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Equilibrium of a Rigid Body
When some external forces (which may be concurrent or parallel) are acting on a stationary body, the body may start moving or may start rotating about any point. But if the body does not start moving
and also does not start rotating about any point, then the body** is said to be in equilibrium.
1.12.1. Principle of Equilibrium. The principle of equilibrium states that, a stationary body which is subjected to coplanar forces (concurrent or parallel) will be in equilibrium if the algebraic
sum of all the external forces is zero and also the algebraic sum of moments of all the external forces about any point in their plane is zero. Mathematically, it is expressed by the equations :
ΣF = 0
ΣM =0
The sign l: is known as sigma which is a Greek letter. This sign represents the algebraic sun of forces or moments.
The equation (1.22) is also known as force law of equilibrium whereas the equation (1.23) is known as moment law of equilibrium.
The forces are generally resolved into horizontal and vertical components. Hence equation (1.22) is written as
Σ F [x] =0
Σ F [y] =0
where ΣF = Algebraic sum of all horizontal components
and ΣF = Algebraic sum of all vertical components.
1.12.2. Equations of Equilibrium for Non-concurrent Force Systems. A non-concurrent force systems will be in equilibrium if the resultant of all forces and moment is zero.
Hence the equations of equilibrium are
ΣF [x] = 0, ΣF [y] = 0 and ΣM = 0
1.12.3. Equations of Equilibrium for Concurrent Force System. For the concurrent forces, the lines of action of all forces meet at a point, and hence the moment of those force about that very point
will be zero or ΣM = 0 automatically.
Thus for concurrent force system, the condition I.M = 0 becomes redundant and only two conditions, i.e., ΣF[ x ]= 0 and ΣF[Y] = 0 are required.
1.12.4. Force Law of Equilibrium. Force law of equilibrium is given by equation (1.22) or by equations (1.24) and (1.25). Let us apply this law to the following important force system:
(i) Two force system (ii) Three force system
(iii) Four force system.
1.12.5. Two Force System. When a body is subjected to two forces, then the body will be in equilibrium if the two forces Fig. 1. 71 are collinear, equal and opposite as shown in Fig. 1.71.
If the two forces acting on a body are equal and opposite but are parallel, as shown in Fig. 1.71(a), then the body will not be in equilibrium. This is due to the fact that the three conditions of
equilibrium will not be satisfied. This is proved as given below :
(i) Here Σ F [y] = 0 as there is no horizontal force acting on the body. Hence first condition of equilibrium is satisfied.
(ii) Also here ΣF[Y] = 0 as F [1] = F[2 ].
Hence second condition of equilibrium is also satisfied.
(iii) ΣM about any point should be ·zero. The resultant moment about point A is given by · Fig. 1.71(a)
M [A] = – F[2] x AB (- ve sign is due to clockwise moment)
But M[A] is not equal to zero. Hence the third condition is not satisfied.
Hence a body will not be in equilibrium under the action of two equal and opposite parallel forces.
Two equal and opposite parallel forces produce a couple and moment of the couple is – F 1 x AB [Fig. 1.71(a)].
1.12.6. Three Force System. The three forces acting on a body which is in equilibrium may be either concurrent or parallel. Let us first consider that the body is in equilibrium when three forces,
acting on the body, are concurrent. This is shown in Fig. 1.72.
(a) When three forces are concurrent. The three concurrent forces F[1]‘ F[2] and F[3] are acting on a body at point 0 and the body is in equilibrium. The resultant of F [1] and F[2] is given by R. If
the force F[3] is collinear, equal and opposite to the resultant R, then the body will be in equilibrium. The forceF[3] which is equal and opposite to the resultant R is known as equilibrant. Hence
for three concurrent forces acting on a body when the body is in equilibrium, the resultant of the two forces should be equal and opposite to the third force.
and F[3] are acting and the body is in equilibrium. If three forces F[1] F[2] and F[3] are acting in the same direction, then there will be a resultant R =F[1] +F[2 ]+ F[3] and body will not be in
equilibrium. The three forces are acting in opposite direction and their magnitude is so adjusted that there is no resultant force and body is in equilibrium. Let us suppose that F2 is acting in
opposite direction as shown in Fig. 1.73.
Now let us apply the three conditions of equilibrium:
(i) ΣF x = 0 as there is no horizontal force acting on the body
(ii) ΣF y.= 0 i.e., F1 + F3 = F2
(iii) ΣM = 0 about any point .
Taking the moments of F[1]F[2] and F[3] about point A,
ΣM[A] = -F[2] x AB +F[3] x AC
(Moment of F3 is anti-clockwise whereas moment of F2 is clockwise)
For equilibrium, Σ.MA should be zero
i.e., -F[2 ]x A[B] + F[3] x AC = 0
If the distances AB and AC are such that the above equation is satisfied, then the body will be in equilibrium under the action of three parallel forces.
1.12.7. Four Force System. The body will be in equilibrium if the resultant force in horizontal direction is zero (i.e., Σ F x = 0), resultant force in vertical direction is zero (i.e., r. F. = 0)
and moment of all forces about any point in the plane of forces is zero (i.e., r. M = 0). >
Problem 1.32. Two forces F1 and F2 are acting on a body and the body is in equilibrium. If the magnitude of the force F1 is 100 N and its acting at 0 long x-axis as shown in Fig. 1. 74, then
determine the magnitude and direction of force F2.
Sol. Given:
Force, F1 = 100 N
The body is in equilibrium under the action of two forcesF1 andF2.
When two forces are acting on a body and the body is in equilibrium, then the two forces should be collinear, equal and opposite.
F [2] = F[1] = 100 N
The force F[2] should pass through 0, and would be acting in the opposite direction of F[1]
Problem 1.33. Three forces F[1], F[2] and F[3] are acting on a body as shown in Fig. 1. 75 and the body is in equilibrium. If the magnitude of force F[3] is 400 N, find the magnitudes of force F[1]
and F[2].
Sol. Given:
Force, F [3] = 400 N
As the body is in equilibrium, the resultant force in
x-direction should be zero and also the resultant force in
y-direction. should be zero.
(i) for Σ F [x] = 0
F[1] cos 30^o + F[2] cos 30^o = 0
F[1] – F[2 ]= 0
ΣF[Y] = 0 , we get
(ii) for Σ F [y] = 0 , we have
F[1] sin 30^o + F[2] sin 30^o - 400 = 0
F[1] x 0.5 + f[2] x 0.5 = 400
or f[1] x 0.5 + f[2] x 0.5 = 400
or f[1] = 400 N
also f[2] = f1 = 400 N.
2nd Method
If three forces are acting on a body at a point and the body is in equilibrium, Lame’s Theorem can be applied.
Using Lame’s theorem,
F[1] / sin 120^o = f[2] / sin 120^ o = 400 / sin 120^ o
F[1] = F[2] = 400 N.
Problem 1.39.Three parallel forces F[1] F[2] and F[3] are acting on a body as shown in
Fig. 1.76 and the body is in equilibrium. Ifforce F[1]= 250 Nand F[3]= 1000 N and the distance
between F[1] and F[2]= 1.0 m, then determine the magnitude offorce F[2] and the distance ofF[2] from
force F[3].
Sol. Given:
Force, F[1 ]=250 N
Force, F[3]=1000 N
Distance AB =1. 0 m
The body is in equilibrium.
Find F 2 and distance BC.
For the equilibrium of the body, the resultant force in the vertical direction should be zero (here there
is no force in horizontal direction).
ForƩF[y =]0, we get
F[1]+ F[3]- F2 = 0
or 250 + 1000 -F[2]=0
or F[2]= 250 + 1000 = 1250 N. Ans.
For the equilibrium of the body, the moment of all forces about any point must be zero:
Taking moments of all forces about point A and considering distance BC = x, we get
or (AC = AB + BC = 1 + x)
or 1250 – 1000 – 1000x = 0
or 250 = 1000x
or X=
Problem 1.40.The five forces F[1], F[2], F[3], F[4] and F[5]are acting at a point on a body as
shown in Fig. 1.77 and the body is in equilibrium. IfF[1]= 18 N, F[2]= 22.5 N, F[3]= 15 Nand F[4]= 30 N, find the force F[5] in magnitude and direction.
Sol. Given:
Forces, F[1]= 18 N, F[2]=2.25N,
F[3]= 15 N andF[4]= 30 N.
The body is in equilibrium. Find force F[5]in magnitude and direction. This problem can
be solved analytically and graphically.
1. Analytical method
Let θ= Angle made by force F[5]with horizontal axis O-X’.
As the body is in equilibrium, the resultant forcein x-direction and y-direction should be zero.
(i) For ƩF[x]= 0, we get
F[1]+ F[2], cos 45^0- F[4]cos 30^0-F[5]cas θ= 0
or 18 + 22.5 x 0.707 – 30 x 0.866 -F[5]cos θ= 0
or 18 + 15.9 – 25.98 -F[5]cos θ= 0
or F[5]cos θ= 18 + 15.9 – 25.98
F[5]case = 7.92 ……(i)
(ii) For ƩF= 0, we get
F[2]sin 45^0+ F[3]- F[4]sin 30^0 – F[5]sin θ= 0
or 22.5 x 0.707 + 15 – 30 x 0.5 -F[5]sin θ= 0
or 15.9 + 15 – 15 – F[5]sin θ= 0
or F[5]sin θ= 15.9
Dividing equation (ii) byequation (i), we get =or tan θ= 2.0075
θ= tan^-1 2.0075 = 63.52°. Ans.
Substituting the value of e in equation (i), we get
F[5]cos 63.52° = 7.92
F5 = = 17.76 N. Ans.
2. Graphical method
(i) First draw a space diagram with given four forces F[1], F[2]F[3]and F[4]at correct angles as shown in Fig. 1.78(a).
(ii) Now choose a suitable scale, say 1 cm= 5 N for drawing a force diagram. Take any point O in the force diagram as shown in Fig. 1.78(b).
(iii) Draw line Oaparallel to force F[1]and cut Oa= F[l]= 18 N to the same scale.
(iv) Froma, draw the line abparallel to F[2]and cut ab= F[2]= 22.5 N
(v) From b, draw the line be parallel to F[3]and cut be = F[3]= 15 N
(vi) From c, draw the line cd parallel to F[4]and cut cd = F[4]= 30 N
(vii) Now join d to O. Then the closing side dOrepresents the force F[5]in magnitude and direction. Now measure the length dO.
By measurement, length dO= 3.55 em.
Force F[5]= Length dOx Scale = 3.55 x 5 = 17.75 N. Ans.
The direction is obtained in the space diagram by drawing the force F[5]parallel to line
Measure the angle θ, which is equal to 63.5°. Or the force F[5]is making an angle of
180 + 63.5 = 243.5° with the force F[1].
Problem 1.41.Fig. 1.78(c) shows the coplanar system of forces acting on a flat plate.
Determine: (i) the resultant and (ii) x and y intercepts of the resultant.
Sol. Given:
Force at A = 2240 N
Angle with x-axis = 63.43°
Force at B = 1805 N
Angle with x-axis = 33.67°
Force at C = 1500 N
Angle with x-axis = 60°
Lengths OA = 4 m
DB = 3m
DC = 2m
OD = 3m.
Each force is resolved into X and Y components as shown in Fig. 1. 78(d).
(i) Force at A= 2240 N
Its X-component = 2240 × cos 63.43° = 1001.9 N
Its Y-component = 2240 × sin 63.43° = 2003.4 N
(ii) Force at B = 1805 N.
Its X-component = 1805 × cos 33.67° = 1502.2 N
Its Y-component = 1805 × sin 33.67° = 1000.7 N
(iii) Force at C = 1500 N.
Its X-component = 1500 × cos 60° = 750 N
Its Y-component = 1500 × sin 60° = 1299 N
The net force along X-axis,
R[x] = ƩLF[x ]= 1001.9 – 1502.2 – 750 = – 1250.3 N
The net force along Y-axis,
R[y]= ƩLF[y]= – 2003.4 – 1000.7 + 1299 = – 1705.1 N
(i) The resultant force is given by,
R= √(R_(x^2 )+R_(y^2 ) )= √((-1250.3)^2+(-1750.1)^2 )
= √(1563250+2907366)=2114.4 N. Ans
The angle made by the resultant with x-axis is given by
The net moment* about point O,
M[o] = 2003.4 × 4 + 1000.7 × 3 -1299 × 2 -1502.2 × 3 – 750 × 3
= 8012.16 + 3002.1- 2598 – 4506.6 – 2250
= 11014.26 – 9354.6 = 1659.55 Nm
As the net moment about O is clockwise, hence the resultant must act towards right of origin O, making an angle = 53.7° with x-axis as shown in Fig. 1. 78(e).The components R[x] and R[y ]are also
negative. Hence this condition is also satisfied.
(ii) Intercepts of resultant on x-axis and y-axis [Refer to Fig. 1.78(e)].
Let x= Intercept of resultant along x-axis.
y = Intercept of resultant along y-axis.
The moment of a force about a point is equal to the sum of the moments of the components of the force about the same point. Resolving the resultant (R) into its component R[x] and R[y] at F.
Moment of R about 0 = Sum of moments of R[x] and R[y] at O
*Considering clockwise moment positive and anti-clockwise moment as negative. At A, the X component of 1001.9 N passes through O and hence has no moment.
But moment of R about O = 1659.66 (M[o] = 1659.66)
1659.66 = Rx × O + R[y] × x
(as R[x] at F passes through O hence it has no moment)
1659.66 = 1705.1 × x
To find y-intercept, resolve the resultant R at G into its component R[x] and R[y].
Moment of R about O = Sum of moments of R[x] and R[y] at O
or 1659.66 = R[x] × y + R[y] × O.
(At G, R[y] passes through O and hence has no moment)
1659.66 = 1250.3 × y
Problem 1.42. A lamp weighing 5 N is suspended from the ceiling by a chain. It is
pulled aside by a horizontal cord until the chain makes an angle of 60° with the ceiling as
shown in Fig. 1.79. Find the tensions in the chain and the cord by applying Lame’s theorem and
also by graphical method.
Sol. Given:
Weight of lamp= 5 N
Angle made by chain with ceiling = 60°
Cord is horizontal as shown in Fig. 1.79.
(i) By Lame’s theorem
Let T[1] = Tension (or pull) in the cord
T[2] = Tension (or pull) in the chain.
Now from the geometry, it is obvious that angles between T[1] and lamp will be 90°, between lamp and T[2] 150° and between T[2] and T[1 ]120°. [Refer to Fig. 1.79(b)].
Applying Lame’s theorem, we get
(ii) By Graphical method
(1) First draw the space diagram at correct angles as shown in Fig. 1.79(b). Now choose
a suitable scale say 1 cm= 1 N for drawing a force diagram as shown in Fig. 1.79(c). Take any
point O in the force diagram.
(2) From O, draw the line Oa vertically downward to represent the weight of the lamp. Cut Oa = 5N.
(3) From a, draw the line ab parallel to T[2.] The magnitude of T[2] is unknown. Now from
0, draw the line Ob horizontally (i.e., parallel to T[1]) cutting the line ab at point b.
(4) Now measure the lengths ab and bO).
Then ab represents T[2] and bO represents T[1]. By measurements, ab= 5.77 cm and bO
= 2.9 cm.
Pull in the cord = bO= 2.9 cm × scale = 2.9 × 1
= 2.9 N. Ans.
Pull in the chain = ab= 5.77 cm × scale = 5.77 × 1
= 5.77 N. Ans.
Problem 1.43.On a horizontal line PQRS 12 cm long, where PQ = QR = RS = 4 cm,
forces of 1000 N, 1500 N, 1000 N and 500 N are acting at P, Q, Rand S respectively, all
downwards, their lines of action making angles of 90°, 60°, 45° and 30° respectively with PS.
Obtain the resultant of the system completely in magnitude, direction and position graphically
and check the answer analytically.
Sol. Given:
PQ = QR = RS = 4 cm
Force at P = 1000 N. Angle with PS = 90°
Force at Q = 1500 N. Angle with QS = 60°
Force at R = 1000 N. Angle with RS = 45°
Force at S = 500 N. Angle with PS = 30°
Graphical method
Draw the space diagram of the forces as shown in Fig. 1.79(d). The procedure is as follows:
(i) Draw a horizontal line PQRS = 12 cm in which take PQ = QR = RS = 4 cm.
(ii) Draw the line of action of forces P, Q, R, S of magnitude 1000 N, 1500 N, 1000 N and
500 N respectively at an angle of 90°, 60°, 45° and 30° respectively with line PS as shown in
Magnitude and Direction of Resultant Force (R*)
To find the magnitude and direction of the resultant force, the force diagram is drawn
as shown in Fig. 1. 79(e) as given above:
(i) Draw the vector ab to represent the force 1000 N to a scale of 1 cm= 500 N. The
vector ab is parallel to the line of action of force P.
(ii) From point b, draw vector bc = 1500 N and parallel to the line of action of force Q. Similarly the vectors, cd = 1000 N and parallel to line of action of force R and de = 500 N and
parallel to the line of action of force S, are drawn.
(iii) Join ae which gives the magnitude of the resultant. Measuring ae, the resultant force is equal to 3770 N.
(iv) To get the line of action of the resultant, choose any point O on force diagram (called
the pole) and join Oa, Ob, Oc, Od and Oe.
(v) Now choose any point X; on the line of action of force P and draw a line parallel to Oa.
(vi) Also from the point X[1] draw another line parallel to Ob, which cuts the line of action
of force Q at X[2]. Similarly from point X[2], draw a line parallel to Oc to cut the line of action of
force R at X[3]. From point X[3], draw a line parallel to Od to cut the line of action of force S atX[4].
(vii) From point X[4], draw a line parallel Oe.
(viii) Produce the first line (i.e., the line from X[l] and parallel to Oa) and the last line (i.e.,
the line from X[4] and parallel to Oe) to interest at.X, Then the resultant must pass through this
(ix) From point X, draw a line parallel to ae which determines the line of action of resultant force. Measure PX. By measurements:
Resultant force, R* = 3770 N
Point of action, PX = 4.20 cm
Direction, θ = 60° 30′ with PS.
Analytical method
In analytical method, all the forces acting can be resolved horizontally and vertically. Resultant of all vertical and horizontal forces can be calculated separately and then the final
resultant can be obtained.
Resolving all forces and considering the system for vertical forces only.
Vertical force at P = 1000 N
Vertical force at Q = 1500 sin 60° = 1299 N
Vertical force at R = 1000 sin 45° = 707 N
Vertical force at S = 500 sin 30° = 250 N
Let Rv*= the resultant of all vertical forces and acting at a distance x cm from P.
