(5)
00
f(z) = L cn(Z- Zo)n
n==-oo
1 1 f(z) dz
Cn = 2rri C (z- zo)n+l
(n = 0, ±1, ±2, ...).
In either of the forms (1) or (4), it is called a Laurent series.
Observe that the integrand in expression (3) can be written f (z) (z - zot-1. Thus
it is clear that when f is actually analytic throughout the disk lz Zol < R2, this
integrand is too. Hence all of the coefficients bn are zero; and, because (Sec. 48)
_1_1 f(z) dz = J<n)(zo)
2rri c (z- z0)n+l nl
(n = 0, 1, 2, ...),
192 SERIES CHAP. 5
expansion (1) reduces to a Taylor series about z0.
If, however, f fails to be analytic at zo but is otherwise analytic in the disk
lz - zol < R2, the radius R1 can be chosen arbitrarily small. Representation (1) is
then valid in the punctured disk 0 < lz- z0 1< R2. Similarly, iff is analytic at each
point in the finite plane exterior to the circle Iz - zoI= R1, the condition of validity is
R1 < lz- zol < oo. Observe that iff is analytic everywhere in the finite plane except
at z0, series (1) is valid at each point of analyticity, or when 0 < lz - z0 1< oo.
We shall prove Laurent's theorem first when zo = 0, in which case the annulus is
centered at the origin. The verification of the theorem when zo is arbitrary will follow
readily.
We start the proof by forming a closed annular region r1 < lzl < r2 that is con-
tained in the domain R1 < lzl < R2 and whose interior contains both the point z and
the contour C (Fig. 75). We let C1 and C2 denote the circles lzl =rt and lzl =r2, re-
spectively, and we assign those two circles a positive orientation. Observe that f is
analytic on C1 and C2, as well as in the annular domain between them.
Next, we construct a positively oriented circle y with center at zand small enough
to be completely contained in theinterioroftheannularregionr1 < lzl < r2, as shown in
Fig. 75. It then follows from the extension ofthe Cauchy-Goursat theorem to integrals
of analytic functions around the oriented boundaries of multiply connected domains
(Theorem 2, Sec. 46) that
{ f(s) ~s _ { f(s) ds -1 f(s) ds = O.
lc2
s- 4 lc1 s- z y s- z
y
FIGURE75
194 SERIES CHAP. 5
These limits are readily established by a method already used in the proof of Taylor's
theorem in Sec. 53. We write lzl = r, so that r1 < r < r2, and let M denote the
maximum value of lf(s)l on C1 and C2. We also note that if sis a point on C2, then
Is- zl > r2 - r; and if sis on C1o lz- sl > r- r 1• This enables us to write
Since (rIr2) < 1 and (rtf r) < 1, it is now clear that both PN(z) and aN(Z) have the
desired property.
Finally, we need only recall Corollary 2 in Sec. 46 to see that the contours used in
integrals (10) may be replaced by the contour C. This completes the proof ofLaurent's
theorem when zo = 0 since, if z is used instead of s as the variable of integration,
expressions (10) for the coefficients an and bn are the same as expressions (2) and (3)
when zo = 0 there.
To extend the proof to the general case in which z0 is an arbitrary point in the
finite plane, we let f be a function satisfying the conditions in the theorem; and, just
as we did in the proof of Taylor's theorem, we write g(z) = f(z +z0). Since f(z) is
analytic in the annulus R1 < !z- zol < R2, the function f(z + zo) is analytic when
R1 < l(z +z0)- zol < R2. That is, g is analytic in the annulus R1 < lzl < R2, which is
centered at the origin. Now the simple closed contour C in the statement ofthe theorem
has some parametric representation z =z(t) (a < t <b), where
(12)
for all t in the interval a < t < b. Hence if r denotes the path
(13) z = z(t)- zo
r is not only a simple closed contour but, in view of inequalities (12), it lies in the
domain R1 < lzl < R2• Consequently, g(z) has a Laurent series representation
(14)
where
(15)
(16) bn = _I_ { g(z) dz
2Jri Jr z-n-;-I
(n =0, 1, 2, ...),
(n = 1, 2, ...).
Representation (1) is obtained if we write f(z +z0) instead of g(z) in equation
(14) and then replace zby z - zo in the resulting equation, as well as in the condition of
validity R1 < lzl < R2. Expression (15) for the coefficients an is, moreover, the same
SEC.j6 EXAMPLES 195
as expression (2) since
rg(z) dz =1b J[z(t)]z'(t) dt = r f(z) dz .
Jr zn+l a [z(t) - zo]n+l Jc (z- zo)n+l
Similarly, the coefficients bn in expression (16) are the same as those in expres-
sion (3).
56. EXAMPLES
The coefficients in a Laurent series are generally found by means other than by
appealing directly to their integral representations. This is illustrated in the examples
below, where it is always assumed that, when the annular domain is specified, a Laurent
series for a given function in unique. As was the case with Taylor series, we defer the
proof of such uniqueness until Sec. 60.
EXAMPLE 1. Replacing z by 1/z in the Maclaurin series expansion
00 zn z z2 z3
ez = L - = 1+ - + - + - + ...
n=O n! 1! 2! 3!
(lzl < oo),
we have the Laurent series representation
00
1 1 1 1
ellz = L =1+ - + - +- +...
n!zn l!z 2!z2 3!z3
n=O
(0 < lzl < oo}.
Note that no positive powers of z appear here, the coefficients of the positive
powers being zero. Note, too, that the coefficient of 1/z is unity; and, according to
Laurent's theorem in Sec. 55, that coefficient is the number
b1 = -
1
-. relfz dz,
21fl lc
where C is any positively oriented simple closed contour around the origin. Since
b1 = 1, then,
Lelfz dz = 2ni.
This method of evaluating certain integrals around simple closed contours will be
developed in considerable detail in Chap. 6.
sEc.s6 ExERCISES 199
7. Write the two Laurent series in powers of z that represent the function
1
/(z) = z(l +z2)
in certain domains, and specify those domains.
00 1
Ans. L(-1)n+lz2n+l +- (0 < lzl < 1);
n=O Z
oo (-l)n+l
"' (1 < lzl < oo).~ z2n+I
n=l
8. (a) Let a denote a real number, where -1 <a < 1, and derive the Laurent series
representation
(Ia I < lzl < oo).
(b) Write z = eiO in the equation obtained in part (a) and then equate real parts and
imaginary parts on each side of the result to derive the summation formulas
00 () 2
"'n () acos -a~a cos n =-------=-
n=I 1- 2a cos() +a 2
and
00 • ()
"n· () asm~a smn = ,
n=I 1- 2a cos()+ a2
where -1 <a < 1. (Compare Exercise 4, Sec. 52.)
9. Suppose that a series
00
L x[n]z-n
n=-oo
converges to an analytic function X(z) in some annulus R1 < lzl < R2• That sum X(z)
is called the z-transform of x[n1(n =0, ±1, ±2, ...).*Use expression (5), Sec. 55, for
the coefficients in a Laurent series to show that if the annulus contains the unit circle
lzl = 1, then the inverse z-transform of X(z) can be written
x[n]=- X(e1
)em dB1 JT( ·e · e
21l" -T(
(n =0, ±1, ±2, ...).
10. (a) Let z be any complex number, and let C denote the unit circle
w =it/J
in the w plane. Then use that contour in expression (5), Sec. 55, for the coefficients
in a Laurent series, adapted to such series about the origin in the w plane, to show
*The z-transfonn arises in studies of discrete-time linear systems. See, for instance, the book by
Oppenheim, Schafer, and Buck that is listed in Appendix 1.
200 SERIES CHAP. 5
that
(0 < lwl < oo).
where
ln(Z)=-1 frr exp[-i(n<J>-zsin4>)]d4>
2Jr -7r
(n =0, ±1, ±2, ...).
(b) With the aid of Exercise 6, Sec. 37, regarding certain definite integrals of even and
odd complex-valued functions of a real variable, show that the coefficients in part
(a) can be written*
1 !orrln(Z) = - cos(n4>- zsin 4>) d4>
1r 0
(n =0, ±1, ±2, ...).
11. (a) Let f (z) denote a function which is analytic in some annular domain about the origin
that includes the unit circle z =eilf> (-Jr < 4> < 1r ). By taking that circle as the path
of integration in expressions (2) and (3), Sec. 55, for the coefficients an and bn in a
Laurent series in powers of z, show that
j(z) = _1 !Jr f(eiif>) d4> + _1 t frr f(ei!/J) ~( ~t/J)n + (e~if>)n] d4>
21f -rr 21r n=l -rr L' e ~
when z is any point in the annular domain.
(b) Write u(B) =Re[f(ei0)], and show how it follows from the expansion in part (a)
that
1 11C 1 00 17[u(B) =- u(4>) d4> +- L u(4>) cos[n(B- 4>)]d4>.
21f -rr 1r n=l -rr
This is one form of the Fourier series expansion of the real-valued function u(e) on
the interval-Jr < e< 1r. The restriction on u(e) is more severe than is necessary in
order for it to be represented by a Fourier series.t
57. ABSOLUTE AND UNIFORM CONVERGENCE
OF POWER SERIES
This section and the three following it are devoted mainly to various properties of
power series. A reader who wishes to simply accept the theorems and any corollaries
there can easily skip their proofs in order to reach Sec. 61 more quickly.
*These coefficients ln(Z) are called Bessel functions of the first kind. They play a prominent role in
certain areas of applied mathematics. See, for example, the authors' "Fourier Series and Boundary
Value Problems," 6th ed., Chap. 8, 2001.
t For other sufficient conditions, see Sees. 31 and 32 of the book cited in the footnote to Exercise 10.
SEC. 57 ABSOLUTE AND UNIFORM CONVERGENCE OF POWER SERIES 201
We recall from Sec. 52 that a series of complex numbers converges absolutely
if the series of absolute values of those numbers converges. The following theorem
concerns the absolute convergence of power series.
Theorem I. Ifa power series
(1)
converges when z =z1 (z1 f= z0), then it is absolutely convergent at each point z in the
open disk iz- zol < R1, where Rt =lzt- zol (Fig. 77).
y
----"' ..../ '
/ •z 'I
I ~Z1
( ~I l
Zo I
I
I
' /
' ..,"'
...... ____
0 X
FIGURE77
We first prove the theorem when zo = 0, and we assume that the series
converges. The terms anz~ are thus bounded; that is,
(n = 0, 1, 2, ...)
forsomepositiveconstant M (see Sec. 52). If lzl < lz11and we let p denote the modulus
lz/z11, we can see that
n
z
lanzn! = ianz71 < Mpn
Z1
(n = 0, 1, 2, ...),
where p < 1. Now the series whose terms are the real numbers Mpn(n =0, 1, 2, ...)
is a geometric series, which converges when p < 1. Hence, by the comparison test for
series of real numbers, the series
202 SERIES CHAP. 5
converges in the open disk lzl < lz11; and the theorem is proved when z0 = 0.
When z0 is any nonzero number, we assume that series (1) converges at z =z1
(z1 i= z0). If we write w = z - z0, series (1) becomes
(2)
00
Lanwn;
n=O
and this series converges at w = z1 - z0. Consequently, since the theorem is known to
be true when zo =0, we see that series (2) is absolutely convergent in the open disk
lwl < lz1 - z0 l. Finally, by replacing w by z- zo in series (2) and this condition of
validity, as well as writing R1 = lz1 - z01, we arrive at the proof of the theorem as it
is stated.
The theorem tells us that the set of all points inside some circle centered at z0
is a region of convergence for the power series (1), provided it converges at some
point other than z0. The greatest circle centered at zo such that series (1) converges at
each point inside is called the circle of convergence of series (1). The series cannot
converge at any point z2 outside that circle, according to the theorem; for if it did, it
would converge everywhere inside the circle centered at zo and passing through z2•
The first circle could not, then, be the circle of convergence.
Our next theorem involves terminology that we must first define. Suppose that
the power series (1) has circle of convergence lz- zoi = R, and let S(z) and SN(z)
represent the sum and partial sums, respectively, of that series:
00 N-1
S(z) =L an(Z- z0)n, SN(Z) = L an(Z- zot (lz- zol < R).
n.=O n=-0
Then write the remainder function
(3) (lz- zol < R).
Since the power series converges for any fixed value of z when lz- zol < R, we
know that the remainder PN(Z) approaches zero for any such z as N tendc; to infinity.
According to definition (2), Sec. 51, of the limit of a sequence, this means that,
corresponding to each positive number e, there is a positive integer N8
such that
(4) IPN(z)l < e whenever N > Ne.
When the choice of Ne depends only on the value of e and is independent of the point
z taken in a specified region within the circle of convergence. the convergence is said
to be uniform in that region
SEC. 57 ABSOLUTE AND UNIFORM CoNVERGENCE oF PowER SERIES 203
Theorem 2. If z1 is a point inside the circle ofconvergence lz - zol = R ofa power
.serzes
00
(5) Lan(Z- zo)n,
n=O
then that series must be uniformly convergent in the closed disk lz- zol < R1, where
R1 = lzt- zol (Fig. 78).
y
,....----..... '/ ....
/
'/
'I
I
I
I Zt 1
I
I II I
I I
I
I
'
0 .....
"' X
FIGURE78.....
--...... --..,...
As in the proof of Theorem 1, we first treat the case in which zo = 0. Given that
z1 is a point lying inside the circle of convergence of the series
(6)
we note that there are points with modulus greater than lz11for which it converges.
According to Theorem 1, then, the series
(7)
converges. Letting m and N denote positive integers, where m > N, we can write the
remainders of series (6) and (7) as
(8)
and
(9)
respectively.
m
PN(Z) = lim "' anznm-+oo L...J
n=N
m
aN= lim "' lanz~l,m-+oo L...J
n=N
204 SERIES
Now, in view of Exercise 3, Sec. 52,
m
IPN(Z)i = lim ""anzn ;m->oo l..-J
n=N
and, when lzl < lzd,
m m m m
I: anzn < I: lanllzln < I: lanllz.ln = I: lanz~l·
n=N n=N n=N n=N
Hence
(10) IPN(z)l < CfN when lzl < lzif.
CHAP. 5
Since aN are the remainders of a convergent series, they tend to zero as N tends to
infinity. That is, for each positive number e, an integer N6
exists such that
(11) aN < e whenever N > Ne.
Because ofconditions (10) and (11), then, condition (4) holds for all points zin the disk
lzl < lztf; and the value of Ne is independent of the choice of z. Hence the convergence
of series (6) is uniform in that disk.
The extension of the proof to the case in which zo is arbitrary is, of course,
accomplished by writing w = z - z0 in series (5). For then the hypothesis of the
theorem is that z1 - z0 is a point inside the circle of convergence lwl =R ofthe series
Since we know that this series converges uniformly in the disk fwl < lz1 - z0 1, the
conclusion in the statement of the theorem is evident.
58. CONTINUITY OF SUMS OF POWER SERIES
Our next theorem is an important consequence of uniform convergence, discussed in
the previous section.
Theorem. A power series
00
(1) Lan(Z- zo)n
n=O
represents a continuous function S(z) at each point inside its circle of convergence
lz- zol = R.
sEc.s8 CONTINUITY OF SUMS OF POWER SERIES 205
Another way to state this theorem is to say that if S(z) denotes the sum of series
(1) within its circle of convergence Iz - zoI= R and if z1 is a point inside that circle,
then, for each positive number s, there is a positive number 8 such that
(2) IS(z)- S(z1)1 < 8 whenever lz- z11< 8,
the number 8 being small enough so that zlies in the domain of definition lz - zol < R
of S(z). [See definition (4), Sec. 17, of continuity.]
To show this, we let SN(Z) denote the sum of the first N terms of series (1) and
write the remainder function
(lz- zol < R).
Then, because
(lz- zol < R),
one can see that
or
If z is any point lying in some closed disk lz- zol < R0 whose radius R0 is greater
than iz1 - zol but less than the radius R of the circle of convergence of series (1) (see
Fig. 79), the uniform convergence stated in Theorem 2, Sec. 57, ensures that there is
a positive integer Ne such that
8
IPN(z)l <- whenever N > Ne.
3
(4)
In particular, condition (4) holds for each point z in some neighborhood lz- z11 < 8
of z1 that is small enough to be contained in the disk lz- zol < R0.
y
I
0 X
----- FIGURE79
206 SERIES CHAP. 5
Now the partial sum SN(Z) is a polynomial and is, therefore, continuous at z1 for
each value of N. In particular, when N = Ne + 1, we can choose our 8 so small that
(5)
s
)SN(Z)- SN(z1)J <- whenever )z- z1) < 8.
3
By writing N = Ne + 1in inequality (3) and using the fact that statements (4) and (5)
are true when N = Ne + 1, we now find that
s s s
)S(z)- S(z1)1 <- +- +- whenever lz- ztl < 8.
3 3 3
This is statement (2), and the corollary is now established.
By writing w = 1/(z- z0), one can modify the two theorems in the previous
section and the theorem here so as to apply to series of the type
00 b
I: (z-nzo)n.
n=l
(6)
If, for instance, series (6) converges at a point z1(z1 t= z0), the series
00
must converge absolutely to a continuous function when
1
lwl< ·
lzt- zol
(7)
Thus, since inequality (7) is the same as lz- zol > lz1 - zol. series (6) must converge
absolutely to a continuous function in the domain exterior to the circle lz- z01= R1,
where R1 = iz1 - z0 ). Also, we know that if a Laurent series representation
00 00 b
/(z) =L a11 (Z- Zo)n +L n
n=O n=l (z - Zo)n
is valid in an annulus R1 < lz - z0).< R2, then both of the series on the right converge
uniformly in any closed annulus which is concentric to and interior to that region of
validity.
59. INTEGRATION AND DIFFERENTIATION OF
POWER SERIES
We have just seen that a power series
00
(1) S(z) =L an(Z - zot
n=O
SEC. 59 INTEGRATION AND DIFFERENTIATION OF POWER SERIES 207
represents a continuous function at each point interior to its circle of convergence. In
this section, we prove that the sum S(z) is actually analytic within that circle. Our
proof depends on the following theorem, which is of interest in itself.
Theorem I. Let C denote any contour interior to the circle of convergence of the
power series (1), and let g(z) be any function that is continuous on C. The series
formed by multiplying each term of the power series by g(z) can be integrated term
by term over C; that is,
(2)
00
1g(z)S(z) dz = L an 1g(z)(z - zo)n dz.
C n=O C
To prove this theorem, we note that since both g(z) and the sum S(z) ofthe power
series are continuous on C, the integral over C of the product
N-l
g(z)S(z) = L an g(z)(z- Zo)n +g(z)PN(Z),
n=O
where PN(z) is the remainder of the given series after N terms, exists. The terms of
the finite sum here are also continuous on the contour C, and so their integrals over
C exist. Consequently, the integral of the quantity g(z)PN(z) must exist; and we may
write
N-1
(3) rg(z)S(z) dz = L an rg(z)(z- zo)n dz + rg(z)PN(Z) dz.
Jc n=O Jc Jc
Now let M be the maximum value of lg(z)l on C, and let L denote the length of
C. In view of the uniform convergence of the given power series (Sec. 57), we know
that for each positive numbers there exists a positive integer N8 such that, for all points
z on C,
IPN(z)l < s whenever N > N8 •
Since N8 is independent of z, we find that
fc g(z)PN(Z) dz <MeL whenever N > N8 ;
that is,
lim { g(z)PN(Z) dz = 0.
N-+oo lc
208 SERIES CHAP. 5
It follows, therefore, from equation (3) that
N-1
1g(z)S(z) dz = lim L an 1g(z)(z- zot dz.
C N-+oo C
n=O
This is the same as equation (2), and Theorem 1 is proved.
If lg(z) I = 1 for each value of z in the open disk bounded by the circle of
convergence of power series (1), the fact that (z - zo)n is entire when n = 0, 1, 2, ...
ensures that
Lg(z)(z - zo)n dz = L(z - z0)n dz =0 (n = 0, 1, 2, ...)
for every closed contour C lying in that domain. According to equation (2), then,
LS(z)dz =0
for every such contour; and, by Morera's theorem (Sec. 48), the function S(z) is
analytic throughout the domain. We state this result as a corollary.
Corollary. The sum S(z) ofpower series (1) is analytic at each point z interior to the
circle ofconvergence ofthat series.
This corollary is often helpful in establishing the analyticity of functions and in
evaluating limits.
EXAMPLE 1. To illustrate, let us show that the function defined by the equations
f(z)= { (sinz)/z whenz f.O,
1 whenz =0
is entire. Since the Maclaurin series expansion
00
7 2n+l
sinz = L(-l)n-""__
n=O (2n + 1)!
represents sin z for every value of z, the series
00
z2n z2 ,.4
L(-1)
11
= 1-- + ::.._- ···
n=O (2n + 1)! 3! 5!
(4)
obtained by dividing each term of that Maclaurin series by z, converges to f(z) when
z :f= 0. But series (4) clearly converges to f(O) when z = 0. Hence f(z) is represented
by the convergent power series (4) for all z: and f is, therefore, an entire function.
SEC. 59 INTEGRATION AND DIFFERENTIATION OF POWER SERIES 209
Note that, since f is continuous at z =0 and since (sin z)/z = f(z) when z =F 0,
(5) lim sin z =lim f(z) = f(O) = 1.
z-+0 z z-+0
This is a result known beforehand because the limit here is the definition of the
derivative of sin zat z =0.
We observed at the beginning of Sec. 54 that the Taylor series for a function f
about a point zo converges to f(z) at each point z interior to the circle centered at zo
and passing through the nearest point z1 where f fails to be analytic. In view of the
above corollary, we now know that there is no larger circle about zo such that at each
point z interior to it the Taylor series converges to f (z). For if there were such a circle,
f would be analytic at z1; but f is not analytic at z1.
We now present a companion to Theorem 1.
Theorem 2. The power series (1) can be differentiated term by term. That is, at each
point z interior to the circle ofconvergence ofthat series,
(6)
00
S'(z) = Lnan(Z- zo)n-l.
n=l
To prove this, let zdenote any point interior to the circle of convergence of series
(1); and let C be some positively oriented simple closed contour surrounding z and
interior to that circle. Also, define the function
(7)
1 1
g(s) = -2.. ( )2
:rrz s -z
at each points on C. Since g(s) is continuous on C, Theorem 1 tells us that
(8)
00
( g(s)S(s) ds = L an ( g(s)(s- Zo)n ds.
~ n=O ~
Now S(s) is analytic inside and on C, and this enables us to write
( g(s)S(s) ds = ~ f S(s) ds = S'(z)
lc 2:rn lc (s- z)2
with the aid of the integral representation for derivatives in Sec. 48. Furthermore,
1 n d 1 1(s - Zo)n d d ng(s)(s- zo) s = -. s = -(z- zo)
c 2:rn c (s - z)2 dz
(n = 0, 1, 2, ...).
210 SERIES CHAP. 5
Thus equation (8) reduces to
which is the same as equation (6). This completes the proof.
EXAMPLE 2. In Example 4, Sec. 54, we saw that
00
1 "'' n n
-=~(-1) (z-1)
z n=O
(lz- 11 < 1).
Differentiation of each side of this equation reveals that
00
1 L n n-1
-- = (-1) n(z- 1)
z2
n=l
(lz- II< 1),
or
(iz- 11 < 1).
60. UNIQUENESS OF SERIES REPRESENTATIONS
The uniqueness of Taylor and Laurent series representations, anticipated in Sees. 54
and 56, respectively, follows readily from Theorem 1 in Sec. 59. We consider first the
uniqueness of Taylor series representations.
Theorem I. Ifa series
00
(1) L an(Z- zo)n
n=O
converges to f (z) at allpoints interior to some circle lz - zol = R, then it is the Taylor
series expansion for f in powers of z - z0.
To prove this, we write the series representation
00
(2) /(z) =L an(Z- zot (lz- z0 1 < R)
n=O
in the hypothesis of the theorem using the index of summation m:
00
f(z) = L arn(Z- zo)rn (lz- zol < R).
m=O
SEC.60 UNIQUENESS OF SERIES REPRESENTATIONS 211
Then, by appealing to Theorem 1 in Sec. 59, we may write
(3)
00
[ g(z)f(z) dz =Lam [ g(z)(z- zo)m dz,
C m=O C
where g(z) is any one of the functions
(4)
1 1
g(z) =- .---.2:rri (z-zo)n+I
(n =0, 1, 2, ...)
and C is some circle centered at zo and with radius less than R.
In view of the generalized form (5), Sec. 48, of the Cauchy integral formula (see
also the corollary in Sec. 59), we find that
(5) { g(z)f(z) dz = _1_ { f(z) dz = f(n)(zo);
lc 2:rri lc (z- z0)n+l n!
and, since (see Exercise 10, Sec. 40)
(6) z z-z dz=- =[
m 1 [ dz { 0
c g( .)( o) 2:rri c (z- zo)n-m+l 1
it is clear that
(7)
Because of equations (5) and (7), equation (3) now reduces to
when m =I= n,
whenm =n,
and this shows that series (2) is, in fact, the Taylor series for f about the point z0.
Note how itfollows from Theorem 1that if series (1) converges to zero throughout
some neighborhood of z0, then the coefficients an must all be zero.
Our second theorem here concerns the uniqueness of Laurent series representa-
tions.
Theorem 2. Ifa series
(8)
converges to f (z) at allpoints in some annular domain about z0, then it is the Laurent
series expansion for f in powers ofz - z0 for that domain.
212 SERIES CHAP. 5
The method of proof here is similar to the one used in proving Theorem 1. The
hypothesis of this theorem tells us that there is an annular domain about zo such that
00
J(z) = L Cn(Z- Zo)n
n=-oo
for each point z in it. Let g(z) be as defined by equation (4), but now allow n to be
a negative integer too. Also, let C be any circle around the annulus, centered at z0
and taken in the positive sense. Then, using the index of summation m and adapting
Theorem 1 in Sec. 59 to series involving both nonnegative and negative powers of
z - z0 (Exercise 10), write
00
1g(z)J(z) dz = L em 1g(z)(z- zo)m dz,
C m=-oo C
or
(9)
00
1 1 J(z)dz "" 1 m-
2
. ( )n+l = ~ em g(z)(z - zo) dz.
1r l C Z - Zo m=-oo C
Since equations (6) are also valid when the integers m and n are allowed to be
negative, equation (9) reduces to
1 r f(z)dz
2ni Jc (z- z0
)n+I =en,
which is expression (5), Sec. 55, for coefficients in the Laurent series for f in the
annulus.
EXERCISES
1. By differentiating the Maclaurin series representation
(lzl < 1),
obtain the expansions
1 00
---.,.2 = L(n+ l)zn
(1- z) n=O
(lzl < 1)
and
2 00
---.,.
3
= L(n + l)(n + 2)zn
(1- z) n=O
(lzl < 1).
SEC.6I MULTIPLICATION AND DIVISION OF POWER SERIES 215
Conclude from these results that if
then
00
1g(z)S(z) dz = L en 1g(z)(z- zo)n dz.
C n=-oo C
11. Show that the function
l
fz(z) = z2 + 1 (z f: ± i)
is the analytic continuation (Sec. 26) of the function
00
ft(z) =L(-l)nz2n (lzl < l)
n=O
into the domain consisting of all points in the z plane except z=± i.
12. Show that the function f2(z) = 1/z2 (z f: 0) is the analytic continuation (Sec. 26) of the
function
00
fl(Z) =L(n+ l)(z + l)n (lz + 11 < 1)
n=O
into the domain consisting of all points in the z plane except z = 0.
61. MULTIPLICATION AND DIVISION OF POWER SERIES
Suppose that each of the power series
(1)
00
L an (z - Zo)n and
n=O
converges within some circle lz- z01= R. Their sums /(z) and g(z), respectively,
are then analytic functions in the disk iz - z0 I < R (Sec. 59), and the product of those
sums has a Taylor series expansion which is valid there:
00
(2) f(z)g(z) =L Cn(Z- zo)n (lz- zol < R).
n=O
SEC. 61 MULTIPLICATION AND DIVISION OF PoWER SERIES 217
and multiplying these two series term by term. To be precise, we may multiply each
term in the first series by 1, then each term in that series by -z, etc. The following
systematic approach is suggested, where like powers of z are assembled vertically so
that their coefficients can be readily added:
1+z+Izz+Iz3+···
2 6
2 1 3 1 4
z - -z - -z - · · ·
2 6
-z-
The desired result is
(5) (lzl < 1).
Continuing to let f(z) and g(z) denote the sums of series (1), suppose that
g(z) ::/= 0 when lz- zol < R. Since the quotient f(z)/g(z) is analytic throughout the
disk lz - zol < R, it has a Taylor series representation
00
f(z) = ~ d ( _ )n
(z) ~ n Z Zo
g n=O
(6) (lz - zol < R),
where the coefficients dn can be found by differentiating f(z)/g(z) successively and
evaluating the derivatives at z= z0• The results are the same as those found by formally
carrying out the division of the first of series (1) by the second. Since it is usually only
the first few terms that are needed in practice, this method is not difficult.
EXAMPLE 2. As pointed out in Sec. 34, the zeros of the entire function sinh z are
the numbers z =nrri (n =0, ±1, ±2, ...). So the quotient
1 1
z2 sinh z- z2(z +z3j3! + z5j5! + ···)'
which can be written
(7)
z2si~hz = z
13
( 1+z2f3! +
1
z4f5! + ...) '
has a Laurent series representation in the punctured disk 0 < lzI < rr. The denominator
of the fraction in parentheses on the right-hand side of equation (7) is a power series
220 SERIES CHAP. 5
7. Let f(z) be an entire function that is represented by a series of the form.
2 3
f (z) =z + a2z + a3z + ··· (lzl < oo).
(a) By differentiating the composite function g(z) = f[/(z)] successively, find the first
three nonzero terms in the Maclaurin series for g(z) and thus show that
f[f(z)] = z +2a2i + 2(a~ +a3)z
3
+ ···
(b) Obtain the result in part (a) in aformal manner by writing
(lzl < oo).
replacing f(z) on the right-hand side here by its series representation, and then
collecting tenns in like powers of z.
(c) By applying the result in part (a) to the function f(z) =sin z, show that
. ( . ) 1 3
sm sm z =z - - z + ···
3
(lzl < oo).
8. The Euler numbers are the numbers En (n = 0, 1, 2, ...) in the Maclaurin series repre-
sentation
00
1 LEn n--- -z
cosh z n=O n!
(lzl < rr/2).
Point out why this representation is valid in the indicated disk and why
(n =0, 1, 2, ...).
Then show that
Eo = 1, E2 = -1, E4 = 5, and E6 = -61.
CHAPTER
6
RESIDUES AND POLES
The Cauchy-Goursat theorem (Sec. 44) states that if a function is analytic at all points
interior to and on a simple closed contour C, then the value of the integral of the
function around that contour is zero. If, however, the function fails to be analytic at a
finite number of points interior to C, there is, as we shall see in this chapter, a specific
number, called a residue, which each of those points contributes to the value of the
integral. We develop here the theory of residues; and, in Chap. 7, we shall illustrate
their use in certain areas of applied mathematics.
62. RESIDUES
Recall (Sec. 23) that a point z0 is called a singular point of a function f iff fails to be
analytic at zobut is analytic at some point in every neighborhood ofz0. A singular point
zo is said to be isolated if, in addition, there is a deleted neighborhood 0 < lz- zol < s
of zo throughout which f is analytic.
EXAMPLE 1. The function
has the three isolated singular points z =0 and z=±i.
221
222 RESIDUES AND POLES CHAP. 6
EXAMPLE 2. The origin is a singular point of the principal branch (Sec. 30)
Log z =In r + i8 (r > 0, -TC < 8 < TC)
of the logarithmic function. It is not, however, an isolated singular point since every
deleted s neighborhood of it contains points on the negative real axis (see Fig. 80) and
the branch is not even defined there.
I
I
I
y
-- - ....../ ....
/ '/ '
£
----~-----~----~----1 0
I
' ' ' - -~
/
/
I
I
I
/
X
EXAMPLE 3. The function
FIGURE SO
1
sin(rc/z)
has the singular points z =0 and z = 1jn (n = ±1, ±2, ...), all lying on the segment,
of the real axis from z = -1 to z = 1. Each singular point except z = 0 is isolated. The
singular point z = 0 is not isolated because every deleted s neighborhood ofthe origin
contains other singular points of the function. More precisely, when a positive number
y
..,.,- -...../ ..../ ....
/ "I
1 £
I
----~-----~~--~----
0 I X
I
' I
" /
' /...... __ _..-"""
FIGURE81
SEC.62 RESIDUES 223
s is specified and m is any positive integer such that m > 1/s, the fact that 0 < 1/m < s
means that the point z = ljm lies in the deleted e neighborhood 0 < lzl < s (Fig. 81).
When zo is an isolated singular point of a function f, there is a positive number
R2 such that f is analytic at each point z for which 0 < iz- z0 ! < R2. Consequently,
f(z) is represented by a Laurent series
00
(1)
~ I! bl b2 bl!
j(z) =L.J an(Z - Zo) + + + ... + . + ...
n=O z - Zo (z - zo)2
(z - zo)n
(0 < lz - zol < R2),
where the coefficients an and bn have certain integral representations (Sec. 55). In
patticular,
bn = _1_ { j(z) dz
2rri Jc (z- z0)-n+I
(n=1,2, ...)
where C is any positively oriented simple dosed contour around z0 and lying in the
punctured disk 0 < lz - zol < R2 (Fig. 82). When n = 1, this expression for bn can be
written
(2) Lj(z) dz = 2rrib1•
The complex number bh which is the coefficient of l/(z- zo) in expansion (1), is
called the residue off at the isolated singular point z0. We shall often use the notation
Res f(z),
z=zo
or simply B when the point zo and the function fare clearly indicated, to denote the
residue b1•
y
0
-------/ '/ ....
/ '/ '/ 'I
I
I
I
I
I
I
'
'' /
' /
' /
------
I
I
/
I
I
I
I
X
FIGURE82
SEC.63 CAUCHY'S RESIDUE THEOREM 225
EXAMPLE 5. Let us show that
(5) Lexp(
2
;) dz = 0,
where Cis the unit circle lzI = 1. Since 1/z2
is analytic everywhere except at the origin,
so is the integrand. The isolated singular point z = 0 is interior to C; and, with the aid
of the Maclaurin series (Sec. 54)
z z z2 z3
e =1+-+-+-+···1! 2! 3!
one can write the Laurent series expansion
(lzl < oo),
(0 < lzl < oo).
The residue of the integrand at its isolated singular point z - 0 is, therefore, zero
(b1=0), and the value of integral (5) is established.
We are reminded in this example that, although the analyticity ofa function within
and on a simple closed contour C is a sufficient condition for the value of the integral
around C to be zero, it is not a necessary condition.
63. CAUCHY'S RESIDUE THEOREM
If, except for afinite number of singular points, a function f is analytic inside a simple
closed contour C, those singular points must be isolated (Sec. 62). The following
theorem, which is known as Cauchy's residue theorem, is a precise statement of the
fact that iff is also analytic on C and if Cis positively oriented, then the value of the
integral of f around C is 2Jri times the sum of the residues of f at the singular points
inside C.
Theorem. Let C be a simple closed contour, described in the positive sense. If a
function f is analytic inside and on C except for a finite number of singular points
Zk (k = 1, 2, ... , n) inside C, then
(1)
n
rf(z) dz = 2ni L Res f(z).
Jc z=zk
k=l
To prove the theorem, let the points Zk (k = 1, 2, ... , n) be centers of positively
oriented circles Ck which are interior to C and are so small that no two of them have
points in common (Fig. 84). The circles Ck• together with the simple closed contour C,
form the boundary ofa closed region throughout which f is analytic and whose interior
SEC.65 THE THREE TYPES oF IsoLATED SINGULAR PoiNTs 231
65. THE THREE TYPES OF ISOLATED SINGULAR POINTS
We saw in Sec. 62 that the theory of residues is based on the fact that if f has an
isolated singular point z0, then f(z) can be represented by a Laurent series
(1)
in a punctured disk 0 < lz- zol < R2. The portion
of the series, involving negative powers of z - z0, is called the principal part of f at
z0. We now use the principal part to identify the isolated singular point zo as one of
three special types. This classification will aid us in the development of residue theory
that appears in following sections.
If the principal part off at zo contains at least one nonzero term but the number
of such terms is finite, then there exists a positive integer m such that
bm f= 0 and bm+1 = bm+2 = · · · = 0.
That is, expansion (1) takes the form
where b111 =I= 0. In this case, the isolated singular point zo is called a pole oforder m.*
A pole of order m = 1is usually referred to as a simple pole.
EXAMPLE 1. Observe that the function
,
z"' - 2z + 3 z(z - 2) + 3 3
2
-
----= =z+ = +(z
z-2 z-2 z 2
2 3
J+z-2
(0 < lz- 21 < oo)
has a simple pole (m = I) at z0 =2. Its residue b1 there is 3.
• Reasons for the terminology pole are suggested on p. 70 of the book by R. P. Boas mentioned in the
footnote in Sec. 64.
232 RESIDUES AND POLES CHAP. 6
EXAMPLE 2. The function
sinh z = _.!._ (z + z3
+ z
5
_ + z
7
+ .. ·) = _.!._ + _!_ . I + _£ + z
3
+ ...
z4 z4 3! 5! 7! z3 3! z 5! 7!
(0 < lzl < oo)
has a pole of order m = 3 at zo = 0, with residue b1= l/6.
There remain two extremes, the case in which all of the coefficients in the
principal part are zero and the one in which an infinite number of them are nonzero.
When all of the bn's are zero, so that
00
(3) f(z) = L an(Z- zo>" = ao + ai(Z- Zo) +az(z- Zo)
2
+ ...
n=O
(0 < lz - zol < Rz),
the point zo is known as a removable singular point. Note that the residue at a remov-
able singular point is always zero. If we define, or possibly redefine, f at zo so that
f(z0) = a0, expansion (3) becomes valid throughout the entire disk lz - zol < R2.
Since a power series always represents an analytic function interior to its circle of
convergence (Sec. 59), it follows that f is analytic at z0 when it is assigned the value
a0 there. The singularity at zo is, therefore, removed.
EXAMPLE 3. The point zo = 0 is a removable singular point of the function
1- cos z 1 [ (· z
2
z
4
z
6
)]
f(z) = =- 1- 1-- +-- -. + ...
· z2 z2 2! 4! 6!
1 z2
z4
=---+--···
2! 4! 6!
(0 < lzl < oo).
When the value f(O) = 1/2 is assigned, f becomes entire.
When an infinite number of the coefficients bn in the principal part are nonzero,
zo is said to be an essential singular point of f. An important result concerning the
behavior of a function near an essential singular point is due to Picard. It states that
in each neighborhood ofan essential singular point, a function assumes every finite
value, with one possible exception, an infinite number oftimes.*
*For a proof of Picard's theorem, see Sec. 51 in Vol. III of the book by Markushevich, cited in
Appendix 1.
SEC.65 EXERCISES 233
EXAMPLE 4. The function
exp(I) = f -1
. _l = 1+ _!_ . I + _!_ . __!_
Z n=On! zn 1! Z 2! z2
(0 < izl < oo)
has an essential singular point at zo =0, where the residue b1is unity. For an illustration
ofPicard's theorem, let us show that exp(l/z) assumes the value -1 an infinite number
of times in each neighborhood of the origin. To do this, we recall from the example in
Sec. 28 that exp z = -1 when z = (2n + l)ni (n = 0, ±1, ±2, ...). This means that
exp(1/z) =-1 when
1 l l
z= ·-=-----
(2n 1)ni t (2n + 1)n
(n = 0, ±1, ±2, ...),
and an infinite number of these points clearly lie in any given neighborhood of the
origin. Since exp(l/z) i= 0 for any value of z, zero is the exceptional value in Picard's
theorem.
In the remaining sections of this chapter, we shall develop in greater depth the
theory of the three types of isolated singular points just described. The emphasis will
be on useful and efficient methods for identifying poles and finding the corresponding
residues.
EXERCISES
1. In each case, write the principal part of the function at its isolated singular point and
determine whether that point is a pole, a removable singular point, or an essential singular
point:
(a) z exp(~}
2
(b) z ;
I+ z
(c) sm z;
z
(d) cos z;
z
1
(e) (2- z)3.
2. Show that the singular point of each of the following functions is a pole. Determine the
order m of that pole and the corresponding residue B.
(a) 1- ~~sh z; (b) 1- ex:c2z); (c) exp(2z) .
" z (z - 1)2
Ans. (a) m =I, B = -1/2; (b) m =3, B =-4/3; (c) m =2, B =2e2.
3. Suppose that a function f is analytic at zo, and write g(z) =f(z)/(z- zo). Show that
(a) if f(zo) i= 0, then zo is a simple pole of g, with residue f(z0);
(b) if f(z0 )- 0, then z0 is a removable singular point of g.
Suggestion: As pointed out in Sec. 53, there is a Taylor series for f(z) about z0
since f is analytic there. Start each part of this exercise by writing out a few terms of
that series.
234 RESIDUES AND POLES
4. Write the function
as
f(z) = </>(z)
(z- ai)3
CHAP. 6
(a> 0)
8a3z2
where </>(z) = 3
.
(z +ai)
Point out why <P (z) has a Taylor series representation about z =ai, and then use it to
show that the principal part of f at that point is
<P"(ai)/2 <P'(ai) <f>(ai)
---+ +--'--'--..:......,.
z - ai (z - ai)2 (z - ai)3
66. RESIDUES AT POLES
ij2
z- az
aj2
(z- ai)2 (z- ai)3 •
When a function f has an isolated singularity at a point z0, the basic method for
identifying zo as a pole and finding the residue there is to write the appropriate Laurent
series and to note the coefficient of 1/(z - z0). The following theorem provides an
alternative characterization of poles and another way of finding the corresponding
residues.
Theorem. An isolated singular point zo ofa function f is a pole oforder m ifand
only if f(z) can be written in the form
(1) f(z) = ¢(z) ,
(z- zo)m
where ¢(z) is analytic and nonzero at z0 . Moreover,
(2) Res f(z) = ¢(zo) if m = 1
z=zo
and
(3) Res f(z) =
¢(m-l)(zo)
if m > 2.
(m- 1)!z=zo
Observe that expression (2) need not have been written separately since, with the
convention that ¢(0)(z0) =¢(z0) and 0! = 1, expression (3) reduces to it when m =1.
SEC. 66 RESIDUES AT POLES 235
To prove the theorem, we first assume that f(z) has the form (1) and recall (Sec.
53) that since ¢(z) is analytic at z0, it has a Taylor series representation
¢(z) =¢(zo) + ¢'(zo) (z- zo) + ¢"(zo) (z
1! 2!
.+.(m-1)( )
)2+ +'f/ Zo ( )m-1
zo · · · z- zo
(m- 1)!
oo .+.(n)( )
+ L '{J
20
(z - zot
n=m n!
in some neighborhood lz- zol < 8 of z0; and from expression (1) it follows that
(
4
) f(z) = ¢(zo) ¢'(zo)/l! + ¢"(zo)/2! + ... + ¢<m-J)(z0)j(m- 1)!
(z- Zo)m (z- zo)m-1 (z- zo)m-2 z- zo
oo .+.(n) ( )
+ '"' 'fJ
20
(z - zot-m
~ n!
n=m
when 0 < lz - zol < 8. This Laurent series representation, together with the fact that
¢(z0) =/:= 0, reveals that zo is, indeed, a pole of order m of f(z). The coefficient of
1j (z - z0) tells us, of course, that the residue of f (z) at zo is as in the statement of the
theorem.
Suppose, on the other hand, that we know only that zo is a pole of order m of f,
or that f (z) has a Laurent series representation
which is valid in a punctured disk 0 < lz- zol < R2. The function ¢(z) defined by
means of the equations
¢(z) = { b:-zo)mf(z)
evidently has the power series representation
00
+ L an (z - Zo)m+n
n=O
when z =/:= z0,
when z = zo
throughout the entire disk lz- zol < R2. Consequently, ¢(z) is analytic in that disk
(Sec. 59) and, in particular, at z0• Inasmuch as ¢(z0) = bm =/:= 0, expression (1) is
established; and the proof of the theorem is complete.
236 RESIDUES AND POLES CHAP. 6
67. EXAMPLES
The following examples serve to illustrate the use of the theorem in the previous
section.
EXAMPLE 1. The function f(z) = (z + 1)/(z2 + 9) has an isolated singular point
at z =3i and can be written as
f(z) = ¢(z)
z - 3i
z+l
where ¢(z) = .
z +3i
Since ¢(z) is analytic at z =3i and ¢(3i) =(3- i)/6 f= 0, that point is a simple pole
of the function f; and the residue there is B1 = (3- i)/6. The point z = -3i is also a
simple pole off, with residue B2 =(3 + i)/6.
EXAMPLE 2. If f(z) = (z3 + 2z)/(z- i)3, then
f(z) = ¢(z) where ,P(z) = z3 + 2z.
(z - i)3
The function ¢(z) is entire, and ¢(i) =i i= 0. Hence f has a pole of order 3 at z =i.
The residue there is
B = ¢"(i) =3i.
2!
The theorem can, of course, be used when branches of multiple-valued functions
are involved.
EXAMPLE 3. Suppose that
where the branch
log z = In r + i() (r > 0, 0 < () < 2n)
of the logarithmic function is to be used. To find the residue off at z = i, we write
f(z) = ¢(z~ (log z)3
where ¢(z) = .
z + iz -l
The function ¢(z) is clearly analytic at z = i; and, since
~(i)= (logi)
3
= (lnl+in/2)
3
=-n
3
-"O
'fJ · 2i 2i 16 r '
the desired residue is B = ¢(i) =-n3/16.
SEC.67 ExAMPLES 237
While the theorem in Sec. 66 can be extremely useful, the identification of an
isolated singular point as a pole of a certain order is sometimes done most efficiently
by appealing directly to a Laurent series.
EXAMPLE 4. If, for instance, the residue of the function
f(z) =sinh z
z4
is needed at the singularity z =0, it would be incorrect to write
j(z) = ,P(z) where ,P(z) =sinh z
z4
and to attempt an application of formula (3) in Sec. 66 with m = 4. For it is necessary
that ¢(z0) i= 0 if that formula is to be used. In this case, the simplest way to find
the residue is to write out a few terms of the Laurent series for f(z), as was done in
Example 2 of Sec. 65. There it was shown that z = 0 is a pole of the third order, with
residue B = 1j6.
In some cases, the series approach can be effectively combined with the theorem
in Sec. 66.
EXAMPLE 5. Since z(ez- 1) is entire and its zeros are
z = 2mri (n = 0, ±1, ±2, ...),
the point z = 0 is clearly an isolated singular point of the function
From the Maclaurin series
1
f(z) = z(ez - 1)
z z2
z3
e2
= 1+ - + - + - + .. ·
1! 2! 3!
we see that
(
z z2
z3
z(e2
- 1) =z - + - + -
ll 2! 3!
Thus
(lzl < oo),
f(z) = ¢(z)
z2
where
1
¢(z)=---~---
1+z/2!+z2/3!+···
(lzl < oo).
SEC.68 ZEROS OF ANALYTIC FUNCTIONS 239
where Cis the circle lzl = 2, described in the positive sense.
Ans. 4JTi.
6. Use the theorem in Sec. 64, involving a single residue, to evaluate the integral of f(z)
around the positively oriented circle lzl =3 when
(a) f(z) (3z +2)2 ; (b) f(z) = z3(1- 3z) ;
z(z - 1)(2z + 5) (1 + z)(l +2z4)
Ans. (a) 9JT i; (b) -3JT i; (c) 2JTi.
68. ZEROS OF ANALYTIC FUNCTIONS
Zeros and poles of functions are closely related. In fact, we shall see in the next section
how zeros can be a source of poles. We need, however, some preliminary results
regarding zeros of analytic functions.
Suppose that a function f is analytic at a point z0 . We know from Sec. 48 that
all of the derivatives f(n)(z) (n = 1, 2, ...) exist at z0. If /(z0) =0 and if there is a
positive integer m such that f(m) (zo) ;f. 0 and each derivative of lower order vanishes
at z0, then f is said to have a zero oforder m at z0. Our first theorem here provides a
useful alternative characterization of zeros of order m.
Theorem 1. A function f that is analytic at a point zo has a zero oforder m there if
and only ifthere is a function g, which is analytic and nonzero at z0, such that
(1) f(z) =(z- zo)mg(z).
Both parts of the proof that follows use the fact (Sec. 53) that if a function is
analytic at a point z0, then it must have a valid Taylor series representation in powers
of z- zo which is valid throughout a neighborhood lz- zol < e of that point.
We start the first part of the proof by assuming that expression (1) holds and
noting that, since g(z) is analytic at z0, it has a Taylor series representation
g'(zo) g"(zo) 2
g(z) = g(zo) + (z - zo) + (z - zo) + ···
1! 2!
in some neighborhood lz- zol < e of z0. Expression (1) thus takes the form
g'(z )
/(z) = g(zo)(z- zo)m + 0
(z- z0)m+l
1!
g"(zo) (z _ zo)m+2 + ...
2!
when lz- zol <e. Since this is actually a Taylor series expansion for f(z), according
to Theorem 1 in Sec. 60, it follows that
(2) /(zo) = /'(zo) = f"(zo) = ···= /(m-l)(z0) =0
and that
(3)
SEC.68 ZEROS OF ANALYTIC FUNCTIONS 241
To prove this, let f be as stated and observe that not all of the derivatives of
f at zo are zero. For, if they were, all of the coefficients in the Taylor series for f
about zo would be zero; and that would mean that f (z) is identically equal to zero in
some neighborhood of z0. So it is clear from the definition of zeros of order m at the
beginning of this section that f must have a zero of some order m at z0• According to
Theorem 1, then,
(4)
where g(z) is analytic and nonzero at z0.
Now g is continuous, in addition to being nonzero, at zo because it is analytic
there. Hence there is some neighborhood lz- zol < c: in which equation (4) holds and
in which g(z) f=. 0 (see Sec. 17). Consequently, f(z) f=. 0 in the deleted neighborhood
0 < lz - z0 1< c:; and the proof is complete.
Our final theorem here concerns functions with zeros that are not all isolated. It
was referred to earlier in Sec. 26 and makes an interesting contrast to Theorem 2 just
above.
Theorem 3. Given a function f and a point z0, suppose that
(i) f is analytic throughout a neighborhood N0 ofz0 ;
(ii) J(z0 ) =0 and J(z) =0 at each point z ofa domain or line segment containing
zo (Fig. 87).
Then f(z) = 0 in N0; that is, f(z) is identically equal to zero throughout N0.
y
----_.. - -'/ ..../ ....
/ ....
I'
'I
I -----..... .... /
I /
I I I
I I I I
I I
~/
No III
I
I
I I
I I I
I
I I
I.... /
...... ____/
I
I
/
0 .... / X
.... /
- _..
------- FIGURE87
We begin the proof with the observation that, under the stated conditions,
f (z) = 0 in some neighborhood N of z0 . For, otherwise, there would be a deleted
neighborhood of z0 throughout which f (z) f=. 0, according to Theorem 2 above; and
that would be inconsistent with the condition that f (z) = 0 everywhere in a domain
or on a line segment containing z0. Since f (z) =0 in the neighborhood N, then, it
242 RESIDUES AND POLES CHAP. 6
follows that all of the coefficients
(n=0,1,2, ...)
in the Taylor series for f (z) about z0 must be zero. Thus f (z) = 0 in the neighborhood
N0, since Taylor series also represents f(z) in N0. This completes the proof.
69. ZEROS AND POLES
The following theorem shows how zeros of order m can create poles of order m.
Theorem 1. Suppose that
(i) two functions p and q are analytic at a point z0 ;
(ii) p(z0) i= 0 and q has a zero oforder mat z0.
Then the quotient p(z)/q(z) has a pole oforder mat z0.
The proof is easy. Let p and q be as in the statement of the theorem. Since q has
a zero of order m at z0, we know from Theorem 2 in Sec. 68 that there is a deleted
neighborhood of z0 in which q(z) i= 0; and so z0 is an isolated singular point of the
quotient p(z)/q(z). Theorem 1 in Sec. 68 tells us, moreover. that
where g is analytic and nonzero at z0; and this enables us to write
(1)
p(z) p(z)jg(z)
q (z) (z - zo)m
Since p(z)1g (z) is analytic and nonzero at z0• it now follows from the theorem in Sec.
66 that zo is a pole of order m of p(z)/q(z).
EXAMPLE 1. The two functions
p(z) =1 and q(z) =z(ez- 1)
are entire; and we know from the example in Sec. 68 that q has a zero of order m = 2
at the point z0 = 0. Hence it follows from Theorem 1 here that the quotient
p(z) 1
q(z) z(ez- 1)
has a pole oforder 2 at that point. This was demonstrated in another way in Example 5,
Sec.67.
SEC.69 ZEROS AND PoLES 243
Theorem 1 leads us to another method for identifying simple poles and finding
the corresponding residues. This method is sometimes easier to use than the one in
Sec. 66.
Theorem 2. Let two functions p and q be analytic at a point z0. If
p(z0) f=- 0, q (z0) = 0, and q'(z0) f=- 0,
then z0 is a simple pole (~{the quotient p(z)/q(z) and
(2) Res p(z) = p(zo) .
z=zo q (z) q'(z0)
To show this, we assume that p and q are as stated and observe that, because of
the conditions on q, the point zo is a zero of order m = 1of that function. According
to Theorem 1in Sec. 68, then,
(3) q(z) =(z- zo)g(z)
where g(z) is analytic and nonzero at z0 . Furthermore, Theorem 1 in this section tells
us that z0 is a simple pole of p(z) jq (z); and equation (1) in its proof becomes
p(z) p(z)/g(z)
q(z) z- zo
Now p(z)/g(z) is analytic and nonzero at z0 , and it follows from the theorem in Sec.
66 that
Res p(z) = p(zo).
z=zo q(z) g(zo)
(4)
But g(z0) = q'(z0), as is seen by differentiating each side of equation (3) and setting
z =z0. Expression (4) thus takes the form (2).
EXAMPLE 2. Consider the function
cos z
f (z) =cot z = . ,
smz
which is a quotient of the entire functions p(z) =cos z and q(z) =sin z. The singu-
larities of that quotient occur at the zeros of q, or at the points
z =nJT (n = 0, ±1, ±2, ...).
Since
p(nrr) =(-It f:-0, q(nrr) =0, and q'(nrr) = (-1)11
f:-0,
246 RESIDUES AND POLES CHAP. 6
6. Show that
1 dz n
c (z2 - 1)2 + 3 = 2./2'
where C is the positively oriented boundary of the rectangle whose sides lie along the
lines x = ±2, y =0, and y = 1.
Suggestion: By observing that the four zeros ofthe polynornialq(z) = (z2
- 1)2
+3
are the square roots of the numbers 1± J3i, show that the reciprocal 1/q (z) is analytic
inside and on C except at the points
~l
zo = and
./2
-zo=
Then apply Theorem 2 in Sec. 69.
7. Consider the function
1
J(z) = [q(z)]2'
-~+i
h
where q is analytic at z0, q(z0) = 0, and q'(z0) ::j:. 0. Show that zo is a pole oforder m= 2
of the function f, with residue
q"(zo)
Bo=- .
[q'(zo)]3
Suggestion: Note that zo is a zero of order m = 1of the function q, so that
q(z) = (z- zo)g(z),
where g(z) is analytic and nonzero at z0. Then write
f(z) = ¢(z)
(z- zo)2
1
where ¢ (z) = 2[g(z)]
The desired form of the residue B0 =¢'(zo) can be obtained by showing that
q'(zo) =g(zo) and q"(z0) =2g'(zo).
8. Use the result in Exercise 7 to find the residue at z = 0 of the function
1
(a) f(z) =csc2
z; (b) f(z) = ( 2
)
2z+z
Ans. (a) 0; (b) -2.
9. Let p and q denote functions that are analytic at a point z0, where p(zo) ::j:. 0 and
q(zo) = 0. Show that if the quotient p(z)jq(z) has a pole of order m at z0, then z0 is
a zero of order m of q. (Compare Theorem 1 in Sec. 69.)
Suggestion: Note that the theorem in Sec. 66 enables one to write
p(z) =
'q(z) (z- zo)m
¢(z)
where ¢(z) is analytic and nonzero at z0. Then solve for q(z).
SEC. 70 BEHAVIOR OFf NEAR ISOLATED SINGULAR POINTS 247
10. Recall (Sec. 10) that a point z0 is an accumulation point of a set S if each deleted
neighborhood ofz0 contains at least one point of S. One form ofthe Bolzano-Weierstrass
theorem can be stated as follows: an infinite set ofpoints lying in a closed bounded region
R has at least one accumulation point in R.* Use that theorem and Theorem 2 in Sec.
68 to show that if a function f is analytic in the region R consisting of all points inside
and on a simple closed contour C, except possibly for poles inside C, and if all the zeros
of f in R are interior to C and are of finite order, then those zeros must be finite in
number.
11. Let R denote the region consisting of all points inside and on a simple closed contour
C. Use the Bolzano-Weierstrass theorem (see Exercise 10) and the fact that poles are
isolated singular points to show that if f is analytic in the region R except for poles
interior to C, then those poles must be finite in number.
70. BEHAVIOR OFf NEAR ISOLATED SINGULAR POINTS
As already indicated in Sec. 65, the behavior of a function f near an isolated singular
point zo varies, depending on whether zo is a pole, a removable singular point, or
an essential singular point. In this section, we develop the differences in behavior
somewhat further. Since the results presented here will not be used elsewhere in the
book, the reader who wishes to reach applications of residue theory more quickly may
pass directly to Chap. 7 without disruption.
Theorem 1. Ifz0 is a pole ofa function f, then
(1) lim f(z) = oo.
z_,..zo
To verify limit (1), we assume that f has a pole of order m at z0 and use the
theorem in Sec. 66. It tells us that
f(z) = c/J(z) ,
(z- zo)m
where c/J(z) is analytic and nonzero at z0. Since
lim (z- z0)m
z_,..zo 0
--=0,
¢(zo)
then, limit (1) holds, according to the theorem in Sec. 16 regarding limits that involve
the point at infinity.
The next theorem emphasizes how the behavior of f near a removable singular
point is fundamentally different from the behavior near a pole.
*See, for example, A. E. Taylor and W. R. Mann. "Advanced Calculus," 3d ed., pp. 517 and 521, 1983.
248 RESIDUES AND POLES CHAP. 6
Theorem 2. If z0 is a removable singular point ofa function f, then f is analytic
and bounded in some deleted neighborhood 0 < lz - zol < £ ofz0.
The proof is easy and is based on the fact that the function f here is analytic
in a disk lz- zol < R2 when f(z0) is properly defined; and f is then continuous in
any closed disk lz- zol <£where£ < R2. Consequently, f is bounded in that disk,
according to Sec. 17; and this means that, in addition to being analytic, f must be
bounded in the deleted neighborhood 0 < lz- z01< £.
The proof of our final theorem, regarding the behavior of a function near an
essential singular point, relies on the following lemma, which is closely related to
Theorem 2 and is known as Riemann's theorem.
Lemma. Suppose that afunction f is analytic and bounded in some deleted neigh-
borhood0 < lz - zol < £ofapoint zo.lff is not analytic at z0, then ithas a removable
singularity there.
To prove this, we assume that f is not analytic at z0. As a consequence, the point
z0 must be an isolated singularity off; and f(z) is represented by a Laurent series
(2)
throughout the deleted neighborhood 0 < lz- zol <£.If C denotes a positively ori-
ented circle lz- zol = p, where p < s (Fig. 88), we know from Sec. 55 that the
coefficients bn in expansion (2) can be written
(3)
y
0
I
I
I
I
/
I
I
/
c
'
""
bn = _1_.1 f(z) dz
2m c (z- z0)-n+I
------/
/
X
(n=l,2, ...).
FIGURE88
SEC. 70 BEHAVIOR OFf NEAR ISOLATED SINGULAR POINTS 249
Now the boundedness condition on f tells us that there is a positive constant M such
that lf(z)l < M whenever 0 < lz- z01<£.Hence it follows from expression (3) that
(n = 1, 2, ...).
Since the coefficients bn are constants and since p can be chosen arbitrarily small, we
may conclude that bn = 0 (n = 1, 2, ...) in the Laurent series (2). This tells us that zo
is a removable singularity of f, and the proof of the lemma is complete.
We know from Sec. 65 that the behavior of a function near an essential singular
point is quite irregular. The theorem below, regarding such behavior, is related to
Picard's theorem in that earlier section and is usually referred to as the Casorati-
Weierstrass theorem. Itstates that, in each deleted neighborhood ofan essential singular
point, a function assumes values arbitrarily close to any given number.
Theorem 3. Suppose that z0 is an essential singularity ofafunction f, and let w0 be
any complex number. Then, for any positive number s, the inequality
(4) lf(z)- wol < e
is satisfied at some point z in each deleted neighborhood 0 < lz - zol < ~ of z0
(Fig. 89).
y v
----/
- ' '/
"/
• '/
f(z)/'I
.,..--- I,.. '
e
/ ~
I
I I
! 8 I Wo I
I
I • I I
z Zo I /
I
' /
/ ' /
', / ' /
.... ,..... __...,..
-----0 X 0 u FIGURE89
The proof is by contradiction. Since zo is an isolated singularity of f, there is a
deleted neighborhood 0 < lz- zol < othroughout which f is analytic; and we assume
that condition (4) is not satisfied for any point z there. Thus lf(z)- w01 >£when
0 < lz- zol <~;and so the function
(5)
1
g(z)=---
f(z)- wo
(0 < lz - zol < 8)
250 RESIDUES AND POLES CHAP. 6
is bounded and analytic in its domain of definition. Hence, according to the above
lemma, z0 is a removable singularity of g; and we let g be defined at zo so that it is
analytic there.
If g(zo) f::- 0, the function j(z), which can be written
1
f(z)=- +wo
g(z)
(6)
when 0 < lz- zol < o, becomes analytic at zo if it is defined there as
1
f(zo) = + wo.
g(zo)
But this means that z0 is a removable singularity of f, not an essential one, and we
have a contradiction.
If g(zo) =0, the function g must have a zero of some finite order m (Sec. 68) at
z0 because g(z) is not identically equal to zero in the neighborhood lz- zol < o. In
view of equation (6), then, f has a pole of order m at zo (see Theorem 1 in Sec. 69).
So, once again, we have a contradiction; and Theorem 3 here is proven.
CHAPTER
7
APPLICATIONS OF RESIDUES
We tum now to some important applications of the theory of residues, which was
developed in the preceding chapter. The applications include evaluation ofcertain types
ofdefinite and improper integrals occurring in real analysis and applied mathematics.
Considerable attention is also given to a method, based on residues, for locating zeros
of functions and to finding inverse Laplace transforms by summing residues.
71. EVALUATION OF IMPROPER INTEGRALS
In calculus, the improper integral of a continuous function f(x) over the semi-infinite
interval x > 0 is defined by means of the equation
(1) r~ j(x)dx= lim {R j(x)dx.
lo R-+oo lo
When the limit on the right exists, the improper integral is said to converge to that
limit. If f(x) is continuous for all x, its improper integral over the infinite interval
-oo < x < oo is defined by writing
(2) foo fo 1R2
f(x) dx = lim f(x) dx + lim f(x) dx;
-oo RJ---*00 -Rt Rz---*00 0
and when both of the limits here exist, integral (2) converges to their sum. Another
value that is assigned to integral (2) is often useful. Namely, the Cauchy principal
251
SEC.71 EVALUATION OF IMPROPER INTEGRALS 253
and we see that integral (1) converges to one half the Cauchy principal value (3) when
that value exists. Moreover, since integral (1) converges and since
10 1R1f(x) dx = f(x) dx,
-R1 0
integral (2) converges to twice the value of integral (1). We have thus shown that when
f(x)(-oo < x < oo) is even and the Cauchy principal value (3) exists, both ofthe
integrals (1) and (2) converge and
(6) P.V.100
.f(x) dx =100
f(x) dx =2 r~ f(x) dx.
-00 -00 h
We now describe a method involving residues, to be illustrated in the next
section, that is often used to evaluate impro er integrals of even rational functions
f(x) = p(x)jq(x), where f(-x) is equal to f(x) and where p(x) and q(x) are
polynomials with real coefficients and no factors in common. We agree that q(z) has
no real zeros but has at least on~£~Z~e the. re~l ~!~J
The method begins with the identification of all of the distinct zeros of the
polynomial q (z) that lie above the real axis. They are, of course, finite in number
(see Sec. 49) and may be labeled ZI> z2, ••• , Zn, where n is less than or equal to the
degree of q(z). We then integrate the quotient
(7) f(z) = p(z)
q(z)
around the positively oriented boundary of the semicircular region shown in Fig. 90.
That simple closed contour consists of the segment of the real axis from z = -:-R to
z =Rand the top half of the circle lzl = R, described counterclockwise and denoted
by CR· It is understood that the positive number R is .large enough that the points
z1, z2, ... , Zn all lie inside the closed path.
y
-R 0 R X
FIGURE90
SEC.72 EXAMPLE 255
reveals that the sixth roots of -1 are
ck=exp[i(: +
2
~n)] (k =0, 1, 2, ... ' 5),
and it is clear that none of them lies on the real axis. The first three roots,
C -eiTC/6 c l
o- , •=,
lie in the upper half plane (Fig. 91) and the other three lie in the lower one. When
R > 1, the points ck (k = 0, 1, 2) lie in the interior of the semicircular region bounded
by the segment z =x (-R < x < R) of the real axis and the upper half CR of the
circle lzl = R from z = R to z = -R. Integrating f(z) counterclockwise around the
boundary of this semicircular region, we see that
(1)
iR f(x) dx + { f(z) dz = 2ni(Bo + Bt + B2),
-R lcR
where Bk is the residue of f(z) at ck (k = 0, 1, 2).
y
-R 0 R X
FIGURE91
With the aid of Theorem 2 in Sec. 69, we find that the points ck are simple poles
off and that
(k=0,1,2).
Thus
2ni(B0 + B1 + B2) =2ni ( ;i -;i + ;i) = ~ ~
and equation (1) can be put in the form
(2)
iR f(x) dx = n - { f(z) dz,
-R 3 lcR
which is valid for all values of R greater than 1.
256 APPLICATIONS OF RESIDUES CHAP. 7
Next, we show that the value of the integral on the right in equation (2) tends to
0 as R tends to oo. To do this, we observe that when lzl = R,
and
So, if z is any point on CR•
where
and this means that
(3)
n R being the length of the semicircle CR. (See Sec. 41.) Since the number
nR3
MRnR = R6 -1
is a quotient of polynomials in R and since the degree of the numerator is less than
the degree of the denominator, that quotient must tend to zero as R tends to oo. More
precisely, if we divide both numerator and denominator by R6
and write
n
R3
MRnR=
1
,
1--
R6
it is evident that M Rn R tends to zero. Consequently, in view of inequality (3),
lim 1f(z) dz = 0.
R--+oo CR
It now follows from equation (2) that
J
R x2 n
lim dx = -,
R--+oo -R x6 + 1 3
or
f
oo x2 n
P.V.
6
dx = -.
-oo X + 1 3
SEC.73 IMPROPER INTEGRALS FROM FOURIER ANALYSIS 259
where a is any real number and A =Ja2 + 1, arises in the theory of case-hardening of
steel by means of radio-frequency heating.* Follow the steps below to derive it.
(a) Point out why the four zeros of the polynomial
q(z) = (z
2
- a)2
+ 1
are the square roots of the numbers a ± i. Then, using the fact that the numbers
and -z0 are the square roots of a+ i (Exercise 5, Sec. 9), verify that ±z0 are the
square roots of a - i and hence that z0 and -z0 are the only zeros of q(z) in the
upper half plane lm z > 0.
(b) Using the method derived in Exercise 7, Sec. 69, and keeping in mind that z6=a+ i
for purposes of simplification, show that the point zo in part (a) is a pole of order 2
of the function f(z) = l/[q(z)]2
and that the residue B1 at zo can be written
81
=_ q
11
(z0) =a- i(2a2
+ 3)
[q'(z0)]3 16A2z0
After observing that q'(-z) = -q'(z) and q"(-z) =q"(z), use the same method to
show that the point -z0 in part (a) is also a pole of order 2 of the function f(z), with
residue
Then obtain the expression
B - { q"(zo) } - -B
2- [q'(zo)J3 - I·
B + B = 1 Im [ -a +i(2a
2
+3) ]
I 2 8A2.
I Zo
for the sum of these residues.
(c) Refer to part (a) and show that lq(z) I > (R- lzol)4
if lzl =R, where R > lzol· Then,
with the aid of the final result in part (b), complete the derivation of the integration
formula.
73. IMPROPER INTEGRALS FROM FOURIER ANALYSIS
Residue theory can be useful in evaluating convergent improper integrals of the form
(1) i:f(x) sin ax dx or 1:f(x) cos ax dx,
*See pp. 359-364 of the book by Brown, Hoyler, and Bierwirth that is listed in Appendix 1.
260 APPLICATIONS OF RESIDUES CHAP. 7
where a denotes a positive constant. As in Sec. 71, we assume that f(x) =p(x)jq(x),
where p(x) and q(x) are polynomials with real coefficients and no factors in common.
Also, q(z) has no real zeros. Integrals of type (1) occur in the theory and application
of the Fourier integral.*
The method described in Sec. 71 and used in Sec. 72 cannot be applied directly
here since (see Sec. 33)
jsin azl
2
= sin2
ax + sinh2
ay
and
Ieos ax!2
= cos2
ax+ sinh2
ay.
More precisely, since
. eay- e-ay
smh ay = ,
2
the moduli !sin azl and jcos az! increase like eay as y tends to infinity. The modification
illustrated in the example below is suggested by the fact that
I:f(x) cos ax dx i !R f(x) sin ax dx =!R f(x)eiax dx,
-R -R
together with the fact that the modulus
jeiazl = jeia(x+iy)l = je-ayeiaxl =e-ay
is bounded in the upper half plane y > 0.
EXAMPLE. Let us show that
(2)
f
oo cos 3x d _ 2rr
X- .
-oo (x2 + 1)2 e3
Because the integrand is even, it is sufficient to show that the Cauchy principal value
of the integral exists and to find that value.
We introduce the function
(3)
1
f(z) = (z2 + 1)2
and observe that the product j(z)ei3z is analytic everywhere on and above the real
axis except at the point z = i. The singularity z =i lies in the interior of the semi-
circular region whose boundary consists of the segment - R < x < R of the real axis
*See the authors' "Fourier Series and Boundary Value Problems," 6th ed., Chap. 7, 2001.
262 APPLICATIONS OF RESIDUES CHAP. 7
74. JORDAN'S LEMMA
In the evaluation of integrals of the type treated in Sec. 73, it is sometimes necessary
to use Jordan's lemma,* which is stated here as a theorem.
Theorem. Suppose that
(i) a function f(z) is analytic at all points z in the upper halfplane y > 0 that are
exterior to a circle lzl = Ro;
(ii) CR denotes a semicircle z = Rei0(0 < () < rr), where R > R0 (Fig. 93);
(iii) for all points z on CR, there is a positive constant MR such that 1/(z)l < MR,
where
lim MR = 0.
R---+oo
Then, for every positive constant a,
(1) lim 1f (z)eiaz dz = 0.
R---+oo CR
y
0 R X
FIGURE93
The proof is based on a result that is known as Jordan's inequality:
(2) {rr e-RsinO de< rr
lo R
(R > 0).
To verify this inequality, we first note from the graphs of the functions y =sine and
y=20/rr when 0 < e< rr/2 (Fig. 94) that sine > 20jrr for all values of e in that
*See the first footnote in Sec. 38.
264 APPLICATIONS OF RESIDUES CHAP. 7
As usual, the existence of the value in question will be established by our actually
finding it.
We write
z zf(z) = - ,
z2 + 2z +2 (z - ZJ)(z- ZJ)
where z1 =-1 +i. The point z1, which lies above the .X axis, is a simple pole of the
function f (z)eiz, with residue
(4)
Hence, when R > v'2 and CR denotes the upper half of the positively oriented circle
lzl = R,
1R xeix dx 1 .2
= 2niB1 - f(z)e12
dz;
-R X + 2x + 2 CR
and this means that
(5)
1R x sinx dx 1 ·2
= Im(2niB1)- Im f(z)e1
z dz.
-R X +2x + 2 CR
Now
(6)
and we note that, when z is a point on CR•
R
lf(z)l < MR where MR = ---=-
(R v'2,)2
and that leizl = e-Y < 1for such a point. By proceeding as we did in the examples in
Sees. 72 and 73, we cannot conclude that the right-hand side of inequality (6), and
hence the left-hand side, tends to zero as R tends to infinity. For the quantity
M n R = n R2
R (R- v'2,)2
does not tend to zero. Limit (1) does, however, provide the desired result.
So it does, indeed, follow from inequality (6) that the left-hand side there tends
to zero as R tends to infinity. Consequently, equation (5), together with expression (4)
SEC. 75 INDENTED PATHS 267
(b) Show that the value of the integral along the arc CR in part (a) tends to zero a'l R
tends to infinity by obtaining the inequality
{ eiz2 dz < R (n:/2 e-R2 sinc/Jdl/>
JcR 2 lo
and then referring to the form (3), Sec. 74, of Jordan's inequality.
(c) Use the results in parts (a) and (b), together with the known integration formula*
1oo -x2 d .jii
e X=-,
0 2
to complete the exercise.
75. INDENTED PATHS
In this and the following section, we illustrate the use of indented paths. We begin with
an important limit that will be used in the example in this section.