= 1000 + 1299 + 707 + 250 = 3256 N
Taking moments of all vertical forces about point P,
Rv* × x = 1299 × 4 + 707 × 8 + 250 × 12 = 13852
Now consider the system for horizontal forces only, horizontal force at P= 0
Horizontal force at Q = 1500 × cos 60° = 750 N
Horizontal force at R = 1000 × cos 45° = 707 N
Horizontal force at S = 500 × cos 30° = 433 N
Resultant of all horizontal forces will be,
R[H]* = 0 + 750 + 707 + 433 = 1890 n
The resultant R* of R[V]* and R[H]* will also pass through point X which is at a distance of 4.25 cm from P.
The resultant will make an angle θ with PS and is given by
θ = tan^-1 1.723 = 59.9°
Thus the resultant of 3764 N makes an angle 59.9° with PS and passing through point
X which is at a distance of 4.25 cm from point P.
This result confirms closely with the values obtained by graphical method.
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double integrals in polar coordinates
June 21st 2010, 09:14 PM #1
Dec 2009
double integrals in polar coordinates
And this is the answer they have given us:
They converted to polar coordinates. I've seen this type of problem being solved without conversion to polar coordinates. I tried to evaluate the integral without converting to polar and I ended
up with a different answer:
$\int^1_0 \int^{-y+1}_{x=0} 2x-2y dx dy$
$\int^1_0 y^2-2y+1dy = \frac{1}{3}-1+1=\frac{1}{3}$
But my answer does not agree with the model answer! Shouldn't we end up with the same result regardless of whether we use polar or rectangular coordinates? I appreciate any help because I have an
exam tomorrow...
Last edited by demode; June 21st 2010 at 09:43 PM.
You left out the r with the jacobian or the r associated with the x or y.
In either case you need two r's in the integral
$x=r\cos\theta$ and $y=r\sin\theta$ while $dxdy=rdrd\theta$
Furthermore, you shoudn't go anywhere near polar, since this is a triangle and your polar bounds are of a circle.
the line x+y=1 converted to polar is hideous.... $r={1\over \sin\theta+\cos\theta}$
June 21st 2010, 10:57 PM #2
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Expanding universe can emerge in remarkably simple way, scientists say
Recipe for a universe: Apply heat and stir.
When soup is heated, it starts to boil. When time and space are heated, an expanding universe can emerge, without requiring anything like a "Big Bang". This phase transition between a boring empty
space and an expanding universe containing mass has now been mathematically described by a research team at the Vienna University of Technology, together with colleagues from Harvard, the MIT and
Edinburgh. The idea behind this result is a remarkable connection between quantum field theory and Einstein's theory of relativity.
A Cookbook for Spacetime
Everybody knows of the transitions between liquid, solid and gaseous phases. But also time and space can undergo a phase transition, as the physicists Steven Hawking and Don Page pointed out in 1983.
They calculated that empty space can turn into a black hole at a specific temperature.
Can a similar process create a whole expanding universe such as ours? Daniel Grumiller from the Vienna University of Technology looked into this, together with colleagues from the USA and Great
Britain. Their calculations show that there is indeed a critical temperature at which an empty, flat spacetime turns into an expanding universe with mass. "The empty spacetime starts to boil, little
bubbles form, one of which expands and eventually takes up all of spacetime", explains Grumiller.
For this to be possible, the universe has to rotate – so the recipe for creating the universe is "apply heat and stir". However, the required rotation can be arbitrarily small. In a first step, a
spacetime with only two spatial dimensions was considered. "But there is no reason why the same should not be true for a universe with three spatial dimensions", says Grumiller.
Daniel Grumiller heats up spacetime - with pencil and paper.
Looking for the Structure of the Universe
Our own universe does not seem to have come into existence this way. The phase-transition model is not meant to replace the theory of the Big Bang. "Today, cosmologists know a lot about the early
universe – we are not challenging their findings. But we are interested in the question, which phase transitions are possible for time and space and how the mathematical structure of spacetime can be
described" says Grumiller.
The new theory is the logical next step after the so called "AdS-CFT correspondence", a conjecture put forward in 1997, which has strongly influenced fundamental physics research ever since. It
describes a peculiar connection between theories of gravity and quantum field theories – two areas which, at first glance, do not have much in common. In certain limiting cases, according to AdS-CFT
correspondence, statements from quantum field theories can be translated into statements concerning gravitational theories and vice versa. This is almost as surprising as the idea of making
statements about a stone falling to the ground by actually calculating the temperature of a hot gas. Two completely different areas are being connected – but it works.
In this kind of correspondence, the quantum field theory is always described in one fewer dimension than the gravitational theory. This is called "holographic principle". Similar to a two dimensional
hologram which can depict a three dimensional object, a quantum field theory with two spatial dimensions can describe a physical situation in three spatial dimensions.
Arjun Bagchi (right) is a physicist from India, currently holding a Lise-Meitner Fellowship, working with Daniel Grumiller (left) on holographic correspondences in flat spacetimes.
A Correspondence Principle for Flat Spacetimes
To do this, the gravitational calculations usually have to be done in an exotic kind of geometry – in so-called "Anti-de-Sitter-spaces", which are quite different from the flat geometry we are used
to. However, it has been suspected for a while, that there may be a similar version of the "holographic principle" for flat spacetimes. But for a long time there haven't been any models showing this.
Last year, Daniel Grumiller and colleagues established such a model (in two spatial dimensions, for simplicity). This led to the current question; phase transitions in quantum field theories are well
known. But for symmetry reasons this would mean that gravitational theories should exhibit phase transitions too.
"At first, this was a mystery for us", says Daniel Grumiller. "This would mean a phase transition between an empty spacetime and an expanding universe. To us, this sounded extremely implausible." But
the calculations showed exactly that. "We are only beginning to understand these remarkable correspondence relations", says Daniel Grumiller. Which new ideas about our own universe can be derived
from this, is hard to say – only spacetime will tell.
More information: "Cosmic evolution from phase transition of 3-dimensional flat space." Arjun Bagchi, Stephane Detournay, Daniel Grumiller, Joan Simon. Phys. Rev. Lett. 111, 181301 (2013)
arXiv:1305.2919. arxiv.org/abs/arXiv:1305.2919
See also: A. Bagchi, S. Detournay and D. Grumiller, Phys. Rev. Lett. 109, 151301 (2012) arxiv.org/abs/arXiv:1208.1658
1.3 / 5 (23) Dec 10, 2013
Amazingly simple. God said "Let there be light", and there was light.
Some things don't have to be complicated.
2.2 / 5 (6) Dec 10, 2013
Space-time differs from ordinary matter less than physicists expected: the three phases of space-time
2 / 5 (10) Dec 10, 2013
Universe is made exclusively by waves, there are no particles.
Waves keep bouncing thanks to the internal pressure of the Universe.
Our perception of time is the slow loss of internal pressure in the Universe.
4.2 / 5 (6) Dec 10, 2013
I'm still stuck in the 18th century. How can a vacuum have a temperature if temperature is defined by the motion of matter? More jiggling=higher temperature.
I guess they mean the activity of the virtual particles.But that still does not get me very far because the total energy is conserved. (Zero)
Or is it?
I can see one sacred cow headed for the slaughterhouse.
Well done the reporter- you made a stout effort at explaining an esoteric subject.
2.3 / 5 (3) Dec 10, 2013
Duplicate, sorry.
3.3 / 5 (3) Dec 10, 2013
@egleton: Also, what does it mean for a vacuum to rotate?
not rated yet Dec 10, 2013
Egleton: Space is almost but not quite empty, in fact its density varies. Think about it - if it was uniformly denser closer to the big bang, then it has to be so.
The average temperature of space is 2.7 Kelvin, look up CMBR on Wikipedia.
2 / 5 (4) Dec 10, 2013
yes funny how they keep coming back to the bible version , i thought there was no such thing as empty space , higgs fields and whatnot
1 / 5 (2) Dec 11, 2013
And the stork brings babies. Learned that at VUT.
1 / 5 (1) Dec 11, 2013
[in case a scientist is lurking this far down the well] Since the basic idea is that our space time is 'projected' from a lower dimension I would like to know if it is an entire dimension that shows
up and if so, which one? Time? Or does this somehow put some "extra" onto three dimensions (of something, wouldn't be space-time) and pulls the fourth a little bit from each of them such that we
can't say specifically which of x,y,z,t came into being?
Also, for those who utterly pan this kind of work, I promise you that the moment you lower the internal rhetoric of your conclusions (you can still blab out loud, but start thinking for
yourself...hesaidhopingtonotoffendbutneedingtobeclear) and put some of that energy to understanding what ~just the main-stream~ physicists are saying these days, it will enrichen your life in
beautiful ways. And, you'll not longer be afraid of the dark -- that goes early.
1 / 5 (1) Dec 11, 2013
If the universe did not yet exists how can it have rotated and heated up? Where di the little bit of energy to heat and stir come from?
3.7 / 5 (3) Dec 11, 2013
If the universe did not yet exists how can it have rotated and heated up? Where di the little bit of energy to heat and stir come from?
From the article:
Our own universe does not seem to have come into existence this way. The phase-transition model is not meant to replace the theory of the Big Bang.
1 / 5 (1) Dec 11, 2013
Sounds interesting, but a deSitter space is a better approximation to our universe than a flat Minkowski one is, which has Zero cosmological constant. The entire AdS-CFT idea has forever been
crippled by demanding that the cosmological constant be negative, when in fact, it is experimentally observed to be positive.
1 / 5 (1) Dec 13, 2013
Egleton: The average temperature of space is 2.7 Kelvin, look up CMBR on Wikipedia.
That is light inside space. Not empty space.
1 / 5 (1) Dec 15, 2013
how can nonexistent spacetime before the beginning of the universe rotate if there is nothing relative to it to measure its rotation?
3 / 5 (2) Dec 15, 2013
How can a vacuum have a temperature if temperature is defined by the motion of matter?
There's several ways to define a temperature (also note that an absolute vacuum is only a theoretical construct)
But that still does not get me very far because the total energy is conserved.
The temperature here is what a black body would settle at if left in outer space (i.e. what temperature it would radiate if it, on average, radiated as much energy as it would receive.) That
temperature would be the current temperature of the CMB
The temperature of the CMB does not originate from the current 'vacuum' or its virtual particles (if it were then the universe would be heating up all the time as you'd be pumping net energy into
it). That temperature is the afterglow of the big bang spread through an expanding spacetime (i.e. red shifted).
3 / 5 (2) Dec 15, 2013
how can nonexistent spacetime before the beginning of the universe rotate if there is nothing relative to it to measure its rotation?
That's the question then, isn't it?. It can rotate if the universe is embedded in a larger structure (branes), as passing other universes can have an effect there and impart net (local) spin to a
However this directly leads to the question how all the other branes got to be there. if by the same mechanism then we need to think about how rotation is not conserved in this larger system...
... because if it is then we'd be back to square one: where does the rotation come from to craete the first brane?.
Unless we can have a phase transition there as well where two parts split off - each with a net rotation but the sum of which is always zero.
1 / 5 (1) Dec 16, 2013
Beware that a gang of Harvard "researchers" has already stole in Nature journals and, with further support of the MIT's ones, in ASC Nano Lett both the ideas and money of taxpayers. There are
numerous swindlers from David H. Koch Inst. for Integrative Cancer Research and Dept of Chemical Engineering, also with Dept of Chemistry and Chem. Biology and School of Eng and Applied Science of
Harvard University at http://issuu.com/...vard_mit & http://issuu.com/...llsens12
Their plagiaristic compilation titled Macroporous nanowire nanoelectronic scaffolds for synthetic tissues (DOI: 10.1038/NMAT3404) and Outside Looking In: Nanotube Transistor Intracellular Sensors
(dx.doi.org/10.1021/nl301623p) was funded by NIH Director's Pioneer Award (1DP1OD003900) and a McKnight Foundation Technol Innovations in Neurosc Award, also a Biotechnology Research Endowment from
the Dept. of Anesthesiology at Children's Hospital Boston and NIH grants GM073626, DE01-3023/6516
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Oil and Gas Operations Stocks, Including Valero, Making Big Moves on August 20, 2012 (NRGY,BRY,COG,SM,TGA,RRC,VLO)
The market has been slipping so far today. The Nasdaq is trading down 0.2%; the S&P 500 is down 0.3%; and the Dow has slipped 0.2%.
The Oil and Gas Operations sector (DIG) is currently lagging behind the overall market, down 1.1%, and its current biggest movers are:
│Company │Market Cap │Percentage Change│
│Inergy (NYSE:NRGY) │$2.6 billion │+4.5% │
│Berry Petroleum (NYSE:BRY) │$2.15 billion │-3.2% │
│Cabot Oil & Gas (NYSE:COG) │$9.03 billion │-3% │
│SM (NYSE:SM) │$3.18 billion │-2.7% │
│TransGlobe Energy Corporation (Nasdaq:TGA) │$762.3 million│+2.2% │
│Range (NYSE:RRC) │$11.32 billion│-2% │
│Valero (NYSE:VLO) │$15.89 billion│+2% │
Broker Summary: Fidelity Online Brokerage Inergy
) has risen 4.5% to hit a current price of $20.60 per share. With 274,607 shares changing hands so far today, the company's volume is 0.7 times its current three-month average. High volume indicates
a lot of investor interest while low volume indicates the opposite. In making a decision about a potential or existing investment, valuation ratios are useful as a basis for seeing whether the stock
price is too high, reasonable, or a bargain. One of the favorite tools of many value investors is analyzing price/book value ratios, as it provides a measure of the underlying value of a company's
assets as compared to the valuation of its equity. NRGY's P/B ratio of 2.49 shows that its share price is higher than its book value. This implies that investors expect management to create more
value from a given set of assets and/or that the market value of the firm's assets is significantly higher than their accounting value. P/B value ratios are particularly useful to value investors,
distressed or "vulture" investors, or any other investors purchasing beaten-down securities but are less useful to investors focused on growth stocks, purchasing IPOs, or investing in technology or
other "asset-lite" companies. SEE:
How Buybacks Warps The Price-To-Book Ratio Berry Petroleum
) is down 3.2% to reach $38.40 per share. This morning, the company is trading a volume of 162,210 shares. Volume is an important indicator because it indicates how significant a price shift is.
Investment valuation ratios provide investors with an estimation, albeit a simplistic one, of the value of a stock. The capitalization ratio is calculated by dividing long-term debt by the sum of
long-term debt and shareholders' equity. The capitalization ratio for BRY is 61%, which is fairly high. A high capitalization ratio is not necessarily bad since higher financial leverage can increase
the return on a shareholder's investment. This ratio is considered to be one of the more meaningful of the "debt" ratios - it delivers the key insight into the use of leverage by a company.
Cabot Oil & Gas
) has fallen 3% and is currently trading at $41.68 per share. The company's volume is currently 507,108 shares for the day, lighter than yesterday's volume of 2.6 million shares. In technical
analysis, trading volume is used to determine the strength of a market indicator. Investors can use valuation ratios as tools to estimate what kind of deal a particular investment is. There are
generally two price/earnings ratios calculated: the first, called the trailing Price/Earnings ratio, is calculated using the previous years actual earnings; the second, called forward Price/Earnings
ratio, is calculated using the next year's estimated earnings. COG's P/E ratio of 81.5 is above the industry average of 14.62. Generally speaking, the higher the P/E ratio, the higher the market
expectations for a company's future performance. From the investor's perspective, a stock with a lower ratio is relatively cheaper than a stock with a higher ratio. SEE:
Understanding The P/E Ratio
Currently trading at $47.40 per share,
) has fallen 2.7%. This morning, the company's volume is 246,510 shares. This is 0.3 times its current daily average. Volume is an important indicator in technical analysis as it is used to measure
the worth of a market move. If the markets have made a strong price move either up or down the perceived strength of that move depends on the volume for that period. The higher the volume during that
price move the more significant the move. Looking at a company's valuation ratios is a good way of getting a basic idea as to its value as an investment. The price/earnings to growth (PEG) ratio
compares a company's P/E ratio to its earnings-per-share growth rate, which tells you whether or not you are getting a good value when purchasing a stock with a high price/earnings ratio (P/E ratio).
PEG ratio for SM is 1.14. While P/E ratios are important indicators of market value, a high P/E in and of itself is not bad because it may indicate a company whose earnings are growing very rapidly,
so many investors look at the PEG ratio in order to get an idea of whether or not a particular P/E ratio is justified by underlying earnings growth.
After an increase of 2.2%,
TransGlobe Energy Corporation
) has reached a current price of $10.63. The company's volume for the day so far is 55,349 shares, in keeping with the average volume over the past three months. If a stock price makes a big move up
or down, volume lets us know the significance of that move. When estimating the value of a particular investment, valuation ratios provide a good basis for assessing the value of an individual stock.
The easy-to-calculate debt ratio is helpful to investors looking for a quick take on the leverage for a company. TGA has a debt ratio of 38%, which is fairly low. In other words, the company is less
sensitive to changes in business or interest rates since less of its cash flow is dedicated to paying off loan expenses. However, one thing to note with this ratio: it isn't a pure measure of a
company's debt (or indebtedness), as it also includes operational liabilities, such as accounts payable and taxes payable.
Falling 2%,
) is currently at a share price of $68.24. The company is currently trading a volume of 565,641 shares. When a stock price moves up or down, watching the volume is a good way of identifying how
significant that shift is. Investors can make use of valuation ratios to estimate whether a stock is fairly valued. The dividend yield is measured by taking the annual dividends per share and
dividing that number by the stock price. RRC has a dividend yield of 0.2%, which is fairly low. This may indicate that the company's stock is overpriced. High dividend yields are generally more
important to value investors, investors in larger companies, and income oriented investors than they are to growth investors, investors in small cap stocks, and investors in new or emerging
companies. SEE:
Investment Valuation Ratios: Dividend Yield
After rising 2%,
) is currently trading at a share price of $29.38. This morning, 3.9 million shares have been traded, on pace to reach yesterday's volume of seven million shares. Volume is also used as a secondary
indicator to help confirm what the price movement is suggesting. Valuation ratios include the price to earnings (P/E) ratio, the price to earnings growth (PEG) ratio, the price to sales (P/S) ratio,
the price to book (P/B) ratio, and the dividend yield. The price/sales ratio is used for spotting recovery situations or for double-checking that a company's growth has not become overvalued. VLO has
a low P/S ratio of 0.1. Coupled with high relative strength in the previous twelve months, a low P/S ratio is one of the most potent combinations of investment criteria. It is important to compare P/
S ratios for companies in the same industry, as ratios can vary quite widely for companies in different industries.
The Bottom Line
On any given day, a particular stock may see positive or negative change in its share price. Daily stock performance should be weighed against historical performance and put in context of the market
overall. Keep in mind that all these ratios should be compared against historical numbers and industry information in order to get a more complete picture.
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MathGroup Archive: June 2006 [00400]
[Date Index] [Thread Index] [Author Index]
Re: Some questions regarding loops and lists.
• To: mathgroup at smc.vnet.net
• Subject: [mg67224] Re: Some questions regarding loops and lists.