Theorem. Suppose that
(i) afunction /(z) has a simple pole at a point z =x0 on the real axis, with a Laurent
series representation in a punctured disk 0 < lz - xol < R2 (Fig. 96) and with
residue B0;
(ii) CP denotes the upperhalfofa circle lz - x0 i = p, where p < R2 and the clockwise
direction is taken.
Then
(1)
y
0
lim 1/(z) dz = -B0 ni.
p-+0 cp
X
FIGURE%
Assuming that the conditions in parts (i) and (ii) are satisfied, we start the proof
of the theorem by writing the Laurent series in part (i) as
f(z) =g(z) + Bo
z -x0
*See the footnote with Exercise 4, Sec. 46.
SEC.77 INTEGRATION ALONG A BRANCH CUT 273
and, by !'Hospital's rule, the product pIn pin the numerator on the far right here tends
to 0 as p tends to 0. So the first of limits (6) clearly holds. Likewise, by writing
n In R
1 1 lnR+n R+R
Re f(z)dz < f(z)dz < 2 2
nR=n ,
eR eR (R -4) (R- ~)-
and using }'Hospital's rule to show that the quotient (In R)/R tends to 0 as R tends to
oo, we obtain the second of limits (6).
Note how another integration formula, namely
(7)
rX! dx n
lo (x2 +4)2 - 32'
follows by equating imaginary, rather than real, parts on each side of equation (4):
(8)
32
Im f f(z) dz- Im f j(z) dz.
leP leR
Formula (7) is then obtained by letting p and R tend to 0 and oo, respectively, since
Im f f(z) dz < f f(z) dz
lcP lep
and Im f j(z) dz
lei?
77. INTEGRATION ALONG A BRANCH CUT
< f f(z) dz .
lei?
Cauchy's residue theorem can be useful in evaluating a real integral when part of the
path of integration of the function f(z) to which the theorem is applied lies along a
branch cut of that function.
EXAMPLE. Let x-a, where x > 0 and 0 < a < 1, denote the principal value of the
indicated power of x; that is, x -a is the positive real number exp(-a In x). We shall
evaluate here the improper real integral
(1)
r)() x-a d
lo X+ l X
(0 <a< 1),
which is important in the study of the gamma function.* Note that integral (1) is
improper notonly because ofits upper limit ofintegration but also because its integrand
has an infinite discontinuity at x =0. The integral converges when 0 < a < 1since the
integrand behaves like x-a near x = 0 and like x-a-l as x tends to infinity. We do not,
*See, for example, p. 4 of the book by Lebedev cited in Appendix 1.
274 APPLICATIONS OF RESIDUES CHAP. 7
however, need to establish convergence separately; for that will be contained in our
evaluation of the integral.
We begin by letting CP and CR denote the circles lzI= p and lzI= R, respectively,
where p < 1 < R; and we assign them the orientations shown in Fig. 99. We then
integrate the branch
(2)
z-a
f(z)=--
z+l
(lzl > 0, 0 < argz < 2n)
of the multiple-valued function z-a j(z + 1), with branch cut arg z = 0, around the
simple closed contour indicated in Fig. 99. That contour is traced out by a point moving
from p to R along the top of the branch cut for f (z), next around CR and back to R,
then along the bottom of the cut top, and finally around CP back top.
y
---
X
FIGURE99
Now 8 = 0 and 8 =2n along the upper and lower "edges," respectively, of the
cut annulus that is formed. Since
f(z) = exp(-a log z) = exp[-a~ln r +W)]
· z + 1 red'+ 1
where z = reUJ, it follows that
J(z) = exp[-a(ln r + iO)] = r-a
r+l r+l
on the upper edge, where z = rei0, and that
exp[-a(In r + i2n)] r-ae-i2arr
f(z)= = - - -
r+l r+l
on the lower edge, where z =rei2rr. The residue theorem thus suggests that
(3) {R r-a dr + { J(z) dz - {R r-ae-i2arr dr + { f(z) dz
JP r + 1 lcR JP r + 1 lcp
= 2ni Res f(z).
z=-1
SEC.78 DEFINITE INTEGRALS INVOLVING SINES AND COSINES 279
evaluate that integral by means of Cauchy's residue theorem once the zeros of the
polynomial in the denominator have been located and provided that none lie on C.
EXAMPLE. Let us show that
(5)
f2
1f de 2n
Jo 1+a sin() - J1 - a2
(-1 <a< 1).
This integration formula is clearly valid when a =0, and we exclude that case in our
derivation. With substitutions (3), the integral takes the form
(6) r 2/a dz,
lc z2 + (2ija)z- 1
where Cis the positively oriented circle lzI= 1. The quadratic formula reveals that the
denominator of the integrand here has the pure imaginary zeros
(
l-JI-a2)·
Z2 = l.
a
So if /(z) denotes the integrand, then
f(z) = 2ja
(z- z1)(z- z2)
Note that, because IaI < 1,
Also, since lz1z21= I, it follows that lz11< 1. Hence there are no singular points on C,
and the only one interior to it is the point z1. The corresponding residue B1 is found
by writing
/(z) = ¢(z)
Z ~ Zl
where f/>(z) = 2
/a .
Z- Z2
This shows that z1 is a simple pole and that
Consequently,
1 2/a . 2n
-=----'---- dz = 2nzB1 = ;
cz2 +(2i/a)z-1 JI-a2
and integration formula (5) follows.
SEC.79 ARGUMENT PRINCIPLE 281
79. ARGUMENT PRINCIPLE
A function f is said to be meromorphic in a domain D if it is analytic throughout
D except for poles. Suppose now that f is meromorphic in the domain interior to
a positively oriented simple closed contour C and that it is analytic and nonzero
on C. The image r of C under the transformation w- f(z) is a closed contour,
not necessarily simple, in the w plane (Fig. 101). As a point z traverses C in the
positive direction, its images w traverses r in a particular direction that determines
the orientation of r. Note that, since f has no zeros on C, the contour r does not pass
through the origin in the w plane.
y v
z
X u
FIGURE 101
Let w and w0 be points on r, where w0 is fixed and ¢0 is a value of arg w0. Then
let arg w vary continuously, starting with the value ¢0, as the point w begins at the point
w0 and traverses r once in the direction of orientation assigned to it by the mapping
w = f (z). When w returns to the point w0, where it started, arg w assumes a particular
value of arg w0, which we denote by ¢ 1• Thus the change in arg w as w describes r
once in its direction of orientation is ¢1 - ¢0. This change is, of course, independent
ofthe point w0 chosen to determine it. Since w = f (z), the number¢1 - ¢0 is, in fact,
the change in argument off(z) as z describes C once in the positive direction, starting
with a point z0; and we write
Llc arg f (z) = ¢J - ¢o·
The value of D-e arg f (z) is evidently an integral multiple of 2n, and the integer
I
-D-e arg f(z)
2n
represents the number oftimes the point w winds around the origin in the w plane. For
that reason, this integer is sometimes called the winding number of r with respect to
the origin w = 0. It is positive if r winds around the origin in the counterclockwise
direction and negative if it winds clockwise around that point. The winding number
is always zero when r does not enclose the origin. The verification of this fact for a
special case is left to the exercises.
282 APPLICATIONS OF RESIDUES CHAP. 7
The winding number can be determined from the number of zeros and poles of
f interior to C. The number of poles is necessarily finite, according to Exercise 11,
Sec. 69. Likewise, with the understanding that f(z) is not identically equal to zero
everywhere else inside C, it is easily shown (Exercise 4, Sec. 80) that the zeros of f
are finite in number and are all of finite order. Suppose now that f has Z zeros and P
poles in the domain interior to C. We agree that f has m0 zeros at a point z0 if it has a
zero of order m0 there; and iff has a pole of order mP at z0, that pole is to be counted
mP times. The following theorem, which is known as the argument principle, states
that the winding number is simply the difference Z - P.
Theorem. Suppose that
(i) a function f(z) is meromorphic in the domain interior to a positively oriented
simple closed contour C;
(ii) f(z) is analytic and nonzero on C;
(iii) counting multiplicities, Z is the number ofzeros and P is the number ofpoles of
f(z) inside C.
Then
(1)
1
-Lie arg f(z) = Z- P.
2n
To prove this, we evaluate the integral of f'(z)/f(z) around C in two different
ways. First, we let z = z(t) (a< t <b) be a parametric representation for C, so that
(2)
f f'(z) dz = fb f'[z(t)]z'(t) dt.
Jc f(z) la f[z(t)]
Since, under the transformation w = f(z), the image r of C never passes through
the origin in the w plane, the image of any point z =z(t) on C can be expressed in
exponential form as w = p (t) exp[i¢ (t)]. Thus
(3) f[z(t)] = p(t)ei<fl(t) (a<t<b);
and, along each of the smooth arcs making up the contour r. it follows that (see
Exercise 5, Sec. 38)
(4) f'[z(t)]z'(t) = !!:_ f[z(t)] = !!:_[p(t)ei<fl(t)] =p'(t)ei<fl(t) +ip(t)ei<fl(t)¢'(t).
dt dt
Inasmuch as p1
(t) and ¢'(t) are piecewise continuous on the interval a< t < b, we
can now use expressions (3) and (4) to write integral (2) as follows:
f f'(z) dz = fb p'(t) dt + i fb ¢1(t) dt =ln p(t)]b + icp(t)]b.
Jc f(z) la p(t) la a a
SEC.79
But
Hence
(5)
ARGUMENT PRINCIPLE 283
p(b) =p(a) and cp(b)- cp(a) = b.c arg f(z).
1;;_f'_(z-'-) dz = i b.c arg f(z).
c f(z)
Another way to evaluate integral (5) is to use Cauchy's residue theorem. To be
specific, we observe that the integrand f'(z)lf(z) is analytic inside and on C except
at the points inside C at which the zeros and poles off occur. If f has a zero of order
m0 at z0, then (Sec. 68)
(6) f(z) = (z- Zo)m0g(z),
where g(z) is analytic and nonzero at z0. Hence
or
(7)
f'(zo) =mo(z- Zo)m0-
1
g(z) + (z- zo)m0 g
1
(z),
f'(z)
f(z)
mo + g'(z).
z- zo g(z)
Since g'(z)I g(z) is analytic at z0, it has a Taylor series representation about that point;
and so equation (7) tells us that f' (z) If (z) has a simple pole at z0, with residue m 0.
If, on the other hand, f has a pole of order mP at z0, we know from the theorem in
Sec. 66 that
(8)
where ¢(z) is analytic and nonzero at z0. Because expression (8) has the same form
as expression (6), with the positive integer m0 in equation (6) replaced by -mP' it is
clear from equation (7) that f' (z)If (z) has a simple pole at z0, with residue -m p·
Applying the residue theorem, then, we find that
(9)
1f'(z) dz =2Jri(Z- P).
c f(z)
Expression (1) now follows by equating the right-hand sides of equations (5) and (9).
EXAMPLE. The only singularity of the function 11z2
is a pole of order 2 at the
origin, and there are no zeros in the finite plane. In particular, this function is analytic
and nonzero on the unit circle z =ei
8 (0 < () < 2n). If we let C denote that positively
oriented circle, our theorem tells us that
_1 b.c arg(_!_) = -2.
2n z2
284 APPLICATIONS OF RESIDUES CHAP. 7
That is, the image r of C under the transformation w = 1/z2
winds around the origin
w =0 twice in the clockwise direction. This can be verified directly by noting that r
has the parametric representation w = e-i28 (0 < e< 2n).
,
80. ROUCHE'S THEOREM
The main result in this section is known as Rouchi's theorem and is a consequence of
the argument principle, just developed in Sec. 79. It can be useful in locating regions
of the complex plane in which a given analytic function has zeros.
Theorem. Suppose that
(i) twofunctions f(z) and g(z) are analytic inside andon a simple closedcontourC;
(ii) lf(z)l > lg(z)l ateachpointon C.
Then f(z) and f(z) + g(z) have the same number ofzeros, counting multiplicities,
inside C.
The orientation of C in the statement of the theorem is evidently immaterial.
Thus, in the proof here, we may assume that the orientation is positive. We begin with
the observation that neither the function f(z) nor the sum f(z) +g(z) has a zero on
C, since
lf(z)l > lg(z)l > 0 and IJ(z) +g(z)l > llf(z)l-lg(z)ll > 0
when z is on C.
If Z1 and Zf+g denote the number of zeros, counting multiplicities, of f(z) and
f(z) +g(z), respectively, inside C, we know from the theorem in Sec. 79 that
1 1
Z1 = 2
n Llc arg f(z) and Zf+g = 2
n Llc arg[f(z) + g(z)].
Consequently, since
it is clear that
(1)
where
[
g(z) J=D-e arg f(z) +de arg 1+ f(z) ,
1
Z f+g = Z1 +-D-e arg F(z),
2n
F(z) = 1+ g(z) .
f(z)
286 APPLICATIONS OF RESIDUES CHAP. 7
that figure; and, with the aid of the theorem in Sec. 79, determine the number of zeros,
counting multiplicities, of f interior to C.
Ans. 6rr; 3.
3. Using the notation in Sec. 79, suppose that r does not enclose the origin w = 0 and that
there is a ray from that point which does not intersect r. By observing that the absolute
value of l::.c arg f(z) must be less than 2n when a point z makes one cycle around C
and recalling that l::.c arg f (z) is an integral multiple of 2rr, point out why the winding
number of r with respect to the origin w =0 must be zero.
4. Suppose that a function f is meromorphic in the domain D interior to a simple closed
contour C on which f is analytic and nonzero, and let D0 denote the domain consisting
of all points in D except for poles. Point out how it follows from the lemma in Sec. 26
and Exercise 10, Sec. 69, that iff(z) is not identically equal to zero in D0, then the zeros
off in D are all of finite order and that they are finite in number.
Suggestion: Note that if a point z0 in D is a zero off that is not of finite order, then
there must be a neighborhood of z0 throughout which j(z) is identically equal to zero.
5. Suppose that a function f is analytic inside and on a positively oriented simple closed
contour C and that it has no zeros on C. Show that iff has n zeros Zk(k = 1, 2, ... , n)
inside C, where each Zk is of multiplicity mk> then
[
! ) II
zj (z .
. dz =2m L mkzk·
c J(z) k=I
[Compare equation (9), Sec. 79 when P =0 there.]
6. Determine the number of zeros, counting multiplicities, of the polynomial
(a) z6
- 5z4
+z3
- 2z; (b) 2z4 - 2z3 +2z2
- 2z +9
inside the circle lzl = 1.
Ans. (a) 4; (b) 0.
7. Determine the number of zeros, counting multiplicities, of the polynomial
(a) z4
+3z3
+6; (b) z4
- 2z3
+9z2
+z 1; (c) z5 + 3z3 +z2 + 1
inside the circle lzl =2.
Ans. (a) 3; (b) 2; (c) 5.
8. Determine the number of roots, counting multiplicities, of the equation
in the annulus 1 < lzl < 2.
Ans. 3.
9. Show that if cis a complex number such that lei > e, then the equation cz11
= ez has n
roots, counting multiplicities, inside the circle lzl = 1.
SEC.80 EXERCISES 287
10. Write f(z) = zn and g(z) = a0 +a1z + ···+an-IZn-I and use Rouche's theorem to
prove that any polynomial
where n > 1, has precisely n zeros, counting multiplicities. Thus give an alternative proof
of the fundamental theorem of algebra (Theorem 2, Sec. 49).
Suggestion: Note that one can let an be unity. Then show that lg(z)l < lf(z)l on
the circle lzl = R, where R is sufficiently large and, in particular, larger than
11. Inequalities (5), Sec. 49, ensure that the zeros of a polynomial
P(z) =ao +a1z + ···+an-IZn-I +anzn
of degree n > I all lie inside some circle lzl = R about the origin. Also, Exercise 4 above
tells us that they are all of finite order and that there is a finite number N of them. Use
expression (9), Sec. 79, and the theorem in Sec. 64 to show that
N=Res P'(l/z),
z=O z2P(ljz)
where multiplicities of the zeros are to be counted. Then evaluate this residue to show
that N = n. (Compare Exercise 10.)
12. Let two functions f and g be as in the statement of Rouche's theorem in Sec. 80, and let
the orientation of the contour C there be positive. Then define the function
<P(t) =_I_ f f'(z) + tg'(z) dz
2rr:i lc f(z) + tg(z)
(0 < t < 1)
and follow the steps below to give another proof of Rouche's theorem.
(a) Point out why the denominator in the integrand ofthe integral defining <l>(t) is never
zero on C. This ensures the existence of the integral.
(b) Lett and t0 be any two points in the interval 0 < t < 1 and show that
I<P(t) _ <P(to)l = It- tol
2rr:
Then, after pointing out why
fg'
_ _.::...;;;:____::.....;:;:__ d z .
i
fg'- f'g
c (j + tg)(f + t0g)
lfg'- f'gl< .:..::._::;:_____.::_~
(f +tg)(f + t0g) - (lfl - lgl)2
at points on C, show that there is a positive constant A, which is independent oft
and t0, such that
I<P(t)- <P(to)l < Alt- t0 1.
Conclude from this inequality that <P(t) is continuous on the interval 0 < t < 1.
288 APPLICATIONS OF RESIDUES CHAP. 7
(c) By referring to equation (9), Sec. 79, state why the value ofthe function <I> is, for each
value of t in the interval 0 < t < 1, an integer representing the number of zeros of
f(z) + tg(z) inside C. Then conclude from the fact that <I> is continuous, as shown
in part (b), that f(z) and f(z) + g(z) have the same number of zeros, counting
multiplicities, inside C.
81. INVERSE LAPLACE TRANSFORMS
Suppose that a function F of the complex variable s is analytic throughout the finite s
plane except for a finite number of isolated singularities. Then let L R denote a vertical
line segment from s = y - i R to s = y + i R, where the constant y is positive and
large enough that the singularities ofF all lie to the left of that segment (Fig. 103). A
new function f ofthe real variable t is defined for positive values oft by means of the
equation
(1) f(t) = ~ lim 1est F(s) ds
2Jrl R~oo LR
(t > 0),
provided this limit exists. Expression (1) is usually written
(2)
1 jy+ioo
f(t) = -. P.V. est F(s) ds
2nz y-ioo
(t > 0)
[compare equation (3), Sec. 71], and such an integral is called a Bromwich integral.
It can be shown that, when fairly general conditions are imposed on the functions
involved, f(t) is the inverse Laplace transform of F(s). That is, if F(s) is the Laplace
transform of f(t), defined by the equation
(3) F(s) = fooo e-st f(t) dt,
y
0 y
_...,y-iR
FIGURE 103
SEC. 81 INVERSE LAPLACE TRANSFORMS 289
then f(t) is retrieved by means of equation (2), where the choice of the positive
number y is immaterial as long as the singularities of F all lie to the left of LR·*
Laplace transforms and their inverses are important in solving both ordinary and partial
differential equations.
Residues can often be used to evaluate the limit in expression (1) when the
function F(s) is specified. To see how this is done, we let sn (n = 1, 2, ... , N) denote
the singularities ofF(s). We then let R0 denote the largest oftheir moduli and consider
a semicircle CR with parametric representation
(4) s = y + Reie -<(J<-
(rr 3n)2- - 2 '
where R > R0 + y. Note that, for each sn,
Hence the singularities all lie in the interior of the semicircular region bounded by CR
and L R (see Fig. 103), and Cauchy's residue theorem tells us that
N
(5) 1e5tF(s)ds=2niL~~s[estF(s)]- {c e5
tF(s)ds.
LR n=1 n J(R
Suppose now that, for allpoints s on CR• there is a positive constant M R such that
IF(s) I< MR• where M R tends to zero as R tends to infinity. We may use the parametric
representation (4) for CR to write
Then, since
we find that
(6)
*For an extensive treatment of such details regarding Laplace transforms, see R. V. Churchill, "Opera-
tional Mathematics," 3d ed., 1972, where transforms F(s) with an infinite number of isolated singular
points, or with branch cuts, are also discussed.
290 APPLICATIONS OF RESIDUES CHAP. 7
But the substitution tf> = () - (:rr/2), together with Jordan's inequality (2), Sec. 74,
reveals that
1
3
rr
12
eRtcose d() = {rt e-Rtsin<P df/> < rr.
rt/2 lo Rt
Inequality (6) thus becomes
(7)
and this shows that
(8) lim [ est F(s) ds =0.
R--+oo lcR
Letting R tend to oo in equation (5), then, we see that the function f (t), defined by
equation (1), exists and that it can be written
(9)
N
f(t) = L Res[e51
F(s)]
S-Sn
n=l
(t > 0).
In many applications ofLaplace transforms, such as the solution of partial differ-
ential equations arising in studies of heat conduction and mechanical vibrations, the
function F (s) is analytic for all values of s in the finite plane except for an infinite
set of isolated singular points sn(n = 1, 2, ...) that lie to the left of some vertical line
Re s = y. Often the method just described for finding f (t) can then be modified in
such a way that the finite sum (9) is replaced by an infinite series of residues:
(10)
00
f(t) = L Res[est F(s)]
S-Sn
n=l
(t > 0).
The basic modification is to replace the vertical line segments L R by vertical line
segments LN (N = 1, 2, ...) from s = y- ibN to s = y +ibN. The circular arcs CR
are then replaced by contours CN (N =1, 2, ...) from y +ibN toy -ibN such that,
for each N, the sum L N +CN is a simple closed contour enclosing the singular points
sb s2, ... , sN. Once it is shown that
(11) lim { est F(s) ds = 0,
N--+oo leN
expression (2) for f (t) becomes expression (lO).
The choice of the contours CN depends on the nature of the function F(s).
Common choices include circular or parabolic arcs and rectangular paths. Also, the
simple closed contour L N +CN need not enclose precisely N singularities. When, for
example, the region between LN + CN and LN+l +CN+l contains two singular points
SEC.82 EXAMPLES 291
of F(s), the pair of corresponding residues of est F(s) are simply grouped together as
a single term in series (10). Since it is often quite tedious to establish limit (11) in any
case, we shall accept it in the examples and related exercises below that involve an
infinite number of singularities.* Thus our use of expression (10) will be only formal.
82. EXAMPLES
Calculation of the sums ofthe residues of est F (s) in expressions (9) and (10), Sec. 81,
is often facilitated by techniques developed in Exercises 12 and 13 of this section. We
preface our examples here with a statement of those techniques.
Suppose that F (s) has a pole of order m at a point s0 and that its Laurent series
representation in a punctured disk 0 < Is -sol < R2 has principal part
Then
(1) Res[estF(s)]=esot[bt+b2t+···+ bm tm-1].
s=so 1! (m -1)!
When the pole s0 is of the form s0 =a+ if3 (f3 =f. 0) and F(s) = F(S) at points of
analyticity of F(s) (see Sec. 27), the conjugate s0 =a - if3 is also a pole of order m.
Moreover,
Res[est F(s)] + Res[est F(s)]
s=so s=so
(2) = 2eat Re{eif3t [bl+ b2 t + ... + bm tm-1]}1! (m-1)!
when tis real. Note that if s0 is a simple pole (m = 1), expressions (1) and (2) become
(3) Res[est F(s)] = esot Res F(s)
s=so s=so
and
(4) ~~~[estF(s)] +~~~[estF(s)] =2eat Re[eif3t ~~~ F(s)J,
respectively.
*An extensive treatment of ways to obtain limit (11) appears in the book by R. V. Churchill that is cited
in the footnote earlier in this section. In fact, the inverse transform to be found in Example 3 in the next
section is fully verified on pp. 220-226 of that book.
298 APPLICATIONS OF RESIDUES CHAP. 7
of the entire function est = esote<s-so)t, show that
Res[estF(s)]=esot[bt+bzt+···+ bm-i tm-2+ bm tm-1],
s=so 1! (m 2)! (m- 1)!
as stated at the beginning of Sec. 82.
13. Let the point s0 =a+ if3 (/3 =!= 0) be a pole of order m of a function F(s), which has a
Laurent series representation
00
F(s) = L a~l~
n=O
in the punctured disk 0 < Is- sol < R2. Also, assume that F(s) =F(S) at points s where
F(s) is analytic.
(a) With the aid of the result in Exercise 6, Sec. 52, point out how it follows that
00 - - -
F(-) "-c- -)n b1 bz bms =~ an S - so + _ _ + _ _ 2
+ .. ·+ _ ____:.:.:.____
n=O s - so (s - so) (S - So)m
when 0 < Is- sol < R2• Then replaces by s here to obtain a Laurent series repre-
sentation for F(s) in the punctured disk 0 < Is- sol < R2, and conclude that s0 is a
pole of order m of F(s).
(b) Use results in Exercise 12 and part (a) above to show that
Res[est F(s)) + Res[es1
F(s)] = 2e011
Re{ei{Jt [b1 + bz t + ... + bm tm-l]}
S=So S=So 1! (m- 1)!
when t is real, as stated at the beginning of Sec. 82.
14. Let F(s) be the function in Exercise 13, and write the nonzero coefficient bm there in
exponential form as bm = rm exp(iem). Then use the main result in part (b) of Exercise
13 to show that when tis real, the sum of the residues of est F(s) at s0 =a+ if3 (/3 =/= 0)
and s0 contains a term of the type
2r 1----'--'m-tm- eat cos(f3t + fJm).
(m- 1)!
Note that if a > 0, the product tm 1e011
here tends to oo as t tends to oo. When the
inverse Laplace transform f(t) is found by summing the residues of e~1
F(s), the term
displayed above is, therefore, an unstable component of f(t) if a > 0; and it is said to
be of resonance type. If m > 2 and a = 0, the term is also of resonance type.
CHAPTER
8
MAPPING BY ELEMENTARY
FUNCTIONS
The geometric interpretation of a function of a complex variable as a mapping, or
transformation, was introduced in Sees. 12 and 13 (Chap. 2). We saw there how the
nature of such a function can be displayed graphically, to some extent, by the manner
in which it maps certain curves and regions.
In this chapter, we shall see further examples of how various curves and regions
are mapped by elementary analytic functions. Applications of such results to physical
problems are illustrated in Chaps. 10 and 11.
83. LINEAR TRANSFORMATIONS
To study the mapping
(1) w= Az.
where A is a nonzero complex constant and z =I= 0, we write A and z in exponential
form:
Then
(2)
and we see from equation (2) that transformation (1) expands or contracts the radius
vector representing z by the factor a =IAI and rotates it through an angle a =arg A
299
300 MAPPING BY ELEMENTARY FUNCTIONS CHAP. 8
about the origin. The image of a given region is, therefore, geometrically similar to
that region.
The mapping
(3) w =z + B,
where B is any complex constant, is a translation by means of the vector representing
B. That is, if
w = u + i v, z = x +i y, and B =b1 +i b2,
then the image of any point (x, y) in the z plane is the point
(4)
in the w plane. Since each point in any given region of the z plane is mapped into the
w plane in this manner, the image region is geometrically congruent to the original
one.
The general (nonconstant) linear transformation
(5) w = Az + B (A :;#: 0),
which is a composition of the transformations
Z = Az (A =1= 0) and w = Z + B,
is evidently an expansion or contraction and a rotation, followed by a translation.
EXAMPLE. The mapping
w- (1 + i)z + 2
transforms the rectangular region shown in the zplane of Fig. 104 into the rectangular
y y v
-1 + 3i 1 +3i
B
1 + 2i
B'
A' A"
0 A X X u
FIGURE 104
w = (l +i)z +2.
SEC.84 THE TRANSFORMATION W = 1/z 301
region shown in the w plane there. This is seen by writing it as a composition of the
transformations
Z =(1 + i)z and w = Z + 2.
Since 1+ i =,J2exp(irc/4), the first of these transformations is an expansion by the
factor ,J2 and a rotation through the angle rcj4. The second is a translation two units
to the right.
EXERCISES
1. State why the transformation w =iz is a rotation of the z plane through the angle rc/2.
Then find the image of the infinite strip 0 < x < 1.
Ans. 0 < v < 1.
2. Show that the transformation w = i z +i maps the half plane x > 0 onto the half plane
v > 1.
3. Find the region onto which the half plane y > 0 is mapped by the transformation
w =(1 +i)z
by using (a) polar coordinates; (b) rectangular coordinates. Sketch the region.
Ans. v > u.
4. Find the image of the half plane y > 1under the transformation w = (1 - i)z.
5. Find the image of the semi-infinite strip x > 0, 0 < y < 2 when w =iz + 1. Sketch the
strip and its image.
Ans. -1 < u < 1, v < 0.
6. Give a geometric description of the transformation w = A(z +B), where A and B are
complex constants and A ::j:. 0.
84. THE TRANSFORMATION w =liz
The equation
1
W=-(1)
z
establishes a one to one correspondence between the nonzero points of the z and the
w planes. Since zz= lzl2
, the mapping can be described by means of the successive
transformations
(2)
1
Z=-
2
z, w=Z.
lzl
302 MAPPING BY ELEMENTARY FUNCTIONS CHAP. 8
The first of these transformations is an inversion with respect to the unit circle
lzI= 1. That is, the image of a nonzero point z is the point Z with the properties
1
IZI= - and arg Z =arg z.
lzl
Thus the points exterior to the circle IzI=1are mapped onto the nonzero points interior
to it (Fig. 105), and conversely. Any point on the circle is mapped onto itself. The
second of transformations (2) is simply a reflection in the real axis.
(3)
y z
0 X
FIGURE 105
If we write transformation (1) as
1
T(z) =-
z
(z "I 0),
we can define T at the origin and at the point at infinity so as to be continuous on the
extended complex plane. To do this, we need only refer to Sec. 16 to see that
(4)
and
(5)
lim T(z) =oo smce
z~o
1
lim =0
z~o T(z)
lim T (z) = 0 since lim T (I)= 0.
z~oo z~o z
In order to make T continuous on the extended plane, then, we write
(6) T(O) =oo, T(oo) =0,
1
and T(z) =-
z
for the remaining values of z. More precisely, equations (6), together with the first of
limits (4) and (5), show that
(7) lim T(z) = T(z0)
z~zo
for every point z0 in the extended plane, including zo =0 and zo = oo. The fact that T
is continuous everywhere in the extended plane is now a consequence of equation (7)
SEC. 85 MAPPINGS BY l/z 303
(see Sec. 17). Because of this continuity, when the point at infinity is involved in any
discussion of the function 1/z, it is tacitly assumed that T(z) is intended.
85. MAPPINGS BY liz
When a point w = u + i v is the image of a nonzero point z = x + i y under the
transformation w = ljz, writing w =z/lzl2
reveals that
X
u - -,----------,-
- x2 + y2'
(1)
-y
v = ----::---=----
x2 + y2
Also, since z = 1jw = wflwl2
,
(2)
u
X-----::---
- u2 + v2'
-v
y = .u2 + v2
The following argument, based on these relations between coordinates, shows that the
mapping w = 1/z transforms circles and lines into circles and lines. When A, B, C,
and Dare all real numbers satisfying the condition B2
+ C2
> 4AD, the equation
(3) A(x
2
+ l) + Bx + Cy + D =0
represents an arbitrary circle or line, where A :f=. 0 for a circle and A = 0 for a line.
The need for the condition B2
+ C2
> 4AD when A :f=. 0 is evident if, by the method
of completing the squares, we rewrite equation (3) as
(
x !!__)
2
( +£)2
= (y'B2+C2-4AD)
2
·
+ 2A + y 2A 2A
When A = 0, the condition becomes B2 + C2 > 0, which means that B and C are not
both zero. Returning to the verification of the statement in italics, we observe that if
x andy satisfy equation (3), we can use relations (2) to substitute for those variables.
After some simplifications, we find that u and v satisfy the equation (see also Exercise
14 below)
(4)
which also represents a circle or line. Conversely, if u and v satisfy equation (4), it
follows from relations (1) that x and y satisfy equation (3).
It is now clear from equations (3) and (4) that
(i) a circle (A :f=. 0) not passing through the origin (D =I= 0) in the z plane is trans-
formed into a circle not passing through the origin in the w plane;
(ii) a circle (A =I= 0) through the origin (D =0) in the z plane is transformed into a
line that does not pass through the origin in the w plane;
304 MAPPING BY ELEMENTARY FUNCTIONS CHAP. 8
(iii) a line (A = 0) not passing through the origin (D f= 0) in the zplane is transformed
into a circle through the origin in the w plane;
(iv) a line (A = 0) through the origin (D = 0) in the z plane is transformed into a line
through the origin in the w plane.
EXAMPLE 1. According to equations (3) and (4), a vertical line x = c1 (c1 f= 0) is
transformed by w = 1/z into the circle -c1(u2 + v2) +u = 0, or
(5)
(
u- _1)2+ v2 = (-1)2.2c1 2c1
which is centered on the u axis and tangent to the v axis. The image of a typical point
(cl> y) on the line is, by equations (1 ),
(
CJ -y )
(u' v) = 2 2' 2 2 ·
c1 + y c1 +y
If c1> 0, the circle (5) is evidently to the right of the v axis. As the point (ch y)
moves up the entire line, its image traverses the circle once in the clockwise direction,
the point at infinity in the extended z plane corresponding to the origin in the w plane.
For if y < 0, then v > 0; and, as y increases through negative values to 0, we see that
u increases from 0 to 1jc1• Then, as y increases through positive values, vis negative
and u decreases to 0.
If, on the other hand, c1 < 0, the circle lies to the left of the v axis. As the point
(cl> y) moves upward, its image still makes one cycle, but in the counterclockwise
direction. See Fig. I06, where the cases c1 = l/3 and c1 =-1/2 are illustrated.
y v
c1 =-! c1 =-!
---- - ----c2 = ~
u
FIGURE
106
w = 1/z.
EXAMPLE 2. A horizontal line y = c2 (c2 =I= 0) is mapped by w = 1/z onto the
circle
(6) u
2
+ v+- = - ,( 1)2 (1)22c2 2c2
SEC.85 EXERCISES 305
which is centered on the v axis and tangent to the u axis. Two special cases are shown
in Fig. 106, where the corresponding orientations of the lines and circles are also
indicated.
EXAMPLE 3. When w = ljz, the half plane x > c1 (c1 > 0) is mapped onto the
disk
(7) (
u- _1)2+ v2 < (-1)2·2c1 2c1
For, according to Example 1, any line x = c (c > c1) is transformed into the circle
(8)
Furthermore, as c increases through all values greater than c1, the lines x = c move
to the right and the image circles (8) shrink in size. (See Fig. 107.) Since the lines
x =c pass through all points in the half plane x > c1 and the circles (8) pass through
all points in the disk (7), the mapping is established.
y
0 X
X= Ct X= C
EXERCISES
v
0 u
FIGURE107
w = 1/z.
1. In Sec. 85, point out how it follows from the first of equations (2) that when w = 1/z,
the inequality x > c1 (c1 > 0) is satisfied ifand only if inequality (7) holds. Thus give an
alternative verification of the mapping established in Example 3 in that section.
2. Show that when c1 < 0, the image of the half plane x < c1 under the transformation
w =1/z is the interior of a circle. What is the image when c1 =0?
3. Show that the image of the half plane y > c2 under the transformation w =1/z is the
interiorofa circle, provided c2 > 0. Find the image when c2 < 0; also find it when c2 = 0.
4. Find the image of the infinite strip 0 < y < 1/(2c) under the transformation w = 1/z.
Sketch the strip and its image.
Ans. u2
+ (v + c)2
> c2
, v < 0.
SEC. 86 LINEAR FRACTIONAL TRANSFORMATIONS 307
86. LINEAR FRACTIONAL TRANSFORMATIONS
The transformation
(1)
az+b
W=---
cz+d
(ad- be ::fo 0),
where a, b, c, and dare complex constants, is called a linearfractional transformation,
or Mobius transformation. Observe that equation (1) can be written in the form
(2) Azw + Bz +Cw + D = 0 (AD- BC ::fo 0);
and, conversely, any equation of type (2) can be put in the form (1). Since this
alternative form is linear in z and linear in w, or bilinear in z and w, another name
for a linear fractional transformation is bilinear transformation.
When e = 0, the condition ad- be# 0 with equation (1) becomes ad ::fo 0; and
we see that the transformation reduces to a nonconstant linear function. When e ::fo 0,
equation (1) can be written
(3)
a be- ad
W=-+---
C e
1
(ad- be ::fo 0).
ez+d
So, once again, the condition ad - be ::fo 0 ensures that we do not have a constant
function. The transformation w = 1/z is evidently a special case of transformation (1)
when c ::fo 0.
Equation (3) reveals that when c ::fo 0, a linear fractional transformation is a
composition of the mappings.
Z =ez +d, W=_!_
z'
a be-adW
W=-+---
c e
(ad- be# 0).
It thus follows that, regardless of whether e is zero or nonzero, any linear fractional
transformation transforms circles and lines into circles and lines because these special
linear fractional transformations do. (See Sees. 83 and 85.)
Solving equation (1) for z, we find that
(4)
-dw+b
z= (ad-be::foO).
cw -a
When a given point w is the image of some point z under transformation (1), the point
zis retrieved by means ofequation (4). If e =0, so that a and d are both nonzero, each,
point in the w plane is evidently the image of one and only one point in the z plane.;
The same is true if e ::fo 0, except when w = aje since the denominator in equation
(4) vanishes if w has that value. We can, however, enlarge the domain of definition
of transformation (1) in order to define a linear fractional transformation T on the
308 MAPPING BY ELEMENTARY FUNCTIONS CHAP. 8
extended z plane such that the point w =ajc is the image of z =oo when c "I 0. We
first write
(5)
We then write
and
T(z) = az + b
cz +d
(ad- be::/= 0).