• From: Bill Rowe <readnewsciv at earthlink.net>
• Date: Tue, 13 Jun 2006 01:07:19 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com
On 6/11/06 at 11:08 PM, thelonias at wildox.net (Eric) wrote:
>The first question is regarding algorithm processing for multiple
>instances of the same form. For example: solution[n_] function is
>created where n_ represents the number of terms in the series-- this
>is not constant, but it does increase. The following steps occur
>for each usage of the function:
>1. Assignment to variable function for plotting purposes. Variable
>changes due to number of terms in each series.
>2. Plotting the function. ( I guess this is only a visual reassurance
>to keep a handle upon the validity of the findroot command.)
>3. using the findroot command to provide the lowest zero of the
>Presently this is done for each variable function (35 presently) by
>hand. Is there a way to automate this process successfully?
It sounds like you are trying to find the smallest root of a function with multiple roots. If so, you might find the RootSearch package Ted Ersek created useful. This package is available on the Wolfram MathSource web page. For example, after loading the RootSearch package
sol = RootSearch[x/2 + Sin[5*x] == 0, {x, -5, 5}]
{{x -> -1.1358415592628808}, {x -> -0.6998127639815517},
{x -> 0.}, {x -> 0.6998127639815517},
{x -> 1.1358415592628808}}
Selecting the smallest root can be done by
and finally visual verification that these are roots and all of the roots can be done with:
Show@Block[{$DisplayFunction = Identity},
{Plot[x/2 + Sin[5 x], {x, -2.5, 2.5}],
ListPlot[{x /. #, x/2 + Sin[5 x] /. #} & /@
sol, PlotStyle -> {PointSize[.015], Red}]}];
>The second question is regarding list appending. There is a set of
>values calculated which are then placed into a data table. This
>table is then used to create a list, and from this list a new list
>is created for graphing purposes. The new list contains two values
>in each element (x,y), instead of the one it was initially created
>with. The new value entered is in the position of x; it is entered
>by hand and is simply the original set of values which were
>calculated. Ideally, I would think there was a way to simply place
>the variables initially calculated into a two dimensional list, use
>the first dimension (the only one with a value initially) to
>calculate the second dimension elements, and then use that list for
>plotting purposes. I hope that gives an idea as to what I am trying
>to accomplish.
It sounds like you want to plot the results of some function on a set of values against the values. That is plot specific points {x, f[x]}. If so, starting with some random points
data = Table[Random[] - .5, {10}];
then squaring each value
results = #^2 & /@ data;
I can plot the results against the original data by:
ListPlot[Transpose@{data, results}];
Alteratively, the last two steps can be combined into a single step with:
ListPlot[{#, #^2} & /@ data];
Note, what you are trying to do would be much clearer if you included Mathematica code showing what you have tried. While attachments are not allowed by the group charter, simply copying and pasting from a Mathematica notebook into your message would suffice. But when you do this, you should convert the cells you copy to InputForm before the copy/paste operation.
To reply via email subtract one hundred and four
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Message Board Basketball Forum - InsideHoops - Money if you can prove I'm Theoo
NauruDude 04-10-2013 04:12 PM
Money if you can prove I'm Theoo
Now there has been rumours about that me, NauruDude, and that prick called Theoo are the one and same guy. These rumours piss me off.
They have been started by morons like millwad who envy me of my writing skills or something.
I'm offering 1111 icelandic kronas if anyone can prove that me and Theoo are the same guy (have the same IP then).
Photo of the money.
GOOD LUCK gangstaz :applause:
CeltsGarlic 04-10-2013 04:13 PM
Re: Money if you can prove I'm Theoo
maad money
Wally450 04-10-2013 04:20 PM
Re: Money if you can prove I'm Theoo
Why 1111?
blablabla 04-10-2013 04:29 PM
Re: Money if you can prove I'm Theoo
InspiredLebowski 04-10-2013 04:31 PM
Re: Money if you can prove I'm Theoo
you got an Ipad 2 Theoo?
kaiteng 04-10-2013 04:43 PM
Re: Money if you can prove I'm Theoo
Originally Posted by blablabla
Seems legit.
BuffaloBill 04-10-2013 04:52 PM
Re: Money if you can prove I'm Theoo
I don't know how much money that is, but it must be a lot because of how nice the guy's beard is.
fiddy 04-10-2013 05:05 PM
Re: Money if you can prove I'm Theoo
~10$ fukk outta here
ripthekik 04-10-2013 05:43 PM
Re: Money if you can prove I'm Theoo
his beard looks like a scrotum.
macmac 04-10-2013 05:58 PM
Re: Money if you can prove I'm Theoo
I know there's at least two people on insidehoops, me and someone else....beyond that, I cannot guarantee anything
Theoo 04-10-2013 06:25 PM
Re: Money if you can prove I'm Theoo
:roll: :roll: sure i chip to these ,a one time good thred by a nauruprude not good enough to make a real name...
ALL YOUR FK** guys retarts called millwad, slikkshoter , your all say "he naurudude he naurudude" well fk** come here to prove!!!! your talk talk and no fk** do anything to make a prove. if your say
afk** thing i well make your fk** ethered to link to these thread. COMONE TELL TO ME , HOW IM NAURUDUDE??? SHOW ME YOUR A FK** PROVES?? that right there is none :facepalm here a money that i chip to:
and if your can prove were same person i well never make posts again!! TALK TO JEFF IP!!! GO HEAD :roll:
Budadiiii 04-10-2013 11:30 PM
Re: Money if you can prove I'm Theoo
Your unduly denial is just making it more obvious.
HarryCallahan 04-10-2013 11:36 PM
Re: Money if you can prove I'm Theoo
Originally Posted by Budadiiii
Your unduly denial is just making it more obvious.
Shut up Godzuki.
Budadiiii 04-10-2013 11:47 PM
Re: Money if you can prove I'm Theoo
Originally Posted by HarryCallahan
Shut up Godzuki.
I have a vague recollection of who the **** Godzuki even is. I can assure you I'm not him, though
FiveRings 04-11-2013 03:54 AM
Re: Money if you can prove I'm Theoo
Theoo, you've been dethroned as the best poster on ISH by that Laker fan 6 for 24 who hails from Mozambique. Step up your game Theoo.
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what is the definition of addition? - WyzAnt Answers
what is the definition of addition?
Tutors, please
to answer this question.
The simple answer is that "adding" is basically "combining amounts." Using the arabic notation we ultimately have to just remember certain things you can find an addition table here:
If you look at the vertical column that has the numeral 6 at the very top in a pinkish box where the number itself is white, there are many numbers in white boxes below it with black text. Now look
at the row where there is a 4 all the way to the left in the pinkish box. If you look at where the row and the column intersect you can see the number 10. This means that 4 + 6 = 10. If you look at
the column that has the number 4 in the pink box and the row that has the number 6 in the pink box, you will find that the intersection is also 10. This means that 6 + 4 = 4 + 6 = 10.
Basically if you start with 4 things, and then add 6 more things then you end up with 10 things. Similarly if you started with 6 things and add 4 more you end up with 10 things still. We say that
addition is commutative for this reason.
There are lots more things that can be said about addition and the definition of addition can get pretty complicated if you get into the nitty gritty of mathematics. If you are trying to add bigger
numbers like 154 + 672 (which equals 826 by the way) that involves understanding what is meant by the number 72, which is actually a little complicated, so don't feel bad about asking this question.
I know that this might have been a bit more complicated than you were looking for so let me and other people on this site know if you need clarification, or have any other questions.
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Homework Help
Posted by nancy on Thursday, June 28, 2012 at 1:54pm.
BBA307: Finance
Exercise 8A
Assignment 8.1
Assignment 8.1: Capital Budgeting Application
You have just graduated and one of your favorite courses was Financial Management. While you were in school, your grandfather died and left you $1 million. You have decided to invest the funds in a
fast-food franchise and have two choices–Franchise L and Franchise S. You only intend to be in business for three years and then sell the franchise. See the cash flows for each year below:
Franchise L
Franchise S
$ 10
$ 70
$ 60
$ 50
$ 80
$ 20
Depreciation, salvage values, net working capital requirements, and tax effects are included in the cash flows. The required rate of return is 10%. You must decide which franchise to invest in.
1.What is each franchise's NPV? Be sure to show your calculations.
2.According to the NPV, which franchise or franchises should be accepted if they are independent? Which should be accepted if they are mutually exclusive?
3.Would the NPV change if the cost of capital changed?
4.What is each franchise's IRR? Be sure to show your calculations.
5.What is the logic behind the IRR method? According to the IRR, which franchises should be accepted if they are independent? Mutually exclusive?
6.Would the franchises' IRR change if the cost of capital changed?
7.Draw the NPV profiles for each franchise. At what discount rate do the profiles cross?
8.Using the NPV profiles above, which franchise or franchises should be accepted if they are independent? Mutually exclusive? Explain. Are your answers correct at any cost of capital less than 23.6%?
9.Which method is best and why?
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New map of E1b1b in Europe and the Middle East
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A Statistical Design for Testing Transgenerational Genomic Imprinting in Natural Human Populations
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PLoS One. 2011; 6(2): e16858.
A Statistical Design for Testing Transgenerational Genomic Imprinting in Natural Human Populations
Vladimir Brusic, Editor^
Genomic imprinting is a phenomenon in which the same allele is expressed differently, depending on its parental origin. Such a phenomenon, also called the parent-of-origin effect, has been recognized
to play a pivotal role in embryological development and pathogenesis in many species. Here we propose a statistical design for detecting imprinted loci that control quantitative traits based on a
random set of three-generation families from a natural population in humans. This design provides a pathway for characterizing the effects of imprinted genes on a complex trait or disease at
different generations and testing transgenerational changes of imprinted effects. The design is integrated with population and cytogenetic principles of gene segregation and transmission from a
previous generation to next. The implementation of the EM algorithm within the design framework leads to the estimation of genetic parameters that define imprinted effects. A simulation study is used
to investigate the statistical properties of the model and validate its utilization. This new design, coupled with increasingly used genome-wide association studies, should have an immediate
implication for studying the genetic architecture of complex traits in humans.
Genomic imprinting arises from a gene when either the maternally or paternally derived copy of it is expressed while the other copy is silenced [1], [2]. Caused by epigenetic modifications such as
DNA methylation established during gametogenesis and maintained throughout somatic development in the offspring, genetic imprinting has been shown to play a pivotal role in regulating the formation,
development, function, and evolution of complex traits and diseases [3], [4], [5], [6], [7], [8], [9], [10]. While most studies of genetic imprinting focus on the epigenetic and molecular mechanisms
of this phenomenon [7], [11], the number and distribution of imprinted genes and their epistatic interactions for quantitative traits are poorly understood, limiting the scope of our inference about
the effects of imprinting genes on the diversity of biological traits or processes. Several authors have started to use genome-wide association and linkage studies to identify the regions of the
genome that contain imprinted sequence variants and further understand the epigenetic variation of complex traits [12], [13], [14], [15].
In a series of recent studies, Cheverud, Wolf, and colleagues categorized genetic imprinting into different types based on the pattern of its expression, i.e., maternal expression, paternal
expression, bipolar dominance, polar overdominance, and polar underdominance [14], [15]. With a three-generation F[16]. By modeling identical-by-descent relationships in multiple related families of
canines, Liu et al. [13] derived a random effect model based on linkage analysis to genome-wide scan for the existence of iQTL that affect canine hip dysplasia. In a recent study, Wang et al. [9]
used reciprocal F
While epigenetic marks resulting in genomic imprinting can be generally stable in an organism's lifetime, they may undergo reprogramming, i.e., a faithful clearing of the epigenetic state established
in the previous generation, in the new generation during gametogenesis and early embryogenesis [17], [18], . However, a growing body of evidence since the early 1980s indicates that genes may escape
such reprogramming and, thus, inherit their imprinting effects into next generations [20], [21], [22], [23], [24], [25]. Two fundamental questions will naturally arise from this discovery: how common
are imprinted genes of this type and how strong is the evidence for their existence in humans and other organisms? If epigenetic changes through imprinted genes can be inherited across generations,
this would significantly alter the way we think about the inheritance of phenotype [26], [27]. Such transgenerational epigenetic inheritance, i.e., modifications of the chromosomes that pass to the
next generation through gametes, may be related with health and diseases with a mechanism for transmitting environmental exposure information that alters gene expression in the next generation(s)
[28]. The identification of imprinted loci displaying transgenerational epigenetic inheritance will be greatly helpful for addressing the two questions mentioned above, in a quest to elucidate the
detailed genetic architecture of complex traits and diseases.
The motivation of this study is to develop a novel strategy for identifying imprinted genes for a quantitative trait and understanding the transgenerational changes of their effects with a
three-generation family design by sampling multiple unrelated nuclear families, each composed of the grandfather, grandmother, father, mother, and grandchildren, from a natural population. This
transgenerational design contains information about how alleles at different loci co-transmit during meiosis from one generation to next and, thus, has been widely used for genetic linkage analysis
[29], [30]. By tracing the inheritance of alleles at a gene(s) from a paternal or maternal parent, this design allows the characterization of parent-of-origin of alleles and provides a powerful way
to estimate genetic imprinting effects. Because only genotypes can be observed, we formulate a mixture model to specify allelic configurations in terms of parental origins of the alleles. The EM
algorithm is implemented to estimate the effects of imprinted genes and their changes across generations. A testing procedure is proposed to study the pattern of transgenerational epigenetic
inheritance. The statistical behavior of the model is examined through simulation studies.
Simulation studies were performed to examine the statistical behavior of the model. A three-generation design is simulated which include a certain number of first-generation families sorted into 9
mating types (as shown in Table S1) according to the genotype frequencies. Assume that the allele frequencies of a gene are 0.6 and 0.4 in a natural population at Hardy-Weinberg equilibrium. Our
simulation will focus on the investigation of the impacts of different sampling strategies and heritabilities on parameter estimation and model power. For a given sample size, two sampling strategies
are simulated, (1) a large family number and small family size, and (2) a small family number and large family size.
The first strategy samples 200 unrelated grandfathers and 200 unrelated grandmothers, who marry to form 200 the first-generation families. Each first-generation family is assumed to have one son who,
as the father, form a second-generation family with the mother from the natural population. There is one child for each second-generation family. This allocation results in a total of 1000 subjects.
All members in the design are typed for the gene, but only the fathers and offspring of the third generation are phenotyped for a normally distributed trait. The second strategy samples 50 unrelated
grandfathers and 50 unrelated grandmothers. In each first-generation family, 3 sons are simulated, forming 150 second-generation families in which 4 children are assumed. This strategy also results
in 1000 subjects.
Different genetic effects of the gene, additive, dominant, and imprinting, are simulated for the second- and third-generations using the designed shown in Tables S2 and S3. Two different heritability
levels, 0.1 and 0.4, are simulated for each generation, from which variances are determined. Table 1 tabulates the estimates of population and quantitative genetic parameters from the
three-generation design. As expected, allele frequency can be very well estimated. The model provides reasonable estimation accuracy and precision for all genetic parameters under different sampling
strategies, even for a modest heritability level. Under both strategies, the model has great power (0.85 or higher) to test the significance of individual genetic effects, additive, dominant, and
imprinting, expressed in different generations. The model is also powerful to detect differences of genetic effects between two consecutive generations. More interesting, the difference of imprinting
effect between different generations, i.e., transgenerational inheritance of genetic imprinting, can be discerned with power 0.80 using our statistical design.
The maximum likelihood estimates (MLEs) of additive (
One major aim of this study is to estimate the change of genetic effects over generation. Although our model has great power to detect the transgenerational change of genetic effects, its false
positive rates should also be assessed. We conducted an additional simulation study to address this issue by simulating a SNP that has the same genetic effects between the two generations. The model
detects a small proportion of simulation replicates (
The haplotype model is also examined through simulation studies. We simulated two SNPs with a recombination fraction of
Table 2 gives the results of simulation for different heritabilities and sample sizes (all subjects used). Overall, all parameters can be estimated reasonably well. As expected, the precision of
parameter estimation increases with heritability and sample size. The additive genetic effects in both generations can well be estimated with a modest sample size (say 400) for a small heritability
(0.1). More sample sizes (say 800) are needed to provide a good estimate for genetic imprinting effects for a small heritability. To well estimate dominant genetic effects, an even larger sample size
(say 2000) is required for the same level of heritability.
Simulation results for transgeneration imprinting effects comparisons.
The traditional view of quantitative trait expression analysis assumes that the maternally and paternally derived alleles of each gene are expressed simultaneously at a similar level. However, this
view is violated by a growing body of evidence that alleles are expressed from only one of the two parental chromosomes [1], [2]. This so-called genetic imprinting or parent-of-origin effect has been
thought to play a pivotal role in regulating the phenotypic variation of a complex trait [3], [4], [6], [8], [9], [12], [13], [14], [15]. With the discovery of more imprinting genes involved in trait
control through molecular and bioinformatics approaches, we will be in a position to elucidate the genetic architecture of quantitative variation for various organisms including humans.
Recent evidence shows that epigenetic inheritance in humans may experience a transgenerational change. This would represent a significant shift in our current understanding of inheritance and disease
aetiology. Despite the development of new technologies that are reducing the time and cost of genotyping by several orders of magnitude [31], [32], the understanding of the underlying genetic events
will be challenging. In this article, we present a computational model for identifying the genomic imprinting effect of genes on quantitative phenotypes and transgenerational change of genomic
imprinting using a multigenerational sampling design for human families. The model formulates a general framework for testing the difference of genetic effects between different generations. By
including multiple SNPs, the model was extended to estimate genomic imprinting and its transgenerational change expressed at the haplotype level. Although several models have been developed to
estimate genomic imprinting for binary disease traits [33], [34], our model is among the first for estimating genetic imprinting operational in regulating the variation of quantitative traits and is
certainly the first of its kind that can discern the transgenerational change of genetic imprinting.
Although no real data were analyzed for the moment, this model presents a conceptual design by which new data can be collected according to the sampling strategy proposed and then analyzed by the
computational algorithm derived. Based on computer simulation, the model should display convincing statistical properties in parameter estimation and test and can be applied to a practical data set.
However, several issues need to be addressed when the model is attempted to solve broader genetic questions. First, the maternal effects that cause parent-of-origin effects of alleles may be
confounded with imprinting effects [35], which should be separated by developing a proper design in order to better study the patterns of gene expression and evolutionary dynamics.
Second, this study assumes the unisex (sons) produced from the first-generation family. One can also assume daughters with no change of the model, allowing the test of genomic imprinting between
mother and offspring. In fact, our model can involve both sexes so that in the second generation sex-specific genetic effects can be characterized. If the sexes in the third generation are
considered, the model can be extended to study the transgenerational changes of gene-sex interactions. Third, it is possible that part of parental genotypes are missing in practice. To infer genomic
imprinting using such data sets, a multi-hierachical mixture model can be derived to estimate the missing parental genotypes based on observed offspring genotypes. Fourth, although a basic premise of
epigenetic processes was that, once established, these marks were maintained through rounds of mitotic cell division and stable for the life of the organism, several recent studies have shown that at
some loci the epigenetic state can be altered by the environment [36]. The questions are how common are genes of this type and how strong is the evidence for their existence in humans? The
development of our design and model will help to address these biological questions of fundamental importance in elucidating the genetic architecture of complex traits.