T (oo) = oo if c = 0
T(oo) =: and r( ~) = oo if c "I 0.
In view of Exercise 11, Sec. 17, this makes T continuous on the extended z plane.
It also agrees with the way in which we enlarged the domain of definition of the
transformation w = ljz in Sec. 84.
When its domain of definition is enlarged in this way, the linear fractional
transformation (5) is a one to one mapping of the extended z plane onto the extended
w plane. That is, T (z 1) "I T (z2) whenever z1 "I z2; and, for each point w in the second
plane, there is a point z in the first one such that T(z) = w. Hence, associated with
the transformation T, there is an inverse transformation r-1, which is defined on the
extended w plane as follows:
T- 1
(w) = z if and only if T(z) = w.
From equation (4), we see that
(6) T-l(W) = -dw + b (ad- be =f: 0).
cw -a
Evidently, r-1 is itself a linear fractional transformation, where
T-oo)=oo if c=O
and
r-1
(:) = oo and r-1
(oo) =- ~ if c =I= 0.
IfT and S are two linear fractional transformations, then so is the composition S[T(z)].
This can be verified by combining expressions of the type (5). Note that, in particular,
r-1
[T(z)] = z for each point z in the extended plane.
There is always a linear fractional transformation that maps three given distinct
points Zb z2, and z3 onto three specified distinct points wb w 2, and w3, respectively.
Verification of this will appear in Sec. 87, where the image w of a point z under such
a transformation is given implicitly in terms of z. We illustrate here a more direct
approach to finding the desired transformation.
314 MAPPING BY ELEMENTARY FUNCTIONS
according to relations (2) and (3), equation (4) can be put in the form
(5)
· z -zow=ela_~
Z- Zt
where a is a real constant and z0 and z1 are (nonzero) complex constants.
CHAP. 8
Next, we impose on transformation (5) the condition that lwl = 1 when z = 1.
This tells us that
or
(1- z1)(1- z1) =(1- z0)(1- ZQ).
But ZtZt = zozo since lztl = lz0 1, and the above relation reduces to
z1 +z1 =zo +zo;
that is, Re z1 = Re z0. It follows that either
z1 = z0 or z1 = zo,
again since lz11= lz0 1. If z1 = z0, transformation (5) becomes the constant function
w = exp(ia); hence z1 =zo.
Transformation (5), with z1 = z0, maps the point zo onto the origin w =0;
and, since points interior to the circle IwI= 1 are to be the images of points above
the real axis in the z plane, we may conclude that Im z0 > 0. Any linear fractional
transformation having the mapping property stated in the first paragraph of this section
must, therefore, be of the form
(6) (lm zo > 0),
where a is real.
It remains to show that, conversely, any linear fractional transformation of the
form (6) has the desired mapping property. This is easily done by taking absolute
values of each side of equation (6) and interpreting the resulting equation,
I 1
- z- zow- -,
z- zo
geometrically. If a point z lies above the real axis, both it and the point zo lie on the
same side of that axis, which is the perpendicular bisector of the line segment joining
z0 and z0 . It follows that the distance lz - zol is less than the distance lz - z01(Fig.
108); that is, lwl < l. Likewise, if z lies below the real axis, the distance lz - zol
is greater than the distance lz - z0 1; and so lwl > 1. Finally, if z is on the real axis,
IwI= 1because then lz - zol = lz - z0 1. Since any linear fractional transformation is
a one to one mapping of the extended z plane onto the extended w plane, this shows
SEC.88 MAPPINGS OF THE UPPER HALF PLANE 315
that transformation (6) maps the halfplane Im z > 0 onto the disk lwl < 1 and the
boundary ofthe halfplane onto the boundary ofthe disk.
Our first example here illustrates the use of the result in italics just above.
EXAMPLE 1. The transformation
(7)
1-Z
W=--
i+z
in Examples 1 in Sees. 86 and 87 can be written
. z- iw =em ___
z -l
Hence it has the mapping property described in italics. (See also Fig. 13 in Appendix
2, where corresponding boundary points are indicated.)
Images of the upper half plane Im z > 0 under other types of linear fractional
transformations are often fairly easy to determine by examining the particular trans-
formation in question.
EXAMPLE 2. By writing z = x +iy and w = u +iv, we can readily show that the
transformation
z-1
w=--
z+l
(8)
maps the half plane y > 0 onto the half plane v > 0 and the x axis onto the u axis. We
first note that when the number z is real, so is the number w. Consequently, since the
image of the real axis y = 0 is either a circle or a line, it must be the real axis v = 0.
Furthermore, for any point w in the finite w plane,
v =Im w =Im (z - l)(z + l) = 2Y
(z + l)(z + 1) lz + 112
(z:F-1).
The numbers y and v thus have the same sign, and this means that points above the
x axis correspond to points above the u axis and points below the x axis correspond
to points below the u axis. Finally, since points on the x axis correspond to points
on the u axis and since a linear fractional transformation is a one to one mapping of
the extended plane onto the extended plane (Sec. 86), the stated mapping property of
transformation (8) is established.
Our final example involves a composite function and uses the mapping discussed
in Example 2.
EXAMPLE 3. The transformation
(9)
z -1
w =Log ,
z+l
316 MAPPING BY ELEMENTARY FUNCTIONS CHAP. 8
where the principal branch of the logarithmic function is used, is a composition of the
functions
z- 1
Z = and w =Log Z.
z+l
(10)
We know from Example 2 that the first of transformations (10) maps the upper
half plane y > 0 onto the upper half plane Y > 0, where z = x +i y and Z = X +i Y.
Furthermore, it is easy to see from Fig. 109 that the second of transformations (10)
maps the halfplane Y > 0 onto the strip 0 < v < n, where w =u +iv. More precisely,
by writing Z = R exp(iE>) and
Log Z = In R + iG (R > 0, -n < E> < n),
we see that as a point Z = R exp(iE>0) (0 < E>0 < n) moves outward from the origin
along the ray E> = G0, its image is the point whose rectangular coordinates in the w
plane are (In R, G0). That image evidently moves to the right along the entire length
of the horizontal line v = G0. Since these lines fill the strip 0 < v < n as the choice of
E>0 varies between G0 =0 to G0 = n, the mapping of the half plane Y > 0 onto the
strip is, in fact, one to one.
This shows that the composition (9) of the mappings (10) transforms the plane
y > 0 onto the strip 0 < v < n. Corresponding boundary points are shown in Fig. 19
of Appendix 2.
FIGURE 109
w=Logz.
EXERCISES
y
0
I
I
I
I
I
I
I
I
I
I
I
I
I
X
1. Recall from Example 1 in Sec. 88 that the transformation
z-z
w=
i+z
v
ni
8 0i____.....,.... _____
0 u
maps the half plane Im z > 0 onto the disk IwI < 1and the boundary of the half plane
onto the boundary of the disk. Show that a point z =x is mapped onto the point
1- x2 . 2x
w = +I '1+x2 1+x2
SEC. 88 EXERCISES 317
and then complete the verification of the mapping illustrated in Fig. 13, Appendix 2, by
showing that segments of the x axis are mapped as indicated there.
2. Verify the mapping shown in Fig. 12, Appendix 2, where
z -1
w=--
z+1
Suggestion: Write the given transformation as a composition of the mappings
i- zW= W=-W.
i +z'
Z = iz,
Then refer to the mapping whose verification was completed in Exercise 1.
3. (a) By finding the inverse ofthe transformation
i-z
w=--
i+z
and appealing to Fig. 13, Appendix 2, whose verification was completed in Exer-
cise 1, show that the transformation
.1- zW=l--
l+z
maps the disk lzl < 1onto the half plane Im w > 0.
(b) Show that the linear fractional transformation
can be written
Z - 7 -1
-"" '
z-2
w=
z
W
.1- z
=l '
I+Z
w=iW.
Then, with the aid of the result in part (a), verify that it maps the disk lz- 11 < 1
onto the left half plane Re w < 0.
4. Transformation (6), Sec. 88, maps the point z = oo onto the point w = exp(ia), which
lies on the boundary of the disk IwI < 1. Show that if 0 < a < 2rr and the points z = 0
and z =1are to be mapped onto the points w = 1and w =exp(ia/2), respectively, then
the transformation can be written
io: z +exp(-iaj2)
w=e .
z +exp(iot/2)
5. Note that when a = n:/2, the transformation in Exercise 4 becomes
iz +exp(irr/4)
w= .
z + exp(irr/4)
Verify that this special case maps points on the x axis as indicated in Fig. 110.
318 MAPPING BY ELEMENTARY FUNCTIONS CHAP. 8
y
FIGUREllO
iz +exp(irr/4)
w = ---"--'---'--
z +exp(irr/4)
6. Show that if Im zo < 0, transformation (6), Sec. 88, maps the lower half plane Im z < 0
onto the unit disk lwl < 1.
7. The equation w = log(z 1) can be written
Z = z - 1, w = log Z.
Find a branch of log Z such that the cut z plane consisting of all points except those on
the segment x > 1of the real axis is mapped by w = log(z - 1) onto the strip 0 < v < 2rr
in the w plane.
89. THE TRANSFORMATION w = sin z
Since (Sec. 33)
sin z = sin x cosh y +i cos x sinh y,
the transformation w =sin z can be written
(1) u = sin x cosh y, v = cos x sinh y.
One method that is often useful in finding images of regions under this transfor-
mation is to examine images of vertical lines x = c1• If 0 < c1 < n J2, points on the
line x =c1 are transformed into points on the curve
(2) u =sin c1cosh y, v =cos c1 sinh y
which is the right-hand branch of the hyperbola
(3)
with foci at the points
v2
---=1
cos2 c1
(-oo < y < oo),
w =±)sin2
c1 +cos2 c1 = ±1.
The second ofequations (2) shows that as a point (c1, y) moves upward along the entire
length of the line, its image moves upward along the entire length of the hyperbola's
branch. Such a line and its image are shown in Fig. 111, where corresponding points
are labeled. Note that, in particular, there is a one to one mapping ofthe top half (y > 0)
of the line onto the top half (v > 0) of the hyperbola's branch. If -n/2 < c1 < 0, the
SEC.89 THE TRANSFORMATION W =sin z 319
y v
F C
E B
_Zf 0 zr X u
2 2
D A FIGURE 111
w =smz.
line x =c1 is mapped onto the left-hand branch of the same hyperbola. As before,
corresponding points are indicated in Fig. Ill.
The line x =0, or the y axis, needs to be considered separately. According to
equations (1), the image of each point (0, y) is (0, sinh y). Hence they axis is mapped
onto the v axis in a one to one manner, the positive y axis corresponding to the positive
v ax1s.
We now illustrate how these observations can be used to establish the images of
certain regions.
EXAMPLE 1. Here we show that the transformation w = sin z is a one to one
mapping of the semi-infinite strip -n/2 < x < n /2, y > 0 in the z plane onto the
upper half v > 0 of the w plane.
To do this, we first show that the boundary of the strip is mapped in a one to one
manner onto the real axis in thew plane, as indicated in Fig. 112. The image of the
line segment B A there is found by writing x = n /2 in equations (1) and restricting y
to be nonnegative. Since u =cosh y and v =0 when x =rr j2, a typical point (rr j2, y)
on BA is mapped onto the point (cosh y, 0) in the w plane; and that image must move
to the right from B' along the u axis as (rr/2, y) moves upward from B. A point (x, 0)
on the horizontal segment DB has image (sin x, 0), which moves to the right from
D' to B' as x increases from x = -rr/2 to x = rr/2, or as (x, 0) goes from D to B.
Finally, as a point (-nj2, y) on the line segment DE moves upward from D, its image
(- cosh y, 0) moves to the left from D'. .
Now each point in the interior -rr/2 < x < rr/2, y > 0 of the strip lies on
one of the vertical half lines x = cl> y > 0 (-rr/2 < c1 < rr/2) that are shown in
E
D
y
M
0
c
L
A
B
1i X
2
v
E' A'
u
FIGURE 112
w = sm z.
320 MAPPING BY ELEMENTARY FUNCTIONS CHAP. 8
Fig. 112. Also, it is important to notice that the images of those half lines are distinct
and constitute the entire half plane v > 0. More precisely, if the upper half L of a line
x = c1 (0 < c1 < rr/2) is thought of as moving to the left toward the positive y axis,
the right-hand branch of the hyperbola containing its image L' is opening up wider
and its vertex (sin c1, 0) is tending toward the origin w = 0. Hence L' tends to become
the positive v axis, which we saw just prior to this example is the image of the positive
y axis. On the other hand, as L approaches the segment BA of the boundary of the
strip, the branch of the hyperbola closes down around the segment B'A' of the u axis
and its vertex (sin c1, 0) tends toward the point w = 1. Similar statements can be made
regarding the half line M and its image M' in Fig. 112. We may conclude that the
image of each point in the interior of the strip lies in the upper half plane v > 0 and,
furthermore, that each point in the half plane is the image of exactly one point in the
interior of the strip.
This completes our demonstration that the transformation w = sin z is a one to
one mapping of the strip -rr/2 < x < rrj2, y > 0 onto the half plane v > 0. The final
result is shown in Fig. 9, Appendix 2. The right-hand half of the strip is evidently
mapped onto the first quadrant of the w plane, as shown in Fig. 10, Appendix 2.
Another convenient way to find the images of certain regions when w = sin z
is to consider the images of horizontal line segments y = c2 ( -rr < x < rr), where
c2 > 0. According to equations (I), the image of such a line segment is the curve with
parametric representation
(4) u =sin x cosh c2, v =cos x sinh c2 (-rr<x<rr).
That curve is readily seen to be the ellipse
(5)
whose foci lie at the points
w = ±Jcosh2
c2 - sinh2
c2 = ±1.
The image of a point (x, c2) moving to the right from point A to point E in Fig. 113
makes one circuit around the ellipse in the clockwise direction. Note that when smaller
values of the positive number c2 are taken, the ellipse becomes smaller but retains the
same foci (±1, 0). In the limiting case c2 = 0, equations (4) become
u =smx, v =0 (-rr <x <rr);
and we find that the interval -rr < x < rr of the x axis is mapped onto the interval
-1 < u < 1 of the u axis. The mapping is not, however, one to one, as it is when
c2 > 0.
The following example relies on these remarks.
SEC.89 THE TRANSFORMATION W =sin Z 321
y v
A B c D E
y= c2 >0
C'
D'
-JC 1C 0 1C 1C
X u
-2 2
A' E'
FIGURE113
w =smz.
EXAMPLE 2. The rectangular region -n/2 < x < n j2, 0 < y < b is mapped by
w =sin z in a one to one manner onto the semi-elliptical region shown in Fig. 114,
where corresponding boundary points are also indicated. For if L is a line segment
y = c2 (-n/2 < x < n/2), where 0 < c2 < b, its image L' is the top half of the ellipse
(5). As c2 decreases, L moves downward toward the x axis and the semi-ellipse L'
also moves downward and tends to become the line segment E'F'A' from w = -1 to
w = 1. In fact, when c2 = 0, equations (4) become
U =Sin X, V = 0
and this is clearly a one to one mapping of the segment E FA onto E'F'A'. Inasmuch
as any point in the semi-elliptical region in the w plane lies on one and only one of
the semi-ellipses, or on the limiting case E'F'A', that point is the image of exactly
one point in the rectangular region in the z plane. The desired mapping, which is also
shown in Fig. 11 of Appendix 2, is now established.
y v
D bi c B
C'
L
E F A E' F' A' B'
-¥ 0 1( X
2
-1 0 1 cosh b u
FIGURE 114
w =smz.
Mappings by various other functions closely related to the sine function are easily
obtained once mappings by the sine function are known.
322 MAPPING BY ELEMENTARY FUNCTIONS
EXAMPLE 3. We need only recall the identity (Sec. 33)
cos z = sin( z + ~)
to see that the transfonnation w = cos z can be written successively as
rr
Z = z + - w = sin Z.
2'
CHAP. 8
Hence the cosine transfonnation is the same as the sine transfonnation preceded by a
translation to the right through rr/2 units.
EXAMPLE 4. According to Sec. 34, the transfonnation w = sinh z can be written
w = -i sin(iz), or
Z = i z, W = sin Z, w =- i W.
It is, therefore, a combination of the sine transformation and rotations through right
angles. The transformation w =cosh zis, likewise, essentially a cosine transformation
since cosh z = cos(iz).
EXERCISES
1. Show that the transformation w = sin z maps the top half (y > 0) of the vertical line
x =c1 (-rr/2 < c1 < 0) in a one to one manner onto the top half (v > 0) ofthe left-hand
branch of hyperbola (3), Sec. 89, as indicated in Fig. 112 ofthat section.
2. Show that under the transformation w =sin z, a line x = c1 (rr/2 < c1 < rr) is mapped
onto the right-hand branch of hyperbola (3), Sec. 89. Note that the mapping is one to
one and that the upper and lower halves ofthe line are mapped onto the lower and upper
halves, respectively, of the branch.
3. Vertical half lines were used in Example 1, Sec. 89, to show that the transformation
w = sin z is a one to one mapping of the open region -rr/2 < x < rrj2, y > 0 onto
the half plane v > 0. Verify that result by using, instead, the horizontal line segments
y- c2 (-rr/2 < x < rr/2), where c2 > 0.
4. (a) Show that under the transformation w =sin z, the images of the line segments
forming the boundary of the rectangular region 0 < x < rr/2, 0 < y < 1are the line
segments and the arc D'E' indicated in Fig. 115. The arc D'E' is a quarter of the
ellipse
(b) Complete the mapping indicated in Fig. 115 by using images of horizontal line
segments to prove that the transformation w = sin z establishes a one to one cor-
respondence between the interior points of the regions ABDE and A'B'D'E'.
SEC. 89 ExERCISES 323
y v
i
E D E'
F c F'
B
A 1C X A' B'C'D'U FIGURE US
2 w=sinz.
5. Verify that the interior of a reetangular region -JC < x < rr, a < y < b lying above the x
axis is mapped by w = sin z onto the interior of an elliptical ring which has a cut along
the segment -sinh b < v < -sinh a of the negative real axis, as indicated in Fig. 116.
Note that, while the mapping of the interior of the rectangular region is one to one, the
mapping of its boundary is not.
A B
-n:
c
7t X
v
FIGURE 116
w=sinz.
6. (a) Show that the equation w =cosh z can be written
Z
. 7t
= lZ +-, w =sin Z.
2
(b) Use the result in part (a), together with the mapping by sin z shown in Fig. 10,
Appendix 2, to verify that the transformation w =cosh z maps the semi-infinite
strip x > 0, 0 < y < rr/2 in the z plane onto the first quadrant u ~ 0, v > 0 of thew
plane. Indicate corresponding parts of the boundaries of the two regions.
7. Observe that the transformation w =cosh z can be expressed as a composition of the
mappings
1
W=Z +-,
z
1
W=-W.
2
Then, by referring to Figs. 7 and 16 in Appendix 2, show that when w = cosh z, the semi-
infinite strip x < 0, 0 < y < rr in the z plane is mapped onto the lower half v < 0 of the
w plane. Indicate corresponding parts of the boundaries.
8. (a) Verify that the equation w =sin z can be written
Z = i ( z + ~). W = cosh Z, w = - W.
324 MAPPING BY ELEMENTARY FUNCTIONS CHAP. 8
(b) Use the result in part (a) here and the one in Exercise 7 to show that the transformation
w =sin z maps the semi-infinite strip -n/2 < x < nf2, y > 0 onto the half plane
v > 0, as shown in Fig. 9, Appendix 2. (This mapping was verified in a different way
in Example 1, Sec. 89.)
90. MAPPINGS BY z2 AND BRANCHES OF z1
12
In Chap 2 (Sec. 12), we considered some fairly simple mappings under the transfor-
mation w = z2, written in the form
(1)
We tum now to aless elementary example and then examine related mappings w = z112,
where specific branches of the square root function are taken.
EXAMPLE 1. Let us use equations (1) to show that the image of the vertical strip
0 < x < 1, y > 0, shown in Fig. 117, is the closed semiparabolic region indicated there.
When 0 < x1 < 1, the point (x1, y) moves up a vertical halfline, labeled L1in Fig.
117, as y increases from y = 0. The image traced out in the uv plane has, according
to equations (1), the parametric representation
(2) 2 2
U =x 1 - y , V =2XJY (0 < y < oo).
Using the second of these equations to substitute for y in the first one, we see that the
image points (u, v) must lie on the parabola
(3)
with vertex at (xf, 0) and focus at the origin. Since v increases with y from v =0,
according to the second of equations (2), we also see that as the point (xb y) moves
up L 1 from the x axis, its image moves up the top half L~ of the parabola from the
u axis. Furthermore, when a number x2 larger than xi> but less than 1, is taken, the
corresponding halfline L 2 has an image L~ that is a half parabola to the right of L~, as
v
c B B' FIGURE 117
1 X
1 u
w =z2
•
SEC. go MAPPINGS BY z2
AND BRANCHES OF z112 325
indicated in Fig. 117. We note, in fact, that the image of the half line BA in that figure
is the top half of the parabola v2
= -4(u - 1), labeled B'A'.
The image of the half line CD is found by observing from equations (1) that a
typical point (0, y), where y > 0, on CD is transformed into the point (-y2, 0) in the
uv plane. So, as a point moves up from the origin along CD, its image moves left from
the origin along the u axis. Evidently, then, as the vertical half lines in the xy plane
move to the left, the half parabolas that are their images in the uv plane shrink down
to become the half line C'D'.
It is now clear that the images of all the half lines between and including CD and
BA fill up the closed semiparabolic region bounded by A'B'C'D'. Also, each point in
that region is the image of only one point in the closed strip bounded by ABCD. Hence
we may conclude that the semiparabolic region is the image of the strip and that there
is a one to one correspondence between points in those closed regions. (Compare Fig.
3 in Appendix 2, where the strip has arbitrary width.)
As for mappings by branches of z112, we recall from Sec. 8 that the values of z112
are the two square roots of zwhen z ::j:; 0. According to that section, if polar coordinates
are used and
z =r exp(i8) (r > 0, -T( < e < rr),
then
(4)
1/2 r::. i(8+2krr)
z =vr exp
2
(k=0,1),
the principal root occurring when k = 0. In Sec. 31, we saw that z1!2 can also be written
(5) z1
1
2
= exp(~ Jog z) (z :f:; 0).
The principal branch Fo(z) of the double-valued function z1
12
is then obtained by
taking the principal branch of log z and writing (see Sec. 32)
Since
Fo(z) = exp(~ Log z) (lzl > 0, -rr < Arg z < rr).
1 1 . i8
- Log z =- (ln r + t 8) = ln Jr + -
2 2 2
when z = r exp(i8), this becomes
(6)
i8
Fo(z) =v'rexp -
2
(r > 0, -T( < e < rr).
326 MAPPING BY ELEMENTARY FUNCTIONS CHAP. 8
The right-hand side of this equation is, of course, the same as the right-hand side of
equation (4) when k = 0 and -rr < 8 < n there. The origin and the ray 8 = n form
the branch cut for F0, and the origin is the branch point.
Images of curves and regions under the transformation w = F0(z) may be ob-
tained by writing w = p exp(i¢), where p = .y'r and¢= 8/2. Arguments are evi-
dently halved by this transformation, and it is understood that w = 0 when z = 0.
EXAMPLE 2. It is easy to verify that w = F0(z) is a one to one mapping of the
quarter disk 0 < r < 2, 0 < (} < n /2 onto the sector 0 < p < _,f2, 0 < ¢ < n /4 in the
w plane (Fig. 118). To do this, we observe that as a point z = r exp(i01) (0 < 01 < n/2)
moves outward from the origin along a radius R1 of length 2 and with angle of
inclination Ob its image w = .y'r exp(iBt/2) moves outward from the origin in the
w plane along a radius Ri whose length is ~ and angle of inclination is 01/2. See
Fig. 118, where another radius R2 and its imageR; are also shown. It is now clear from
the figure that if the region in the z plane is thought of as being swept out by a radius,
starting with D A and ending with DC, then the region in the w plane is swept out by
the corresponding radius, starting with D'A' and ending with D'C'. This establishes a
one to one correspondence between points in the two regions.
v
2
X D'
C'
A' u
FIGURE 118
w = F0(z).
EXAMPLE 3. The transformation w = F0(sin z) can be written
Z=sinz, w=F0(Z) (IZI > 0, -n < Arg Z < rr).
As noted at the end of Example 1 in Sec. 89, the first transformation maps the semi-
infinite strip 0 < x < n /2, y > 0 onto the first quadrant X > 0, Y > 0 in the Z plane.
The second transformation, with the understanding that F0 (0) = 0, maps that quadrant
onto an octant in the w plane. These successive transformations are illustrated in Fig.
119, where corresponding boundary points are shown.
When -rr < 8 <nand the branch
log z = ln r +i (e + 2rr)
SEC. 90
y
D
c
MAPPINGS BY z2
AND BRANCHES OF zlf2 327
FIGURE119
w = F0(sin z).
of the logarithmic function is used, equation (5) yields the branch
(7) F()
r.: i(8+2n)
1 z = v r exp - -
2
-- (r > 0, -JT < e < JT)
ofz1
12, which corresponds to k = 1in equation (4). Since exp(in) =-1, it follows that
F1(z) =-F0(z). The values ±F0(z) thus represent the totality of values of zl!2 at all
points in the domain r > 0, -n < 8 < n. If, by means of expression (6), we extend
the domain of definition of F0 to include the ray 8 =n and if we write F0(0) =0,
then the values ±Fo(z) represent the totality of values of z1
12 in the entire z plane.
Other branches of z112 are obtained by using other branches of log zin expression
(5). A branch where the ray() =a is used to form the branch cut is given by the equation
(8)
·e
fcAz) = ~ exp _l
2
(r > 0, a < () < a +2n).
Observe that when a =-n, we have the branch F0(z) and that when a = n, we have
the branch F1(z). Just as in the case of F0, the domain of definition of fa can be
extended to the entire complex plane by using expression (8) to define fa at the nonzero
points on the branch cut and by writing fa(O) = 0. Such extensions are, however, never
continuous in the entire complex plane.
Finally, suppose that n is any positive integer, where n > 2. The values of z11n are
the nth roots of z when z 1- 0; and, according to Sec. 31, the multiple-valued function
z1ln can be written
1/ ( 1 ) i(e +2kn)
(9) z n = exp n log z = yr;: exp n (k =0, 1, 2, ... , n - 1),
where r = lzl and 8 = Arg z. The case n = 2 has just been considered. In the general
case, each of the n functions
(10) F ( )
nC i (8 +2kJT)
k z =vrexp----
n
(k = 0, 1, 2, ... , n - 1)
is a branch of z1
1n, defined on the domain r > 0, -n < 8 < n. When w = pei¢, the
transformation w = Fk(Z) is a one to one mapping of that domain onto the domain
p > 0,
(2k - 1)n (2k + 1)n
-'----- < ¢ < .
n n
328 MAPPING BY ELEMENTARY FUNCTIONS CHAP. 8
These n branches of z11n yield then distinct nth roots of z at any point z in the domain
r > 0, -rr < e < rr. The principal branch occurs when k =0, and further branches
of the type (8) are readily constructed.
EXERCISES
1. Show. indicating corresponding orientations, that the mapping w =z2 transforms lines
y =c2 (c2 > 0) into parabolas v2 = 4c~(u + c~). all with foci at w = 0. (Compare
Example 1, Sec. 90.)
2. Use the result in Exercise 1to show that the transformation w = z2 is a one to one mapping
of a strip a < y < b above the x axis onto the closed region between the two parabolas
v2
= 4a2
(u +a2
), v2
=4b2
(u +b2
).
3. Point out how it follows from the discussion in Example 1, Sec. 90, that the transfor-
mation w = z2 maps a vertical strip 0 < x < c, y > 0 of arbitrary width onto a closed
semiparabolic region, as shown in Fig. 3, Appendix 2.
4. Modify the discussion in Example 1. Sec. 90, to show that when w =z2, the image of
the closed triangular region formed by the lines y = ± x and x = 1is the closed parabolic
region bounded on the left by the segment -2 < v < 2 of the v axis and on the right by
a portion of the parabola v2 = -4(u - 1). Verify the corresponding points on the two
boundaries shown in Fig. 120.
y v
2
A X
A'
-2
C'
l u
F'IGURE 120
1
w =z-.
5. By referring to Fig. 10, Appendix 2, show that the transformation w = sin2
z maps the
strip 0 < x < 7( f2, y > 0 onto the half plane v > 0. Indicate corresponding parts of the
boundaries.
Suggestion: Sec also the first paragraph in Example 3, Sec. 12.
6. Use Fig. 9, Appendix 2, to show that if w =(sin z)114, where the principal branch of
the fractional power is taken, the semi-infinite strip -7( /2 < x < n j2, y > 0 is mapped
onto the part of the first quadrant lying between the line v =u and the u axis. Label
corresponding parts of the boundaries.
SEC.91 SQUARE ROOTS OF POLYNOMIALS 329
7. According to Example 2, Sec. 88, the linear fractional transformation
Z=z-1
z+1
maps the x axis onto the X axis and the half planes y > 0 and y < 0 onto the halfplanes
Y > 0 and Y < 0, respectively. Show that, in particular, it maps the segment -1 < x < 1
of the x axis onto the segment X < 0 of the X axis. Then show that when the principal
branch of the square root is used, the composite function
( )
l/2
w =zl/2 = z- 1
z+1
maps the z plane, except for the segment 1 < x < 1of the x axis, onto the half plane
u >0.
8. Determine the image of the domain r > 0, -:rr < e < :rr in the z plane under each of
the transformations w = Fk(z) (k = 0, 1, 2, 3), where Fk(z) are the four branches ofzl!4
given by equation (10), Sec. 90, when n = 4. Use these branches to determine the fourth
roots of i.
91. SQUARE ROOTS OF POLYNOMIALS
We now consider some mappings that are compositions of polynomials and square
roots of z.
EXAMPLE 1. Branches of the double-valued function (z- z0) 112 can be obtained
by noting that it is a composition of the translation Z = z - zo with the double-valued
function z112. Each branch of z112 yields a branch of (z- z0) 112. When Z = ReiB,
branches of Z1/
2
are
'()
zl/2 =.JRexp l
2
Hence if we write
(R > 0, a < () <a+ 2:rr).
R = lz- zol. e = Arg(z- zo). and 0 = arg(z- zo).
two branches of (z zo)1
12
are
(1)
and
(2)
~n ie
Go(z) = v r< exp -
2
'()
go(z) = .JRexp _z
2
(R > 0, -n < E> < n)
(R > 0, 0 < () < 2n).
The branch of z112 that was used in writing G0(z) is defined at all points in the
Z plane except for the origin and points on the ray Arg Z = n. The transformation
SEC.91 SQUARE ROOTS OF POLYNOMIALS 331
The branch of (z + 1)112 given by equation (2) is
where
i02
fz(z) =Fz exp -
2
(r2 > 0, 0 < 02 < 2n),
r2 = lz + 11 and 02 = arg(z + 1).
The product of these two branches is, therefore, the branch f of (z2 - 1)112 defined
by the equation
(4)
i (01 + 02)
f (z) =Jrli2 exp ,
2
where
rk > 0, 0 < ()k < 2n (k=l,2).
As illustrated in Fig. 122, the branch f is defined everywhere in the z plane except on
the ray r2 > 0, 02 = 0, which is the portion x > -1 of the x axis.
The branch f of (z2 - 1)1
12 given in equation (4) can be extended to a function
(5)
i(01 +112)
F(z) =Jr1i2exp ,
2
where
rk > 0, 0 < (Jk < 2n (k = 1, 2) and rr + r2 > 2.
As we shall now see, this function is analytic everywhere in its domain of definition,
which is the entire z plane except for the segment -1 < x < 1 of the x axis.
Since F(z) =f(z) for all z in the domain of definition ofF except on the ray
r1 > 0, e1 = 0, we need only show that F is analytic on that ray. To do this, we form
the product of the branches of (z - 1)1
12 and (z + 1)112 which are given by equation
(1). That is, we consider the function
i(E>r + E>2)
G(z) =Jrli2exp ,
2
y
FIGURE 122
332 MAPPING BY ELEMENTARY FUNCTIONS CHAP. 8
where
rl = lz II, rz = lz +II, el = Arg(z- I), e2 = Arg(z + 1)
and where
(k =I, 2).
Observe that G is analytic in the entire z plane except for the ray r1 > 0, E>1 = rr.
Now F(z) = G(z) when the point z lies above or on the ray r1 > 0, e1 = 0; for then
(}k = E>k(k = 1, 2). When z lies below that ray, (}k = E>k +2rr (k = 1, 2). Consequently,
exp(iOk/2) = -exp(iE>k/2); and this means that
i((h +Oz) ( Wt) ( Wz) i(E>t + E>z)exp = exp - exp - = exp .
2 2 2 2
So again, F(z) = G(z). Since F(z) and G(z) are the same in a domain containing the
ray r1 > 0, e1 = 0 and since G is analytic in that domain, F is analytic there. Hence
F is analytic everywhere except on the line segment P2P1 in Fig. 122.
The function F defined by equation (5) cannot itself be extended to a function
which is analytic at points on the line segment P2P1; for the value on the right in
equation (5) jumps from i#2 to numbers near -i#2 as the point z moves
downward across that line segment. Hence the extension would not even be continuous
there.
The transformation w = F (z) is, as we shall see, a one to one mapping of the
domain Dz consisting of all points in the z plane except those on the line segment
P2P1 onto the domain Dw consisting of the entire w plane with the exception of the
segment -1 < v < I of the v axis (Fig. 123).
Before verifying this, we note that if z =iy (y > 0), then
r1 =r2 >1 and 01 +02 =rr;
hence the positive y axis is mapped by w =F(z) onto that part of the v axis for which
v > 1. The negative y axis is, moreover, mapped onto that part of the v axis for which
v < -1. Each point in the upper half y > 0 of the domain Dz is mapped into the upper
half v > 0 of the w plane, and each point in the lower half y < 0 of the domain Dz
y v
z w
Dw l
p2 P2
-1 1 X o• u
I
I
-~
FIGURE 123
w = F(z).
SEC.91 SQUARE ROOTS OF POLYNOMIALS 333
is mapped into the lower half v < 0 of thew plane. The ray r1 > 0, 01 =0 is mapped
onto the positive real axis in thew plane, and the ray r2 > 0, 02 = n is mapped onto
the negative real axis there.
To show that the transformation w = F(z) is one to one, we observe that if
F(z 1) = F(z2), then zi -1 =z~ -1. From this, it follows that z1 =z2 or z1 =-z2•
However, because of the manner in which F maps the upper and lower halves of the
domain Dz, as well as the portions of the real axis lying in Dz, the case z1 =-z2 is
impossible. Thus, if F(z 1) = F(z2), then z1 =z2; and F is one to one.
We can show that F maps the domain Dz onto the domain Dw by finding a
function H mapping Dw into Dz with the property that if z = H(w), then w = F(z).
This will show that, for any point w in Dw, there exists a point z in Dz such that
F(z) =w; that is, the mapping F is onto. The mapping H will be the inverse of F.
To find H, we first note that if w is a value of (z2 - 1)112 for a specific z, then
w2 =z2 - 1; and zis, therefore, a value of (w2 + 1)112 for that w. The function H will
be a branch of the double-valued function
(w =I= ±i).
Followingourprocedureforobtaining the function F(z), we write w- i = p1exp(i¢1)
and w +i = p2 exp(ic/>2). (See Fig. 123.) With the restrictions
n 3n
Pk > 0, -l < cl>k < 2 (k = 1, 2) and P1 +P2 > 2,
we then write
(6)
i (1'/>1 +1'/>2)
H(w)=~exp ,
2
the domain of definition being Dw. The transformation z = H(w) maps points of Dw
lying above or below the u axis onto points above or below the x axis, respectively. It
maps the positive u axis into that part of the x axis where x > 1and the negative u axis
into that part of the negative x axis where x < -1. If z = H (w), then z2 =w 2 + 1;
and so w2 =z2 - 1. Since z is in Dz and since F(z) and -F(z) are the two values of
(z2
- 1)112
for a point in Dz, we see that w = F(z) or w = -F(z). But it is evident
from the manner in which F and H map the upper and lower halves of their domains
of definition, including the portions of the real axes lying in those domains, that
w = F(z).
Mappings by branches of double-valued functions
(7)
334 MAPPING BY ELEMENTARY FUNCTIONS CHAP. 8
where A = - 2z0 and B = z6 = zt, can be treated with the aid of the results found for
the function F in Example 2 and the successive transformations
(8)
z- z0
z = -----=-,
Zt
W = czz- l)t;z, wW = Zt •
EXERCISES
1. The branch F of(z2 - 1)112 inExample2, Sec. 91, was defined in terms ofthe coordinates
r1, r2 , ()1, ()2• Explain geometrically why the conditions r1 > 0, 0 < 111 +112 < Jr describe
the quadrant x > 0, y > 0 of the z plane. Then show that the transformation w = F(z)
maps that quadrant onto the quadrant u > 0, v > 0 of the w plane.
Suggestion: To show that the quadrant x > 0, y > 0 in the z plane is described, note
that 111 +e2 = Jr at each point on the positive yaxis and that 01 +02 decreases as a point
z moves to the right along a ray e2 =c (0 < c < Jr/2).
2. For the transformation w = F (z) ofthe first quadrant ofthe zplane onto the first quadrant
of the w plane in Exercise 1, show that
1 I ? 2 1 I
u = ,fj_V r1r2 +x-- y - 1 and v = .fiYr 1r2 x2 + y2 + 1,
where
Crtrz)2
= (x
2
+ i + 1)
2
- 4x
2
,
and that the image of the portion of the hyperbola x2 - y2 = 1in the first quadrant is the
ray v =u (u > 0).
3. Show that in Exercise 2 the domain D that lies under the hyperbola and in the first
quadrant of the z plane is described by the conditions r1 > 0, 0 < e1+e2 < Jr/2. Then
show that the image of Dis the octant 0 < v < u. Sketch the domain D and its image.
4. Let F be the branch of (z2 - 1)112 defined in Example 2, Sec. 91, and let zo =r0 exp(i()0)
be a fixed point, where r0 > 0 and 0 < e0 < 2Jr. Show that a branch Fo of (z2
- z5)1
12
whose branch cut is the line segment between the points zo and -z0 can be written
Fo(z) = z0 F(Z), where Z = z/zo.
5. Write z - 1=r1 exp(ie1) and z + 1 = r2 exp(i8z), where
0 < el < 2Jr and - 7r < e2 < 7r'
to define a branch of the function
(b) (z-1)1/2.
z+l
In each case, the branch cut should consist of the two rays e1 = 0 and 8 2 = rr.
6. Using the notation in Sec. 91, show that the function
(
z 1)112
/?;1 i(01- 02)w = = - exp _.:__,_______,::.:..
z + 1 r2 2
SEC.92 RIEMANN SURFACES 335
is a branch with the same domain ofdefinition Dz and the same branch cut as the function
w = F(z) in that section. Show that this transformation maps Dz onto the right half plane
p > 0, -rr:/2 < q, < rr: j2, where the point w = 1is the image of the point z =oo. Also,
show that the inverse transformation is
1+w2
z=---
1- w2
(Compare Exercise 7, Sec. 90.)
(Rew>O).
7. Show that the transformation in Exercise 6 maps the region outside the unit circle lzl = 1
in the upper halfofthe zplane onto the region in the first quadrant ofthew plane between
the line v = u and the u axis. Sketch the two regions.
8. Write z =r exp(i8), z- 1= r1 exp(i81), and z 1= r2 exp(i82), where the values
of all three arguments lie between -rr: and rr:. Then define a branch of the function
[z(z2 - 1)]112 whose branch cut consists of the two segments x < -1 and 0 < x < 1 of
the x axis.
92. RIEMANN SURFACES
The remaining two sections of this chapter constitute a briefintroduction to the concept
of a mapping defined on a Riemann surface, which is a generalization of the complex
plane consisting of more than one sheet. The theory rests on the fact that at each point
on such a surface only one value of a given multiple-valued function is assigned. The
material in these two sections will not be used in the chapters to follow, and the reader
may skip to Chap. 9 without disruption.
Once a Riemann surface is devised for a given function, the function is single-
valued on the surface and the theory of single-valued functions applies there. Complex-
ities arising because the function is multiple-valued are thus re1ieved by a geometric
device. However, the description of those surfaces and the arrangement of proper con-
nections between the sheets can become quite involved. We limit our attention to fairly
simple examples and begin with a surface for log z.
EXAMPLE 1. Corresponding to each nonzero number z, the multiple-valued func-
tion
(1) log z = In r + if1
has infinitely many values. To describe log zas a single-valued function, we replace the
zplane, with the origin deleted, by a surface on which a new point is located whenever
the argument of the number z is increased or decreased by 2n, or an integral multiple
of2n.
We treat the zplane, with the origin deleted, as a thin sheet R0 which is cut along
the positive half of the real axis. On that sheet, let() range from 0 to 2n. Let a second
sheet R1 be cut in the same way and placed in front of the sheet R0. The lower edge of
the slitin R0 is then joined to the upper edge of the slit in R1• On R1, the angle() ranges
336 MAPPING BY ELEMENTARY FuNCTIONS CHAP. 8
from 2n to 4n; so, when z is represented by a point on R1, the imaginary component
of log zranges from 2n to 4n.
A sheet R2 is then cut in the same way and placed in front of R1• The lower edge
of the slit in R1 is joined to the upper edge of the slit in this new sheet, and similarly
for sheets R3, R4 , .... A sheet R_1on which() varies from 0 to -2n is cut and placed
behind R0, with the lower edge of its slit connected to the upper edge of the slit in R0;
the sheets R_2, R_3, ••• are constructed in like manner. The coordinates r and() of a
point on any sheet can be considered as polar coordinates ofthe projection of the point
onto the original z plane, the angular coordinate () being restricted to a definite range
of 2n radians on each sheet.
Consider any continuous curve on this connected surface of infinitely many
sheets. As a point zdescribes that curve, the values oflog zvary continuously since(), in
addition tor, varies continuously; and log znow assumes just one value corresponding
to each point on the curve. For example, as the point makes a complete cycle around
the origin on the sheet R0 over the path indicated in Fig. 124, the angle changes from
0 to 2n. As it moves across the ray () =2n, the point passes to the sheet R1 of the
surface. As the point completes a cycle in R1o the angle () varies from 2n to 4rr; and,
as it crosses the ray() = 4n, the point passes to the sheet R2.
yl
I
-I-.
/'I~
1 R0 1 ,
- - ...1 - - ::;<? - _,_ -1--.--
0 1 I X
' I ;I' ..... L_...,....
I
I
FIGURE 124
The surface described here is a Riemann surface for log z. It is a connected surface
of infinitely many sheets, arranged so that log z is a single-valued function of points
on it.
The transformation w =log z maps the whole Riemann surface in a one to one
manner onto the entire w plane. The image of the sheet R0 is the strip 0 < v < 2n (see
Example 3, Sec. 88). As a point zmoves onto the sheet R1 over the arc shown in Fig.
125, its image w moves upward across the line v =2n, as indicated in that figure.
h
v
21ti
01 I X
'-L.#RI 0
0 uI
FIGURE 125l
SEC.92 RIEMANN SURFACES 337
Note that log z, defined on the sheet RI• represents the analytic continuation (Sec.
26) of the single-valued analytic function
f(z) =In r +i8 (0 < () < 2.rr)
upward across the positive real axis. In this sense, log z is not only a single-valued
function of all points z on the Riemann surface but also an analytic function at all
points there.
The sheets could, ofcourse, be cut along the negative real axis, or along any other
ray from the origin, and properly joined along the slits to form other Riemann surfaces
for log z.
EXAMPLE 2. Corresponding to each point in the z plane other than the origin, the
square root function
(2)
has two values. A Riemann surface for zll2 is obtained by replacing the z plane with
a surface made up of two sheets Ro and R., each cut along the positive real axis and
with R1 placed in front of R0. The lower edge of the slit in R0 is joined to the upper
edge of the slit in RI, and the lower edge of the slit in RI is joined to the upper edge
of the slit in R0.
As a point zstarts from the upper edge ofthe slit in R0 and describes a continuous
circuit around the origin in the counterclockwise direction (Fig. 126), the angle ()
increases from 0 to 2.rr. The point then passes from the sheet R0 to the sheet R., where
() increases from 2.rr to 4.rr. As the point moves still further, it passes back to the sheet
R0, where the values of() can vary from 4.rr to 6.rr or from 0 to 2.rr, a choice that does
not affect the value of zi/2, etc. Note that the value of z112 at a point where the circuit
passes from the sheet R0 to the sheet R I is different from the value of zI/2 at a point
where the circuit passes from the sheet R1 to the sheet R0.
We have thus constructed a Riemann surface on which zll2 is single-valued for
each nonzero z. In that construction, the edges of the sheets R0 and R1 are joined in
pairs in such a way that the resulting surface is closed and connected. The points where
two of the edges are joined are distinct from the points where the other two edges are
joined. Thus it is physically impossible to build a model of that Riemann surface. In
-I-.
/'I".
t R0 1 ,
_.J __ ~-.1...-f--1---
01 1 X
' I ;I
' ..... L-"'
I
I
FIGURE 126
338 MAPPING BY ELEMENTARY FUNCTIONS CHAP. 8
visualizing a Riemann surface, it is important to understand how we are to proceed
when we arrive at an edge of a slit.
The origin is a special point on this Riemann surface. It is common to both sheets,
and a curve around the origin on the surface must wind around it twice in order to be
a closed curve. A point of this kind on a Riemann surface is called a branch point.
The image of the sheet R0 under the transformation w = z1
12
is the upper half
of the w plane since the argument of w is ()/2 on R0, where 0 < () j2 < rr. Likewise,
the image of the sheet R1 is the lower half of the w plane. As defined on either sheet,
the function is the analytic continuation, across the cut, of the function defined on the
other sheet. In this respect, the single-valued function z1
12
of points on the Riemann
surface is analytic at all points except the origin.
EXERCISES
1. Describe the Riemann surface for log z obtained by cutting the z plane along the negative
real axis. Compare this Riemann surface with the one obtained in Example 1, Sec. 92.
2. Determine the image under the transformation w =log z of the sheet Rn, where nis an
arbitrary integer, of the Riemann surface for log z given in Example 1, Sec. 92.
3. Verify that, under the transformation w = z112, the sheet R1 of the Riemann surface for
z112 given in Example 2, Sec. 92, is mapped onto the lower half of the w plane.
4. Describe the curve, on aRiemann surface for zl/2
, whose image is the entire circle Iw I= 1
under the transformation w =zl/2•
5. Let C denote the positively oriented circle lz- 21 = 1on the Riemann surface described
in Example 2, Sec. 92, for z1
12
, where the upper half of that circle lies on the sheet R0
and the lower half on R1. Note that, for each point z on C, one can write
z112 = Jrei012 where 4rr - 1r < e< 4rr + 1r.
2 2
State why it follows that
[ z1
12
dz = 0.
Generalize this result to fit the case ofthe other simple closed curves that cross from one
sheet to another without enclosing the branch points. Generalize to other functions, thus
extending the Cauchy-Goursat theorem to integrals of multiple-valued functions.
93. SURFACES FOR RELATED FUNCTIONS
We consider here Riemann surfaces for two composite functions involving simple
polynomials and the square root function.
SEC.93 SURFACES FOR RELATED FUNCTIONS 339
EXAMPLE 1. Let us describe a Riemann surface for the double-valued function
(1) f(z) = (z2 _ 1)112 =Jrii2exp i(01 + fh),
2
where z- 1=r1 exp(i01) and z + 1=r2 exp(i02). A branch of this function, with the
line segment P1P2 between the branch points z =±1 as a branch cut (Fig. 127), was
described in Example 2, Sec. 91. That branch is as written above, with the restrictions
rk > 0, 0 < Ok < 2rr (k = 1, 2) and r1 + r2 > 2. The branch is not defined on the
segment P1Pz.
y
X
FIGURE 127
A Riemann surface for the double-valued function (1) must consist of two sheets
of R0 and R1. Let both sheets be cut along the segment P1P2• The lower edge of the
slit in R0 is then joined to the upper edge of the slit in RI> and the lower edge in R1 is
joined to the upper edge in R0.
On the sheet R0, let the angles 01 and 02 range from 0 to 2rr. If a point on the
sheet R0 describes a simple closed curve that encloses the segment P1P2 once in the
counterclockwise direction, then both 01 and 02 change by the amount 2rr upon the
return of the point to its original position. The change in (01 +02)/2 is also 2rr, and
the value of f is unchanged. If a point starting on the sheet R0 describes a path that
passes twice around just the branch point z = 1, it crosses from the sheet R0 onto the
sheet R1and then back onto the sheet R0 before it returns to its original position. In this
case, the value of01 changes by the amount 4rr, while the value of 02 does not change
at all. Similarly, for a circuit twice around the point z = -1, the value of 02 changes
by 4rr, while the value of 01 remains unchanged. Again, the change in (01 +02) /2 is
2rr; and the value of f is unchanged. Thus, on the sheet R0, the range of the angles 01
and 02 may be extended by changing both 01 and 02 by the same integral multiple of
2rr or by changing just one of the angles by a multiple of 4rr. In either case, the total
change in both angles is an even integral multiple of 2rr.
To obtain the range of values for 01and 02 on the sheet R1, we note that if a point
starts on the sheet R0 and describes a path around just one of the branch points once, it
crosses onto the sheet R1 and does not return to the sheet R0. In this case, the value of
one of the angles is changed by 2rr, while the value of the other remains unchanged.
Hence on the sheet R1one angle can range from 2rr to 4rr, while the other ranges from
0 to 2rr. Their sum then ranges from 2rr to 4rr, and the value of (01 + 02) /2, which is
the argument of f (z), ranges from rr to 2rr. Again, the range of the angles is extended
340 MAPPING BY ELEMENTARY FUNCTIONS CHAP. 8
by changing the value of just one of the angles by an integral multiple of 4rr or by
changing the value of both angles by the same integral multiple of 2rr.
The double-valued function (1) may now be considered as a single-valued
function of the points on the Riemann surface just constructed. The transformation
w = f (z) maps each of the sheets used in the construction of that surface onto the
entire w plane.
EXAMPLE 2. Consider the double-valued function
(2) f(z) = [z(z2 - 1)]112 =.jrr1
r2
exp i(() +()1 +()2)
2
(Fig. 128). The points z = 0. ±I are branch points of this function. We note that if the
point z describes a circuit that includes all three of those points, the argument off(z)
changes by the angle 3rr and the value of the function thus changes. Consequently, a
branch cut must run from one of those branch points to the point at infinity in order to
describe a single-valued branch off. Hence the point at infinity is also a branch point,
as one can show by noting that the function f (1/z) has a branch point at z = 0.
Let two sheets be cut along the line segment L2 from z = -1 to z =0 and along
the part L1 of the real axis to the right of the point z = 1. We specify that each of the
three angles (), ()1> and ()2 may range from 0 to 2rr on the sheet R0 and from 2rr to
4rr on the sheet R1• We also specify that the angles corresponding to a point on either
sheet may be changed by integral multiples of 2rr in such a way that the sum of the
three angles changes by an integral multiple of 4rr. The value of the function f is,
therefore, unaltered.
A Riemann surlace for the double-valued function (2) is obtained by joining the
lower edges in R0 of the slits along L 1 and L2 to the upper edges in R1 of the slits
along L 1 and L2, respectively. The lower edges in R1 of the slits along L 1 and L2 are
then joined to the upper edges in R0 of the slits along L 1 and L2, respectively. It is
readily verified with the aid of Fig. 128 that one branch of the function is represented
by its values at points on R0 and the other branch at points on R1.
y
FIGURE 128
EXERCISES
1. Describe a Riemann surface for the triple-valued function w = (z- 1)1
!3, and point out
which third of the w plane represents the image of each sheet of that surface.
SEC.93 EXERCISES 341
2. Corresponding to each point on the Riemann surface described in Example 2, Sec. 93,
for the function w = j(z) in that example, there is just one value of w. Show that,
corresponding to each value of w, there are, in general, three points on the surface.
3. Describe a Riemann surface for the multiple-valued function
( )
1/2
J(z) = z ~ 1
4. Note that the Riemann surface described in Example 1, Sec. 93, for (z2 1)112 is also a
Riemann surface for the function
g(z) = z+ (z2- 1)1/2.
Let fo denote the branch of (z2 - 1)112 defined on the sheet R0, and show that the branches
g0 and g1 of g on the two sheets are given by the equations
1
go(z) = = z + fo(z).
8I(Z)
5. In Exercise 4, the branch fo of (z2
- 1)1
12
can be described by means of the equation
( i01)( i02)fo(z) =Jrl.F2 exp
2 exp
2 ,
where el and e2 range from 0 to 2n and
z- 1=r 1 exp(it11), z + 1 =rz exp(il12).
Note that 2z =r1 exp(ie1) + r2 exp(i82), and show that the branch g0 of the function
g(z) = z+ (z2 - 1)112 can be written in the form
1( ifh ie2 )
2
go(z) = - Jrlexp - +y'F2 exp - .
2 2 2
Find g0(z)g0(z), and note that r1 +r2 > 2 and cos[(e1 - e2)/2] > 0 for all z, to prove that
lgo(z)I > I. Then show that the transformation w = z + (z2 - 1)112maps the sheet R0 of
the Riemann surface onto the region lwl > 1, the sheet R1onto the region lwl < 1, and the
branch cut between the points z= ±1onto the circle Iw I= 1. Note that the transformation
used here is an inverse of the transformation
CHAPTER
9
CONFORMAL MAPPING
In this chapter, we introduce and develop the concept ofa conformal mapping, with em-
phasis on connections between such mappings and harmonic functions. Applications
to physical problems will follow in the next chapter.
94. PRESERVATION OF ANGLES
Let C be a smooth arc (Sec. 38), represented by the equation
z =z(t) (a<t<b),
and let f (z) be a function defined at all points z on C. The equation
w = f[z(t)] (a<t<b)
is a parametric representation ofthe imager of C under the transformation w = f(z).
Suppose that C passes through a point zo =z(t0) (a < t0 <b) at which f is
analytic and that f'(z0) =J 0. According to the chain rule given in Exercise 5, Sec.
38, if w(t) = j[z(t)], then
(1) w'(t0) = f'[z(t0)].z:'(t0);
and this means that (see Sec. 7)
(2) arg w'(t0) = arg f'[z(to)] +arg z'(to).
343
344 CoNFORMAL MAPPING CHAP. 9
Statement (2) is useful in relating the directions of C and r at the points zo and
wo = f(zo), respectively.
To be specific, let lfro denote a value of arg f'(z0), and let 00 be the angle of
inclination of a directed line tangent to C at zo (Fig. 129). According to Sec. 38, 00 is
a value of arg z'(t0); and it follows from statement (2) that the quantity
is a value of arg w'(t0) and is, therefore, the angle of inclination of a directed line
tangent tor at the point w0 = f(z0). Hence the angle of inclination of the directed
line at w0 differs from the angle of inclination of the directed line at zo by the angle
ofrotation
(3)
y v
0 X
1/lo =arg !'(zo).
['
u
FIGURE 129
¢o = 1/lo +Bo.
Now let C1 and C2 be two smooth arcs passing through z0, and let 01 and 02 be
angles of inclination of directed lines tangent to C1 and C2, respectively, at z0. We
know from the preceding paragraph that the quantities
are angles of inclination of directed lines tangent to the image curves r 1 and r 2,
respectively, at the point w0 = f(z0). Thus ¢2 -qy1 = 02 - 01; that is, the angle ¢2 - ¢1
from r 1 to r 2 is the same in magnitude and sense as the angle 02 -01 from C1 to C2.
Those angles are denoted by a in Fig. 130.
Because of this angle-preserving property, a transformation w = f(z) is said to
be conformal at a point z0 iff is analytic there and f' (zo) f= 0. Such a transformation
y v
0 X 0 u FIGURE 130
SEC.94 PRESERVATION OF ANGLES 345
is actually conformal at each point in a neighborhood of z0. For f must be analytic in
a neighborhood of Zo (Sec. 23); and, since r is continuous at zo (Sec. 48), it follows
from Theorem 2 in Sec. 17 that there is also a neighborhood of that point throughout
which f'(z) f= 0.
A transformation w = f(z), defined on a domain D, is referred to as a conformal
transformation, or conformal mapping, when it is conformal at each point in D. That
is, the mapping is conformal in D if f is analytic in D and its derivative f' has no
zeros there. Each of the elementary functions studied in Chap. 3 can be used to define
a transformation that is conformal in some domain.
EXAMPLE 1. The mapping w = ez is conformal throughout the entire zplane since
(eZ)' =ez f= 0 for each z. Consider any two lines x = c1 andy= c2 in the z plane, the
first directed upward and the second directed to the right. According to Sec. 13, their
images under the mapping w = ez are a positively oriented circle centered at the origin
and a ray from the origin, respectively. As illustrated in Fig. 20 (Sec. 13), the angle
between the lines at their point ofintersection is a right angle in the negative direction,
and the same is true of the angle between the circle and the ray at the corresponding
point in the w plane. The conformality of the mapping w = ez is also illustrated in
Figs. 7 and 8 of Appendix 2.
EXAMPLE 2. Consider two smooth arcs which are level curves u (x, y) = c1 and
v(x, y) = c2 of the real and imaginary components, respectively, of a function
f(z) = u(x, y) +iv(x, y),
and suppose that they intersect at a point z0 where f is analytic and f'(zo) f= 0. The
transformation w = f(z) is conformal at z0 and maps these arcs into the lines u = c1
and v = c2, which are orthogonal at the point w0 = f(z0). According to our theory,
then, the arcs must be orthogonal at z0. This has already been verified and illustrated
in Exercises 7 through 11 of Sec. 25.
A mapping that preserves the magnitude of the angle between two smooth arcs
but not necessarily the sense is called an isogonal mapping.
EXAMPLE 3. The transformation w = z, which is a reflection in the real axis,
is isogonal but not conformal. If it is followed by a conformal transformation, the
resulting transformation w = f mis also isogonal but not conformal.
Suppose that f is not a constant function and is analytic at a point z0. If, in
addition, f' (zo) = 0, then zo is called a critical point of the transformation w = f (z).
EXAMPLE 4. The point z = 0 is a critical point of the transformation
W = 1+z2
,
346 CONFORMAL MAPPING CHAP. 9
which is a composition of the mappings
Z =z2
and w = 1+ Z.
A ray () =a from the point z =0 is evidently mapped onto the ray from the point
w = 1 whose angle of inclination is 2a. Moreover, the angle between any two rays
drawn from the critical point z = 0 is doubled by the transformation.
More generally, it can be shown that if zo is a critical point of a transformation
w = f(z), there is an integer m(m > 2) such that the angle between any two smooth
arcs passing through z0 is multiplied by m under that transformation. The integer m is
the smallest positive integer such that f(m)(zo) :f=. 0. Verification of these facts is left
to the exercises.
95. SCALE FACTORS
Another property of a transformation w = f(z) that is conformal at a point zo is
obtained by considering the modulus of f'(z0). From the definition of derivative and a
property of limits involving moduli that was derived in Exercise 7, Sec. 17, we know
that
(1) 1/'(zo)l = lim f(z)- f(zo) = lim lf(z)- f(zo)l.
z-+zo Z - ZQ Z-+Zo lz - Zol
Now lz - zol is the length of a line segmentjoining zo and z, and If(z) - f (zo) Iis the
length of the line segmentjoining the points f(z0) and f(z) in thew plane. Evidently,
then, if zis near the point z0, the ratio
lf(z)- f(zo)l
lz- zol
of the two lengths is approximately the number I!'(z0) 1. Note that If'(z0) Irepresents
an expansion if it is greater than unity and a contraction if it is less than unity.
Although the angle of rotation arg f'(z) (Sec. 94) and the scale factor 1/'(z)l
vary, in general, from point to point, it follows from the continuity of f' that their
values are approximately arg J'(zo) and 1/'(zo)l at points z near z0. Hence the image
of a small region in a neighborhood of zo conforms to the original region in the sense
that it has approximately the same shape. A large region may, however, be transformed
into a region that bears no resemblance to the original one.
EXAMPLE. When f (z) =z2
, the transformation
w = /(z) = x2
- i +i2xy
SEC.95 SCALE FACTORS 347
is conformal at the point z = 1+i, where the half lines
y = x (x > 0) and x = 1 (x > 0)
intersect. We denote those half lines by C1 and C2, with positive sense upward, and
observe that the angle from C1 to C2 is rc/4 at their point of intersection (Fig. 13I).
Since the image of a point z = (x, y) is a point in the w plane whose rectangular
coordinates are
u =x2
-l and v =2xy,
the half line C1 is transformed into the curve r 1 with parametric representation
(2) u = 0, v = 2x2
(0 < x < oo).
Thus f 1 is the upper half v > 0 of the v axis. The half line C2 is transformed into the
curve r 2 represented by the equations
(3) (0 < y < oo).
Hence r2 is the upper half of the parabola v2
= -4(u- 1). Note that, in each case,
the positive sense of the image curve is upward.
y
0
v
X
FIGURE 131
~
w =z-.
If u and v are the variables in representation (3) for the image curve r 2, then
dv _ dvfdy
du dufdy
2
-2y
2
v
In particular, dv/ du = -I when v = 2. Consequently, the angle from the image curve
r 1 to the image curve r2 at the point w = f (I +i) =2i is rcf4, as required by the
conformality of the mapping at z = I +i. As anticipated, the angle of rotation rc/4 at
the point z = 1+i is a value of
arg[J'(l +i)] =arg[2(1 +i)] = rc +2mr
4
The scale factor at that point is the number
(n = 0, ±1, ±2, ...).
lf'(l + i)l = 12(1 + i)l = 2../2.
348 CONFORMAL MAPPING CHAP. 9
To illustrate how the angle of rotation and the scale factor can change from point
to point, we note that they are 0 and 2, respectively, at the point z= 1since f'(1) =2.
See Fig. 131, where the curves C2 and r2 are the ones just discussed and where the
nonnegative x axis C3 is transformed into the nonnegative u axis r3.
96. LOCAL INVERSES
A transformation w = f (z) that is conformal at a point zo has a local inverse there. That
is, if w0 = f(z0), then there exists a unique transformation z = g(w), which is defined
and analytic in a neighborhood N of w0, such that g(w0) =zo and f[g(w)] =w for
all points w in N. The derivative of g (w) is, moreover,
(1) ' I
g (w) = f'(z)
We note from expression (1) that the transformation z =g(w) is itself conformal at
wo.
Assuming that w = f (z) is, in fact, conformal at z0, let us verify the existence
of such an inverse, which is a direct consequence of results in advanced calculus.* As
noted in Sec. 94, the conformality of the transformation w =f(z) at zo implies that
there is some neighborhood of z0 throughout which f is analytic. Hence if we write
z =x +iy, zo =x0 +iy0 , and f(z) == u(x, y) +iv(x, y),
we know that there is a neighborhood of the point (x0, y0) throughout which the
functions u (x, y) and v (x, y) along with their partial derivatives of all orders, are
continuous (see Sec. 48).
Now the pair of equations
(2) u =u(x, y), v =v(x, y)
represents a transformation from the neighborhood just mentioned into the uv plane.
Moreover, the determinant
which is known as the Jacobian of the transformation, is nonzero at the point (x0, y0).
For, in view of the Cauchy-Riemann equations ux = vy and uy = -vx, one can write
J as
*The results from advanced calculus to be used here appear in, for instance, A. E. Taylor and W. R.
Mann, "Advanced Calculus,'' 3d ed., pp. 241-247, 1983.
SEC. 96 LoCAL INVERSES 349
and f'(z0) f=. 0 since the transformation w = f(z) is conformal at z0. The above con-
tinuity conditions on the functions u(x, y) and v(x, y) and their derivatives, together
with this condition on the Jacobian, are sufficient to ensure the existence of a local
inverse of transformation (2) at (x0, Yo). That is, if
(3) u0 =u(x0, y0) and v0 = v(x0, y0),
then there is a unique continuous transformation
(4) x = x(u, v), y = y(u, v),
defined on aneighborhood N ofthe point (u0, v0) and mapping that point onto (x0, y0),
such that equations (2) hold when equations (4) hold. Also, in addition to being
continuous, the functions (4) have continuous first-order partial derivatives satisfying
the equations
(5)
1
Yv = -ux
J
throughout N.
If we write w u +iv and w0 =u0 +iv0 , as well as
(6) g(w) =x(u, v) + iy(u, v),
the transformation z =g (w) is evidently the local inverse ofthe original transformation
w = f(z) at z0. Transformations (2) and (4) can be written
u +iv = u(x, y) +iv(x, y) and x + iy = x(u, v) + iy(u, v);
and these last two equations are the same as
w = f(z) and z =g(w),
where g has the desired properties. Equations (5) can be used to show that g is analytic
in N. Details are left to the exercises, where expression (1) for g'(w) is also derived.
EXAMPLE. We saw in Example I, Sec. 94, that if /(z) =ez, the transformation
w =f(z) is conformal everywhere in the z plane and, in particular, at the point
zo =2ni. The image of this choice of zo is the point w0 = 1. When points in the w
plane are expressed in the form w =p exp(i4J), the local inverse at zo can be obtained
by writing g(w) =log w, where log w denotes the branch
log w =In p +i4J (p > 0, n < e< 3n)
of the logarithmic function, restricted to any neighborhood of w0 that does not contain
the origin. Observe that
g(l) =In 1+ i2rr =2rri
350 CONFORMAL MAPPING CHAP. 9
and that, when w is in the neighborhood,
f[g(w)] =exp(log w) =w.
Also,
1 d 1 1
g (w) = - log w = - = - -
dw w expz
in accordance with equation (1).
Note that, if the point zo = 0 is chosen, one can use the principal branch
Log w =In p +icp (p > 0, -rc < c/J < rc)
of the logarithmic function to define g. In this case, g(l) =0.
EXERCISES
1. Determine the angle ofrotation at the point z =2 +i when the transfonnation is w =z2,
and illustrate it for some particular curve. Show that the scale factor ofthe transfonnation
at that point is 2.J5.
2. What angle of rotation is produced by the transfonnation w = 1/zat the point
(a) z = I; (b) z =i?
Ans. (a) n; (b) 0.
3. Show that under the transfonnation w = 1/z, the images of the lines y = x - 1 and
y =0 are the circle u2 +v2 - u - v =0 and the line v =0, respectively. Sketch all four
curves, determine corresponding directions along them, and verify the confonnality of
the mapping at the point z = 1.
4. Show that the angle ofrotation at a nonzero point zo = r0 exp(i00) under the transfonna-
tion w = zn (n = 1, 2, ...) is (n - l)e0. Detennine the scale factor of the transformation
at that point.
A
n-I
ns. nr0 .
5. Show that the transfonnation w =sin z is confonnal at all points except
JT
z=-+nn
2
(n =0, ±1, ±2, ...).
Note that this is in agreement with the mapping of directed line segments shown in Figs.
9, 10, and 11 of Appendix 2.
6. Find the local inverse of the transformation w = z2
at the point
(a) zo = 2; (b) zo = -2; (c) Zo =-i.
Ans. (a) wll2 = ,JPeicfJ/2 (p > 0, -n < ¢; < n);
(c) wlf2 = ,JPei¢12 (p > 0, 2n < ¢ < 4n).
7. In Sec. 96, it was pointed out that the components x(u, v) and y(u, v) of the inverse
function g(w) defined by equation (6) are continuous and have continuous first-order
SEC. 97 HARMONIC CONJUGATES 351
partial derivatives in the neighborhood N. Use equations (5), Sec. 96, to show that the
Cauchy-Riemann equations Xu= Yv• Xv = -yu hold inN. Then conclude that g(w) is
analytic in that neighborhood.
8. Show that if z = g(w) is the local inverse of a conformal transformation w = f(z) at a
point z0, then
I 1
g (w) = f'(z)
at points w in the neighborhood N where g is analytic (Exercise 7).
Suggestion: Start with the fact that f[g(w)] = w, and apply the chain rule for
differentiating composite functions.
9. Let C be a smooth arc lying in a domain D throughout which a transformation w =f (z)
is conformal, and let r denote the image of C under that transformation. Show that r is
also a smooth arc.
10. Suppose that a function f is analytic at z0 and that
for some positive integer m(m > 1). Also, write w0 = f(z0).
(a) Use the Taylor series for f about the point zo to show that there is a neighborhood
of zo in which the difference f(z)- w0 can be written
f(m)(z )
f(z)- wo = (z- zo)m 0
[1 +g(z)],
m!
where g(z) is continuous at zo and g(z0) = 0.
(b) Let r be the image of a smooth arc C under the transformation w =f (z), as shown
in Fig. 129 (Sec. 94), and note that the angles of inclination 80 and ¢0 in that figure
are limits of arg(z- zo) and arg[f(z) - w0], respectively, as z approaches zo along
the arc C. Then use the result in part (a) to show that 80 and ¢0 are related by the
equation
(c) Let a denote the angle between two smooth arcs C1 and C2 passing through zo.
as shown on the left in Fig. 130 (Sec. 94). Show how it follows from the relation
obtained in part (b) that the corresponding angle between the image curves r 1 and
r2 at the point w0 = f (zo) is ma. (Note that the transformation is conformal at z0
when m = 1and that zo is a critical point when m > 2.)
97. HARMONIC CONJUGATES
We saw in Sec. 25 that if a function
j(z) = u(x, y) + iv(x, y)
352 CONFORMAL MAPPING CHAP. 9
is analytic in a domain D, then the real-valued functions u and v are harmonic in that
domain. That is, they have continuous partial derivatives of the first and second order
in D and satisfy Laplace's equation there:
(1) Uxx +u yy = 0, V.u + Vyy = 0.
We had seen earlier that the first-order partial derivatives ofu and v satisfy the Cauchy-
Riemann equations
(2)
and, as pointed out in Sec. 25, v is called a harmonic conjugate of u.
Suppose now that u(x, y) is any given harmonic function defined on a simply
connected (Sec. 46) domain D. In this section, we show that u(x, y) always has a
harmonic conjugate v(x, y) in D by deriving an expression for v(x, y).
To accomplish this, we first recall some important facts about line integrals in
advanced calculus.* Suppose that P(x, y) and Q(x, y) have continuous first-order
partial derivatives in a simply connected domain D of the xy plane, and let (x0, y0)
and (x, y) be any two points in D. If Py = Qx everywhere in D, then the line integral
LP(s, t) ds + Q(s, t) dt
from (x0, Yo) to (x, y) is independent of the contour C that is taken as long as the
contour lies entirely in D. Furthermore, when the point (x0, y0) is kept fixed and (x, y)
is allowed to vary throughout D, the integral represents a single-valued function
l
(x,y)
F(x, y) = P(s, t) ds + Q(s, t) dt
(xo.Yo)
(3)
of x andy whose first-order partial derivatives are given by the equations
(4) Fx(x, y) = P(x, y), Fy(x, y) = Q(x, y).
Note that the value of F is changed by an additive constant when a different point
(xo, y0) is taken.
Returning to the given harmonic function u(x, y), observe how it follows from
Laplace's equation uxx + Uyy = 0 that
(-uy)y = (ux)x
everywhere in D. Also, the second-order partial derivatives of u are continuous in D;
and this means that the first-order partial derivatives of -uy and Ux are continuous
*See, for example, W. Kaplan, "Advanced Mathematics for Engineers," pp. 546-550, 1992.
SEC.98 TRANSFORMATIONS OF HARMONIC FUNCTIONS 353
there. Thus, if (x0, y0) is a fixed point in D, the function
l
(x ,y)
v(x, y) = -u1(s, t) ds + us(s, t) dt
(xo.Yo)
(5)
is well defined for all (x, y) in D~ and, according to equations (4),
(6)
These are the Cauchy-Riemann equations. Since the first-order partial derivatives of
u are continuous, it is evident from equations (6) that those derivatives of v are also
continuous. Hence (Sec. 21) u(x, y) +iv(x, y) is an analytic function in D; and vis,
therefore, a harmonic conjugate of u.
The function v defined by equation (5) is, of course, not the only harmonic
conjugate of u. The function v(x, y) + c, where c is any real constant, is also a
harmonic conjugate of u. [Recall Exercise 2, Sec. 25.]
EXAMPLE. Consider the function u (x, y) = x y, which is harmonic throughout the
entire xy plane. According to equation (5), the function
1
(x,y)
v(x,y)= -sds+tdt
{0,0)
is aharmonic conjugate of u(x, y). The integral here is readily evaluated by inspection.
It can also be evaluated by integrating first along the horizontal path from the point
(0, 0) to the point (x, 0) and then along the vertical path from (x, 0) to the point (x, y).
The result is
1 2 1 2
v(x,y)=-
2
x +
2
y,
and the corresponding analytic function is
. .
l 2 2 l 2
j(z) = xy- -(x - y ) = --z .
2 2
98. TRANSFORMATIONS OF HARMONIC FUNCTIONS
The problem of finding a function that is harmonic in a specified domain and satisfies
prescribed conditions on the boundary of the domain is prominent in applied mathe-
matics. If the values of the function are prescribed along the boundary, the problem
is known as a boundary value problem of the first kind, or a Dirichlet problem. If the
values of the normal derivative of the function are prescribed on the boundary, the
boundary value problem is one of the second kind, or a Neumann problem. Modifica-
tions and combinations of those types of boundary conditions also arise.