Sampling Strategies
Suppose there is a natural human population at Hardy–Weinberg equilibrium (HWE) from which a panel of three-generation families, each composed of the grandfather, grandmother, father, mother, and
grandchildren, are sampled. Each member in a family is typed for single nucleotide polymorphisms (SNPs) from the human genome. Consider a quantitative trait affected by a SNP with two alleles A in a
frequency of a in a frequency of Table S1). Given a cross type, the genotypes of sons or daughters can be inferred. Here we first assume one sex (say son) in the second generation, although both
sexes can be considered. The sons from a family serve as the father to mate with the females as the mother derived from a natural population, with genotypes,
According to this design, the grandfathers and grandmothers are founders whose parents are unknown. Alleles of sons from a first-generation family can be traced directly or indirectly, but the
females used to generate the second-generation family are the founders with the unknown origin of alleles. For this reason, we will measure the phenotype for sons from the first-generation families
and grandchildren from the second-generation families. This design will allow us to characterize imprinting effects of a gene in the second- and third-generations.
Genetic Models
There are three genotypes,
The difference in the genetic architecture of a complex trait between two different generations is described as
By testing whether these differences are equal to zero jointly or individually, we can determine the transgenerational changes of the pattern of genetic control. If a significant imprinting effect is
detected, we can test the type of genetic imprinting, i.e., parental or maternal dominance, by incorporating the imprinting models of Cheverud et al. [14].
The grandfather and grandmother in the first generation from a natural population constitutes Table S1 (
It is not difficult to derive the maximum likelihood estimate of allele frequency from the three-generation family design as
The male individuals from the first generation are typed for the marker, with four distinct configurations,
Since offspring genotypes depend on parental genotypes, the log-likelihood of paternal and offspring parameters given marker (M) and phenotypic (
where Table S1 has a particular configuration is calculated by
In the M step, the genotypic values of configurations and variance are calculated by
The EM algorithm can also be implemented to estimate genetic parameters
In the M step, the genotypic values of configurations and variance are calculated by
Hypothesis Tests
It is imperative to know whether there exists a significant association between a specific SNP and a complex trait and how a significant SNP triggers an additive, dominant, or imprinting effect on
the trait. To test for the overall significant association of SNP genotype and trait phenotype, we generate the following hypotheses:
The log-likelihood ratio under the null and alternative hypotheses is calculated. Since the null hypothesis contains a nuisance parameter, allele frequency, this log-likelihood ratio test statistic
may have an unclear distribution. For this reason, the critical threshold for claiming the existence of a significant SNP is determined from permutation tests [37]. If our interest is in testing
whether there is an additive, dominant, or imprinting effect, the null hypothesis should be
The transgenerational changes of different genetic effects can also be tested. The null hypotheses used to test whether the additive, dominant, and imprinting effects display significant changes from
one generation to next are expressed as
Haplotyping Model
Recent molecular surveys suggest that the human genome contains many discrete haplotype blocks that are sites of closely located SNPs [38], [39], [40]. Each block may have a few common haplotypes
which account for a large proportion of chromosomal variation. Between adjacent blocks are there large regions, called hotspots, in which recombination events occur with high frequencies. Several
algorithms have been developed to identify a minimal subset of SNPs, i.e., tagging SNPs, that can characterize the most common haplotypes [41]. The number and type of tagging SNPs within each
haplotype block can be determined prior to association studies. In this section, we will derive a model for detecting the association between haplotypes constructed by alleles at a set of SNPs and
complex traits.
For the simplicity of our description, consider two SNPs A (with two alleles A and a) and B (with two alleles B and b). They form four haplotypes [42]. Risk and non-risk haplotypes from the maternal
and paternal parents generate four composite diplotypes, [43] and Wang et al. [44] proposed a two- and three-SNP model for estimating and testing genetic imprinting effects in a natural population,
respectively. Wu et al.'s procedure [45] allows the choice of an optimal number and combination of risk haplotypes within a multiallelic model framework. Here, we adopted Cheng et al.'s two-SNP model
to estimate haplotype imprinting genetic effects and their transgenerational change.
In this example, four haplotypes AABB (coded as 1), AABb (coded as 2), …, aabb (coded as 9), which are actually observed. Each subject must bear one of these genotypes, and the parents in each family
will be one of 9 Tables S2 and S3 show the structure and frequencies of mother by father genotype combinations under random mating and their offspring genotype frequencies in the second and third
generation, respectively. For a double heterozygote AaBb, its observed genotype may be derived from two possible diplotypes,
A similar likelihood (5) cane be formulated for haplotype models. A complicated EM algorithm is derived to estimate haplotype frequencies using the parental information. Let
In the M step, estimate the haplotype frequencies and recombination fraction by
In the M step, the equations for estimating additive, dominant, imprinting effects expressed in paternal and offspring generations are also derived. The E and M steps are iterated until the estimates
converge to a stable value. These stable values are the maximum likelihood estimates (MLEs) of parameters. The estimated haplotype frequencies and recombination fraction are embedded into a mixture
model for estimating genotypic values and variances for different generations.
Supporting Information
Table S1
A three-generation family design used to study transgenerational inheritance.
Table S2
A three-generation family design showing how to produce the second generation by mating different genotypes of grandfathers and grandmothers sampled from a natural population.
Table S3
A three-generation family design showing how to produce the second generation by mating different genotypes of grandfathers and grandmothers sampled from a natural population.
Competing Interests: The authors have declared that no competing interests exist.
Funding: This work is supported by the Changjiang Scholars Award, “One-thousand Person Plan” Award, National Natural Science Foundation of China (Grant No.30900854) and Beijing Forestry University
Young Scientist Fund (Grant No. Blx2w8003). The funders had no role in study design, data collection and analysis, decision to publish, or preparation of the manuscript.
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This discussion is precisely why I am alarmed about the new Core. At the high school level I really don't see much in the way of what I am supposed to teach. I get that students need to be problem
solvers, persistent, etc. and am on board with that, but unless we know more about the skills needed to solve these problems, students will be continually penalized by the system.
Here is an example on the Algebra 2/Trig exam that actually involved a skill I KNEW needed to be taught. Students were required to solve an absolute value inequality and graph the solution set on a
number line drawn in the booklet. Many of my students did exactly that and did it correctly. Yet they lost one or two points because the rubric REQUIRED that the answer also be written as an
inequality with the word OR, in addition to graphing it on the number line. NOwhere did it state that it had to be done algebraically. It could have been done graphically.
The new Core is too vague and will lead to way more unpleasant surprises. And if the test questions are not released, we will never know what we failed to cover.
I would like to see something similar to the Acorn book for Advanced Placement Calculus issued by the College Board. Without mandating when or how, it conveys to a teacher exactly what needs to be
Eleanor Pupko
From: bobbi@alumni.nd.edu
Subject: Re: Geometry #21
Date: Thu, 18 Aug 2011 13:06:46 -0400
To: nyshsmath@mathforum.org
Yes, it would because in that case a trap could be a parallelogram.
I remember the back-and-forth of a few years ago that you mention because in my first years of teaching (1962+) I had a Regents geom class and an honors geom. The textbook for the Regents geom had
the usual def of a trapezoid while the book for the honors class had the other def. that you mentioned.
Bobbi E
I believe isosceles trapezoid is correct. Although I do recall an earlier conversation on this list from several years ago where a few people defined a trapezoid as a quadrilateral with AT LEAST
one pair of parallel sides where most of us used EXACTLY one pair of parallel sides. This would make a difference, I think.
Liz Waite
-----Original Message-----
From: Jonathan <
To: nyshsmath <
Sent: Thu, Aug 18, 2011 10:24 am
Subject: Geometry #21
The diagonals of a quadrilateral are congruent but do not bisect eac other. The quadrilateral is:
Iso trapezoid
But none of these are necessarily correct (kite, anyone?)
Sent from my iPhone
Hi All-
Sorry this is a bit late, but I have been away for the last week. I hope you are all enjoying some time off over the summer.
I know most of you have seen the news around the release of the PARCC Model Content Frameworks for public comment. For those of you who have not, read on.
· The Model Content Frameworks in Mathematics and English language arts/literacy were released for public review on August 3^rd, after several rounds of feedback from the PARCC states. This
public review period is an opportunity for an even wider group of interested parties to provide feedback on all parts of the frameworks, including the introductions and the grade level
analyses, which contain suggested areas of emphasis and priority.
By following the this link,
, you will be able to review the draft Model Content Frameworks and provide your feedback through an on-line survey.
All feedback is due to PARCC by Wednesday, August 17^th.
During this public review period, PARCC hopes that teachers in particular will provide their feedback on the draft Model Content Frameworks. While teachers have helped to develop the
frameworks to this point, the feedback of a broader group of educators is critical. The Model Content Frameworks are being shared directly with NCTM, NCSM, NCTE, AFT, and NEA, as well as
others, so these organizations can share them with their members, as well.
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The Dold-Thom theorem for infinity categories?
up vote 17 down vote favorite
Let $\mathcal{M}$ denote the category of finite sets and monomorphisms, and let $\mathcal T$ denote the category of based spaces. For a based space $X \in \mathcal T$, one has a canonical funtor $S_X
: \mathcal M \rightarrow \mathcal T$ defined by $\{n\} \mapsto X^n$. The definition on morphisms is to insert basepoints on the factors which are not in the image of a given monomorphism.
As is well know, the homotopy groups of $\mathrm{colim} S_X = SP^\infty X$ give the homology of $X$ (this is the Dold-Thom theorem), and the homotopy groups of $\mathrm{hocolim} S_X = SP^\infty_h X$
given the stable homotopy of $X$.
Is there a model for $SP^\infty X$, the ordinary infinite symmetric product, as a homotopy colimit as opposed to a categorical colimit?
The motivation for this question comes from thinking about $\infty$-categories. In an $\infty$-category, one does not really have a good notion (at least not one that I am aware of) of strict
categorical colimits. So I'm wondering if there is, nonetheless, some easily defined functor on the $\infty$-category of spaces which will let us calculate ordinary homology. In short, is there any $
\infty$-categorical analog of the Dold-Thom theorem?
Update: Following up on André's remark it seems using the orbit category is heading in the right direction, at least for the $n$-th approximations. I'll just quickly sketch what I have so far:
Let $\mathcal O(\Sigma_n)$ denote the orbit category. The objects are the homogeneous (discrete) spaces $\Sigma_n/H$ (with left actions) as $H$ runs over all the subgroups of $\Sigma_n$, and the
morphisms are the $\Sigma_n$-equivariant maps. There is a canonical functor $$\Sigma_n \rightarrow \mathcal O(\Sigma_n)^{op}$$ where we regard $\Sigma_n$ as a category with one object as usual.
Given a $\Sigma_n$ space $X$, right Kan extension along this inclusion produces a $\mathcal O(\Sigma_n)^{op}$ diagram $\tilde X$ defined by $$\tilde X(\Sigma_n/H) = X^H$$ It turns out that the above
inclusion is final so that it induces an isomorphism of colimits. Hence $\mathrm{colim}_{\mathcal O(\Sigma_n)} \tilde X \cong X_{\Sigma_n}$, i.e., the coinvariants. It's also not hard to see that the
undercategories are copies of $B\Sigma_n$, hence not contractible, so we don't expect an equivalence of homotopy colimits, which is good.
On the other hand, I can now show that when $X$ is discrete, the canonical map $$\mathrm{hocolim} \tilde X \rightarrow \mathrm{colim} \tilde X$$ is an equivalence. My methods here do not generalize
to all spaces, so if someone has a reference for why this is true in general, that would be much appreciated. (I think something like this must appear in May's book on equivariant homotopy theory if
it's true, but I did not have it available this weekend.)
The remaining part would be to let $n \rightarrow \infty$, but somehow this seems like it should not be too bad. (Something like: make a functor $\mathcal M \rightarrow \mathcal Cat$ by $n \mapsto \
mathcal O(\Sigma_n)$. Take the Grothendieck construction. Some natural diagram on this category might give the right answer.)
homotopy-theory infinity-categories at.algebraic-topology
1 I only know how to go the other direction. There's a model category with a monoidal Quillen equivalence to the category of spaces so that the infinite symmetric product gives the stable homotopy
groups rather than the homology. (This is forthcoming work of Sagave and Schlichtkrull.) I'm not sure what really makes symmetric products of ordinary spaces behave as nicely as they do. – Tyler
Lawson Sep 3 '10 at 20:01
2 This is an interesting question, and I'd also like to know an answer! – Charles Rezk Sep 3 '10 at 23:24
Let's look at the n-th finite approximation of the space you care about, i.e., the n-th symmetric power of X. That's the strict coinvariants of S_n acting on X^n. If you want a homotopy invariant
1 description of those coinvariants, you should start with a model of X^n that lives in the oo-category of S_n-spaces. By Elmendorff's thm, that category is equivalent to the oo-category of O_{S_n}
- spaces (diagrams indexed y the orbit category of S_n). I forget what the homotopy invariant desrption of strict coinvariants is in terms of O_{S_n} - spaces... – André Henriques Sep 4 '10 at
@André: After mulling it over a bit last night, I came to a similar conclusion about somehow getting orbit categories into the picture, so this seems like a promising thing to look at. I'll report
back if I find anything. – Eric Finster Sep 4 '10 at 9:02
@Eric: "As is well know...the homotopy groups of hocolim $SX = SP^\inft_h X$ give the stable homotopy of X." Isn't this a rather recent theorem of Christian Schlichtkrull? (Algebr. Geom. Topol. 7
(2007), 1963--1977) – Dan Ramras Sep 5 '10 at 17:16
show 1 more comment
1 Answer
active oldest votes
It so happens that Emmanuel Dror Farjoun is visiting the EPFL this week. I figured I'd ask him about this problem at lunch today. What a coincidence! He proved exactly this statement
using the exact same techniques. In fact, the construction of $SP^n$ as a homotopy colimit is the subject of Chapter 4 in "Cellular Spaces, Null Spaces, and Homotopy Localization,"
Lecture Notes in Mathematics, 1622.
It turns out, the idea works more generally so that we can always replace strict colimits with homotopy colimits: define an orbit on a category $\mathcal C$ to be a functor $O : \
up vote 5 mathcal C \rightarrow \mathcal Set$ such that $\mathrm{colim}_{\mathcal C} O \cong *$. There is a category of such functors which we call the orbit category of $\mathcal C$, denoted $\
down vote mathcal O(\mathcal C)$. The Yoneda embedding factors through $\mathcal O(\mathcal C)$, and the right Kan extension along this inclusion always results in a "free" diagram.
I still want to play around with the construction a bit to see if there are any wrinkles with $n \rightarrow \infty$, and if I can use this to give easy calculations of homology in the
$\infty$-category $\mathcal S$, but I think it's safe to say at this point that answer to my question is yes.
add comment
Not the answer you're looking for? Browse other questions tagged homotopy-theory infinity-categories at.algebraic-topology or ask your own question.
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Thirty first British Mathematical Colloquium
This was held at University College, London: 4 - 6 April 1979
The enrolment was 413. The chairman was C A Rogers and the secretary was C B Thomas.
Minutes of meetings, etc. are available by clicking on a link below
General Meeting Minutes for 1979
Committee Meeting Minutes for 1979
The plenary speakers were:
Bauer, H Korovkin approximation and convexity
Browder, W Topologies of varieties
Roth, K F Irregularities of distribution
The morning speakers were:
Adams, J F Finite H-spaces and algebras over the Steenrod algebra
Blyth, T S Semigroups and lattices
Brown, A L Chebyshev sets and approximation theory
Cassels, J W S Embedding fields in p-adic fields
Chatters, A W Non-commutative rings with chain conditions
Chillingworth, D R J Bifurcation and catastrophes
Craw, I G Commutative Banach algebras and the topology of the maximal ideal space
Crossley, J N Logic and the natural numbers
Davies, E B Asymptotic analysis for one-parameter semigroups
Heath-Brown, R Differences between successive primes
Johnson, D L Finite groups of deficiency zero
Lloyd, K Graphs motivated by chemical mechanisms
Marstrand, J M Packing curves and surfaces in Euclidean space
Mislin, G The Euler class and group representations
Shiel-Small, T Linear operators in the theory of analytic functions
Walters, P Ergodic theory
Willmore, T J Mean value theorems in Riemannian manifolds
Wilson, R M The theory of combinatorial designs
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differentiating cosh
April 10th 2008, 10:42 AM #1
Mar 2008
differentiating cosh
I have to differentiate the function above. I think you should use the chain rule but it doesnt seem to be working for me. The answer is 7tanh(7x+9). But i dont know where the the tanh comes
from. Can anybody help?
$f(x) = \ln( \cosh(7x+9))$
$f'(x) = \frac{7 \sinh(7x+9)}{\cosh(7x+9)}$
did you manage to get this far ?
if so all you need to know is that $\tanh x = \frac{\sinh x}{\cosh x}$
April 10th 2008, 10:49 AM #2
Super Member
Oct 2007
London / Cambridge
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Math Success for Everyone
As teachers we sometimes miss interesting, candid comments students make. Paying more attention to what students say adds life to our classrooms and highlights interesting ideas. Sometimes the most
important thing a teacher can do is listen carefully and reflect back what students are saying. We forget how important this is, but it is helpful to remember.
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Dissipative Effect and Tunneling Time
Advances in Mathematical Physics
Volume 2011 (2011), Article ID 138358, 13 pages
Research Article
Dissipative Effect and Tunneling Time
Physics and Applied Mathematics Unit, Indian Statistical Institute, 203 Barrackpore Trunk Road, Kolkata 700 108, India
Received 15 March 2011; Revised 7 June 2011; Accepted 10 June 2011
Academic Editor: Yao-Zhong Zhang
Copyright © 2011 Samyadeb Bhattacharya and Sisir Roy. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
The quantum Langevin equation has been studied for dissipative system using the approach of Ford et al. Here, we have considered the inverted harmonic oscillator potential and calculated the effect
of dissipation on tunneling time, group delay, and the self-interference term. A critical value of the friction coefficient has been determined for which the self-interference term vanishes. This
approach sheds new light on understanding the ion transport at nanoscale.
1. Introduction
Caldeira and Laggett [1] started a systematic investigation of the quantum dissipative system and quantum tunneling in dissipative media. After that influential work, many authors have discussed the
dissipative tunneling in numerous papers but not with profound illustration in the aspect of tunneling time. In fact, the proper definition of tunneling time has been debated for decades and it is
yet to have definite answer [2]. Hauge and Støveng [3] mentioned seven different definitions of tunneling time of which the dwell time and the phase time or group delay are well accepted by the
community. Winful [4] studied a general relation between the group delay and the dwell time. In case of quantum dissipative system Caldeira and Laggett used the path integral technique to study the
dissipative quantum tunneling. Brouard et al. [5] made an important clarification of the existence of many tunneling times and the relations among them in a comprehensive framework. Ford et al. [6]
investigated the dissipative quantum tunneling using quantum Langevin equation. The quantum Langevin equation is nothing but the Heisenberg equation of motion for the coordinate operator of a
particle with certain mass, under a particular potential. This is the macroscopic description of a quantum particle interacting with a bath. The interaction with the bath corresponds to energy loss;
in other words, it is the signature of dissipation. The memory function present in the equation describes the interaction with the bath. The nature of the dissipation is contained in the memory
function. In a recent work [7] one of the present authors (S. Roy) investigated the transport of ions in biological system and constructed a nonlinear Schrödinger equation where the transport of ion
occurs at nanoscale. Here, the authors proposed a particular type of memory kernel associated to the non-Markovian behaviour of ions so as to understand the observational findings. The mechanism of
ion transport at nanoscale is becoming an important area of research. Considering the memory kernel for ion transport at nanoscale, we will discuss the tunneling phenomena within quantum Langevin
framework. Here we consider a parabolic potential barrier of the form . The transmittance is calculated using the method devised by Ford et al. [6]. The misty aspect of tunneling time and various
approaches towards it are critically analyzed. We will also try to figure out a process to calculate the tunneling time for a dissipative medium. In Section 2 we will briefly review quantum Langevin
equation for convenience. Then, we will discuss various concepts related to tunneling time, that is, phase time, dwell time, and so forth, and the effect of dissipation in Section 3. The
self-interference effect is discussed within this framework for dissipative systems. This method has been applied to understand the transport of ion in the biological domain which is much relevant at
the nanoscale in Section 4. Possible implications are indicated in Section 5.
2. Quantum Langevin Equation
We begin with the discussion on the tunneling of the ions through the dissipative potential barrier. The theory of dissipative quantum tunneling is pioneered by Caldeira and Laggett [1], where they
treated the problem through the technique of path integral. But we will address how quantum Langevin equation can be used to discuss dissipative quantum tunneling using the approach developed by Ford
et al. [6, 8, 9]. Here it is easy to incorporate non-Markovian and strong coupling effects using suitable memory function. The memory function present in the quantum Langevin equation describes the
interaction with the bath. At first we will briefly discuss the general theory with a general memory function. Then, we will deal with a specific memory function which we have at our hands [7], in
our problem of potassium ion transfer through ion channels. We consider an inverted harmonic oscillator potential barrier and see how the transmission coefficient is modified for inclusion of the
memory function. The quantum Langevin equation has the form where the dot and the prime, respectively, describ the derivative with respect to and . This is nothing but the Heisenberg equation of
motion for the coordinate operator of a particle of mass , under a potential . Here the coupling with the bath is described by two terms, the random force with mean zero and a mean force
characterized by the memory function . The autocorrelation of is given by In this expression is nothing but the Fourier transformation of . The coupling to the bath is given by the function .
Now this function has three important mathematical properties corresponding to some very important physical principles [5, 7]. (1)The condition states that the function is analytic in the upper half
plane. This is a consequence of causality.(2)The second condition is the “positivity condition” stated as This is a consequence of the 2nd law of thermodynamics.(3)The third condition is the reality
condition stated as
This follows from the fact that is a Hermitian operator. These properties are very elaborately explained by Ford and his coauthors. Based on these three properties, the function can be specified to
belong in a restricted class of functions, having a general representation in the upper half plane where is a positive constant, which can be absorbed in the particle mass.
Ford and his collaborators have considered harmonic oscillator potential as a simple example. Under this potential, the quantum Langevin equation takes the form This equation can be solved by the
method of Fourier transformation. We get the Fourier transformation of the coordinate operator as where is called the susceptibility and expressed as We consider an inverted harmonic oscillator
potential of the form , where is the width of the barrier. In the absence of dissipation we have an exact expression for the transmittance by the WKB approximation method If dissipation is included,
the tunneling frequency will be changed. Then the expression will be modified by replacing the frequency in the nondissipative case by that in the dissipative case.
In case of the inverted harmonic oscillator potential the susceptibility takes the form The normal mode frequencies of this coupled system are the poles of the susceptibility. However, there is an
isolated imaginary normal mode frequency corresponding to a pole of the susceptibility, which is classically forbidden and can be interpreted as the tunneling frequency . The determining equation for
is Putting the expression of from (2.6), we get Since the left-hand side of (2.13) is a monotonically increasing function, the value of will always be less than .
Let us consider a simple frictional coefficient “”. Under which the frequency determining equation becomes Since must be real and positive, we take the positive solution of this quadratic equation
Considering , we get ).
Then by replacing by in the expression of transmittance, we get the transmittance for dissipative medium: where and are the transmittance with and without dissipation, respectively.
Now the presence of dissipation can be incorporated in the potential function. The potential barrier without dissipation is .
The potential barrier with dissipation can be expressed as . Relating these two, we get So we can say that dissipation reduces the potential function. The dissipative contribution is included in the
tunneling frequency . We will use this fact to calculate the tunneling time and incorporate effect of dissipation in it.
3. Tunneling Time
In case of quantum mechanical tunneling through a barrier, it is well known how to calculate the probability of tunneling, the escape rate, and the lifetime in initial well. But the question is if
there is a time analogous to classical time spent in the barrier region how long does it take a particle to tunnell through a barrier? The subject of this so-called tunneling time or traversal time
has been covered by many authors in numerous independent approaches [3, 10–15]. Brouard et al. [5] have discussed various aspects of tunneling time in a very systematic approach. The very elegant
review of Hauge and Støveng [3] lists at least seven different types of tunneling time of which the phase time (group delay) and dwell time are considered well established.
3.1. Relation between Phase Time and Dwell Time
The group delay or phase time measures the delay between appearance of a wave packet at the beginning and the end of the barrier. By the method of stationary phase, it is given by the energy
derivative of the transmission phase shift: Here , where is the length of the barrier. Similarly the group delay for reflection is given by where is the reflection phase shift. The total group delay
is defined as the total group delay: where and are the transmission and reflection coefficients, respectively. In case of symmetric barriers .
Regardless of transmission or reflection, the dwell time is a measure of the time spent by a particle in the barrier region . It is given by the expression where is the wave function corresponding to
energy and is the flux of the incident particles. This equation gives us the time that the incident flux has to be turned on, to provide the accumulated particle storage in the barrier. Winful [4]
has discussed that delay time and dwell time are related by a linear relation: where is called the self-interference term given by the expression The self-interference term comes from the overlap of
incident and reflected waves in front of the barrier. This term is of considerable importance at low energies, when the particle spends most of its time dwelling in front of the barrier, interfering
with itself. In the relation given by (3.5), the self-interference term is disentangled and given by a separate expression in (3.6).
3.2. Calculation of Dwell Time and Self-Interference Term in Dissipative Case
The delay time can be calculated considering the effect of dissipation. In order to do that, first we will calculate the dwell time for the parabolic barrier we have taken as a model potential.
Following Er-Juan and Qi-Qing [16] we begin with the parabolic potential barrier and then subdivide the potential into infinitesimal rectangular barrier elements and then summing up the individual
dwell times spent by the particles inside the barrier elements, and the dwell time of the parabolic barrier is calculated.
Let us take a rectangular potential The solutions for the three regions can be written as After some manipulations Now the transmission coefficient can be found as and the probability of transmission
is For thin barrier we have . Then the transmission coefficient and probability of transmission, respectively, become and . The dwell time is defined as . Substituting these values, we get
Considering such barrier as a succession of the adjacent thin rectangular barrier elements in width ,
The corresponding dwell time for the th barrier The dwell time for the entire barrier becomes Since the length of each barrier is very small, the summation can be replaced by integral: where , is a
kind of average height approximation for the potential barrier . Our model potential is the inverse harmonic oscillator potential of the form . Therefore , where .
Therefore, the integrant becomes
Now . Finally putting the value of , we get where .
The effect of dissipation is included in , where , is the coefficient of friction.
Now we have to calculate the self-interference term ). To calculate this term, we have to know the reflection coefficient for the concerned potential.
By the method of WKB approximation the reflection coefficient can be easily shown as Therefore, where For small , Therefore, the self-interference term is found to be So from (3.17) and (3.22), we
find the complete expression of the group delay for the case of a particle tunneling through a barrier of inverse harmonic oscillator potential:
3.3. Effect of Dissipation in Self-Interference Term
Now we will estimate the effect of dissipation in the self-interference term. The effect of dissipation is included in the frequency term . From (3.20) we can write Hence, we can write where and .
Therefore, and are both positive terms. Therefore, the denominator of the 2nd term on the right-hand side will reduce due to them and the 2nd term will increase. It will reduce the whole term. So the
presence of dissipative term reduces the self-interference effect.
Now if we take , then Therefore, it is evident that, for a critical value of dissipation coefficient given by (3.27), the self-interference term vanishes. Let us now apply the approach to a
biological phenomena, that is, transport of potassium ion through ion channels.
4. Potassium Ion Transfer through Ion Channel
Here we will consider a special case for potassium ion transport through ion channel. From the above discussion we can emphasize on the fact that the memory function is the all important function in
this theory. It signifies the nature of the dissipative medium. By choosing this memory function properly, we can determine the tunneling coefficients and tunneling times for various dissipative
media. Now at this very moment, our memory function representing the potassium ion transfer through ion channel comes into play. Ion channels are transmembrane protein structures that selectively
allow given ion species to travel across the cell membrane. Zhou et al. [17] demonstrate that the channel protein transiently stabilizes three states, two within the selectivity filter and one within
the water basket towards the intracellular side of the selectivity filter. Experimental evidence indicates that the selectivity filter is devoid of water molecules other than single water molecule
between K ions [18]. The memory kernel of our specific problem [7] can be written as
The oscillatory ionic dynamics in ion channels is proposed to occur at the limit of the weak non-Markovian approximation associated with a time reversible Markov process, at the selectivity filter.
This reversible stochastic process belongs to a different time scale to that governing diffusion across the rest of the channel, which is determined by the glue-like properties of water at the water
basket. The framework of stochastic mechanics provides a model for such dissipative force in terms of quantum theory. That channel ionic permeation can be associated with nonlinear Schrödinger
equation which addresses the issue of de-coherence and time scale considerations. Now the memory kernel contains both Markovian and non-Markovian contributions that allows a continuous change from
Markovian to non-Markovian dynamics and enables identification of both the terms. The non-Markovian process has two time scales and whose contributions are dominated by the parameters and . The first
term contains the Markovian contribution. It is also clear that is the weak non-Markovian limit.
Averaging over and taking the Fourier transformation over the memory kernel, we get The determining equation of the tunneling frequency is given by (2.13). Under the present circumstances, the
equation becomes Here we consider an approximation . That is, . That is, the time scales and are very small. Since is a finite positive quantity, it is very small compared to the inverse of and .
Taking up to the first order of the binomial terms, we get Neglecting the second-order terms of and , we get Let . So (4.5) may be written in the same form of (2.14), The tunneling coefficient is
found to be
In case of weak non-Markovian limit (), we neglect the part, and This is similar to (2.17).
For the strong non-Markovian case, we get If we put this in the tunneling time expression, we get the group delay for this specific case of potassium ion transfer too.
The expression of delay time is given by (3.23), where .
The effect of dissipation is included in . So we get
In this case of potassium ion channel with the memory kernel as given in (4.1), the expression of the delay time will be with
So the delay time or phase time will depend on the parameter values , , , , and .
5. Possible Implications
It is evident from the above analysis that the effect of dissipation on group delay can be estimated directly in terms of the frictional coefficient. It is also possible to express the
self-interference term in terms of the friction coefficient (), and we can estimate the critical value of for which the interference term vanishes. The chosen biological example indicates that the
present approach may play an important role in understanding the ion transport at nanoscale which will be considered in subsequent papers. We are also interested in the numerical estimation of
tunneling time and the effect of dissipation on it. For that purpose, currently we have the required data for electron tunneling through water. But when one considers electron tunneling through
water, the electron-phonon interaction must also play an important role. This interaction will contribute in the potential in a considerable manner. But at nanoscale water behaves more like frozen
ice [19]. In that frozen water configuration, the electron transfer through water is only weakly affected by electron-phonon interaction [20]. But currently we do not have sufficient data for that
numerical calculation. We hope to present this thorough numerical estimation of tunneling times in subsequent papers.
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congruence modulo m
January 22nd 2007, 04:36 PM #1
Mar 2006
congruence modulo m
Hello, can anyone help me with these two problems?? Thank you so much in advance.
1) Prove: If x ≡ y (mod m), then (x, m) = (y, m)
2) Show that if n > 4 is not prime, then (n-1)! ≡ 0 (mod n).
Let $x,y,m>0$.
I presume it means $\gcd (x,m)=\gcd (y,m)$.
We know that,
$m|(x-y)$ thus, $x-y=km$ for some $k$.
Let $\gcd(x,m)=d_1$.
Let $\gcd(y,m)=d_2$.
We will prove it by trichtonomy.
Assume $d_1>d_2$.
But that cannot be because,
And $d_1|x$ and $d_1|km$.
Thus, $d_1|(x-km)$ thus $d_1|y$. And we also know that $d_1|m$.
Thus, $\gcd(y,m)ot = d_2$ because $d_2<d_1$. And hence it is not "greatest".
By similar reasoning. We can show $d_1<d_2$ leads to contradiction.
Thus, by trichtonomy,
There are two possibilities.
1)n is not a square.
2)n is a square.
If n is not a square then it has a non-trivial proper factorization $n=ab$ where $2\leq a,b \leq n-1$.
Where $a,b$ are distinct.
Thus among the factors of $(n-1)!$:
We can find its factors $a,b$.
When $n$ is a square there are 2 possibilities.
1)n is not a square of a prime.
2)n is a square of a prime.
If n is not a square of a prime we know that $n=c^2=cc=cab$ where $2\leq a,b\leq c-1$ because it is not a prime. Thus, it has a factorization in the form $n=ab$ where $a,b$ are distinct and the
same argument applies.
If n is a square of a prime then we have a minor problem. For example if n=4=2^2 it does not work. Which is why the initial conditions of the problem says n>4. Thus we know that $n=p^2$ and in no
other way. We know that $1,2,...,p,...n-1$ contains one factor. But what about other another factor of $p$? It turns out that when $n>4$ the factor $2p$ appears among $1,2,...,n-1$. Thus, there
is another number that has a factor of $p$. The reason why is because $p=\sqrt{n}\leq (n-1)/2$ for $n>4$thus, $2p\leq n-1$ and is hence among one of the factors.
thanks theperfecthacker for the quick reply!
There are two possibilities.
1)n is not a square.
2)n is a square.
If n is not a square then it has a non-trivial proper factorization $n=ab$ where $2\leq a,b \leq n-1$.
Where $a,b$ are distinct.
Thus among the factors of $(n-1)!$:
We can find its factors $a,b$.
When $n$ is a square there are 2 possibilities.
1)n is not a square of a prime.
2)n is a square of a prime.
If n is not a square of a prime we know that $n=c^2=cc=cab$ where $2\leq a,b\leq c-1$ because it is not a prime. Thus, it has a factorization in the form $n=ab$ where $a,b$ are distinct and the
same argument applies.
If n is a square of a prime then we have a minor problem. For example if n=4=2^2 it does not work. Which is why the initial conditions of the problem says n>4. Thus we know that $n=p^2$ and in no
other way. We know that $1,2,...,p,...n-1$ contains one factor. But what about other another factor of $p$? It turns out that when $n>4$ the factor $2p$ appears among $1,2,...,n-1$. Thus, there
is another number that has a factor of $p$. The reason why is because $p=\sqrt{n}\leq (n-1)/2$ for $n>4$thus, $2p\leq n-1$ and is hence among one of the factors.
Hmm, you have:
2 <= a,b <= n - 1, shouldn't it be: 1 < a < b < n - 1? Maybe you got it mixed up for when it is a square. The proof I was given was:
Either n is a perfect square, n = a^2 in which case 2 < a < 2a <= n−1 and hence a and 2a are among the numbers {3,4, . . . ,n−1} or n is not a perfect square, but still composite, with n = ab, 1
< a < b < n−1.
Does this work? It contradicts a few of your results.
Hmm, you have:
2 <= a,b <= n - 1, shouldn't it be: 1 < a < b < n - 1? Maybe you got it mixed up for when it is a square. The proof I was given was:
Either n is a perfect square, n = a^2 in which case 2 < a < 2a <= n−1 and hence a and 2a are among the numbers {3,4, . . . ,n−1} or n is not a perfect square, but still composite, with n = ab, 1
< a < b < n−1.
Does this work? It contradicts a few of your results.
No. I said proper nontrivial factorization. Meaning the obvious factors, 1 and the number itself are excluded.
Sorry to be a pain, but I clearly understand your proof now TPF. What I don't understand is how the proof I was given (below) proves the claim; how does that tie it all together showing that n is
composite if n > 4 and that n will then divide (n - 1)!
Either n is a perfect square, n = a^2 in which case 2 < a < 2a <= n−1 and hence a and 2a are among the numbers {3,4, . . . ,n−1} or n is not a perfect square, but still composite, with n = ab, 1
< a < b < n−1.
Sorry to be a pain, but I clearly understand your proof now TPF. What I don't understand is how the proof I was given (below) proves the claim; how does that tie it all together showing that n is
composite if n > 4 and that n will then divide (n - 1)!
Either n is a perfect square, n = a^2 in which case 2 < a < 2a <= n−1 and hence a and 2a are among the numbers {3,4, . . . ,n−1} or n is not a perfect square, but still composite, with n = ab, 1
< a < b < n−1.
How do you know that 2a<=n-1?
And hence among the factors.
See I acutally used an inequality to justify that.
January 22nd 2007, 04:45 PM #2
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Category theory
Universal constructions
Limits and colimits
A sketch is a small category equipped with a subset of its limit cones and colimit cocones.
A limit-sketch is a sketch with just limits and no colimits specified.
A model of a sketch is a Set-valued functor preserving the specified limits and colimits.
A category is called sketchable if it is the category of models of a sketch.