The domains most frequently encountered in the applications are simply con-
nected; and, since a function that is harmonic in a simply connected domain always
354 CONFORMAL MAPPING CHAP. 9
has a harmonic conjugate (Sec. 97), solutions of boundary value problems for such
domains are the real or imaginary parts of analytic functions.
EXAMPLE 1. In Example 1, Sec. 25, we saw that the function
T(x, y) =e-Y sinx
satisfies a certain Dirichlet problem for the strip 0 < x < rr, y > 0 and noted that it
represents a solution ofa temperature problem. The function T (x, y), which is actually
harmonic throughout the xy plane, is evidently the real part of the entire function
-ieiz = e-Y sin x - ie-Y cos x.
It is also the imaginary part of the entire function eiz.
Sometimes a solution of a given boundary value problem can be discovered by
identifying it as the real or imaginary part of an analytic function. But the success of
that procedure depends on the simplicity of the problem and on one's familiarity with
the real and imaginary parts of a variety of analytic functions. The following theorem
is an important aid.
Theorem. Suppose that an analytic function
(1) w = f(z) =u(x, y) + iv(x, y)
maps a domain Dz in the z plane onto a domain Dw in the w plane. If h(u, v) is a
harmonic function defined on Dw, then the function
(2) H(x, y) =h[u(x, y), v(x, y)]
is harmonic in Dz·
We first prove the theorem for the case in which the domain Dw is simply
connected. According to Sec. 97, that property of Dw ensures that the given harmonic
function h(u, v) has a harmonic conjugate g(u, v). Hence the function
(3) <P(w) = h(u, v) + ig(u, v)
is analytic in Dw. Since the function f(z) is analytic in Dz, the composite function
<P[j(z)] is also analytic in Dz. Consequently, the real part h[u(x, y), v(x, y)] of this
composition is harmonic in Dz. ·
If Dw is not simply connected, we observe that each point w0 in Dw has a
neighborhood lw- w0i < e lying entirely in Dw. Since that neighborhood is simply
connected, a function ofthe type (3) is analytic in it. Furthermore, since f is continuous
at a point z0 in Dz whose image is w0, there is aneighborhood lz - zoI< c5 whose image
is contained in the neighborhood Iw - w0I< e. Hence it follows that the composition
SEC.99 TRANSFORMATIONS OF BOUNDARY CoNDITIONS 355
<l>[f(z)] is analytic in the neighborhood lz- zol < 8, and we may conclude that
h[u(x, y), v(x, y)] is harmonic there. Finally, since w0 was arbitrarily chosen in Dw
and since each point in Dz is mapped onto such a point under the transformation
w = f(z), the function h[u(x, y), v(x, y)] must be harmonic throughout Dz.
The proof of the theorem for the general case in which Dw is not necessarily
simply connected can also be accomplished directly by means of the chain rule for
partial derivatives. The computations are, however, somewhat involved (see Exercise
8, Sec. 99).
EXAMPLE 2. The function h(u, v) = e-v sin u is harmonic in the domain Dw con-
sisting ofall points in the upper halfplane v > 0 (see Example 1). Ifthe transformation
is w =z2, then u(x, y) =x 2 - y2 and v(x, y) = 2xy; moreover, the domain Dz in the
z plane consisting of the points in the first quadrant x > 0, y > 0 is mapped onto the
domain Dw, as shown in Example 3, Sec. 12. Hence the function
H(x, y) = e-2xy sin(x2 - yl)
is harmonic in Dz.
EXAMPLE 3. Consider the function h(u, v) = Im w = v, which is harmonic in
the horizontal strip -n/2 < v < n j2. We know from Example 3, Sec. 88, that the
transformation w = Log z maps the right half plane x > 0 onto that strip. Hence, by
writing
Log z = In Jx2 + y2 + i arctan Y,
X
where -n/2 < arctan t < n j2, we find that the function
H(x, y) =arctan y
X
is harmonic in the half plane x > 0.
99. TRANSFORMATIONS OF BOUNDARY CONDITIONS
The conditions that a function or its normal derivative have prescribed values along
the boundary of a domain in which it is harmonic are the most common, although not
the only, important types of boundary conditions. In this section, we show that certain
of these conditions remain unaltered under the change of variables associated with a
conformal transformation. These results wi11 be used in Chap. 10 to solve boundary
value problems. The basic technique there is to transform a given boundary value
problem in the xy plane into a simpler one in the uv plane and then to use the theorems
of this and the preceding section to write the solution of the original problem in terms
of the solution obtained for the simpler one.
356 CONFORMAL MAPPING CHAP. 9
Theorem. Suppose that a transfonnation
(1) w = f(z) = u(x, y) + iv(x, y)
is conformal on a smooth arc C, and let r be the image ofC under that transfonnation.
If, along r, a function h(u, v) satisfies either ofthe conditions
(2)
dh
h =h0 or - =0,
dn
where h0 is a real constant and dhjdn denotes derivatives normal tor, then, along
C, the function
(3) H(x, y) =h[u(x, y), v(x, y)]
satisfies the corresponding condition
(4) H=ho or dH =0
dN '
where dHjdN denotes derivatives normal to C.
To show that the condition h = h0 on r implies that H = h0 on C, we note from
equation (3) that the value of Hat any point (x, y) on Cis the same as the value of
hat the image (u, v) of (x, y) under transformation (1). Since the image point (u, v)
lies on r and since h =ho along that curve, it follows that H =h0 along C.
(5)
Suppose, on the other hand, that dh/ dn = 0 on r. From calculus, we know that
dh
-=(grad h)· n,
dn
where grad h denotes the gradient of h at a point (u, v) on r and n is a unit vector
normal tor at (u, v). Since dhjdn = 0 at (u, v), equation (5) tells us that grad his
orthogonal ton at (u, v). That is, grad his tangent tor there (Fig. 132). But gradients
are orthogonal to level curves; and, because grad h is tangent to r, we see that r is
orthogonal to a level curve h(u, v) = c passing through (u, v).
y v
c
gradh
r
gradH
0 X 0 u FIGURE 132
SEC.99 TRANSFDRMATIONS OF BOUNDARY CONDITIONS 357
Now, according to equation (3), the level curve H(x, y) = c in the z plane can
be written
h[u(x, y), v(.x, y)] =c;
and so it is evidently transformed into the level curve h(u, v) = c under transformation
(1). Furthermore, since Cis transformed into rand r is orthogonal to the level curve
h(u, v) = c, as demonstrated in the preceding paragraph, it follows from the confor-
mality of transformation (1) on C that C is orthogonal to the level curve H (x, y) = c
at the point (x, y) corresponding to (u, v). Because gradients are orthogonal to level
curves, this means that grad His tangent to Cat (x, y) (see Fig. 132). Consequently,
if N denotes a unit vector normal to C at (x, y), grad H is orthogonal to N. That is,
(6)
Finally, since
(grad H)· N = 0.
dH
=(grad H)· N,
dN
we may conclude from equation (6) that dHjdN = 0 at points on C.
In this discussion, we have tacitly assumed that grad h f=. 0. If grad h = 0, it
follows from the identity
Igrad H (x, y) I = Igrad h(u, v) ll.f'(z) I,
derived in Exercise lO(a) below, that grad H =0; hence dhjdn and the corresponding
normal derivative dH1dN are both zero. We also assumed that
(i) grad h and grad H always exist;
(ii) the level curve H(.x, y) =cis smooth when grad h f=. 0 at (u, v).
Condition (ii) ensures that angles between arcs are preserved by transformation
(1) when it is conformal. In all of our applications, both conditions (i) and (ii) will be
satisfied.
EXAMPLE. Consider, for instance, the function h(u, v) = v + 2. The transforma-
tion
is conformal when z f=. 0. It maps the half line y = x (x > 0) onto the negative u
axis, where h = 2, and the positive x axis onto the positive v axis, where the normal
derivative hu is 0 (Fig. 133). According to the above theorem, the function
H (x, y) = x 2
- y2
+2
must satisfy the condition H = 2 along the half line y = x (x > 0) and Hy = 0 along
the positive x axis, as one can verify directly.
358 CONFORMAL MAPPING CHAP. 9
y v
C'
A
h =0u
B H =0y c X A' h=2 B' u
FIGURE 133
A boundary condition that is not ofone ofthe two types mentioned in the theorem
may be transformed into a condition that is substantially different from the original
one (see Exercise 6). New boundary conditions for the transformed problem can be
obtained for a particular transformation in any case. It is interesting to note that, under
a conformal transformation, the ratio of a directional derivative of H along a smooth
arc C in the z plane to the directional derivative of h along the image curve r at the
corresponding point in thew plane is lf'(z)l; usually, this ratio is not constant along
a given arc. (See Exercise 10.)
EXERCISES
1. Use expression (5), Sec. 97, to find a harmonic conjugate of the harmonic function
u(x, y) = x 3 - 3xy2. Write the resulting analytic function in terms of the complex
variable z.
2. Let u(x, y) be harmonic in a simply connected domain D. By appealing to results in
Sees. 97 and 48, show that its partial derivatives of all orders are continuous throughout
that domain.
3. The transformation w = exp z maps the horizontal strip 0 < y < :rc onto the upper half
plane v > 0, as shown in Fig. 6 of Appendix 2; and the function
,., ,., 2
h(u, v) = Re(w~) = u~- v
is harmonic in that half plane. With the aid of the theorem in Sec. 98, show that the
function H(x, y) = e2x cos 2y is harmonic in the strip. Verify this result directly.
4. Under the transformation w =exp z, the image of the segment 0 < y < :rc of the y axis
is the semicircle u2 +v2
=1, v > 0. Also, the function
h(u, v) = Re(2- w + ~) = 2- u + 2
u 2w u +v
is harmonic everywhere in the w plane except for the origin; and it assumes the value
h =2 on the semicircle. Write an explicit expression for the function H(x, y) defined
in the theorem of Sec. 99. Then illustrate the theorem by showing directly that H =2
along the segment 0 < y < :rc of the y axis.
SEC. 99 EXERCISES 359
5. The transformation w = z2
maps the positive x and y axes and the origin in the z plane
onto the u axis in the w plane. Consider the harmonic function
h(u, v) = Re(e-w) = e-u cos v,
and observe that its normal derivative hv along the u axis is zero. Then illustrate the
theorem in Sec. 99 when f(z) =z2 by showing directly that the normal derivative of the
function H (x, y) defined in that theorem is zero along both positive axes in the z plane.
(Note that the transformation w = z2 is not conformal at the origin.)
6. Replace the function h(u, v) in Exercise 5 by the harmonic function
h(u, v) = Re(~2iw + e-w) = 2v e-u cos v.
Then show that hv = 2 along the u axis but that Hy =4x along the positive x axis and
Hx =4y along the positive y axis. This illustrates how a condition of the type
dh
- = ho :f::. 0
dn
is not necessarily transformed into a condition of the type dHjdN =h0.
7. Show that if a function H (x, y) is a solution of a Neumann problem (Sec. 98), then
H (x, y) + A, where A is any real constant, is also a solution of that problem.
8. Suppose that an analytic function w = f(z) = u(x, y) + iv(x, y) maps a domain Dz in
the z plane onto a domain Dw in thew plane; and let a function h(u, v), with continuous
partial derivatives of the first and second order, be defined on Dw. Use the chain rule for
partial derivatives to show that if H(x, y) = h[u(x, y), v(x, y)], then
Hxx(X, y) + Hyy(x, y) = [huu(u, v) + hvv(U, v)]lf'(z)l
2
.
Conclude that the function H(x, y) is harmonic in Dz when h(u, v) is harmonic in
Dw. This is an alternative proof of the theorem in Sec.98, even when the domain Dw is
multiply connected.
Suggestion: In the simplifications, it is important to note that since f is analytic,
the Cauchy-Riemann equations ux = Vy, uy = -vx hold and that the functions u and v
both satisfy Laplace's equation. Also, the continuity conditions on the derivatives of h
ensure that hvu = huv·
9. Let p(u, v) be a function that has continuous partial derivatives of the first and second
order and satisfies Poisson's equation
Puu(u, v) + Pvv(u, v) =<l>(u, v)
in a domain Dw of the w plane, where <I> is a prescribed function. Show how it follows
from the identity obtained in Exercise 8 that if an analytic function
w = f(z) =u(x, y) + iv(x, y)
maps a domain Dz onto the domain Dw, then the function
P(x, y) = p[u(x, y), v(x, y)]
360 CONFORMAL MAPPING CHAP. 9
satisfies the Poisson equation
Pxx(x, y) + Pyy(x, y) =<P[u(x, y), v(x, y)JI/'(z)l 2
in Dz.
10. Suppose that w =f (z) =u(x, y) +iv(x, y) is a conformal mapping of a smooth arc C
onto a smooth arc r in thew plane. Let the function h(u, v) be defined on r, and write
H(x, y) =h[u(x, y), v(x, y)].
(a) From calculus, we know that the x and y components of grad H are the partial
derivatives Hx and Hy, respectively; likewise, grad h has components hu and hv.
By applying the chain rule for partial derivatives and using the Cauchy-Riemann
equations, show that if (x, y) is a point on C and (u, v) is its image on r, then
lgrad H(x, y)l = lgrad h(u, v)ll/'(z)l.
(b) Show that the angle from the arc C to grad Hat a point (x, y) on Cis equal to the
angle from r to grad hat the image (u, v) of the point (x, y).
(c) Let s and a denote distance along the arcs C and r, respectively; and let t and 't'
denote unit tangent vectors at a point (x, y) on C and its image (u, v), in the direction
of increasing distance. With the aid of the results in parts (a) and (b) and using the
fact that
dH dh
- = (grad H) · t and - = (grad h) · t',
ds da
show that the directional derivative along the arc r is transformed as follows:
dH = dh lf'(z)l.
ds da
CHAPTER
10
APPLICATIONS OF
CONFORMAL MAPPING
We now use conformal mapping to solve a number of physical problems involving
Laplace's equation in two independent variables. Problems in heat conduction, elec-
trostatic potential, and fluid flow will be treated. Since these problems are intended to
illustrate methods, they will be kept on a fairly elementary level.
100. STEADY TEMPERATURES
In the theory of heat conduction, the.flux across a surface within a solid body at a point
on that surface is the quantity of heat flowing in a specified direction normal to the
surface per unit time per unit area at the point. Flux is, therefore, measured in such
units as calories per second per square centimeter. It is denoted here by <l>, and it varies
with the normal derivative of the temperature T at the point on the surface:
<l>=-KdT
dN
(1) (K > 0).
Relation (1) is known as Fourier's law and the constant K is called the thermal
conductivity of the material of the solid, which is assumed to be homogeneous.*
The points in the solid are assigned rectangular coordinates in three-dimensional
space, and we restrict our attention to those cases in which the temperature T varies
*The law is named for the French mathematical physicist Joseph Fourier (1768-1830). A translation of
his book, cited in Appendix I, is a classic in the theory of heat conduction.
361
362 APPLICATIONS OF CoNFORMAL MAPPING CHAP. 10
with only the x and y coordinates. Since T does not vary with the coordinate along
the axis perpendicular to the xy plane, the flow of heat is, then, two-dimensional and
parallel to that plane. We agree, moreover, that the flow is in a steady state~ that is, T
does not vary with time.
It is assumed that no thermal energy is created or destroyed within the solid.
That is, no heat sources or sinks are present there. Also, the temperature function
T (x, y) and its partial derivatives of the first and second order are continuous at each
point interior to the solid. This statement and expression (1) for the flux of heat are
postulates in the mathematical theory ofheat conduction, postulates that also apply at
points within a solid containing a continuous distribution of sources or sinks.
Consider now an element of volume that is interior to the solid and that has the
shape of a rectangular prism of unit height perpendicular to the xy plane, with base
L).x by L).y in that plane (Fig. 134). The time rate offlow ofheat toward the right across
the left-hand face is - K Tx(x, y)L).y; and, toward the right across the right-hand face,
it is -KTx(x + L).x, y)L).y. Subtracting the first rate from the second, we obtain the
net rate of heat loss from the element through those two faces. This resultant rate can
be written
or
(2)
if L).x is very small. Expression (2) is, of course, an approximation whose accuracy
increases as L).x and L).y are made smaller.
y
~X
r=J~y
(x,y)
FIGURE 134
X
In like manner, the resultant rate ofheat loss through the otherfaces perpendicular
to the xy plane is found to be
(3)
Heat enters or leaves the element only through these four faces, and the temperatures
within the element are steady. Hence the sum of expressions (2) and (3) is zero; that
IS,
(4)
SEC. 101 STEADY TEMPERATURES IN A HALF PLANE 363
The temperature function thus satisfies Laplace's equation at each interior point of the
solid.
In view of equation (4) and the continuity of the temperature function and its
partial derivatives, T is a harmonicfunction ofx andy in the domain representing the
interior of the solid body.
The surfaces T (x, y) =c1, where c1 is any real constant, are the isotherms within
the solid. They can also be considered as curves in the xy plane; then T(x, y) can be
interpreted as the temperature at a point (x, y) in a thin sheet of material in that plane,
with the faces of the sheet thermally insulated. The isotherms are the level curves of
the function T.
The gradient ofT is perpendicularto the isotherm at each point, and the maximum
flux at a point is in the direction ofthe gradientthere. IfT (x, y) denotes temperatures in
a thin sheet and ifSis a harmonic conjugate ofthe function T, then a curve S (x, y) =c2
has the gradient of T as a tangent vector at each point where the analytic function
T (x, y) + iS(x, y) is conformal. The curves S(x, y) = c2 are called lines offlow.
Ifthe normal derivative d TIdN is zero along any part ofthe boundary ofthe sheet,
then the flux of heat across that part is zero. That is, the part is thermally insulated and
is, therefore, a line of flow.
The function T may also denote the concentration of a substance that is diffusing
through a solid. In that case, K is the diffusion constant. The above discussion and the
derivation of equation (4) apply as well to steady-state diffusion.
101. STEADY TEMPERATURES IN A HALF PLANE
Let us find an expression for the steady temperatures T(x, y) in a thin semi-infinite
plate y > 0 whose faces are insulated and whose edge y = 0 is kept at temperature
zero except for the segment -1 < x < 1, where it is kept at temperature unity (Fig.
135). The function T (x, y) is to be bounded; this condition is natural if we consider
the given plate as the limiting case of the plate 0 < y < Yo whose upper edge is kept
at a fixed temperature as Yo is increased. In fact, it would be physically reasonable to
stipulate that T (x, y) approach zero as y tends to infinity.
The boundary value problem to be solved can be written
(1) (-OO<X<OO,y>O),
y v
C' T=l B'
C'{_ D~: ls'A T=O B T=l T=O DX T=O u
FIGURE 135
z- I(r1 :rr 3:rr)w =log-- - > 0, -- < 81 - 82 < - .
z + 1 r2 2 2
364 APPLICATIONS OF CONFORMAL MAPPING CHAP. 10
(2) T(x O) = { 1 when lxl < 1,
' 0 when lxI > 1;
also, IT(x, y) I < M where M is some positive constant. This is a Dirichlet problem
for the upper half of the xy plane. Our method of solution will be to obtain a new
Dirichlet problem for a region in the uv plane. That region will be the image of the
half plane under a transformation w = f(z) that is analytic in the domain y > 0 and
that is conformal along the boundary y =0 except at the points (±1, 0), where it is
undefined. It will be a simple matter to discover a boundedharmonic function satisfying
the new problem. The two theorems in Chap. 9 will then be applied to transform the
solution ofthe problem in the uv plane into a solution ofthe original problem in the xy
plane. Specifically, a harmonic function of u and v will be transformed into a harmonic
function of x andy, and the boundary conditions in the uv plane will be preserved on
corresponding portions ofthe boundary in the xy plane. There should be no confusion
if we use the same symbol T to denote the different temperature functions in the two
planes.
Let us write
z- 1=r1 exp(i81) and z + 1=r2 exp(i82),
where 0 < ek < n (k =1, 2). The transformation
(3)
z- 1 r 1 •
w =log = ln- + z(81 - 82)
z + 1 r 2
is defined on the upper half plane y > 0, except for the two points z = ± 1, since
0 < 81 - 82 < n in the region. (See Fig. 135.) Now the value of the logarithm is the
principal value when 0 < e1 - 82 < n, and we recall from Example 3 in Sec. 88 that
the upper half plane y > 0 is then mapped onto the horizontal strip 0 < v < n in the
w plane. As already noted in that example, the mapping is shown with corresponding
boundary points in Fig. 19 of Appendix 2. Indeed, it was that figure which suggested
transformation (3) here. The segment of the x axis between z =-1 and z = 1, where
81 - 82 =n, is mapped onto the upper edge ofthe strip; and the rest ofthe x axis, where
81 - 82 = 0, is mapped onto the lower edge. The required analyticity and conformality
conditions are evidently satisfied by transformation (3).
A bounded harmonic function of u and v that is zero on the edge v =0 of the
strip and unity on the edge v = n is clearly
(4)
1
T= -v;
J'(
it is harmonic since it is the imaginary part of the entire function (ljn)w. Changing
to x and y coordinates by means of the equation
(5) z-1 (z- 1)w=ln +iarg ,
z+l z+l
SEC. 102 A RELATED PROBLEM 365
we find that
[
(z- l)(Z + 1)] [x2
+ y
2
- 1+ i2y]v=arg =arg ,
(z + l)(z + 1) (x + 1)2 + y2
or
v =arctan( 2
2
Y2 ) .
X + y -1
The range of the arctangent function here is from 0 to n since
arg ( z - 1)= ei - e2
z + 1
and 0 < 01 - 02 < n. Expression (4) now takes the form
(6) T = _!_ arctan (
2
2
Y
2
)
n x+y-1
(0 <arctan t < n).
Since the function (4) is harmonic in the strip 0 < v < n and since transformation
(3) is analytic in the half plane y > 0, we may apply the theorem in Sec. 98 to conclude
that the function (6) is harmonic in that half plane. The boundary conditions for the
two harmonic functions are the same on corresponding parts of the boundaries because
they are ofthe type h = h0, treated in the theorem of Sec. 99. The bounded function (6)
is, therefore, the desired solution of the original problem. One can, of course, verify
directly that the function (6) satisfies Laplace's equation and has the values tending to
those indicated on the left in Fig. 135 as the point (x, y) approaches the x axis from
above.
The isotherms T(x, y) = c1 (0 < c1 < 1) are arcs of the circles
x2 + (y -cot nc1)
2 = csc2net>
passing through the points (±1, 0) and with centers on they axis.
Finally, we note that since the product of a harmonic function by a constant is
also harmonic, the function
T = To arctan(
2
2
Y
2
)
n x+y-1
(0 <arctan t < n)
represents steady temperatures in the given half plane when the temperature T = 1
along the segment -1 < x < 1 of the x axis is replaced by any constant temperature
T =T0.
102. A RELATED PROBLEM
Consider a semi-infinite slab in the three-dimensional space bounded by the planes
x = ±n/2 and y = 0 when the first two surfaces are kept at temperature zero and the
SEC. 102 A RELATED PROBLEM 367
Since the denominator here reduces to sinh2
y cos2
x, the quotient can be put in the
form
2 cos x sinh y
sinh2
y - cos2 x
2(cosxl sinh y)
--'----'---=---,. =tan 2a,
1- (cos xl sinh y)2
where tan a = cos xI sinh y. Hence T = (2ITC )a; that is
(6) T = -2
arctan(-~-os_x_)
TC smh y ( 0 <arctan t < TC)·- - 2
This arctangent function has the range 0 to TC12 since its argument is nonnegative.
Since sin z is entire and the function (5) is harmonic in the half plane v > 0, the
function (6) is harmonic in the strip -TC12 < x < TC12, y > 0. Also, the function (5)
satisfies the boundary condition T = 1when iu I < 1and v = 0, as well as the condition
T =0 when lui > 1and v =0. The function (6) thus satisfies boundary conditions (2)
and (3). Moreover, IT(x, y) I < 1throughout the strip. Expression (6) is, therefore, the
temperature formula that is sought.
The isotherms T (x, y) = c1 (0 < c1 < 1) are the portions of the surfaces
cos x =tan( TC;
1
) sinh y
within the slab, each surface passing through the points (±TC12, 0) in the xy plane. If
K is the thermal conductivity, the flux of heat into the slab through the surface lying
in the plane y =0 is
2K
-KTy(x, 0) =- -
TC COS X
The flux outward through the surface lying in the plane x = TC12 is
-KT (TC ) - _2K_
X 2'y - • hTCSlll y
(y > 0).
The boundary value problem posed in this section can also be solved by the
method ofseparation ofvariables. That method is more direct, but it gives the solution
in the form of an infinite series.*
*A similar problem is treated in the authors' "Fourier Series and Boundary Value Problems," 6th ed.,
Problem 7, p. 142, 2001. Also, a short discussion of the uniqueness of solutions to boundary value
problems can be found in Chap. lO of that book.
368 APPLICATIONS OF CONFORMAL MAPPING CHAP. 10
103. TEMPERATURES IN A QUADRANT
Let us find the steady temperatures in a thin plate having the form of a quadrant if a
segment at the end of one edge is insulated, if the rest of that edge is kept at a fixed
temperature, and if the second edge is kept at another fixed temperature. The surfaces
are insulated, and so the problem is two-dimensional.
The temperature scale and the unit of length can be chosen so that the boundary
value problem for the temperature function T becomes
(1)
(2)
(3)
(x > 0, y > 0),
{
Ty(x, 0) = 0 when 0 < x < 1,
T(x,O)=l whenx>1,
T(O, y) = 0 (y > 0),
where T (x, y) is bounded in the quadrant. The plate and its boundary conditions
are shown on the left in Fig. 137. Conditions (2) prescribe the values of the normal
derivative ofthe function T over a part ofa boundary line and the values ofthe function
itself over the rest of that line. The separation of variables method mentioned at the end
of Sec. 102 is not adapted to such problems with different types of conditions along
the same boundary line.
As indicated in Fig. 10 of Appendix 2, the transformation
(4) z=smw
is a one to one mapping of the semi-infinite strip 0 < u < rr/2, v > 0 onto the quadrant
x > 0, y > 0. Observe now that the existence of an inverse is ensured by the fact
that the given transformation is both one to one and onto. Since transformation (4) is
conformal throughout the strip except at the point w = rr/2, the inverse transformation
must be conformal throughout the quadrant except at the point z = 1. That inverse
transformation maps the segment 0 < x < 1of the x axis onto the base of the strip and
the rest of the boundary onto the sides of the strip as shown in Fig. 137.
Since the inverse of transformation (4) is conformal in the quadrant, except when
z = 1, the solution to the given problem can be obtained by finding a function that is
y
v---
T=O
v
T=O T=l
C' B'
cjaa~J!t>rr~uu
2 FIGURE 137
SEC. 103 TEMPERATURES IN A QUADRANT 369
harmonic in the strip and satisfies the boundary conditions shown on the right in Fig.
137. Observe that these boundary conditions are of the types h =h0 and dhfdn =0
in the theorem of Sec. 99.
The required temperature function T for the new boundary value problem is
clearly
(5)
2
T= -u,
rr
the function (2/rr)u being the real part of the entire function (2/rr)w. We must now
express T in terms of x and y.
To obtain u in terms of x andy, we first note that, according to equation (4),
(6) x =sin u cosh v, y =cos u sinh v.
When 0 < u < rr/2, both sin u and cos u are nonzero; and, consequently,
(7)
Now it is convenient to observe that, for each fixed u, hyperbola (7) has foci at the
points
z = ±Jsin2
u +cos2 u = ±1
and that the length of the transverse axis, which is the line segment joining the two
vertices, is 2 sin u. Thus the absolute value of the difference of the distances between
the foci and a point (x, y) lying on the part of the hyperbola in the first quadrant is
J<x + 1)2+ y2 - J<x- 1)2+ y2 = 2 sin u.
It follows directly from equations (6) that this relation also holds when u = 0 or
u = rr/2. In view of equation (5), then, the required temperature function is
(8)
2 . [ J(x + 1)2 + y2 _ J(x _ 1)2 + y2]
T =- arcsm ,
rr 2
where, since 0 < u < rr/2, the arcsine function has the range 0 torr/2.
If we wish to verify that this function satisfies boundary conditions (2), we must
remember that J(x - 1)2 denotes x- 1 when x> 1 and 1 - xwhen 0 < x< 1, the
square roots being positive. Note, too, that the temperature at any point along the
insulated part of the lower edge of the plate is
T(x, 0) = ~ arcsin x
rr
(0 <X< 1).
It can be seen from equation (5) that the isotherms T(x, y) = c1 (0 < c1 < 1)
are the parts of the confocal hyperbolas (7), where u = rrc!l2, which lie in the first
370 APPLICATIONS OF CONFORMAL MAPPING CHAP. IO
quadrant. Since the function (2/n)v is a harmonic conjugate of the function (5), the
lines of flow are quarters of the confocal ellipses obtained by holding v constant in
equations (6).
EXERCISES
1. In the problem of the semi-infinite plate shown on the left in Fig. 135 (Sec. 101), obtain
a harmonic conjugate of the temperature function T(x, y) from equation (5), Sec. 101,
and find the lines of flow of heat. Show that those lines of flow consist of the upper half
ofthey axis and the upper halves ofcertain circles on either side of that axis, the centers
of the circles lying on the segment AB or CD of the x axis.
2. Show that if the function T in Sec. 101 is not required to be bounded, the harmonic
function (4) in that section can be replaced by the harmonic function
T =Im ( ~ w+ A cosh w) = ~ v + A sinh u sin v,
where A is an arbitrary real constant. Conclude that the solution ofthe Dirichlet problem
for the strip in the uv plane (Fig. 135) would not, then, be unique.
3. Suppose that the condition that T be bounded is omitted from the problem for temper-
atures in the semi-infinite slab of Sec. 102 (Fig. 136). Show that an infinite number of
solutions are then possible by noting the effect of adding to the solution found there the
imaginary part of the function A sin z. where A is an arbitrary real constant.
4. Use the function Log z to find an expression for the bounded steady temperatures in a
plate having the form of a quadrant x > 0, y > 0 (Fig. 138) ifits faces are perfectly insu-
lated and its edges have temperatures T(x, 0) =0 and T(O, y) =1. Find the isotherms
and lines of flow, and draw some of them.
Ans. T = ~ arctan(~).
y
T=l
T=O X
FIGURE138
5. Find the steady temperatures in a solid whose shape is that ofa long cylindrical wedge if
its boundary planes() = 0 and() = 00 (0 < r < r0) are kept at constant temperatures zero
and T0, respectively, and if its surface r =r0 (0 < e < e0) is perfectly insulated (Fig.
139).
Ans. T = To arctan ( Y) .
00 x
372 APPLICATIONS OF CONFORMAL MAPPING
Suggestion: This problem can be transformed into the one in Exercise 4.
2 (tanh y)Ans. H = - arctan .
rc tan x
y
H= 1 H=O
H=Orc x
2 FIGURE 142
CHAP. 10
9. Derive an expression for temperatures T (r, 8) in a semicircular plate r < 1, 0 < e< rc
with insulated faces if T =1 along the radial edge e=0 (0 < r < 1) and T =0 on the
rest of the boundary.
Suggestion: This problem can be transformed into the one in Exercise 8.
Ans. T = ~ arctan(
1
- r cot
8
).
rc l+r 2
10. Solve the boundary value problem for the plate x > 0, y > 0 in the z plane when the
faces are insulated and the boundary conditions are those indicated in Fig. 143.
Suggestion: Use the mapping
. ·-l lZ
w----
- z- lzl2
to transform this problem into the one posed in Sec. 103 (Fig. 137).
T=I
T=O X
FIGURE 143
11. The portions x < 0 (y =0) and x < 0 (y =rc) of the edges ofan infinite horizontal plate
0 < y < rc are thermally insulated, as are the faces of the plate. Also, the conditions
T(x, 0) = 1 and T(x, rr) = 0 are maintained when x > 0 (Fig. 144). Find the steady
temperatures in the plate.
Suggestion: This problem can be transformed into the one in Exercise 6.
12. Consider a thin plate, with insulated faces, whose shape is the upper half of the region
enclosed by an ellipse with foci (±1, 0). The temperature on the elliptical part of its
SEC. 104 ELECTROSTATIC POTENTIAL 373
rei T = 0
T=l X
FIGURE 144
boundary is T = 1. The temperature along the segment -1 < x < 1ofthe x axis is T = 0,
and the rest of the boundary along the x axis is insulated. With the aid of Fig. 11 in
Appendix 2, find the lines of flow of heat.
13. According to Sec. 50 and Exercise 7 of that section, if f(z) =u(x, y) + iv(x, y) is
continuous on a closed bounded region R and analytic and not constant in the interior of
R, then the function u(x, y) reaches its maximum and minimum values on the boundary
of R, and never in the interior. By interpreting u(x, y) as a steady temperature, state a
physical reason why that property of maximum and minimum values should hold true.
104. ELECTROSTATIC POTENTIAL
In an electrostatic force field, the field intensity at a point is a vector representing the
force exerted on a unit positive charge placed at that point. The electrostatic potential
is a scalar function of the space coordinates such that, at each point, its directional
derivative in any direction is the negative of the component of the field intensity in that
direction.
For two stationary charged particles, the magnitude of the force of attraction or
repulsion exerted by one particle on the other is directly proportional to the product
of the charges and inversely proportional to the square of the distance between those
particles. From this inverse-square law, it can be shown that the potential at a point
due to a single particle in space is inversely proportional to the distance between the
point and the particle. In any region free of charges, the potential due to a distribution
of charges outside that region can be shown to satisfy Laplace's equation for three-
dimensional space.
If conditions are such that the potential V is the same in all planes parallel to
the xy plane, then in regions free of charges Vis a harmonic function ofjust the two
variables x andy:
Vxx(x, y) + Vyy(x, y) =0.
The field intensity vector at each point is parallel to the xy plane, with x and y
components - Vx(x, y) and - Vy(x, y), respectively. That vector is, therefore, the
negative of the gradient of V (x, y).
A surface along which V (x, y) is constant is an equipotential surface. The
tangential component ofthe field intensity vector at a point on a conducting surface is
zero in the static case since charges are free to move on such a surface. Hence V (x, y)
is constant along the surface of a conductor, and that surface is an equipotential.
374 APPLICATIONS OF CONFORMAL MAPPING CHAP. 10
If U is a harmonic conjugate of V, the curves U(x, y) =c2 in the xy plane are
calledflux lines. When such a curve intersects an equipotential curve V (x, y) =c1 at
a point where the derivative of the analytic function V(x, y) +i U(x, y) is not zero,
the two curves are orthogonal at that point and the field intensity is tangent to the flux
line there.
Boundary value problems for the potential V are the same mathematical problems
as those for steady temperatures T; and, as in the case of steady temperatures, the
methods of complex variables are limited to two-dimensional problems. The problem
posed in Sec. 102 (see Fig. 136), for instance, can be interpreted as that of finding the
two-dimensional electrostatic potential in the empty space
Jr Jr
--<X<- y > 0
2 2'
boundedby the conducting planes x =± n/2 and y =0, insulated at their intersections,
when the first two surfaces are kept at potential zero and the third at potential unity.
The potential in the steady flow of electricity in a plane conducting sheet is also
a harmonic function at points free from sources and sinks. Gravitational potential is a
further example of a harmonic function in physics.
105. POTENTIAL IN A CYLINDRICAL SPACE
A long hollow circular cylinder is made out of a thin sheet of conducting material,
and the cylinder is split lengthwise to form two equal parts. Those parts are separated
by slender strips of insulating material and are used as electrodes, one of which is
grounded at potential zero and the other kept at a different fixed potential. We take the
coordinate axes and units of length and potential difference as indicated on the left in
Fig. 145. We then interpret the electrostatic potential V (x, y) over any cross section of
the enclosed space that is distant from the ends of the cylinder as a harmonic function
inside the circle x2
+ y2 = 1in the xy plane. Note that V = 0 on the upper half of the
circle and that V =1on the lower half.
y v
V=O
1 X V=1 1 V=O u
FIGURE 145
A linear fractional transformation that maps the upper half plane onto the interior
of the unit circle centered at the origin, the positive real axis onto the upper half ofthe
circle, and the negative real axis onto the lower half of the circle is verified in Exercise
SEC. 105 POTENTIAL IN A CYLINDRICAL SPACE 375
1, Sec. 88. The result is given in Fig. 13 of Appendix 2; interchanging z and w there,
we find that the inverse of the transfonnation
(1)
I - W
z=--
i + w
gives us a new problem for V in a half plane, indicated on the right in Fig. 145.
Now the imaginary part of the function
(2)
1 1 i
- Log w = - In p + -¢ (p > 0, 0 < rp < TC)
TC TC TC
is a bounded function of u and v that assumes the required constant values on the two
parts ¢ =0 and ¢ =TC of the u axis. Hence the desired harmonic function for the half
plane is
(3) V = ~ arctan(:).
where the values of the arctangent function range from 0 to TC.
The inverse of transfonnation (1) is
.1- z
w = l '
l+z
(4)
from which u and v can be expressed in terms of x andy. Equation (3) then becomes
(5) V = _!.. arctan(
1
x
2
y
2
)
TC 2y
(0 < arctan t < TC).