Relation to accessible and locally representable categories
From the discussion there we have that
• an accessible category is equivalently:
□ a full subcategory of a presheaf category that’s closed under $\kappa$-filtered colimits for some $\kappa$
□ the category of models of a sketch
• a locally presentable category is equivalently:
□ a reflective full subcategory of a presheaf category that’s closed under $\kappa$-filtered colimits for some $\kappa$
□ the category of models of a limit sketch
□ an accessible category with all small limits
□ an accessible category with all small colimits
We can “break in half” the difference between the two and define
• a locally multipresentable category to be equivalently:
□ a multireflective full subcategory of a presheaf category that’s closed under $\kappa$-filtered colimits for some $\kappa$
□ the category of models of a limit and coproduct sketch
□ an accessible category with all small connected limits
□ an accessible category with all small multicolimits
• a weakly locally presentable category to be equivalently:
□ a weakly reflective full subcategory of a presheaf category that’s closed under $\kappa$-filtered colimits for some $\kappa$
□ the category of models of a limit and epi sketch
□ an accessible category with all small products
□ an accessible category with all small weak colimits
An overview of the theory is given in
An extensive treatment of the links between theories, sketches and models can be found in
• Michael Makkai, Robert Paré, Accessible categories: The foundations of categorical model theory Contemporary Mathematics 104. American Mathematical Society, Rhode Island, 1989.1989.
That not only every sketchable category is accessible but that conversely every accessible category is sketchable is due to
Revised on March 6, 2014 08:33:04 by
Tim Porter
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A-level Physics (Advancing Physics)/Electron Behaviour as a Quantum Phenomenon
From Wikibooks, open books for an open world
So far, we have considered how quantum physics applies to photons, the quanta of light. In reality, every other particle is also a quantum, but you only need to know about photons and electrons.
The image on the right shows what happens when you fire electrons through a pair of slits: it arrives in lumps, but you get fringes due to superposition as well. The electrons are behaving as both
waves and particles. Actually, they are behaving as quanta. The equations describing quantum behaviour in electrons are similar to those describing it in photons.
Frequency and Kinetic Energy[edit]
We know that, for photons:
$f = \frac{E}{h}$
In suggesting that electrons are quanta, we assume that they must have a frequency at which the phasors representing them rotate. We also know that h is a constant; it does not change. So, when
adapting the above equation to apply to electrons, all we need to adapt is E. In electrons, this energy is their kinetic energy. If the electron has some form of potential energy, this must first be
subtracted from the kinetic energy, as this portion of the energy does not affect frequency. So:
$f = \frac{E_{kinetic} - E_{potential}}{h}$
De Broglie Wavelength[edit]
If electrons exhibit some wavelike properties, they must also have a 'wavelength', known as the de Broglie wavelength, after its discoverer. This is necessary in order to work out a probability
distribution for the position of an electron, as this is the distance the electron travels for each phasor arrow rotation. The de Broglie wavelength λ is given by the equation:
$\lambda = \frac{h}{p} = \frac{h}{mv}$,
where h = Planck's constant, p = momentum, m = mass of electron = 9.1 x 10^-31kg and v = velocity of electron.
Potential Difference and Kinetic Energy[edit]
Potential difference is what causes electrons to move. You already know how power is related to charge, voltage and time:
$P = \frac{QV}{t}$
Since power is the rate at which work is done:
$W = QV\,$
We know that the charge on an electron equals -1.6 x 10^-19, and that work done is energy, so:
$E_{kinetic} = 1.6 \times 10^{-19} \times V$
Energy, in the SI system of units, is measured in Joules, but, sometimes it is measured in electronvolts, or eV. 1 eV is the kinetic energy of 1 electron accelerated by 1V of potential difference.
So, 1 eV = 1.6 x 10^-19 J.
1. An electron moves at 30,000 ms^-1. What is its de Broglie wavelength?
2. What is its frequency?
3. What is its kinetic energy, in eV?
4. Given that it is travelling out of an electron gun, what was the potential difference between the anode and the cathode?
5. An electron is accelerated by a potential difference of 150V. What is its frequency?
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How to measure volumes in a drum... [Archive] - Biodiesel & SVO Discussion Forums Archive - Live Forums at biodieseldiscussion.com
03-08-2006, 06:56 PM
I think I have this figured out for a drum... doing so in a carboy shouldn't be as hard.
I'm hoping someone can check my logic and math. I'm looking for an easy way to measure how much VO, etc I have in a 55 gallon drum.
A drum is 34" tall.
It can hold 55 gallons.
That works to 1.617647 gallons per vertical inch.
That equals 6.12346 Liters per inch.
That equals 2.4108 L / cm.
I'm not really asking you to look up the conversion, just to confirm that my
logic makes sense.
Therefore I could use a yardstick or some other method and figure out what volume I have.
Whaddya think?
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Next | Prev | Top | Index | JOS Index | JOS Pubs | JOS Home | Search
A matrix is defined as a rectangular array of numbers, e.g.,
which is a rows, and columns of the matrix. For example, the general
Either square brackets or large parentheses may be used to delimit the matrix. The ^H.1 of a matrix square.
The transpose of a real matrix
A complex matrix complex numbers. The transpose of a complex matrix is normally defined to include conjugation. The conjugating transpose operation is called the Hermitian transpose. To avoid
confusion, in this tutorial, without conjugation, while conjugating transposition will be denoted by
Subsections Next | Prev | Top | Index | JOS Index | JOS Pubs | JOS Home | Search [How to cite this work] [Order a printed hardcopy] [Comment on this page via email]
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FOM: Re: F.O.M./pure math; general intellectual interest
Harvey Friedman friedman at math.ohio-state.edu
Wed Dec 17 06:07:38 EST 1997
Lou writes:
>Okay, I had forgotten this old discussion where I mentioned number theory
>in connection with complexity (but even there it was in reaction to
>earlier claims by you concerning fom & complexity). Anyway, this is
>not worth our energies to spell out further, we do not have a real
>disagreement on this matter, I think.
>Otherwise I am not conceding anything. Sure, "suspect" is vague, so is
>"general intellectual interest", and perhaps even more, the significance
>of "general intellectual interest".
So you are refusing to explain "suspect?" I'll try again to get you to do
this. I claim that of these three, you are the only one on the fom list of
about 245 subscribers who would entertain the idea that they are of even
remotely comparable vagueness. If you don't explain what "suspect" is, I
might have to, and continue the brawl from there.
If you don't explain "suspect" you are just being silly and grossly unfair.
I have done a lot of preliminary explanation of "general intellectual
interest." You haven't even done any preliminary explaining of "suspect."
>I find P=NP a very interesting
>problem, and it sure has to do with our basic intuitions and experience
>with algorithms. As to how it would play for a fairly general audience
>compared to abc, I wouldn't be able to predict.
You are also the only one on the fom list of about 245 subscribers who
would assert (falsely) that "you wouldn't be able to predict." Let's have a
contest. We will design a meeting with selected participants and test the
matter. We could do this at Illinois or at OSU. Then we wouldn't have to
pay travel expenses for the judges.
>Both are the kind of
>mathematical questions that can be discussed in a Scientific American
>article (maybe this has already happened).
I subscribe to the Scientific American, and they have gotten away from
articles on things like abc. Want to test this? We can write to editors of
Scientific American testing out a possible submission for appropriateness
for the readership. Have you carefully listened to the video on Wiles? It
is extremely well done, and of great human interest. Do you think that a
single mathematical idea was communicated successfully compared to tapes on
other matters? It is clear that Scientific American published this article
because of the human interest involved, and the Mt. Everest aspect.
Historians - I seem to recall several articles touching on P=NP directly
and indirectly, but none on abc. Don't trust me on this. Is it mentioned in
the recent article on FLT?
>Anyway, if P=NP is false
>(as seems plausible), and even if it's true (with probably huge
>constants), and one of these is established, this would surely
>increase our theoretical understanding of algorithms in a big way,
>but there is no reason to think it would have direct practical
>(technological) consequences. In discussing P=NP for a general
>audience it seems to me this should be pointed out.
This is completely misleading. The actual situation is absolutely perfect
for a crucial mathematical problem. Almost nobody thinks that P=NP, and I
think that if P=NP, then it is very likely that we misunderstand algorithms
so much that the solution will have substantial practical consequences for
algorithm design. On the other hand, since notP=NP, there is the question
of the practical significance of that. Well, the right proof of that can be
expected to lead to rigorous proofs of the security of cryptosystems. This
is one of the motivations behind eliminating number theory in the design of
cryptosystems. Otherwise, one is faced with yet more work beyond a good
proof of notP=NP. The general point to be made to general audiences is
that, e.g., P=NP is intimately tied up with cryptosystems and everything
else, and that before good proofs of notP=NP are found, we simply cannot
begin to do what we need to do in order to know that cryptosystems are
secure. Beyond that, it is perhaps not known to be directly practical, but
is the ultimate in "theoretical practicalness," in the sense that it would
justify a lot of present ways of doing algorithms virtually everywhere as
best possible. This is always a great kind of "application." And the sense
of best possible will be more and more in tune with practical
considerations as the theory surrounding notP=NP is developed further and
further. What is so intriguing is that notP=NP is obviously an enormous
intellectual logjam. And everybody cares about this.
> As to abc, it seems to express a very basic and at the same time
>deep property of another kind of basic objects, namely natural
>numbers (their behavior under addition and prime factorization).
>(A special case says: there is a positive constant C such that if
>a+b=c, where a and b are positive integers with gcd=1, then
>c < C.P^2, where P is the product of the distinct prime factors of abc.
>Reference: S. Lang, Old and new conjectured diophantine inequalities,
>Bull. AMS 23, 37-74, 1990.)
This is a purely technical problem. P=NP is a deep conceptual problem as
well as a technical problem. Yours may even be a great technical problem.
If I recall it is related to FLT, and also involved in the kind of
elementary number theory that is relevant to current number theoretic
approaches to complexity(!). But the only way you can get anybody to care
about it outside math is to claim some relevance to something like P=NP or
cryptosystems - along with other math with relevance. Otherwise it has the
appearance of being an imbred occupation of a bunch of harmless genuises
who are best left alone to themselves. Its relevance to things like P=NP
and cryptosystems are the only way I see of getting any of the following to
scientific philosophy, statistics, finance, computer security, complex
systems, complexity theory, database theory, algorithm design,
cryptography, logic, artificial intelligence, robotics, networking,
hardware design, control theory, operations research, neural nets, quantum
computing, expert systems, code verification, proof checking, statistical
mechanics, linguistics, dynamical systems, signal processing, etcetera; and
many, many more.
And it is among many pieces of math that are so relevant. How are you going
to get such people to see any intrinsic interest in such a problem? They
have no problem seeing intrinsic interest to things like P=NP and
More information about the FOM mailing list
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limits pass papers
\[ \lim_{x \to +\4} = \frac{\sqrt[2]{x}}{x^2 - 5x + 4} .\]
So... $\ \lim_{x \to \4} = \frac{\sqrt[2]{x}}{x^2 - 5x + 4} .\]$ You forgot the TEX tags. The overall limit is indeterminate, although from the right side, denoted by $\lim_{x \to 4+}$, it should be
infinity. (Note the plus is on the other side of the four) The easiest way to do this is to graph it then approximate the value by plugging in an obscure number such as 4.0000000000000001 instead of
4, this should give you some idea.
Last edited by Xeritas; March 18th 2012 at 04:23 PM.
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How do you place students into calculus courses?
August 29, 2011 11:30 AM Subscribe
Attention math teachers: what processes does your college/university use for placement into Calculus courses?
I teach at a mid-sized open-admissions research university in the US; total student body is about 6000 (about 1100 first-year students).
Currently, we use Accuplacer to place students anywhere from developmental arithmetic into Calculus I, but we have no method to accurately place students into Calc I vs Calc II vs Calc III (should we
be so lucky) except if they've gotten adequate scores on an AP test. Many of our students who start in Calc I have already taken Calculus in high school; some of them probably should re-take it at
the college level, but probably some shouldn't. (And I'm not sure how good a job Accuplacer does at placing students, either. I don't think we have the data.)
Placing students into Calculus should be a reasonably solved problem. What do folks use? I've heard of the MAA-sponsered one that uses MapleTA; is it any good? (We're not a Maple campus,
particularly.) An out-of-the-box solution would be easier than having to roll my own. Computer-administered/graded placement tests would be ideal, but ideally not multiple-choice. (although that may
be the option.)
posted by leahwrenn to Education (6 answers total) 2 users marked this as a favorite
It seems like the AP test should be a pretty good indicator.
At my school, a 5 on the AB/BC test got you out of a quarter of math, a 3 or a 4 put you in the normal classes. Anything less (or no score) and you had to take college calculus over the summer to be
admitted in the fall.
This seems like the exact kind of test you would want to check in on them. The challenge is mapping scores to your classes (which is easier than making a test from scratch I would think).
If students have not taken the test but wish to place, give them a retired test, there are plenty to be had out there (my AP teacher in high school had access to them for sure). Part of it is
multiple choice and computer gradable and part of it is not...
posted by milqman at 12:46 PM on August 29, 2011
As far as I know, ours were written in-house at some point (and the computerised). There's a 'college algebra' one and a 'calculus' one. (I've never seen them. I gathered this from google and finding
the website where undergrads are supposed to log in to take the placement tests.) They don't use the placement test for students who have taken an AP calculus exam. I think most of the students
who've taken calculus before are being placed by the AP exam. (I think using AP scores for placement is fairly successful. I want to say the MAA tracked this at some point, but I can't find the
articles. I'm not sure how successful the actual placement tests are.)
Berkeley has an
online placement test
if you wish to play with it and
information for choosing a first course
. It doesn't place people into anything below pre-calculus, though. However, I don't know that anyone actually uses this placement test. (I don't think anyone I knew did. It didn't apply in my
situation ) It's certainly not required.
I wonder if the easiest and perhaps best thing to do would be to rig up a placement test using old final exams from the various courses, choosing, I don't know, five of the harder questions from each
course's exam (but excluding the questions that turned out to be disasters, if you have that information). That might get you something that works for everything up to calculus and then you can
either do separate calculus placement exams or rely on AP scores. (Are you likely to have many students who didn't take AP Calculus, but should skip a semester or more of calculus?)
posted by hoyland at 1:09 PM on August 29, 2011 [3 favorites]
Sorry, I wasn't clear. I am not worried about placing students who actually took an AP test. I am concerned about placing students who have had some calculus in high school but did not take an AP
In particular, I am especially interested in determining which students who claim to have passed a year of high school calculus but who did not take an AP test should be placed into Calculus II, and
which should retake Calculus I (and, conceivably, who really need to retake precalculus---it happens).
posted by leahwrenn at 1:43 PM on August 29, 2011
UVM requires students to take a placement test that, I believe, was written inhouse and computerized with something called WebAssign. I know nothing about it, but based on my memory of it from a few
years ago, it used steadily more difficult questions, and went beyond the trig stuff in that Berkeley link. It included integrals and stuff that (after taking Calc 2) I still don't recognize. The
results are some sort of number. They actually don't let you register for a class below the number you test into, but they do allow you to register up if you feel you can handle it or whatever. The
link (though you can't take the test without being a student, I think) is http://www.uvm.edu/~cems/mathstat/?Page=MRT/default.php
Perhaps email one of the professors in the dept?
posted by papayaninja at 2:46 PM on August 29, 2011
My very large university used to conduct their own placement test through the Math department but now uses a test through
. I'm not sure if it's customized or right out of the box.
posted by Squeak Attack at 3:10 PM on August 29, 2011
My school had three Calculus courses (at least, three that all the science majors had to take.) If you got a 4 or 5 on the B/C test you could skip the first and second classes. If you scored at the
highest level of the placement test, you were put into the first one. There were something like 16 levels you could test into, ranging from seriously developmental (they reviewed single-digit
arithmetic) to college algebra & trigonometry.
Then, if the student wanted to move up beyond that placement they just gave them a modified version of the comprehensive
final exam
for each course they wanted to skip. You could get out of all three main Calc classes plus the fourth class (which almost no one needed) with this kind of testing. I knew a few kids who just got out
of the three main classes that way.
The website that explains all of this is
. I personally took precalc in high school, got a 610 Math on the SAT, but decided to take the accelerated algebra/trig courses rather than risk bombing out in the first Calculus class. This was an
experience shared by several hundred other freshmen that quarter.
posted by SMPA at 4:47 PM on August 29, 2011
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OK Corral: Local versus non-local QM
1. I'd like to point out that I began with the exact equation that Bell used [1964; equation (3)]. I get the identical result also: - a.b'.
In physics it is important to understand what physical quantity the terms in an equation stand for. In Bell's paper
represent possible angles of the stern-gerlach device used to measure the spin of the two particles, and these measurements will always yield one of two results, which in the "Formulation" paragraph
on p. 1 of
the paper you refer to
he labels as +1 and -1. The "expectation value" refers to the average expected value of the product of the two measurements, which would be:
yields +1, angle
yields +1) +
yields -1, angle
yields -1) +
yields +1, angle
yields -1) +
yields -1, angle
yields +1)
This experiment is
one where the result of each measurement is an arbitrary real number between -1 and +1, and where the expectation value is the average value of the product of these two real numbers, as you seem to
assume in your example. Again, Bell is assuming that each measurement always yields one of two results which are assigned values +1 and -1, so when you multiply the two values you always get the
result -1 or +1 on any given trial; the expectation value refers to the average this product over many trials.
As I pointed out in a previous post, if you assume that each experimenter has a device which projects the vector
onto their own angle (either
), like
, and then this continuous value is used to determine the
) + 1/2 that the experimenter will get a +1 result on that trial or a -1 result, then it actually does work out that the expectation value for the product of their results will end up being
as you had in your attempted proof. But again, in this case you don't have a guarantee that when they pick the same angle they always get opposite results on a given trial, so Bell's theorem would
only rule out inequalities which don't include this assumption, like the CHSH inequality.
2. NB: At the moment I have limited my derivation to that which I offered: A wholly LOCAL and CLASSICAL derivation of the EPRB correlation. That is, I have derived the limit to which your derivation
must tend in accord with Bohr's Correspondence Principle.
I don't see how the correspondence principle would imply that the expectation value for an experiment in which each experimenter can get any result between +1 and -1 on a given trial would be
identical to the expectation value for an experiment in which each experimenter can only get one of two results, either +1 or -1. Is this what you're claiming here?
3. I hope we might agree on the following important point: Since the space-like experimental results were derived by me in terms of high-school maths, AND without any reference whatsoever to
non-locality, there must be an equivalent QM derivation equally devoid of non-locality.
If you were indeed able to reproduce the result that the expectation value is -cos(
) in a purely classical experiment, where on each trial each experimenter gets either +1 or -1 and the expectation value is for the average of the products of their two answers, and your classical
experiment obeyed the conditions of Bell's theorem like the source not having foreknowledge of the detector settings, then yes, this would show that QM was compatible with local hidden variables. The
problem is you didn't do this--you seem to assume that each experiment can yield a continuous spectrum of values rather than just +1 or -1, and even if you make the assumption I mentioned above where
the probability of getting +1 is (1/2)*(
) + 1/2, so that the expectation value is indeed just
, there seems to be an error in your "high school math", since this is not equal to -cos(
4. So: May I ask you again to provide your fundamental derivation of the EPRB correlation (ie, from first principles; and preferably in the terms of the OP), beginning with Bell's equation (just as I
You're looking for a derivation of why
quantum mechanics
predicts that the expectation value is -
? Why would this be useful, since here we are just trying to figure out whether this expectation value can be reproduced in a classical experiment?
5. I request this of you because you are a PF MENTOR and because SCIENCE ADVISER DrC has not been able to derive it and SCIENCE ADVISER JesseM is a bit confused on my mathematics (but I will sort
that out soon: noting for now that there are no errors in my maths, so far as I can see from my high-school text on vector-analysis).
Yes, please state whatever theorems from your vector textbook you are making use of in your proof. But in the meantime, could you please check the math on my example of
= 0 degrees,
= 60 degrees, and
= 90 degrees? Do you disagree that in this case, -
= - cos(90)*cos(30) = - (0)*(0.866) = 0, while - cos(
) = - cos(60) = -0.5? If you agree with my math on this example, then it seems clear there must be an error in your proof somewhere, unless I misunderstood what you claimed to have proved.
6. Your derivation will not be wasted as I am keen to learn. HOWEVER: If you will not be providing this important derivation; could you please point to where I might find a detailed version;
preferably one that complies with your own local interpretation of QM?
Have you ever studied the basics of QM? Derivations of probabilities and expectation values have nothing to do with one's interpretation, they basically just involve finding state vector representing
the quantum state of the system, expanding it into a weighted sum of eigenvectors of the operator representing the variable you want to measure (energy, for example), and then the square of the
complex amplitude for a given eigenvector represents the probability that you'll get a given value when you measure that variable (the value corresponding to a particular eigenvector is just the
eigenvalue of that vector). And of course, once you know the probability for each possible value, the expectation value is just the sum of each value weighted by its probability. If you're not
familiar with the general way probabilities and expectation values are derived in QM, then a specific derivation of the expectation value for the spins of two entangled electrons won't make much
sense to you. And like I said, the derivation itself would have nothing to say about locality or nonlocality, it's just when you apply Bell's theorem to the predictions of QM that you see they are
not compatible with local hidden variables.
by the way, if you are familiar with calculations in QM, you can look at
this page
for a nearly complete derivation. What they derive there is that if q represents the angle between the two detectors, then the probability that the two detectors get the same result (both spin-up or
both spin-down) is [tex]sin^2 (q/2)[/tex], and the probability they get opposite results (one spin-up and one spin-down) is [tex]cos^2 (q/2)[/tex]. If we represent spin-up with the value +1 and
spin-down with the value -1, then the product of their two results when they both got the same result is going to be +1, and the product of their results when they got different results is going to
be -1. So, the expectation value for the product of their results is:
[tex](+1)*sin^2 (q/2) + (-1)*cos^2 (q/2) = sin^2 (q/2) - cos^2 (q/2)[/tex]
Now, if you look at the page on trigonometric identities
, you find the following identity:
[tex]cos(2x) = cos^2 (x) - sin^2 (x)[/tex]
So, setting 2x = q, this becomes:
[tex]cos(q) = cos^2 (q/2) - sin^2 (q/2)[/tex]
Multiply both sides by -1 and you get:
[tex]sin^2 (q/2) - cos^2 (q/2) = - cos (q)[/tex]
This fills in the final steps to show that the expectation value for the product of their results will be the negative cosine of the angle between their detectors.
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Posts about CS on The Lumber Room
Posts Tagged ‘CS’
This seems counter-intuitive, at least to me, but if you can find an intuitive explanation for it, please comment!
Fact: Let G be a cycle (connected graph where every vertex has degree 2: you know what I mean) of length n (i.e., n vertices and n edges). Start a random walk on G from any node u. Then the
probability that v is the last node visited (i.e., the random walk hits all other nodes before hitting v) is the same for every v other than u!
[Such a property is obviously true for the complete graph, and apparently it's true only for cycles and complete graphs.]
I do not know a probabilist’s proof; what I could think of is a distinctly ComputerSciencey (Dynamic Programming / recurrence relation) proof.
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Year 10 GCSE Paper question varies inversely as the square
July 10th 2008, 03:50 AM #1
Jul 2008
Year 10 GCSE Paper question varies inversely as the square
The shutter speed, S, of a camera varies inversely as the square of the aperture setting, f.
When f= 8, S= 125
a) find a formula for S in terms of f.
b) Hence, or otherwise calculate the value of S when f= 4
thanks for any time anybody spends on this question
If x varies inversely to y, then:
$x = \frac{k}{y}$
Where k is the constant value for any values of (x,y).
If x varies inversely to the square of y, then:
$x = \frac{k}{y^2}$
Find the constant, and then replace in the equation:
$S = \frac{k}{f^2}$
Use this equation to find (b).
July 10th 2008, 03:57 AM #2
Super Member
Jun 2008
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st: RE: Stata 12 Announcement
Notice: On March 31, it was announced that Statalist is moving from an email list to a forum. The old list will shut down at the end of May, and its replacement, statalist.org is already up and
[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
st: RE: Stata 12 Announcement
From "Dwamena, Ben" <bdwamena@med.umich.edu>
To "statalist@hsphsun2.harvard.edu" <statalist@hsphsun2.harvard.edu>
Subject st: RE: Stata 12 Announcement
Date Mon, 27 Jun 2011 00:57:40 +0000
Hurray to Stata 12 especially regarding contour plots and rocreg, automatic memory management, Constrasts and pairwise comparisons
From: owner-statalist@hsphsun2.harvard.edu [owner-statalist@hsphsun2.harvard.edu] on behalf of William Gould, StataCorp LP [wgould@stata.com]
Sent: Sunday, June 26, 2011 7:07 PM
To: statalist@hsphsun2.harvard.edu
Subject: st: Stata 12 Announcement
Following long tradition, we are informing Statalist first:
Stata 12 begins shipping Monday, July 25.
Orders are now being accepted at http://www.stata.com.
Below are some highlights.
Automatic memory management
Automatic memory management means that you no longer have to
-set memory- and never again will you be told that there is no
room because you set too little! Stata automatically adjusts its
memory usage up or down according to current requirements.
The memory manager is tunable. You can set a maximum if you wish.
Old do-files can still -set memory-. Stata merely responds, "-set
memory- ignored".
We have tested the memory manager on systems with 1 TB (the largest
currently available), and it is designed to scale to even more
Import Excel files, export PDFs, and new interface features
Importing Excel files is easy. And the new Import Preview Tool
lets you see the file's contents and adjust import settings before
you import it.
You can now directly export PDFs of graphs and logs.
Stata's windows are now laid out to fit wider screens better. You
can still get back the old layout from Edit -> Preferences.
A new Properties window -- always available -- lets you manage
your variables, including their names, labels, value labels,
notes, formats, and storage types.
The Viewer is now tabbed, and it has buttons at the top to access
dialogs, to jump within the document, and to jump to Also See
The Data Editor also has a new Properties window; has another tool
that lets you Hide, Show, Filter, and Reorder the variables; and
has the new Clipboard Preview tool, which lets you see and prepare
your raw data before pasting.
Structural equation modeling (SEM)
-sem- is a new estimation command, itself the subject of
an entire manual.
If you are new to SEM, you should be interested if you fit linear
regressions, multivariate regressions, seemingly unrelated
regressions, or simultaneous systems, or if you're interested in
generalized method of moments (GMM). And if you think you are
still not interested, take a look anyway. SEM is a remarkably
flexible framework.
If you know about SEM, you will be more interested in path
analysis models, single- and multiple-factor measurement models,
MIMIC models, latent growth models, correlated uniqueness models,
and more, all of which can be fit by -sem-. You will also be
interested in -sem-'s standardized and unstandardized coefficients,
direct and indirect effects, goodness-of-fit statistics,
modification indices, predicted values and factor scores, and
groupwise analysis with tests of invariance.
You can use the GUI or command language to specify your model.
The command language is a variation on standard path notation.
You can type
. sem (L1 -> m1 m2 m3)
(L2 -> m4 m5)
(L1 -> L2)
In -sem-, lowercase names refer to variables in the data and
uppercase names are latent variables. The above corresponds to
m1 = a1 + b1*L1 + e1
m2 = a2 + b2*L1 + e2
m3 = a3 + b3*L1 + e3
m4 = a4 + b4*L2 + e4
m5 = a5 + b5*L2 + e5
L2 = c1 + d1*L1 + e6
Maximum likelihood (ML) and asymptotic distribution free (ADF)
estimation methods are provided. ADF is generalized method of
moments (GMM). Robust estimates of standard errors and SEs for
clustered samples are available, as is full support for survey
data via the -svy:- prefix. Missing at random (MAR) data are
supported via FIML.
Survey, cluster robust, and mixed models
-xtmixed- now supports sampling weights and robust and cluster-
robust standard errors for use with survey data, although you do
*NOT* use the -svy:- prefix as you might have expected.
That is because multilevel models with survey data differ from
standard models in that sampling weights need to be specified at
each modeling level rather than just at the observation level.
Sampling weights must reflect selection probability conditional on
selection at the next highest level.
Thus, -xtmixed- expects you to specify a weight for each level in
your model and warns you if you do not.
Multiple imputation
-mi impute- now supports
1. Chained equations.
Chained equations are used to impute missing values when
variables may be of different types and missing-value
patterns are arbitrary. The first variable could be
imputed using logit, the second using linear regression,
and the third using multinomial logistic regression.
2. Conditional imputation.
Conditional imputation is customized imputation within
group when group itself might be imputed. You can
restrict imputation of number of pregnancies to females
even when female itself contains missing values and so is
being imputed.
3. Imputation by groups.
Australians could have their missing values imputed using
data from other Australians only.
-mi estimate- now
1. Supports panel-data and multilevel models, so you can use
-mi- with -xtreg- or -xtmixed-.
2. Allows you to measure the amount of simulation error in
your final model, so you can decide whether you need more
-mi predict- and -mi predictnl- create linear and nonlinear
predictions in the original (m=0) data, and not just for complete
observations but also for observations with missing values.
Time series
Check out the
1. New estimators for
a. GARCH
b. ARFIMA
c. UCM
2. New postestimation command -psdensity- to estimate the
spectral density of a stationary process using the
parameters of a previously estimated parametric model.
3. New command -tsfilter-, which filters a series to keep only
selected periodicities (frequencies) and which can be used
to separate a series into trend and cyclical components.
Multivariate GARCH deals with models of time-varying volatility in
multiple series. These models allow the conditional covariance
matrix of the dependent variables to follow a flexible dynamic
structure and the conditional mean to follow a
vector-autoregressive (VAR) structure.
ARFIMA is a generalization of the ARMA and ARIMA models. ARMA
models assume short memory. ARIMA models assume shocks are
permanent. ARFIMA provides the middle ground. ARFIMA stands for
autoregressive, fractionally integrated moving average.
UCM stands for unobserved component model and decomposes a series
into trend, seasonal, cyclic, and idiosyncratic components after
controlling for optional exogenous variables.
Business calendars
There is a new %t format: %tb. The b stands for business
calendars. Business calendars allow you to define your own
calendars so that dates display correctly and lags and leads work
as they should.
You could create file lse.stbcal that records the days the London
Stock Exchange is open (or closed) and then Stata would understand
format %tblse just as it understands the usual date format %td.
Once you define a calendar, Stata deeply understands it. You can,
for instance, easily convert between %tblse and %td values.
Constrasts and pairwise comparisons
We were tempted to call this "Stata for Experimentalists" except
that the features are useful to Stata users of all disciplines.
Contrasts, pairwise comparisons, and margins plots are about
understanding and communicating results from your model. How does
a covariate affect the response? Is the effect nonlinear? Does
the effect depend on other covariates?
New commands -contrast-, -pwcompare-, and -marginsplot- join
1. -contrast- compares effects of factor variables and their
interactions. It can perform ANOVA-style tests of main
effects, simple effects, interactions, and nested effects.
It also decomposes these effects into comparisons against
reference categories, comparisons of adjacent levels,
comparisons against the grand mean, orthogonal
polynomials, and such.
In addition to predefined standard contrasts, user-defined
contrasts are also supported. Consider
. contrast ar.educ
The -ar.- out front is one of the new, predefined contrast
operators. -ar.- stands for "adjacent, reversed", and
-contrast ar.educ- compares adjacent levels of education,
for instance, high school to some college, some college to
college graduate, etc.
2. -pwcompare- performs all (or subsets) of the pairwise
comparisons. This can be done for all levels of a single
factor variable or for interactions or interactions with
continuous variables.
3. -margins- now allows the new contrast operators and has a
-pwcompare- option to perform pairwise comparisons.
4. -marginsplot- graphs results from -margins-.
ROC adjusted for covariates
New command -rocreg- is like regression for ROC. You can model
how sensitivity and specificity depend on covariates, and you
can draw graphs.
Contour plots
You just have to see one. Visit
There's more. For instance -rename- has a new syntax that allows
you to rename groups of variables.
. rename (vara varb varc) (varc varb vara)
swaps the names around.
. rename jan* *1
renames all variables starting with jan to instead end in 1.
. rename v# stat#
renames v1 to be stat1, v2 to be stat2, and so on.
. rename v# v(##)
renames v1 to be v01, v2 to be v02, ...
. rename (a b c) v#, addnumber
rename a to be v1, b to be v2, and c to be v3.
. rename v# (a b c)
does the reverse.
There really is a lot more. See http://www.stata.com/stata12.
-- Bill
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
Electronic Mail is not secure, may not be read every day, and should not be used for urgent or sensitive issues
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
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Nicholas Sze of Yahoo Finds Two-Quadrillionth Digit of Pi
16301986 story
Posted by
from the I'm-a-good-guesser-in-binary-too dept.
gregg writes
"A researcher has calculated the 2,000,000,000,000,000th digit of pi — and a few digits either side of it. Nicholas Sze, of technology firm Yahoo, determined that the digit — when expressed in binary
— is 0."
This discussion has been archived. No new comments can be posted.
• Re:So, what is the digit in decimal? (Score:3, Informative)
by Anonymous Coward on Thursday September 16, 2010 @06:50PM (#33605494)
We only know how to calculate it in binary (or any base that is a power of 2). You can't convert to decimal without know all the rest of the digits.
Parent Share
• Re:You fail math forever (Score:2, Informative)
by voutasaurus (1895138) on Thursday September 16, 2010 @06:56PM (#33605540)
What they should have said is: The two quadrillionth digit in the binary expansion of pi is 0.
Parent Share
• Re:You fail math forever (Score:3, Informative)
by Penguinshit (591885) on Thursday September 16, 2010 @07:04PM (#33605634) Homepage Journal
Parent Share
• Re:So, what is the digit in decimal? (Score:5, Informative)
by Haxamanish (1564673) on Thursday September 16, 2010 @07:09PM (#33605692)
We only know how to calculate it in binary (or any base that is a power of 2). You can't convert to decimal without know all the rest of the digits.
Parent is correct, digits of pi can be calculated independently in base 2, 4, 8, 16 or 2^n since the 1990s [maa.org]. So, it is possible to calculate the 2,000,000,000,000,000th number of pi
without calculating the digits before that one. Now, if we want to calculate the digit in decimal (or converse the binary digit to decimal), we need to calculate all of the two-quadrillion
digits. Knowing this digit is in itself not very interesting.
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• Re:The interesting thing about this article is how (Score:3, Informative)
by DerekLyons (302214) <fairwater&gmail,com> on Thursday September 16, 2010 @07:13PM (#33605718) Homepage
The interesting thing about this article is how they calculated the digits. They broke the problem up into small pieces and had them calculated in parallel. This approach isn't something
that's new or all the unique, but what is is applied to is. Most mathematical calculations are done in a near linear fashion, not in parallel. So for them to be able to do this is a big step
forward in how we approach these types of problem in the future.
At least with regards to calculating Pi, it's isn't particularly new. They first used this parallel method back in the 1980's.
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• Re:how do they do it (Score:3, Informative)
by Surt (22457) on Thursday September 16, 2010 @07:20PM (#33605772) Homepage Journal
Regardless of what actually happened, there isn't any computation that requires keeping data in memory rather than hard disk. Memory is just faster, if you need more space for the computation,
you can always actually use the 100 disks.
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• Re:Oh yeah? (Score:1, Informative)
by Anonymous Coward on Thursday September 16, 2010 @07:22PM (#33605778)
You're wrong, because TFA is discussing the binary representation of pi. It's either a 1 or a 0.
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• Re:Last Digit? (Score:4, Informative)
by by (1706743) (1706744) on Thursday September 16, 2010 @08:58PM (#33606442)
Pi is NOT irrational! It is transcendental. Look it up!
http://en.wikipedia.org/wiki/Transcendental_number [wikipedia.org] :
All real transcendental numbers are irrational, since all rational numbers are algebraic.
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• Re:an so are an infinite other digits in that numb (Score:4, Informative)
by Tacvek (948259) on Friday September 17, 2010 @02:06AM (#33607916) Journal
The hexadecimal digit extraction formula for PI (that allows you to skip calculating the previous hex digits) is already known. It can calulcuate the N'th hexadecimaldigit of Pi without
calculating most of the previous digits: http://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula [wikipedia.org]
A slower generalized version that can extract the n'th digit of Pi in any base (including decimal) has also been found: http://web.archive.org/web/19990116223856/www.lacim.uqam.ca/plouffe/Simon/
articlepi.html [archive.org]
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• Re:an so are an infinite other digits in that numb (Score:3, Informative)
by u38cg (607297) <calum@callingthetune.co.uk> on Friday September 17, 2010 @03:46AM (#33608338) Homepage
The thing that I find funny, is that had they used the Bailey-Borwein-Plouffe formula, [wikipedia.org] they could have saved themselves some very considerable computing resources.
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MathGroup Archive: February 1996 [00341]
[Date Index] [Thread Index] [Author Index]
Re: Constant term in polynomial?
• Subject: [mg3333] Re: Constant term in polynomial?
• From: bruck at pacificnet.net (Ronald Bruck)
• Date: 28 Feb 1996 10:12:10 -0600
• Approved: usenet@wri.com
• Distribution: local
• Newsgroups: wri.mathgroup
• Organization: University of Southern California
• Sender: daemon at wri.com
In article <4gmhdc$gmv at dragonfly.wolfram.com>, bruck at mtha.usc.edu (Ronald
Bruck) wrote:
:Arrgh, I feel stupid asking this question, but I can't think how to do it:
:how do I find the constant term in a polynomial in several variables in
:Mathematica? For example, the "7" in 7 + 3 x y + y^2 ?
:First[7 + 3 x y + y^2] will work for this one, since the 7 is present and
:appears first in the FullForm representation. But it won't work in
:First[3 x y + y^2], which returns 3 x y.
:OK, so I can build a command which computes Variables[First[expr]], and
:if that's empty, returns 0; otherwise returns First[expr]. Also clunky
:IMHO, but it seems the most workable--unless there's some trap I'm missing?