The function (5) is the potential function for the space enclosed by the cylindrical
electrodes since it is hannonic inside the circle and assumes the required values on the
semicircles. If we wish to verify this solution, we must note that
lim arctan t =0 and lim arctan t =TC.
t........Y.O
t<O
The equipotential curves V (x, y) = c1 (0 < c1 < 1) in the circular region are arcs
of the circles
with each circle passing through the points (±1, 0). Also, the segment of the x axis
between those points is the equipotential V (x, y) = 1/2. A harmonic conjugate U of
V is -(1/TC) In p, or the imaginary part of the function -{i jTC) Log w. In view of
equation (4), U may be written
1 1 - '7
U =--In "·
rr 1+z
376 APPLICATIONS OF CONFORMAL MAPPING CHAP. IO
From this equation, it can be seen that the flux lines U (x, y) =c2 are arcs of circles
with centers on the x axis. The segment of the y axis between the electrodes is also a
flux line.
EXERCISES
1. The harmonic function (3) of Sec. 105 is bounded in the half plane v > 0 and satisfies
the boundary conditions indicated on the right in Fig. 145. Show that if the imaginary
part of Aew, where A is any real constant, is added to that function, then the resulting
function satisfies all of the requirements except for the boundedness condition.
2. Show that transformation (4) ofSec. 105 maps the upperhalfofthe circular region shown
on the left in Fig. 145 onto the first quadrant of the w plane and the diameter CE onto the
positive v axis. Then find the electrostatic potential V in the space enclosed by the half
cylinder x2 + y2 = 1, y > 0 and the plane y = 0 when V = 0 on the cylindrical surface
and V = 1 on the planar surface (Fig. 146).
2 ( 1 x2 2)Ans. V = - arctan - - y .
lf 2y
y
V=O
-1 V= 1 1 X FIGURE 146
3. Find the electrostatic potential V (r, 0) in the space 0 < r < 1, 0 < e< n /4, bounded by
the half planes e= 0 and e=lf /4 and the portion 0 < e< lf/4 of the cylindrical surface
r = 1, when V =1on the planar surfaces and V =0 on the cylindrical one. (See Exercise
2.) Verify that the function obtained satisfies the boundary conditions.
4. Note that all branches of log z have the same real component, which is harmonic
everywhere except at the origin. Then write an expression for the electrostatic potential
V (x, y) in the space between two coaxial conducting cylindrical surfaces x2 +y2 = 1
and x2 +y2 = rji (r0 :j:. 1) when V = 0 on the first surface and V = I on the second.
ln(x
2
+i>Ans. V = .2ln r0
5. Find the bounded electrostatic potential V (x, y) in the space y > 0 bounded by an infinite
conducting plane y = 0 one strip (-a < x <a, y = 0) of which is insulated from the
rest of the plane and kept at potential V = 1, while V = 0 on the rest (Fig. 147). Verify
that the function obtained satisfies the boundary conditions.
Ans. V =_!.arctan(
2
la;
2
) (0 <arctan t < rr).
rr x +y -a
SEC. 105 EXERCISES 377
y
a
V=O V=l V=O X FIGURE 147
6. Derive an expression for the electrostatic potential in the semi-infinite space indicated in
Fig. 148, bounded by two half planes and a half cylinder, when V = 1on the cylindrical
surface and V = 0 on the planar surfaces. Draw some of the equipotential curves in the
xy plane.
Ans. V =~arctan( 2
Zy
1
).
TC X y2
y
FIGURE 148
7. Find the potential V in the space between the planes y = 0 and y = rc when V = 0 on
the parts of those planes where x > 0 and V = 1 on the parts where x < 0 (Fig. 149).
Check the result with the boundary conditions.
Ans. V = _.!_ arctan(-~-in-'-y-)
rc smhx
(0 <arctan t < rr).
y
V=l rri V= 0
I
I J_
V=l V=O X
FIGURE 149
8. Derive an expression for the electrostatic potential V in the space interior to a long
cylinder r = 1when V =0 on the first quadrant (r = 1, 0 < () < rr/2) of the cylindrical
surface and V = 1 on the rest (r = 1, rc/2 < () < 2rr) of that surface. (See Exercise 5,
Sec. 88, and Fig. 110 there.) Show that V =3/4 on the axis of the cylinder. Check the
result with the boundary conditions.
9. Using Fig. 20 of Appendix 2, find a temperature function T (x, y) that is harmonic in the
shaded domain of the xy plane shown there and assumes the values T = 0 along the arc
SEC. 106 1vo-DIMENSIONAL FLUID FLOW 379
V=O V=O X FIGURE 151
106. TWO-DIMENSIONAL FLUID FLOW
Harmonic functions play an important role in hydrodynamics and aerodynamics.
Again, we consider only the two-dimensional steady-state type of problem. That is,
the motion of the fluid is assumed to be the same in all planes parallel to the xy plane,
the velocity being parallel to that plane and independent of time. It is, then, sufficient
to consider the motion of a sheet of fluid in the x y plane.
We let the vector representing the complex number
v =p + iq
denote the velocity of a particle of the fluid at any point (x, y); hence the x and y
components of the velocity vector are p(x, y) and q (x, y), respectively. At points
interior to a region of flow in which no sources or sinks of the fluid occur, the
real-valued functions p(x, y) and q(x, y) and their first-order partial derivatives are
assumed to be continuous.
The circulation of the fluid along any contour C is defined as the line integral
with respect to arc length a of the tangential component VT(x, y) of the velocity
vector along C:
(1) fc VT(x, y) da.
The ratio of the circulation along C to the length of C is, ther~fore, a mean speed of
the fluid along that contour. It is shown in advanced calculus that such an integral can
be written*
(2) fc VT(x, y) da = fc p(x, y) dx +q(x, y) dy.
When C is a positively oriented simple closed contour lying in a simply connected
domain of flow containing no sources or sinks, Green's theorem (see Sec. 44) enables
*Properties of line integrals in advanced calculus that are used in this and the following section are to
be found in, for instance, W. Kaplan, "Advanced Mathematics for Engineers," Chap. 10, 1992,
380 APPLICATIONS OF CONFORMAL MAPPING CHAP. 10
us to write
fc p(x, y) dx +q(x, y) dy =JL[qx(x, y)- Py(x, y)] dA,
where R is the closed region consisting of points interior to and on C. Thus
(3) fc Vr(x, y) da = JL[qx(x, y)- Py(x, y)] dA
for such a contour
A physical interpretation of the integrand on the right in expression (3) for the
circulation along the simple closed contour C is readily given. We let C denote a circle
ofradius r which is centered at a point (x0 , y0) and taken counterclockwise. The mean
speed along C is then found by dividing the circulation by the circumference 2rrr,
and the corresponding mean angular speed of the fluid about the center of the circle is
obtained by dividing that mean speed by r:
l !11 . . .]d-
2
-[qx(x, y)- Py(x, y) A.
nr R 2
Now this is also an expression for the mean value of the function
(4)
1
w(x, y) =
2
[qx(x, y)- Py(x, y)]
over the circular region R bounded by C. Its limit as r tends to zero is the value of
w at the point (x0, y0). Hence the function w (x, y), called the rotation of the fluid,
represents the limiting angular speed of a circular element of the fluid as the circle
shrinks to its center (x, y), the point at which w is evaluated.
If w (x, y) = 0 at each point in some simply connected domain, the flow is
irrotational in that domain. We consider only irrotational flows here, and we also
assume that the fluid is incompressible andfreefrom viscosity. Under our assumption
of steady irrotational flow of fluids with uniform density p, it can be shown that the
fluid pressure P(x, y) satisfies the following special case of Bernoulli's equation:
p 1
- + -IV12
=constant.
p 2
Note that the pressure is greatest where the speed IVI is least.
Let D be a simply connected domain in which the flow is irrotational. According
to equation (4), Py = qx throughout D. This relation between partial derivatives implies
that the line integral
fc p(s, t) ds +q(s, t) dt
SEC. 107 THE STREAM FUNCTION 381
along a contour C lying entirely in D and joining any two points (x0, y0) and (x, y)
in D is actually independent of path. Thus, if (x0, y0) is fixed, the function
l
(x ,y)
(j>(x, y) = p(s, t) ds +q(s, t) dt
(xo.Yol
(5)
is well defined on D; and, by taking partial derivatives on each side of this equation,
we find that
(6) 4>x(x, y) =p(x, y), 4>y(x, y) =q(x, y).
From equations (6), we see that the velocity vector V = p + iq is the gradient of
4>; and the directional derivative of 4> in any direction represents the component of the
velocity of flow in that direction.
The function (j>(x, y) is called the velocity potential. From equation (5), it is
evident that 4> (x, y) changes by an additive constant when the reference point (x0, Yo)
is changed. The level curves (j>(x, y) = c1 are called equipotentials. Because it is the
gradient of 4> (x, y), the velocity vector V is normal to an equipotential at any point
where V is not the zero vector.
Just as in the case of the flow of heat, the condition that the incompressible fluid
enter or leave an element of volume only by flowing through the boundary of that
element requires that 4> (x, y) must satisfy Laplace's equation
4>xx(x, y) +4>yy(X, y) =0
in a domain where the fluid is free from sources or sinks. In view of equations (6)
and the continuity of the functions p and q and their first-order partial derivatives, it
follows that the partial derivatives of the first and second order of 4> are continuous in
such a domain. Hence the velocity potential 4> is a hannonic function in that domain.
107. THE STREAM FUNCTION
According to Sec. I06, the velocity vector
(1) V =p(x, y) + iq(x, y)
for a simply connected domain in which the flow is irrotational can be written
(2) V = 4>x(x, y) +i(j>y(x, y) =grad (j>(x, y),
where 4> is the velocity potential. When the velocity vector is not the zero vector, it
is normal to an equipotential passing through the point (x, y). If, moreover, 'if!(x, y)
denotes a harmonic conjugate of4> (x, y) (see Sec. 97), the velocity vector is tangent to
a curve 'if/(x, y) = c2. The curves 1/f(x, y) = c2 are called the streamlines of the flow,
and the function 1/1 is the streamfunction. In particular, a boundary across which fluid
cannot flow is a streamline.
382 APPLICATIONS OF CONFORMAL MAPPING
The analytic function
F(z) =!/J(x, y) +i1/f(x, y)
is called the complex potential of the flow. Note that
F'(z) =!/Jx(X, y) +i1frx(x, y),
or, in view of the Cauchy-Riemann equations,
F'(z) = !/Jx(x, y)- i!/Jy(x, y).
Expression (2) for the velocity thus becomes
(3) V = F'(z).
The speed, or magnitude of the velocity, is obtained by writing
lVI = IF'(z)l.
CHAP. IO
According to equation (5), Sec. 97, if!fJ is harmonic in a simply connected domain
D, a harmonic conjugate of !fJ there can be written
J
(x,y)
1/f(x,y)= - -!/J1(s,t)ds+!/J8
(s,t)dt,
(xo.Yo)
where the integration is independent of path. With the aid of equations (6), Sec. 106,
we can, therefore, write
(4) 1/f(x,y)= L-q(s,t)ds+p(s,t)dt,
where Cis any contour in D from (x0 , y0) to (x, y).
Now it is shown in advanced calculus that the right-hand side of equation (4)
represents the integral with respect to arc length a along C of the normal component
VN(x, y) of the vector whose x andy components are p(x, y) and q(x, y), respec-
tively. So expression (4) can be written
(5) 1/f(x, y) = LVN(s, t) da.
Physically, then, 1/1(x, y) represents the time rate of flow of the fluid across C. More
precisely, 1/1(x, y) denotes the rate of flow, by volume, across a surface of unit height
standing perpendicular to the xy plane on the curve C.
EXAMPLE. When the complex potential is the function
(6) F(z) = Az,
SEC. 108 FLOWS AROUND A CORNER AND AROUND A CYLINDER 383
where A is a positive real constant,
(7) tf>(x, y) =Ax and 1/f(x, y) = Ay.
The streamlines 1/f(x, y) =c2 are the horizontal lines y =c2/ A, and the velocity at
any point is
V=F'(z)=A.
Here a point (x0, y0) at which 1/f(x, y) =0 is any point on the x axis. If the point
(x0, y0) is taken as the origin, then 1/f(x, y) is the rate offlow across any contour drawn
from the origin to the point (x, y) (Fig. 152). The flow is uniform and to the right. It
can be interpreted as the uniform flow in the upper half plane bounded by the x axis,
which is a streamline, or as the uniform flow between two parallel lines y = y1 and
Y = Y2·
Yl
I
(x,y) 1
--.---.--:r:- V
s:~ : X FIGURE 152
The stream function 1/1 characterizes a definite flow in a region. The question of
whether just one such function exists corresponding to a given region, except possibly
for a constant factor or an additive constant, is not examined here. In some of the
examples to follow, where the velocity is uniform far from the obstruction, or in Chap.
11, where sources and sinks are involved, the physical situation indicates that the flow
is uniquely determined by the conditions given in the problem.
A harmonic function is not always uniquely determined, even up to a constant
factor, by simply prescribing its values on the boundary of a region. In this example,
the function 1/f(x, y) = Ay is harmonic in the half plane y > 0 and has zero values
on the boundary. The function 1/11(x, y) = Bex sin y also satisfies those conditions.
However, the streamline 1/11(x, y) =0 consists not only of the line y =0 but also of
the lines y =nrr(n =1, 2, ...). Here the function F1(z) = Bez is the complex potential
for the flow in the strip between the lines y =0 and y =rr, both lines making up the
streamline 1/f (x, y) =0; if B > 0, the fluid flows to the right along the lower line and
to the left along the upper one.
108. FLOWS AROUND A CORNER
AND AROUND A CYLINDER
In analyzing a flow in the xy, or z. plane, it is often simpler to consider a corresponding
flow in the uv, or w, plane. Then, if 4> is a velocity potential and 1/f a stream function
for the flow in the uv plane, results in Sees. 98 and 99 can be applied to these harmonic
384 APPLICATIONS OF CONFORMAL MAPPING CHAP. 10
functions. That is, when the domain offlow Dw in the uv plane is the image ofa domain
Dz under a transformation
w = f(z) =u(x, y) + iv(x, y),
where f is analytic, the functions
q>[u(x, y), v(x, y)] and 1/f[u(x, y), v(x, y)]
are harmonic in Dz. These new functions may be interpreted as velocity potential and
stream function in the xy plane. A streamline or natural boundary 1/f(u, v) = c2 in the
uv plane corresponds to a streamline or natural boundary 1/f[u(x, y), v(x, y)] = c2 in
the xy plane.
In using this technique, it is often most efficient to first write the complex potential
function for the region in the w plane and then obtain from that the velocity potential
and stream function for the corresponding region in the xy plane. More precisely, if
the potential function in the uv plane is
F(w) = q>(u, v) + i1/f(u, v),
then the composite function
F[f(z)] =q>[u(x, y), v(x, y)] + i1/f[u(x, y), v(x, y)]
is the desired complex potential in the xy plane.
In order to avoid an excess ofnotation, we use the same symbols F, 4>, and 1/1 for
the complex potential, etc., in both the xy and the uv planes.
EXAMPLE 1. Consider a flow in the first quadrant x > 0, y > 0 that comes in
downward parallel to they axis but is forced to turn a corner near the origin, as shown in
Fig. 153. To determine the flow, we recall (Example 3, Sec. 12) that the transformation
w = z2
= x2
-l + i2xy
maps the first quadrant onto the upper half of the uv plane and the boundary of the
quadrant onto the entire u axis.
From the example in Sec. 107, we know that the complex potential for a uniform
flow to the right in the upper half ofthe w plane is F = Aw, where A is a positive real
y
0 x FIGURE 153
SEC. 108 FLOWS AROUND A CORNER AND AROUND A CYLINDER 385
constant. The potential in the quadrant is, therefore,
(1)
and it follows that the stream function for the flow there is
(2) 1/1 =2Axy.
This stream function is, of course, harmonic in the first quadrant, and it vanishes on
the boundary.
The streamlines are branches of the rectangular hyperbolas
2Axy =c2.
According to equation (3), Sec. 107, the velocity of the fluid is
V =2Az =2A(x- iy).
Observe that the speed
lVI = 2Ay'x2 + y2
of a particle is directly proportional to its distance from the origin. The value of the
stream function (2) at a point (x, y) can be interpreted as the rate of flow across a line
segment extending from the origin to that point.
EXAMPLE 2. Let a long circular cylinder of unit radius be placed in a large body
of fluid flowing with a uniform velocity, the axis of the cylinder being perpendicular
to the direction offlow. To determine the steady flow around the cylinder, we represent
the cylinder by the circle x2 + y2 = 1and let the flow distant from it be parallel to the
x axis and to the right (Fig. 154). Symmetry shows that points on the x axis exterior
to the circle may be treated as boundary points, and so we need to consider only the
upper part of the figure as the region of flow.
The boundary of this region of flow, consisting of the upper semicircle and the
parts of the x axis exterior to the circle, is mapped onto the entire u axis by the
transformation
X
1
w =z + -.
z
FIGURE 154
386 APPLICATIONS OF CONFORMAL MAPPING CHAP. IO
The region itself is mapped onto the upper half plane v > 0, as indicated in Fig. 17,
Appendix 2. The complex potential for the corresponding uniform flow in that half
plane is F =A w, where A is a positive real constant. Hence the complex potential for
the region exterior to the circle and above the x axis is
(3)
The velocity
(4)
approaches A as lzl increases. Thus the flow is nearly uniform and parallel to the x
axis at points distant from the circle, as one would expect. From expression (4), we
see that V (z) = V (z); hence that expression also represents velocities of flow in the
lower region, the lower semicircle being a streamline.
According to equation (3), the stream function for the given problem is, in polar
coordinates,
(5) 1/1 = A (r -:)sin e.
The streamlines
A (r -:) sin e= Cz
are symmetric to they axis and have asymptotes parallel to the x axis. Note that when
c2 =0, the streamline consists of the circle r =I and the parts of the x axis exterior
to the circle.
EXERCISES
1. State why the components ofvelocity can be obtained from the stream function by means
of the equations
p(x, y) =1/ly(x, y), q(x, y) =-1/lx(x, y).
2. At an interior point of a region of flow and under the conditions that we have assumed,
the fluid pressure cannot be less than the pressure at all other points in a neighborhood
of that point. Justify this statement with the aid of statements in Sees. 106, 107, and 50.
3. For the flow around a corner described in Example 1, Sec. 108, at what point ofthe region
x > 0, y > 0 is the fluid pressure greatest?
4. Show that the speed of the fluid at points on the cylindrical surface in Example 2, Sec.
108, is 2AIsineIand also that the fluid pressure on the cylinder is greatest at the points
z ± 1and least at the points z = ± i.
SEC. I08 ExERCISES 387
5, Write the complex potential for the flow around a cylinder r = r0 when the velocity V
at a point z approaches a real constant A as the point recedes from the cylinder.
6. Obtain the stream function
1{r = Ar4 sin 4(}
for a flow in the angular region r > 0, 0 < (} < :rr/4 (Fig. 155), and sketch a few of the
streamlines in the interior of that region.
x FIGURE 155
7. Obtain the complex potential F = A sin z for a flow inside the semi-infinite region
-:rr/2 < x < :rrf2, y > 0 (Fig. 156). Write the equations of the streamlines.
_li
2
y
1T X
2 FIGURE 156
8. Show that if the velocity potential is 4> =A ln r (A > 0) for flow in the region r > r0,
then the streamlines are the halflines (} = c (r > r0) and the rate of flow outward through
each complete circle about the origin is 2:rrA, corresponding to a source of that strength
at the origin.
9. Obtain the complex potential
F= A(z2
+ 2
;)
for a flow in the region r > 1, 0 < (} < :rr/2. Write expressions for V and 1{r. Note how
the speed IVI varies along the boundary of the region, and verify that 1{r(x, y) = 0 on the
boundary.
10. Suppose that the flow at an infinite distance from the cylinder of unit radius in Example
2, Sec. 108, is uniform in a direction making an angle a with the x axis; that is,
lim V = Aeia
JzJ-+oo
(A> 0).
388 APPLICATIONS OF CONFORMAL MAPPING CHAP. 10
Find the complex potential.
(
. 1 . )Ans. F =A ze-ux + ~ e1
a .
11. Write
and
where
The function (z2 - 4)112 is then single-valued and analytic everywhere except on the
branch cut consisting of the segment of the x axis joining the points z =± 2. We know,
moreover, from Exercise 13, Sec. 85, that the transformation
1
z=w+-
w
maps the circle IwI= 1onto the line segment from z =-2 to z = 2 and that it maps the
domain outside the circle onto the rest of the z plane. Use all of the observations above
to show that the inverse transformation, where lwl > 1for every point not on the branch
cut, can be written
1
w=-[z+(z2
2
4)1
12
] = Jrlexp :..I+ Jr2 exp ~ .1( ·e ·e )
2
4 2 2
The transformation and this inverse establish a one to one correspondencebetween points
in the two domains.
12. With the aid of the results found in Exercises 10 and 11, derive the expression
F = A[z cos a- i(z2
- 4)1
12
sin a]
for the complex potential of the steady flow around a long plate whose width is 4 and
whose cross section is the line segment joining the two points z = ±2 in Fig. 157,
assuming that the velocity of the fluid at an infinite distance from the plate is A exp(ia).
The branch of (z2 - 4)112 that is used is the one described in Exercise 11, and A > 0.
13. Show that if sin a -f=- 0 in Exercise 12, then the speed of the fluid along the line segment
joining the points z = 2 is infinite at the ends and is equal to AIcos aIat the midpoint.
14. For the sake of simplicity, suppose that 0 < a < n /2 in Exercise 12. Then show that the
velocity of the fluid along the upper side of the line segment representing the plate in
Fig. 157 is zero at the point x =2 cos a and that the velocity along the lower side of the
segment is zero at the point x = -2 cos a.
SEC. 108 EXERCISES 389
FIGURE 157
15. A circle with its center at a point x0 (0 < x0 < 1) on the x axis and passing through the
point z = 1is subjected to the transformation
1
w = z+ -.
z
Individual nonzero points z can be mapped geometrically by adding the vectors
z = rei0
and
1 1 -ie
- = -e .
z r
Indicate by mapping some points that the image ofthe circle is a profile ofthe type shown
in Fig. 158 and that points exterior to the circle map onto points exterior to the profile.
This is a special case of the profile of a Joukowski airfoil. (See also Exercises 16 and 17
below.)
16. (a) Show that the mapping of the circle in Exercise 15 is conformal except at the point
z = -1.
-2
(b) Let the complex numbers
~z
t = lim
Az-+0 IAzl
~w
and r = lim
Aw-+0 IAwl
represent unit vectors tangent to a smooth directed arc at z=-1 and that arc's image,
respectively, under the transformation w =z + (1/z). Show that T = -t
2
and hence
that the Joukowski profile in Fig. 158 has a cusp at the point w =-2, the angle
between the tangents at the cusp being zero.
v
/.....- ----"""'Jl:.
/ / '/ / ~
I / '- W
I / I
I / I
I
I
'"....
--~
/ I
/
----
I
/
/
/
/
u
FIGURE 158
390 APPLICATIONS OF CONFORMAL MAPPING CHAP. IO
17. Find the complex potential for the flow around the airfoil in Exercise 15 when the velocity
V of the fluid at an infinite distance from the origin is a real constant A. Recall that the
inverse of the transformation
1
w=z+-
z
used in Exercise 15 is given, with z and w interchanged, in Exercise 11.
18. Note that under the transformation w = e2
+z, both halves, where x 2: 0 and x < 0, of
the line y =1r are mapped onto the halfline v =n(u < -1). Similarly, the line y =-n is
mapped onto the halfline v = -rr(u < -1); and the strip -n < y < n is mapped onto the
w plane. Also, note that the change ofdirections, arg(dw/dz), under this transformation
approaches zero as x tends to -co. Show that the streamlines of a fluid flowing through
the open channel formed by the half lines in the w plane (Fig. 159) are the images of the
lines y =c2 in the strip. These streamlines also represent the equipotential curves of the
electrostatic field near the edge of a parallel-plate capacitor.
u
FIGURE 159
CHAPTER
11
THE SCHWARZ-CHRISTOFFEL
TRANSFORMATION
In this chapter, we construct a transformation, known as the Schwarz-Christoffel
transformation, which maps the x axis and the upper half of the z plane onto a given
simple closed polygon and its interior in the w plane. Applications are made to the
solution of problems in fluid flow and electrostatic potential theory.
109. MAPPING THE REAL AXIS ONTO A POLYGON
We represent the unit vector which is tangent to a smooth arc C at a point zo by the
complex number t, and we let the number r denote the unit vector tangent to the image
r of Cat the corresponding point w0 under a transformation w = f(z). We assume
that f is analytic at z0 and that f'(z0) =f=. 0. According to Sec. 94,
(1) arg r =arg f'(z0) +arg t.
In particular, if C is a segment of the x axis with positive sense to the right, then t = 1
and arg t =0 at each point z0 =x on C. In that case, equation (1) becomes
(2) arg r = arg j'(x).
If f'(z) has a constant argument along that segment, it follows that arg r is constant.
Hence the image r of C is also a segment of a straight line.
Let us now construct a transformation w =f (z) that maps the whole x axis onto
a polygon ofn sides, where x1, x2, ... , Xn-l• and oo are the points on that axis whose
391
392 THE SCHWARZ-CHRISTOFFEL TRANSFORMATION CHAP. II
images are to be the vertices of the polygon and where
The vertices are the points wj = f(x1) (j = 1, 2, ... , n- 1) and wn = f(oo). The
function f should be such that arg J'(z) jumps from one constant value to another at
the points z =x J as the point z traces out the x axis (Fig. 160).
y v
I
u
FIGURE 160
If the function f is chosen such that
(3) f'(z) = A(z ~ xl)-k1(z- Xz)-k2 • • • (z- Xn_ 1)-kn-I,
where A is a complex constant and each kj is a real constant, then the argument of
f'(Z) changes in the prescribed manner as z describes the real axis; for the argument
of the derivative (3) can be written
(4) arg f'(z) = arg A- k1 arg(z- x1)
- k2 arg(z- xz) ~ · · ·- kn-1 arg(z ~ xn-1).
When z =x and x < xI>
arg(z- x1) = arg(z x2) = ···=arg(z- Xn-1) =Jr:.
When x 1 < x < x2, the argument arg(z- x 1) is 0 and each of the other arguments is
Jr. According to equation (4), then, arg J'(z) increases abruptly by the angle k1n as
z moves to the right through the point z =x1. It again jumps in value, by the amount
k2n, as z passes through the point x2, etc.
In view of equation (2), the unit vector r is constant in direction as z moves from
x1_1 to x1; the point w thus moves in that fixed direction along a straight line. The
direction of r changes abruptly, by the angle k1n, at the image point w1 of x1• as
shown in Fig. 160. Those angles k1;r: are the exterior angles of the polygon described
by the point w.
The exterior angles can be limited to angles between -;r: and ;r:, in which case
-1 < k1 < 1. We assume that the sides of the polygon never cross one another and
that the polygon is given a positive, or counterclockwise, orientation. The sum of the
SEC. IIO SCHWARZ-CHRISTOFFEL 'TRANSFORMATION 393
exterior angles of a closed polygon is, then, 2n; and the exterior angle at the vertex
wn, which is the image of the point z =oo, can be written
Thus the numbers kj must necessarily satisfy the conditions
(5) -l<kj<1 (j=1,2, ... ,n).
Note that kn = 0 if
(6) kt +kz + ... +kn-1 =2.
This means that the direction of r does not change at the point Ww So Wn is not a
vertex, and the polygon has n 1sides.
The existence of a mapping function f whose derivative is given by equation (3)
will be established in the next section.
110. SCHWARZ-CHRISTOFFEL TRANSFORMATION
In our expression (Sec. 109)
(1)
for the derivative of a function that is to map the x axis onto a polygon, let the factors
(z- xj)-kj represent branches of power functions with branch cuts extending below
that axis. To be specific, write
(2)
where ej =arg(z - xj) and j = 1, 2, ... , n - 1. Then f'(z) is analytic everywhere
in the half plane y > 0 except at then- 1branch points xj.
If zo is a point in that region of analyticity, denoted here by R, then the function
(3) F(z) = 1zf'(s) ds
zo
is single-valued and analytic throughout the same region, where the path of integration
from z0 to z is any contour lying within R. Moreover, F'(z) = f'(z) (see Sec. 42).
To define the function Fat the point z = x1 so that it is continuous there, we note
that (z- x1)-k' is the only factor in expression (1) that is not analytic at x 1• Hence, if
¢ (z) denotes the product ofthe rest ofthe factors in that expression, </J(z) is analytic at
x1and is represented throughout an open disk lz- x11 < R1 by its Taylor series about
394 THE SCHWARZ-CHRISTOFFEL TRANSFORMATION CHAP. II
x1. So we can write
or
(4)
where 1/J is analytic and, therefore, continuous throughout the entire open disk. Since
1- k1 > 0, the last term on the right in equation (4) thus represents a continuous
function of z throughout the upper half of the disk, where Im z > 0, if we assign it
the value zero at z = x 1• It follows that the integral
of that last term along a contour from Z1 to z, where Z 1 and the contour lie in the half
disk, is a continuous function of z at z = x1. The integral
along the same path also represents a continuous function of z at x 1 if we define the
value of the integral there as its limit as z approaches x 1 in the half disk. The integral
of the function (4) along the stated path from Z1 to z is, then, continuous at z =x1; and
the same is true of integral (3) since it can be written as an integral along a contour in
R from zo to Z 1 plus the integral from Z 1 to z.
The above argument applies at each of the n - 1points xJ to make F continuous
throughout the region y > 0.
From equation (1), we can show that, for a sufficiently large positive number R,
a positive constant M exists such that if lm z > 0, then
(5)
I M
If (z)l < lzl2-kn whenever lzl > R.
Since 2 - kn > 1, this order property of the integrand in equation (3) ensures the
existence of the limit of the integral there as z tends to infinity; that is, a number
wn exists such that
(6) lim F(z) = W11
z-+oo
(Imz>O).
Details of the argument are left to Exercises 1 and 2.
SEC. IIO SCHWARZ-CHRISTOFFEL TRANSFORMATION 395
Our mapping function, whose derivative is given by equation (1 ), can be written
f(z) =F(z) + B, where B is a complex constant. The resulting transformation,
(7) w =A ls- x1)-k1(s- x2)-k2 • • • (s- Xn_ 1)-kn-J ds + B,
zo
is the Schwarz-Christoffel transformation, named in honor of the two German math-
ematicians H. A. Schwarz (1843-1921) and E. B. Christoffel (1829-1900) who dis-
covered it independently.
Transformation (7) is continuous throughout the half plane y > 0 and is con-
formal there except for the points xj. We have assumed that the numbers kj satisfy
conditions (5), Sec. 109. In addition, we suppose that the constants xj and kj are such
that the sides of the polygon do not cross, so that the polygon is a simple closed con-
tour. Then, according to Sec. 109, as the point z describes the x axis in the positive
direction, its image w describes the polygon P in the positive sense; and there is a
one to one correspondence between points on that axis and points on P. According to
condition (6), the image wn of the point z =oo exists and wn = Wn + B.
If z is an interior point of the upper half plane y > 0 and x0 is any point on the
x axis other than one of the x j, then the angle from the vector t at x0 up to the line
segmentjoining x0 and zis positive and less than 1f (Fig. 160). At the image w0 of x0,
the corresponding angle from the vector r to the image of the line segment joining x0
and zhas that same value. Thus the images of interior points in the half plane lie to the
left ofthe sides ofthe polygon, taken counterclockwise. A proofthat the transformation
establishes a one to one correspondence between the interior points of the half plane
and the points within the polygon is left to the reader (Exercise 3).
Given a specific polygon P, let us examine the number of constants in the
Schwarz-Christoffel transformation that must be determined in order to map the x
axis onto P. For this purpose, we may write zo =0, A = 1, and B =0 and simply
require that the x axis be mapped onto some polygon P' similar to P. The size and
position of P' can then be adjusted to match those of P by introducing the appropriate
constants A and B.
The numbers kj are all determined from the exterior angles at the vertices of P.
The n - 1constants xj remain to be chosen. The image of the x axis is some polygon
P' that has the same angles as P. But if P' is to be similar to P, then n - 2 connected
sides must have a common ratio to the corresponding sides of P; this condition is
expressed by means of n - 3 equations in the n - 1real unknowns xj. Thus two ofthe
numbers xj• or two relations between them, can be chosen arbitrarily, provided those
n - 3 equations in the remaining n - 3 unknowns have real-valued solutions.
When a finite point z =Xn on the x axis, instead ofthe point at infinity, represents
the point whose image is the vertex Wn, it follows from Sec. 109 that the Schwarz-
Christoffel transformation takes the form
(8) w =A iz(s- x1)-kl(s- x2)-kz · · · (s- Xn)-kn ds + B,
zo
396 THE SCHWARZ-CHRISTOFFEL TRANSFORMATION CHAP. I I
where k1 + k2 + ···+ kn =2. The exponents ki are determined from the exterior
angles of the polygon. But, in this case, there are n real constants xJ that must satisfy
then - 3 equations noted above. Thus three ofthe numbers x1, or three conditions on
those n numbers, can be chosen arbitrarily in transformation (8) of the x axis onto a
given polygon.
EXERCISES
1. Obtain inequality (5), Sec. 110.
Suggestion: Let R be larger than any of the numbers lxjl(j = 1, 2, ... , n- 1).
Note that if R is sufficiently large, the inequalities lzl/2 < lz- x1I < 21zl hold for each
x1 when lzl > R. Then use equation (1), Sec. 110, along with conditions (5), Sec. 109.
2. Use condition (5), Sec. 110, and sufficient conditions for the existence of improper
integrals of real-valued functions to show that F(x) has some limit Wn as x tends to
infinity, where F(z) is defined by equation (3) in that section. Also, show that the integral
of j'(z) over each arc of a semicircle lzl = R (Im z > 0) approaches 0 as R tends to oo.
Then deduce that
lim F(z) = Wn
z~oo
(Imz > 0),
as stated in equation (6) of Sec. 110.
3. According to Sec. 79, the expression
N = _1_1 g'(z) dz
2rri c g(z)
can be used to determine the number (N) of zeros of a function g interior to a positively
oriented simple closed contour C when g(z) =fo 0 on C and when C lies in a simply
connected domain D throughout which g is analytic and g'(z) is never zero. In that
expression, write g(z) = f(z)- w0, where f(z) is the Schwarz-Christoffel mapping
function (7), Sec. 110, and the point w0 is either interior to or exterior to the polygon P
that is the image of the x axis; thus f (z) =fo w0. Let the contour C consist of the upper
half of a circle IzI= R and a segment - R < x < R of the x axis that contains all n - 1
points x1, except that a small segment about each point xJ is replaced by the upper half
of a circle Iz - xJI=pi with that segment as its diameter. Then the number of points z
interior to C such that f (z) = w0 is
Nc = _1_1 f'(z) dz.
2rri c f(z)- wo
Note that f(z)- w0 approaches the nonzero point Wn w0 when lzl =Rand R tends
to oo, and recall the order property (5), Sec. 110, for If'(z) 1. Let the p1 tend to zero, and
prove that the number of points in the upper half of the z plane at which f (z) = w0 is
N = _1_ lim 1R f'(x) dx.
2rri R~oo -R f(x)- Wo
SEC. III TRIANGLES AND RECTANGLES 397
Deduce that since
1 dw . JR f'(x)
~--= hm dx,
p w- w0 R--.oo -R f(x)- Wo
N = 1if w0 is interior to P and that N = 0 if w0 is exterior to P. Thus show that the
mapping of the half plane Im z > 0 onto the interior of P is one to one.
111. TRIANGLES AND RECTANGLES
The Schwarz-Christoffel transformation is written in terms of the points xj and not in
terms of their images, the vertices of the polygon. No more than three of those points
can be chosen arbitrarily; so, when the given polygon has more than three sides, some
ofthe points xj must be determined in order to make the given polygon, or any polygon
similarto it, be the image ofthe x axis. The selection ofconditions for the determination
of those constants, conditions that are convenient to use, often requires ingenuity.
Another limitation in using the transformation is due to the integration that is
involved. Often the integral cannot be evaluated in terms of a finite number of elemen-
tary functions. In such cases, the solution of problems by means of the transformation
can become quite involved.
If the polygon is a triangle with vertices at the points Wt. w2, and w3 (Fig. 161),
the transformation can be written
(1) w =A lz(s- x1)-k1(s- x2)-k2 (s - x3)-k3 ds + B,
zo
where k1 +k2 +k3 = 2. In terms of the interior angles 01,
(j = 1, 2, 3).
Here we have taken all three points xJ as finite points on the x axis. Arbitrary values
can be assigned to each ofthem. The complex constants A and B, which are associated
with the size and position ofthe triangle, can be determined so that the upper halfplane
is mapped onto the given triangular region.
y v
X
FIGURE 161
398 THE SCHWARZ-CHRISTOFFEL TRANSFORMATION CHAP. I I
If we take the vertex w3 as the image of the point at infinity, the transformation
becomes
(2) w =A ls- x1)-k1(s- x2)-k2 ds + B,
zo
where arbitrary real values can be assigned to x1 and x2.
The integrals in equations (1) and (2) do not represent elementary functions unless
the triangle is degenerate with one or two of its vertices at infinity. The integral in
equation (2) becomes an elliptic integral when the triangle is equilateral or when it is
a right triangle with one of its angles equal to either rr/3 or rrf4.
EXAMPLE 1. For an equilateral triangle, k1 =k2 =k3 =2f3. It is convenient to
write x1 =-1, x2 = 1, and x3 = oo and to use equation (2), where z0 = 1, A= 1, and
B = 0. The transformation then becomes
(3) w -1s+ 1)-2
1s- 1)-2
/
3
ds.
The image of the point z = 1 is clearly w = 0; that is, w2 = 0. If z = 1in this
integral, one can write s = x, where -1 < x < L Then
x + 1 > 0 and arg(x + 1) =0,
while
lx - 11 = 1 - x and arg(x - 1) = rr.
Hence
(4)
(
rri) (
1
2dx
= exp 3 Jo (1 - x2)213.
With the substitution x = .;i, the last integral here reduces to a special case of the one
used in defining the beta function (Exercise 7, Sec. 77). Let b denote its value, which
is positive:
(5) h= t 2dx = f-t;2(t-t)-213dt=B(I.I).
lo (1 - x 2)213 lo 2 3
The vertex w1 is, therefore, the point (Fig. 162)
(6)
rri
w1 =hexp -.
3
SEC. I 12 DEGENERATE POLYGONS 401
It is left to the exercises to show that
(12) W3 =b, w4 =b+ic.
The position and dimensions of the rectangle are shown in Fig. 163.
y
v
lC
WI w4
-I I -b b
XI x2 0 x3 x4 X
w2 0 w3 u
FIGURE 163
112. DEGENERATE POLYGONS
We now apply the Schwarz-Christoffel transformation to some degenerate polygons
for which the integrals represent elementary functions. For purposes of illustration,
the examples here result in transformations that we have already seen in Chap. 8.
EXAMPLE 1. Let us map the half plane y > 0 onto the semi-infinite strip
TC TC
--<u<- v>O.
2- - 2'
We consider the strip as the limiting form of a triangle with vertices wb w2, and w3
(Fig. 164) as the imaginary part of w3 tends to infinity.
y
-I I
X 1r u
2 FIGURE 164
The limiting values of the exterior angles are
and
We choose the points x1=-1, x2 =1, and x3 =oo as the points whose images are the
vertices. Then the derivative of the mapping function can be written
dw = A(z + 1)-lf2(z- 1)-1/2 = A'(l- z2) 1/2
dz
402 THE SCHWARZ-CHRISTOFFEL TRANSFORMATION CHAP. II
Hence w = A' sin-I z + B. If we write A' = 1/a and B = bIa, it follows that
z =sin(aw- b).
This transformation from the w to the zplane satisfies the conditions z= -1 when
w = -n/2 and z= 1when w = 1r/2 ifa = 1and b = 0. The resulting transformation is
z= srn w,
which we verified in Sec. 89 as one that maps the strip onto the half plane.
EXAMPLE 2. Consider the strip 0 < v < 1r as the limiting form of a rhombus with
vertices at the points w1 =ni, w2, w3 =0, and w4 as the points w2 and w4 are moved
infinitely far to the left and right, respectively (Fig. 165). In the limit, the exterior
angles become
We leave x1 to be determined and choose the values x2 = 0, x3 = 1, and x4 = oo. The
derivative of the Schwarz-Christoffel mapping function then becomes
thus
y
X
dw = A(z
dz
w =A Logz +B.
-- ------- -..... _ (
(
tK- -,.
W - -- w) 2 ---- --- 4
-- -- u
FIGURE 165
Now B =0 because w =0 when z = 1. The constant A must be real because the
point w lies on the real axis when z =x and x > 0. The point w =ni is the image of
the point z = x~> where x 1 is a negative number; consequently,
ni =A Logx1 =A ln lxtl +Ani.
By identifying real and imaginary parts here, we see that lx11= 1 and A= 1. Hence
the transformation becomes
w =Log z;
also, x1 = -1. We already know from Example 3 in Sec. 88 that this transformation
maps the half plane onto the strip.
SEC. I 12 EXERCISES 403
The procedure used in these two examples is not rigorous because limiting values
ofangles and coordinates were not introduced in an orderly way. Limiting values were
used whenever it seemed expedient to do so. But, if we verify the mapping obtained,
it is not essential that we justify the steps in our derivation of the mapping function.
The formal method used here is shorter and less tedious than rigorous methods.
EXERCISES
1. In transformation (1), Sec. Ill, write B = z0 = 0 and
3rri
A=exp-,
4
X1 = -1, x2 =0,
3
kl = -,
4
1
kz = -,
2
to map the x axis onto an isosceles right triangle. Show that the vertices of that triangle
are the points
Wt =bi, w2 =0, and
where b is the positive constant
Also, show that
where B is the beta function.
2. Obtain expressions (12) in Sec. 111 for the rest of the vertices of the rectangle shown in
Fig. 163.
3. Show that when 0 < a < 1in equations (8) and (9), Sec. Ill, the vertices of the rectangle
are those shown in Fig. 163, where b and c now have values
b =loa !g(x)! dx, c =11
!g(x)i dx.
4. Show that the special case
of the Schwarz-Christoffel transformation (7), Sec. 110, maps the x axis onto the square
with vertices
w2 =0, W3=b, w4 = b +ib,
404 THE SCHWARZ-CHRISTOFFEL TRANSFORMATION CHAP. II
where the (positive) number b is given in terms of the beta function:
5. Use the Schwarz-Christoffel transformation to arrive at the transformation
v
(0 < m < 1),
which maps the halfplane y > 0 onto the wedge IwI> 0, 0 < arg w < mrr and transforms
the point z = 1 into the point w = 1. Consider the wedge as the limiting case of the
triangular region shown in Fig. 166 as the angle a there tends to 0.
u FIGURE 166
6. Refer to Fig. 26, Appendix 2. As the point z moves to the right along the negative real
axis, its image point w is to move to the right along the entire u axis. As z describes
the segment 0 < x < 1 of the real axis, its image point w is to move to the left along
the half line v =rri (u > 1); and, as z moves to the right along that part of the positive
real axis where x > 1, its image point w is to move to the right along the same half line
v =rri (u > 1). Note the changes in direction of the motion of w at the images of the
points z= 0 and z = 1. These changes suggest that the derivative of a mapping function
should be
f'(z) = A(z- 0)-1
(z- 1),
where A is some constant; thus obtain formally the mapping function,
w =rri +z- Log z,
which can be verified as one that maps the halfplane Rc z > 0 as indicated in the figure.
7. As the point z moves to the right along that part of the negative real axis where x < -1,
its image point is to move to the right along the negative real axis in the w plane. As
z moves on the real axis to the right along the segment - 1 < x < 0 and then along the
segment 0 < x < 1, its image point w is to move in the direction of increasing v along
the segment 0 < v < 1 of the v axis and then in the direction of decreasing v along the
same segment. Finally, as z moves to the right along that part of the positive real axis
where x > 1, its image point is to move to the right along the positive real axis in the
w plane. Note the changes in direction of the motion of w at the images of the points
z =-1, z = 0, and z = 1. A mapping function whose derivative is
J'(z) =A(z + 1) 1
;
2
(z- 0) 1
(z- 1) 1
12
SEC. 112 EXERCISES 405
where A is some constant, is thus indicated. Obtain formally the mapping function
w =Jz2 - 1,
where 0 < arg Jz2 - 1 < rr. By considering the successive mappings
Z - 2- z , W =Z - 1, and w =.JW,
verify that the resulting transformation maps the right half plane Re z> 0 onto the upper
half plane Im w > 0, with a cut along the segment 0 < v < 1of the v axis.
8. The inverse of the linear fractional transformation
Z= z -z
i+z
maps the unit disk IZI < 1conformally, except at the point Z = -1, onto the half plane
Im z > 0. (See Fig. 13, Appendix 2.) Let Zj be points on the circle IZI = 1 whose
images are the points z = xj (j = 1, 2, ... , n) that are used in the Schwarz-Christoffel
transformation (8), Sec. 110. Show formally, without determining the branches of the
power functions, that
dw , -k -k -k
dZ =A (Z- Zt) 1(Z- Z2) 2 ••• (Z- Zn) n,
where A' is a constant. Thus show that the transformation
maps the interiorofthe circle IZI = 1onto the interior ofa polygon, the vertices of the
polygon being the images of the points Zj on the circle.
9. In the integral ofExercise 8, let the numbers Zj (j =1, 2, ... , n) be the nth roots ofunity.
Write w =exp(2ni In) and Z1 =1, Z2 =w, ... , Zn =wn-l. Let each of the numbers
kj (j = 1, 2, ... , n) have the value 2/n. The integral in Exercise 8 then becomes
'1z dSw=A +B.
0 (Sn- 1)2/n
Show that when A' = 1and B = 0, this transformation maps the interior of the unit circle
IZI= 1onto the interior of a regular polygon ofn sides and that the center of the polygon
is the point w = 0.
Suggestion: The image of each of the points Zj (j = 1, 2, ... , n) is a vertex of
some polygon with an exterior angle of 2nIn at that vertex. Write
w - {I __d_S----=-_
l - Jo (Sn - 1)2/n,
where the path of the integration is along the positive real axis from Z =0 to Z = 1and
the principal value of the nth root of (Sn - 1)2 is to be taken. Then show that the images
406 THE SCHWARZ-CHRISTOFFEL TRANSFORMATION CHAP. II
fth · t Z Z n-l th · t n-l · lo e pom s 2 =w, ... , n =w are e pom s wwJ> ... , w wh respective y.
Thus verify that the polygon is regular and is centered at w = 0.
113. FLUID FLOW IN A CHANNEL THROUGH A SLIT
We now present a further example of the idealized steady flow treated in Chap. l0, an
example that will help show how sources and sinks can be accounted for in problems
of fluid flow. In this and the following two sections, the problems are posed in the uv
plane, rather than the xy plane. That allows us to refer directly to earlier results in this
chapter without interchanging the planes.
Consider the two-dimensional steady flow of fluid between two parallel planes
v = 0 and v = rc when the fluid is entering through a narrow slit along the line in the
first plane that is perpendicular to the uv plane at the origin (Fig. 167). Let the rate of
flow of fluid into the channel through the slit be Q units of volume per unit time for
each unit of depth of the channel, where the depth is measured perpendicular to the
uv plane. The rate of flow out at either end is, then, Qj2.
y
X
FIGURE 167
The transformation w =Log z is a one to one mapping of the upper half y > 0 of
the z plane onto the strip 0 < v < rc in thew plane (see Example 2 in Sec. 112). The
inverse transformation
(1)
maps the strip onto the half plane (see Example 3, Sec. 13). Under transformation (1),
the image of the u axis is the positive halfof the x axis, and the image ofthe line v =rc
is the negative half of the x axis. Hence the boundary of the strip is transformed into
the boundary of the half plane.
The image of the point w = 0 is the point z = 1. The image of a point w = u0,
where u0 > 0, is a point z = x0, where x0 > 1. The rate of flow of fluid across a curve
joining the point w = u0 to a point (u, v) within the strip is a stream function l/f(u, v)
for the flow (Sec. 107). If u 1 is a negative real number, then the rate of flow into the
channel through the slit can be written
Now, under a conformal transformation, the function 1/1 is transformed into a function
of x and y that represents the stream function for the flow in the corresponding region
SEC. 113 FLUID FLOW IN A CHANNEL THROUGH A SLIT 407
of the z plane; that is, the rate of flow is the same across corresponding curves in the
two planes. As in Chap. 10, the same symbol1fr is used to represent the different stream
functions in the two planes. Since the image ofthe point w = u1is a point z = x1, where
0 < x1 < 1, the rate of flow across any curve connecting the points z =x0 and z =x1
and lying in the upper half of the zplane is also equal to Q. Hence there is a source at
the point z =1equal to the source at w = 0.
The above argument applies in general to show that under a conformal transfor-
mation, a source or sink at a given point corresponds to an equal source or sink at the
image ofthat point.
As Re w tends to -oo, the image of w approaches the point z =0. A sink of
strength Q/2 at the latter point corresponds to the sink infinitely far to the left in the
strip. To apply the above argument in this case, we consider the rate of flow across a
curve connecting the boundary lines v =0 and v =n of the left-hand part of the strip
and the rate of flow across the image of that curve in the z plane.
The sink at the right-hand end of the strip is transformed into a sink at infinity in
the z plane.
The stream function 1fr for the flow in the upper half of the z plane in this case
must be a function whose values are constant along each of the three parts of the x
axis. Moreover, its value must increase by Q as the point z moves around the point
z = 1from the position z =x0 to the position z =x1, and its value must decrease by
Q/2 as z moves about the origin in the corresponding manner. We see that the function
1fr = ~ [Arg(z - 1) - ~ Arg zJ
satisfies those requirements. Furthermore, this function is harmonic in the half plane
Im z > 0 because it is the imaginary component of the function
F = Q [Log(z- 1)- I Log z] = Q Log(z1
12
- z-1
12
).
1l" 2 1l"
The function F is a complex potential function for the flow in the upper half ofthe
zplane. Since z = ew, a complex potential function F (w) for the flow in the channel is
F(w) = Q Log(ew/2 - e-w/2).
1l"
By dropping an additive constant, one can write
(2) F(w) = ; Log(sinh ~).
We have used the same symbol F to denote three distinct functions, once in the z plane
and twice in the w plane.
-=-:-:--:-
The velocity vector F'(w) is given by the equation
(3)
Q w
V =- coth -.
2rr 2
408 THE SCHWARZ-CHRISTOFFEL TRANSFORMATION CHAP. II
From this, it can be seen that
lim V=R.lul-+oo 2:rr
Also, the point w = :rri is a stagnation point; that is, the velocity is zero there. Hence
the fluid pressure along the wall v =:rr of the channel is greatest at points opposite
the slit.
The stream function '1/J(u, v) for the channel is the imaginary component of the
function F(w) given by equation (2). The streamlines '1/J(u, v) =c2 are, therefore, the
curves
This equation reduces to
(4)
v u
tan - = c tanh -,
2 2
where c is any real constant. Some of these streamlines are indicated in Fig. 167.
114. FLOW IN A CHANNEL WITH AN OFFSET
To further illustrate the use of the Schwarz-Christoffel transformation, let us find the
complex potential for the flow of a fluid in a channel with an abrupt change in its
breadth (Fig. 168). We take our unit of length such that the breadth of the wide part of
the channel is :rr units; then h:rr, where 0 < h < 1, represents the breadth of the narrow
part. Let the real constant V0 denote the velocity of the fluid far from the offset in the
wide part; that is,
lim V = V0 ,
U-+ -00
where the complex variable V represents the velocity vector. The rate of flow per unit
depth through the channel, or the strength of the source on the left and of the sink on
the right, is then
(1) Q- :rrVo.
y
u
FIGURE 168
SEC. I 14 FLOW IN A CHANNEL WITH AN OFFSET 409
The cross section of the channel can be considered as the limiting case of the
quadrilateral with the vertices wl> w2, w3, and w4 shown in Fig. 168 as the first and
last of these vertices are moved infinitely far to the left and to the right, respectively.
In the limit, the exterior angles become
As before, we proceed formally, using limiting values whenever it is convenient to do
so.lfwe write x1 =0, x3 =1, x4 =oo and leave x2 to be determined, where 0 < x2 < 1,
the derivative of the mapping function becomes
(2) dw = Az-'(z- x2)-1/2(z- 1)1/2.
dz
To simplify the determination of the constants A and x2 here, we proceed at once
to the complex potential of the flow. The source of the flow in the channel infinitely
far to the left corresponds to an equal source at z = 0 (Sec. 113). The entire boundary
of the cross section of the channel is the image of the x axis. In view of equation (1),
then, the function
(3) F = V0 Log z = V0 In r + i V0e
is the potential for the flow in the upper half of the z plane, with the required source
at the origin. Here the stream function is 1/f = v0e. It increases in value from 0 to
V0n over each semicircle z = Reie(0 < e< n ), where R > 0, as evaries from 0 ton.
[Compare equation (5), Sec. 107, and Exercise 8, Sec. 108.]
The complex conjugate of the velocity V in the w plane can be written
V(w) = dF = dF dz .
dw dz dw
Thus, by referring to equations (2) and (3), we can see that
(4) V (w) = Vo z - x2
( )
1/2
A z -1
At the limiting position of the point wh which corresponds to z = 0, the velocity
is the real constant V0. It therefore follows from equation (4) that
Vo
Vo=-F2·
A
At the limiting position of w4, which corresponds to z = oo, let the real number V4
denote the velocity. Now it seems plausible that as a vertical line segment spanning
the narrow part of the channel is moved infinitely far to the right, V approaches V4 at
each point on that segment. We could establish this conjecture as a fact by first finding
w as the function of z from equation (2); but, to shorten our discussion, we assume
410 THE SCHWARZ-CHRISTOFFEL TRANSFORMATION
that this is true, Then, since the flow is steady,
n hV4 = n V0 = Q,
or V4 = V0/ h. Letting z tend to infinity in equation (4), we find that
Vo Vo-=-
h A
Thus
(5) A =h,
and
(6) V(w) = Vo (z-h2)I/2
h z -1
CHAP. II
From equation (6), we know that the magnitude lVI of the velocity becomes
infinite at the comer w3 of the offset since it is the image of the point z = 1. Also,
the comer w2 is a stagnation point, a point where V = 0. Along the boundary of the
channel, the fluid pressure is, therefore, greatest at w2 and least at w3•
To write the relation between the potential and the variable w, we must integrate
equation (2), which can now be written
(7) dw = h ( z - 1 )
112
dz z z- h2
By substituting a new variables, where
z- h2 2
--=s,
z -1
one can show that equation (7) reduces to
dw _ 2h( 1 _ 1 )
ds - 1- s2 h2 - s2 ·
Hence
hL
l+s L h+s
w = og - og
1-s h-s
(8)
The constant of integration here is zero because when z = h2, the quantity s is zero
and so, therefore, is w.
In terms of s, the potential F of equation (3) becomes
h2- s2
F = V0 Log 2 ;
1-s
SEC. 115 ELECTROSTATIC POTENTIAL ABOUT AN EDGE OF A CONDUCTING PLATE 411
consequently,
2 exp(FI V0) - h2
s = ___;:__:__;__..:::.;___
exp(FI V0) - 1
(9)
By substituting s from this equation into equation (8), we obtain an implicit relation
that defines the potential F as a function of w.
115. ELECTROSTATIC POTENTIAL ABOUT AN EDGE
OF A CONDUCTING PLATE
Two parallel conducting plates of infinite extent are kept at the electrostatic potential
V =0, and a parallel semi-infinite plate, placed midway between them, is kept at the
potential V = 1. The coordinate system and the unit of length are chosen so that
the plates lie in the planes v = 0, v = :n:, and v = :n:12 (Fig. 169). Let us determine
the potential function V (u, v) in the region between those plates.
y
V=O
-1 1
-----,........ --
------;;:w2_______ v= 1
..::::- -21 -
w4 ----- --------- ...... _--- ---- ----V=O FIGURE169
X
The cross section of that region in the uv plane has the limiting form of the
quadrilateral bounded by the dashed lines in Fig. 169 as the points w1 and w3 move
out to the right and w4 to the left. In applying the Schwarz-Christoffel transformation
here, we let the point x4, corresponding to the vertex w4, be the point at infinity. We
choose the points x1 =-1, x3 = 1and leave x2 to be determined. The limiting values
of the exterior angles of the quadrilateral are
k2rr = -rr,
Thus
dw =A(z+l)-l(z-xz)(z-1)-l=A (z-xz) =A (l+xz + 1-xz).
dz z2 - 1 2 z + 1 z - 1
and so the transformation of the upper half of the z plane into the divided strip in the
w plane has the form
(1)
A
w = -[(1 +x2) Log(z + 1) + (1- x2) Log(z- 1)] +B.
2
Let A1, A2 and B1, B2 denote the real and imaginary parts ofthe constants Aand
B. When z = x, the point w lies on the boundary of the divided strip; and, according
412 THE SCHWARZ-CHRISTOFFEL TRANSFORMATION CHAP. II
to equation (1),
. AI+ iA2 .
u + zv = {(I+ xz)[ln lx + 11 + z arg(x + 1)]
2
(2)
+ (1- x2)[ln lx- 11 + i arg(x- 1)]} + B1 + iB2.
To determine the constants here, we first note that the limiting position of the
line segment joining the points wi and w4 is the u axis. That segment is the image of
the part of the x axis to the left of the point x1 =-1; this is because the line segment
joining w3 and w4 is the image of the part of the x axis to the right of x3 = 1, and
the other two sides of the quadrilateral are the images of the remaining two segments
of the x axis. Hence when v =0 and u tends to infinity through positive values, the
corresponding point x approaches the point z = -1 from the left. Thus
arg(x + 1) =rr, arg(x - 1) = rr,
and ln lx + II tends to -oo. Also, since -1 < x2 < 1, the real part ofthe quantity inside
the braces in equation (2) tends to oo. Since v = 0, it readily follows that A2 =0;
for, otherwise, the imaginary part on the right would become infinite. By equating
imaginary parts on the two sides, we now see that
Hence
(3) A2 =0.
The limiting position of the line segment joining the points w1 and w2 is the half
line v = 11: j2 (u > 0). Points on that half line are images of the points z = x, where
-1 < x < x2; consequently,
arg(x + 1) = 0, arg(x - 1) =n.
Identifying the imaginary parts on the two sides of equation (2), we thus arrive at the
relation
(4)
rr AI
- = -(1- x2)rr + B2.
2 2
Finally, the limiting positions of the points on the line segment joining w3 to w4
are the points u +11: i, which are the images ofthe points x when x > 1. By identifying,
for those points, the imaginary parts in equation (2), we find that
rr = B2•
Then, in view of equations (3) and (4),
A1 =-1, x2 =0.
414 THE SCHWARZ-CHRISTOFFEL TRANSFORMATION CHAP. I I
2. Explain why the solution of the problem of flow in a channel with a semi-infinite
rectangular obstruction (Fig. 171) is included in the solution of the problem treated in
Sec. 114.
FIGURE 171
3. Refer to Fig. 29, Appendix 2. As the point z moves to the right along the negative part
of the real axis where x < -1, its image point w is to move to the right along the half
line v =h (u < 0). As the point z moves to the right along the segment -1 < x < l ofthe
x axis, its image point w is to move in the direction of decreasing v along the segment
0 < v < h of the v axis. Finally, as zmoves to the right along the positive part of the real
axis where x > 1, its image point w is to move to the right along the positive real axis.
Note the changes in the direction of motion of w at the images of the points z = -1 and
z= 1. These changes indicate that the derivative of a mapping function might be
dw =A (z+ 1)
112
,
dz z- l
where A is some constant. Thus obtain formally the transformation given with the figure.
Verify that the transformation, written in the form
w = ~{(z + l)I/2(z- l)I/2 +LogLz + (z + l)If2(z- l)l/2]}
:rr
where 0 < arg(z ± 1) < :rr, maps the boundary in the manner indicated in the figure.
4. Let T(u, v) denote the bounded steady-state temperatures in the shaded region of the
w plane in Fig. 29, Appendix 2, with the boundary conditions T (u, h) = l when u < 0
and T = 0 on the rest (B'C'D') of the boundary. Using the parameter a (0 <a < :rr/2),
show that the image of each point z =i tan a on the positive y axis is the point
w = ~ [In(tan a+ sec a) +i (; + sec a)J
(see Exercise 3) and that-the temperature at that point w is
a
T(u, v) =-
:rr
5. Let F(w) denote the complex potential function for the flow of a fluid over a step in
the bed of a deep stream represented by the shaded region of the w plane in Fig. 29,
Appendix 2, where the fluid velocity V approaches a real constant v0 as lwl tends to
infinity in that region. The transformation that maps the upper half of the z plane onto
that region is noted in Exercise 3. Use the chain rule
d F dF dz
- - -
dw dz dw
SEC. I IS EXERCISES 415
to show that
and, in terms of the points z =x whose images are the points along the bed of the stream,
show that
Note that the speed increases from !Vol along A'B' until! VI =oo at B', then diminishes
to zero at C', and increases toward !Vol from C' to D'; note, too, that the speed is !Vol at
the point
.(1 1)w=tl+Jl'h,
between B' and C'.
CHAPTER
12
INTEGRAL FORMULAS
OF THE POISSON TYPE
In this chapter, we develop a theory that enables us to obtain solutions to a variety of
boundary value problems where those solutions are expressed in terms of definite or
improper integrals. Many of the integrals occurring are then readily evaluated.
116. POISSON INTEGRAL FORMULA
Let C0 denote a positively oriented circle, centered at the origin, and suppose that a
function f is analytic inside and on C0. The Cauchy integral formula (Sec. 47)
(1) j(z) = ~ { f(s) ds
2m Jc0 s- z
expresses the value off at any point zinteriorto C0 in terms ofthe values off at points
s on C0. In this section, we shall obtain from formula (1) a corresponding formula for
the real part of the function f; and, in Sec. 117, we shall use that result to solve the
Dirichlet problem (Sec. 98) for the disk bounded by C0.
We let r0 denote the radius of C0 and write z = r exp(i8), where 0 < r < r0
(Fig. 172). The inverse of the nonzero point z with respect to the circle is the point z1
lying on the same ray from the origin as z and satisfying the condition lz111zl =r~;
thus, if s is a point on C0 ,
(2)
2 2 -
ro ro ss
Zt =- exp(iO) =-=- = -=-·r z z
417
SEC. 117 DIRICHLET PROBLEM FOR A DISK 419
Hence, if u is the real part of the analytic function f, it follows from formula (4) that
(6)
1 lo2rr (rJ- r2)u(ro, ¢)
u(r, 6) =- d¢
2rr o rJ - 2r0r cos(¢ - 6) + r 2
(r < r0).
This is the Poisson integral fonnula for the harmonic function u in the open disk
bounded by the circle r = r0 .
Formula (6) defines a linear integral transformation of u(r0, ¢) into u(r, e). The
kernel of the transformation is, except for the factor 1/(2rr), the real-valued function
r2- r2
P(r0 ,r,¢-e)= 2
° ,
r0 - 2r0r cos(¢ -e) + r2
(7)
which is known as the Poisson kernel. In view of equation (5), we can also write
(8)
and, since r < ro, it is clear that P is a positive function. Moreover, since z/(i- Z)
and its complex conjugate z/(s - z) have the same real parts, we find from the second
of equations (3) that
(9) (
s z ) (s+ z)P(r0,r,¢-0)=Re + =Re .
s-z s-z s-z
Thus P(r0 , r, ¢-e) is a harmonic function of r and() interior to C0 for each fixed s
on C0• From equation (7), we see that P(r0, r, ¢-())is an even periodic function of
¢ - (), with period 2rr; and its value is 1when r = 0.
The Poisson integral formula (6) can now be written
(10)
1 lo2rru(r, 0) =- P(ro, r, ¢- O)u(r0, ¢) d¢
2rr o
(r < r0).
When j(z) = u(r' e)= 1, equation (10) shows that p has the property
(11)
1 lo2rr- P(r0, r, ¢ - e) d¢ = 1
2rr o
(r < r0).
We have assumed that f is analytic not only interior to C0 but also on C0 itself
and that u is, therefore, harmonic in a domain which includes all points on that circle.
In particular, u is continuous on C0. The conditions will now be relaxed.
117. DIRICHLET PROBLEM FOR A DISK
Let F be a piecewise continuous function of e on the interval 0 < () < 2rr. The Poisson
integral transform of F is defined in terms of the Poisson kernel P(r0, r, ¢- 6),
SEC. II7 DIRICHLET PROBLEM FOR A DISK 421
An antiderivative of P(1, r, 1/1) is
(4)
f P(l, r, 1/1) dlf! =2 arctan(
1
+r tan 1/1),
1- r 2
the integrand here being the derivative with respect to 1/J of the function on the right.
So it follows from expression (3) that
(
1+ r 2n - e) (1+r n - e)n V (r, ()) =arctan tan - arctan tan .
1-r 2 1-r 2
After simplifying the expression for tanfn V (r, 11)] obtained from this last equation
(see Exercise 3, Sec. 118), we find that
(5) 1 ( 1- r
2
)V (r, B) = - arctan .
n 2r sm ()
(0 <arctan t < n),
where the stated restriction on the values of the arctangent function is physically
evident. When expressed in rectangular coordinates, the solution here is the same as
solution (5) in Sec. 105.
We tum now to the proof that the function U defined in equation (1) satisfies the
Dirichlet problem for the disk r < r0, as asserted just prior to this example. First of
all, U is harmonic inside the circle r =r0 because P is a harmonic function of r and
() there. More precisely, since F is piecewise continuous, integral (1) can be written
as the sum of a finite number of definite integrals each of which has an integrand that
is continuous in r, e, and¢. The partial derivatives of those integrands with respect
to r and eare also continuous. Since the order of integration and differentiation with
respect tor and ecan, then, be interchanged and since P satisfies Laplace's equation
r2
Prr + rP, + Pee = 0
in the polar coordinates r and 11 (Exercise 5, Sec. 25), it follows that U satisfies that
equation too.
In order to verify limit (2), we need to show that if F is continuous at (}, there
corresponds to each positive number s a positive number osuch that
(6) IU(r, 11)- F(B)I < s whenever
We start by referring to property (11), Sec. 116, of the Poisson kernel and writing
U(r, 11)- F(l1) = -
1
(:n: P(r0, r, ¢- 11)[F(¢) - F(B)] d¢.
2n Jo
For convenience, we let F be extended periodically, with period 2n, so that the
integrand here is periodic in ¢ with that same period. Also, we may assume that
0 < r < r0 because of the nature of the limit to be established.
426 INTEGRAL FORMULAS OF THE POISSON TYPE CHAP. I2
l(h,e- eo)
1
h
' 'I
I
I
I
I
I
I
0 eo e0 +h 27f ()
FIGURE176
With the aid of a mean value theorem for definite integrals, show that
where 00 < c < 00 +h, and hence that
1
21!"
lim P (ro, r, </> - e)I(h, </> - 00) d<f> =P(ro. r, 0 - Oo)
h-->0 0
h>O
(r < r0).
Thus the Poisson kernel P(r0 , r, e- e0) is the limit, as h approaches 0 through positive
values, of the harmonic function inside the circle r =r0 whose boundary values are
represented by the impulse function 27f l(h, e- eo).
5. Show that the expression in Exercise 8(b), Sec. 56, for the sum of a certain cosine series
can be written
00 1 2
~ -a1+2 L.. an cos nO = 2n=1 1- 2a cos e+a
(-l<a<l).
Then show that the Poisson kernel has the series representation (10), Sec. 117.
6. Show that the series in representation (10), Sec. 117, for the Poisson kernel converges
uniformly with respect to <f>. Then obtain from formula (I) of that section the series
representation (11) for U (r, 0) there.*
7. Use expressions (11) and (12) in Sec. 117 to find the steady temperatures T(r, e) in a
solid cylinder r.::: r0 of infinite length if T(r0 , 0) =A cos e. Show that no heat flows
across the plane y =0.
A A
Ans. T = -r cos()= -x.
ro ro
*This result is obtained when r0 = 1 by the method of separation of variables in the authors' "Fourier
Series and Boundary Value Problems," 6th ed., Sec. 48, 2001.
SEC. 120 DIRICHLET PROBLEM FOR A HALF PLANE 429
and
(6) /(z) = ~ 1oo (t- x)f(t) dt
m -oo It- zl2
(y > 0).
If f(z);;::;; u(x, y) +iv(x, y), it follows from formulas (5) and (6) that the har-
monic functions u and v are represented in the halfplane y > 0 in terms ofthe boundary
values of u by the formulas
(7) ( )- _! 1oo yu(t, 0) d - _! 1oo yu(t, 0) d
U X, y - t- t
TC -oo It- zl2
rc -oo (t- x)2
+ y2
(y > 0)
and
(8) ( ) l100
(x-t)u(t,O)d
vx,y =- t
rc -oo (t-x)2+y2
(y > 0).
Formula (7) is known as the Schwarz integral formula, or the Poisson integral
formula for the half plane. In the next section, we shall relax the conditions for the
validity of formulas (7) and (8).
120. DIRICHLET PROBLEM FOR A HALF PLANE
Let F denote a real-valued function ofx that is bounded for all x and continuous except
for at most a finite number of finite jumps. When y > s and lxI < 1/s, where s is any
positive constant, the integral
l(x' y) =1oo F(t) dt
-oo (t - X)2 + y2
converges uniformly with respectto x andy, as do the integrals ofthe partial derivatives
of the integrand with respect to x and y. Each of these integrals is the sum of a finite
number of improper or definite integrals over intervals where F is continuous; hence
the integrand of each component integral is a continuous function oft, x, and y when
y > s. Consequently, each partial derivative of l(x, y) is represented by the integral
of the corresponding derivative of the integrand whenever y > 0.
We write U (x, y) = yI(x, y)Irc. Thus U is the Schwarz integral transform of F,
suggested by the second of expressions (7), Sec. 119:
(1) U(x, y) = _! 1oo yF(t) dt
n -oo (t-x)2+y2
(y > 0).
Except for the factor 1/n, the kernel here is y/It- ze.lt is the imaginary component
of the function 1/(t- z), which is analytic in z when y > 0. It follows that the kernel
is harmonic, and so it satisfies Laplace's equation in x and y. Because the order of
differentiation and integration can be interchanged, the function (1) then satisfies that
equation. Consequently, U is harmonic when y > 0.
434 INTEGRAL FoRMULAS OF THE POISSON TYPE
Now, according to equations (1) and (2) of Sec. 117,
Hence
(5)
1 12:rrlim - P(r0, r, Q>- 8)G(Q>) dQ> = G(8).
r--Ho 2n 0
''"O
lim Ur(r, 8) = G(8)
r-+ro
r<ro
for each value of 8 at which G is continuous.
CHAP. 12
When G is piecewise continuous and satisfies condition (4), the formula
r 12:rr(6) U(r, 8) =- __Q_ ln[rJ - 2r0r cos(Q> - 8) +r 2
] G(Q>) dQ> +U0 (r < r0),
2n o
therefore, solves the Neumann problem for the region interior to the circle r = r0,
where G(8) is the normal derivative ofthe harmonic function U (r, 8) at the boundary
in the sense ofcondition (5). Note how it follows from equations (4) and (6) that, since
In rJ is constant, U0 is the value of U at the center r=0 of the circler = r0•
The values U (r, 8) may represent steady temperatures in a disk r < r0 with insu-
lated faces. In that case, condition (5) states that the flux ofheat into the disk through its
edge is proportional to G(8). Condition (4) is the natural physical requirement that the
total rate offlow ofheat into the disk be zero, since temperatures do not vary with time.
A corresponding formula for a harmonic function H in the region exterior to the
circle r =r0 can be written in terms of Q as
(7)
1 12:rrH(R, l/1) = -- Q(r0, R, Q> -lfr)G(Q>) dQ> +Ho
2n o
(R > r0),
where H0 is a constant. As before, we assume that G is piecewise continuous and that
condition (4) holds. Then
and
(8)
H0 = lim H(R, l/1)
R--+oo
lim HR(R, l/1) = G(lfr)
R->ro
R>ro
for each l/1 at which G is continuous. Verification of formula (7), as well as special
cases of formula (3) that apply to semicircular regions, is left to the exercises.
Turning now to a half plane, we let G(x) be continuous for all real x, except
possibly for a finite number of finite jumps, and let it satisfy an order property
(9) (a> 1) |
refresher or introduction to contemporary Fourier Analysis, this book starts from the beginning and assumes no specific background. Readers gain a solid foundation in basic concepts and rigorous mathematics through detailed, user-friendly explanations and worked-out examples, acquire deeper understanding by working through a variety of exercises, and broaden their applied perspective by reading about recent developments and advances in the subject. Features over 550 exercises with hints (ranging from simple calculations to challenging problems), illustrations, and a detailed proof of the Carleson-Hunt theorem on almost everywhere convergence of Fourier series and integrals of L p functions—one of the most difficult and celebrated theorems in Fourier Analysis. A complete Appendix contains a variety of miscellaneous formulae. L p Spaces and Interpolation. Maximal Functions, Fourier transforms, and Distributions. Fourier Analysis on the Torus. Singular Integrals of Convolution Type. Littlewood-Paley Theory and Multipliers. Smoothness and Function Spaces. BMO and Carleson Measures. Singular Integrals of Nonconvolution Type. Weighted Inequalities. Boundedness and Convergence of Fourier Integrals. For mathematicians interested in harmonic analysis. |
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