Yep, there's a trap -- First[3 x y] returns 3! So I ended up implementing
it as follows:
ConstantCoefficient[poly_] := Module[{c},
If[Head[poly] === Plus,
c = First[poly];
If[Variables[c] == {}, Return[c], Return[0]],
(* else head != plus *)
If[Variables[poly] == {}, Return[poly], Return[0]]
This won't recognize E + x as having constant term E, but fortunately my
polynomials all have rational coefficients.
It does seem there ought to be a built-in function to do this.
Someone e-mailed me a suggestion to use poly /. x_^e_ -> 0. This almost
works (it doesn't recognize the x in 7 + x as a power of x); it does
seem rather dangerous to apply the rule x_ -> 0!
The suggestion reminds me of another problem: suppose I want to work with
multivariable polynomials up through a certain degree p. So when, e.g., I
multiply two such polys, I need to get rid of terms like x^p y. This can
be done by applying rules like
(x^m_ y^n_ /; Positive[m+n-p]) -> 0
(although on one machine I have to use
(q__ x^m_ y^n_ /; Positive[m+n-p]) -> 0,
don't ask me why), with the annoyance that we also have to build rules for
x^m_ y -> 0 and x y^m_ -> 0 and x^m_ -> 0 and y^n_ -> 0! Is there some
more elegant way to do this--as well as faster--preferably one which works
for general variables, not just x and y? This came up when I was solving
an integral equation for a function of two variables, and used
undetermined coefficients.
--Ron Bruck
Now 100% ISDN from this address
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2 Lines Stuffs.
May 8th, 2012, 09:21 PM #1
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So, i'm doing this program that takes 2 lines of input, the first line being a number that signifies how many numbers are going to be in the next line. I then have to find which number in the
second line occurs the most often. Sounds simple. However, it is giving me problems. I am thinking with the whole read 2 lines thing. I thought I knew how, but my method of doing it didnt work as
I thought it should, and ended up in not knowing when to stop collecting data from the scanner in the loop, even though I told it when. Buttt anyway, I looked for solutions online and came upon
BufferedReader, which I know nothing about but looked it up and gave it a try. Now I am stuck with this mess of code that I cannot make sense of. The number it prints for occurring the most is 0,
even when 0 is not in the second line. I do not know where to go from here as I cant find where the problem lies. I am really hoping for someone to just suggest a different solution to read the
input. Unless you can fix this. I'll shut up and post the code now.
PS: even though it says if it reads "e" to stop, it really stops after you hit enter twice. If I change this it does not work the same. I have no idea why.
import java.util.*;
import java.io.*;
public class SandwichCounting
public static void main(String [] args) throws IOException
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Please enter the total number of orders, hit enter, then list the order numbers with spaces inbetween each one, then press enter twice when done.: ");
int num = in.read();
int [] list = new int [num];
int n = 0;
int[] number = new int [num];
int [] ammount = new int [num];
while (in.ready())
if (in.readLine().equalsIgnoreCase("e"))
list[n] = in.read();
for(int i = 0; i < num -1; i++)
for(int k = i+1; k < num -1; k++)
if(list[i] == list[k])
number[i] = list[i];
for (int p = 0; p < num; p++)
if(number[i] == list[p])
int order = 0;
int temp = ammount[0];
for (int i = 0; i < ammount.length; i++)
if(ammount[i] > temp)
temp = ammount[i];
order = number[i];
System.out.println("The most popular sandwich was #" + order);
Couldn't you simply just store the entire line in a String object, tokenize the string, parse each token into an integer, then determine the number of times each number occurs?
Possibly so, but how do you tokenize a string? Ive only been doing Java since January, so there's a lot I don't know about! Haha.
I'll go ahead and post this; this is the code I used at first and what I like to use, but it does not work like I need it to, as you'll see if you run it.
import java.util.*;
public class SandwichCounting
public static void main(String [] args)
Scanner in = new Scanner(System.in);
System.out.println("Please enter the total number of orders, hit enter, then list the order numbers with spaces inbetween each one, then press enter twice when done.: ");
int num = Integer.parseInt(in.next());
String [] list1 = new String [num];
int n = 0;
int[] number = new int [num];
int [] ammount = new int [num];
while (in.hasNextLine())
if (in.next().equals(""))
list1[n] = in.next();
int [] list = new int [num];
for (int i = 0; i < num; i++)
list[i] = Integer.parseInt(list1[i]);
for(int i = 0; i < num -1; i++)
for(int k = i+1; k < num -1; k++)
if(list[i] == list[k])
number[i] = list[i];
for (int p = 0; p < num; p++)
if(number[i] == list[p])
int order = 0;
int temp = ammount[0];
for (int i = 0; i < ammount.length; i++)
if(ammount[i] > temp)
temp = ammount[i];
order = number[i];
System.out.println("The most popular sandwich was #" + order);
Simple. Use the StringTokenizer class
StringTokenizer (Java 2 Platform SE v1.4.2)
Thanks a ton! Even though that page said StringTokenizers were outdated and to use the String.split method, I took your advice and got it to read. and semi-work! It reads the lines now, but
perhaps you can help me with potentially making a new design to actually do what the program needs to do. The system im using doesnt seem to be working. Which involves trying to go through and
put the numbers that do occur more than once into an array, then figuring out how many times they occur. Then I have to print the number that occurs the most. I cant say I know whats wrong with
mine, but after some println()s, it just seems like this system isnt going to work. Can you suggest another?
import java.util.*;
public class SandwichCounting
public static void main(String [] args)
Scanner in = new Scanner(System.in);
System.out.println("Please enter the total number of orders, hit enter, then list the order numbers with spaces inbetween each one, then press enter when done.: ");
int num = Integer.parseInt(in.next());
String line = "";
Scanner in2 = new Scanner(System.in);
int[] number = new int [num];
int [] ammount = new int [num];
line = in2.nextLine();
StringTokenizer st = new StringTokenizer(line);
int [] list = new int [num];
for (int i = 0; st.hasMoreTokens(); i++)
list[i] = Integer.parseInt(st.nextToken());
for(int i = 0; i < num; i++)
for(int k = i+1; k < num; k++)
if(list[i] == list[k])
number[i] = list[i];
for (int p = 0; p < num; p++)
if(number[i] == list[p])
int order = 0;
int temp = ammount[0];
for (int i = 0; i < ammount.length; i++)
if(ammount[i] > temp)
temp = ammount[i];
order = number[i];
System.out.println("The most popular sandwich was #" + order);
while (in.ready())
if (in.readLine().equalsIgnoreCase("e"))
list[n] = in.read();
Why are you using in.ready() as a conditional, and why break it when "in.readLine().equalsIgnoreCase("e")"???
Perhaps using num as a maximum, and count with a for loop?
Use a while loop to get the tokens instead.
ArrayList<Integer> numbers = new ArrayList<Integer>();
while(st.hasMoreTokens()) {
Then you could store the values, along with their count in a Map
That makes sense. I will try that as soon as I get a chance tomorrow. And squeakbox, ready was the conditional you have to use for a BufferedReader, and I intended for the user to enter e when
done, since it didnt work when I tried to use enter. Even though I know the input is going to only be two lines. But anyway, that is old code. Ive changed the program a good bit since then,
please look at my last post to get the most recent code.
Hmm... ive never used a map before, but I looked it up and it seems a bit weird. Could you give me a basic explanation as to how to do what I need to do with it?
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Transposition of a
Next | Prev | Up | Top | Index | JOS Index | JOS Pubs | JOS Home | Search
Above, we found the transfer function of the general state-space model to be
By the rules for transposing a matrix, the transpose of this equation gives
The system transpose of the system matrices.
When there is only one input and output signal (the SISO case), scalar, as is
That is, the transfer function of the transposed system is the same as the untransposed system in the scalar case. It can be shown that transposing the state-space representation is equivalent to
transposing the signal flow graph of the filter [75]. The equivalence of a flow graph to its transpose is established by Mason's gain theorem [49,50]. See §9.1.3 for more on this topic.
Next | Prev | Up | Top | Index | JOS Index | JOS Pubs | JOS Home | Search [How to cite this work] [Order a printed hardcopy] [Comment on this page via email]
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Sunday FunctionSunday Function
…you’ve got to ask yourself one question: ‘Do I feel lucky?’ Well, do ya, punk?
- Dirty Harry
The laws of probability, like most of the mathematical rules that govern the world, are a relatively recent discovery. Ancient people like the Romans loved to gamble as much as we do, and they had at
least some idea of how certain kinds of odds worked, but they’d probably have been flummoxed by many of the mathematical tools we use today to study chance. But then again, how many people at your
average casino understand how to calculate the probabilities that govern the flow of their money? Well, probably more than in the general population. But I bet it’s still not that many.
Because there’s so much money involved, problems involving gambling are very popular in introductory probability classes. Take, for instance, the Texas Lottery. The odds of winning are supposedly 1
in 25,827,165. Millions of people buy tickets for each drawing, so what is the probability of having n winners given that N tickets are purchased?
The exact answer is given by the binomial distribution. However, the binomial distribution involves taking factorials of its parameters, and when the parameters are numbers like 25,827,165 that
becomes pretty much impossible. But there’s a very good approximation we can use that makes our life a lot easier in those cases where you have a low-probability event repeated many times. It’s the
Poisson distribution, named after the early 19th century French mathematician Siméon-Denis Poisson. It’s our Sunday Function:
Here n is the number of successes and lambda is the expected number of successes. The expected number of successes is just the number of trials N times the probability of any given trial being
successful. Here “successes” is a term of art meaning just “the low-probability even happens”. Its connotation works well when we’re talking about winning the lottery, but not necessary in other
cases – you could just as well model something like train derailments as a Poisson process, after all.
The population of Texas is about 24,782,302 according to Wikipedia. So let’s pretend every one of them buys a single ticket and they don’t collude in picking their numbers. Multiplying that by the
probability of winning gives an expected number of wins: λ = 0.959544. Use this as our parameter and plot:
This plots the probability of having n winners, for n on the x axis.* As expected, it falls off rapidly. If more people played – say 40 million – the bump would be shifted over to the right as it
became more likely there’d be more winners. As it is, under these conditions the probabilities of n winners are:
0 – 38.3%
1 – 36.8%
2 – 17.6%
3 – 5.6%
4 – 1.4%
5 – 0.3%
And so on, rapidly heading toward zero. In practice I’m sure a lot less than the entire population is buying tickets, so most days the prize ought to roll over or otherwise go unclaimed. As of now
the jackpot is about $35 million, but I think I’ll be saving my money.
*As a commenter points out, while the function is defined for all real n, the Poisson distribution itself is only valid for positive integer n. Plugging in n = 3.14159… wouldn’t tell you anything
meaningful since it’s not possibly to have anything other than a whole number of winners. The reason I’ve plotted the function continuously instead of just at integral values is so that the overall
behavior of the function itself – particularly the location of the maximum – is most clear. From there it’s no difficult thing to understand that in reality it’s usually just the integers we’re
interested in.
1. #1 Aniko January 18, 2010
Matt, your plot has a problem: the Poisson density is defined only for integer n’s, while your plot implies otherwise.
2. #2 Matt Springer January 18, 2010
The function is defined for all real n, even if the distribution isn’t. The main point is to get a feel for how the function behaves, which is more difficult if we only look at positive integer
n. That said, your point about the Poisson distribution itself is correct and I’ll clarify in the entry.
3. #3 Paul Murray January 18, 2010
It’s interesting that this is a continuous function and yet the sum of a set of particular values of it (the integers >= 0) will sum to 1. And for any positive real value of lamba.
4. #4 Clark January 19, 2010
Personally, I’m going to find a lottery where that maximum falls exactly at 1 success. Because when I win the lottery, I don’t want to have to share it with anyone else.
5. #5 Anonymous Coward January 19, 2010
Clark at #4 raises an interesting point: how do you minimize the chances of sharing the winnings?
I don’t think it occurs when the maximum is at 1. I think it occurs in the limit that lambda ->0, which gives a maximum at 0.
This is kinda what you do in quantum optics if you have a coherent source of light, and you want a single photon in your pulse but no multiple-photon pulses. You attenuate the crap out of it
(lambda->0). Unfortunately, this means most of the time you get zero photons.
6. #6 Tim Gaede January 19, 2010
There’s an elegant way to program your computer to use extreme cases of the binomial (or a hypergeometric, or Poisson) without causing an overflow error. You simply keep track of the logarithm of
the probability while looping. This involves adding instead of multiplying, subtracting instead of dividing, and multiplying instead of raising to a power. You can take the inverse-logarithm of
the result when you are finished looping.
7. #7 Anonymous January 23, 2010
“But then again, how many people at your average casino understand how to calculate the probabilities that govern the flow of their money? Well, probably more than in the general population. But
I bet it’s still not that many.”
Are people that understand statistics more likely to gamble or less? You seem to think more, but I would say less.
Once you know that all games are statistically weighted so that you’ll lose money, why play? Unless you have developed some “system”, I think knowing the odds makes it less entertaining or
appealing. That combined with the knowledge that gambling disproportionally affects the poor, and the poor are more likely undereducated, it makes me think less people than average at a casino
understand probability. There are probably a lot of people who have memorized odds for games such as roulette, craps and blackjack (or maybe not even odds but just the proper actions to take in
specific situations), but I doubt all of them understand how to determine what the probability of a result is.
8. #8 MPL February 7, 2010
As an interesting complication, it’s known that lottery number selections are far from uniform—numbers encoding birthdates, “lucky numbers”, etc are selected far above chance rates.
So in actual practice, the distribution of number of winners will be very similar to the Poisson equation, but skewed a bit to the right. On occasion, as I recall, there have been cases where a
truly exceptional number of people win simultaneously, because the number drawn had some sort of significance.
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quadratic formula
September 2nd 2009, 02:30 PM
quadratic formula
i'm trying to use the quadratic formula for $3x^2 - 12x + 5 = 0$
and got it to $x = \frac {12 \pm \sqrt {-12^2 - 60}}{6}$ not sure if i'm going about it right so would like someone to point me in the right direction if possible
September 2nd 2009, 02:39 PM
Yes you are very correct.
But my teacher insisted on being picky so in that sentiment I'll just say that your notation, instead of http://www.mathhelpforum.com/math-he...dac4037a-1.gif
Should be
The addition of brackets removes confusion, that's what my teacher used to say.
September 2nd 2009, 02:44 PM
ah thanks, the answer the book gives is $x = 2 \pm \sqrt {\frac{7}{3}}$ so i can't see how you'd arrive at that from the answer i was at, could you show me the way to get to it please? there are
also meant to be rackets under the root sign around the 7/3 like so: (7/3) but i don't know how to write it out using the brackets like that
September 2nd 2009, 02:57 PM
$a = 3$
$b = -12$
$c = 5$
$x = \frac{12 \pm \sqrt{(-12)^2 - 4(3)(5)}}{2(3)}$
$x = \frac{12 \pm \sqrt{144 - 60}}{6}$
$x = \frac{12 \pm \sqrt{84}}{6}$
$x = \frac{12 \pm \sqrt{4 \cdot 21}}{6}$
$x = \frac{12 \pm 2\sqrt{21}}{6}$
$x = \frac{6 \pm \sqrt{21}}{3} = 2 \pm \frac{\sqrt{21}}{3}$
note that $\sqrt{\frac{7}{3}} = \frac{\sqrt{7}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{21}}{3}$
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Linear Algebra: From the Beginning
Aptly subtitled “From the Beginning,” this text is a self-contained book that works both as a first-semester text for linear algebra and as a guided introduction for those who desire on their own to
gain entry into the land of linear algebra with no more of a passport than a working concept of vectors. The additional phrase on the cover, “for scientists and engineers,” resonates with the
roll-up-your-sleeves approach that dives into practical mechanics, but the book is largely devoid of applications examples beyond basic curve-fitting, digital signal processing, and a few other
things. While the exercises are present to keep the reader on track, someone making the voyage alone will find that the online solutions manual is no longer available, but may appreciate the new
version professionally edited and published by W.H. Freeman and Co. (ISBN-10: 1-4292-0428-1).
Based upon their working experience with undergraduate students, the authors seek to teach linear algebra in a way that minimizes the required mathematical sophistication. This makes the text good
for readers that want to learn the subject on their own. Introductions to mathematical reasoning are made by bringing in geometric perspectives, building up higher-dimensional theorems from
2-dimensional ones, etc. However, this is an introductory text, in no way dense with pure mathematics and theory. As a study aid, the book prefaces the beginning of the chapters with a listing of
“very important formulas” covering matrix basics referred to throughout the text.
In sections packed with plenty of examples and exercises the authors describe vectors, matrices and linear transformations, the solution set of a linear system, the image of a line translation,
determinants, the Eigenvalue problem and Eigenvectors, and abstract vector spaces. Although intended for classroom use, this could also serve as a self-study text on practical approaches to solving
well-defined equations and to elucidate, within the problem-solving context, the mysteries of the dot product in higher dimensions, the discrete Fourier transform and other approaches to
diagonalization, and the Jordan canonical form.
Tom Schulte is working on a PhD in Applied Mathematics at Oakland University and this year checked the box for the SIAM Activity Group on Linear Algebra when renewing his SIAM membership.
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Lahey - LF64 Fortran 8.1
LF64 v8.1 Delivers!
Important features include ...
• Intel EM64T and AMD AMD64 64-bit optimizations
• Unsurpassed global compile-time and runtime diagnostics
• Automatic Parallelization
• OpenMP Support
• Wisk, Winteracter Starter Kit - Graphics package
• Small and medium memory models
• BLAS and LAPACK v4.0 routines - thread-safe
• Fujitsu's SSL2 math library - thread-safe
• Fujitsu's FDB debugger
• Automake, automatic make utility
• COMPLEX constants (Fortran 2003) supported
LF64 v8.1 is available in two configurations, Express and PRO:
LF64 Express includes the powerful Lahey/Fujitsu Fortran 95 optimizing compiler, command line debugger, online documentation, and free e-mail support. Suggested retail price, $249.
LF64 PRO adds auto-parallelization, OpenMP compatibility, the Winteracter Starter Kit, WiSK, for creating Windows GUIs and displaying graphics, thread-safe BLAS and LAPACK, Polyhedron's Automake
utility, and the Fujitsu SSL2 math library (thread-safe for parallel applications). Suggested retail price, $695.
LF64 Performance
Polyhedron Software (www.polyhedron.com) ran their suite of Fortran benchmarks on a Pentium D820 dual core 2.8 GHz processor, with 2 x 1MB L2 cache and 800 MHz FSB, and running SUSE 9.3 Linux.
Specifying the switch --fast, the benchmarks ran an average of 21.8% faster than they did when built with LF95 Linux v6.2. Try LF64 v8.1 on your code today!
LF64 Optimizations
Basic Optimization
• Constant folding
• Common subexpression elimination
• Copy propagation
• Strength Reduction
• Algebraic simplifications
• Dead code elimination
• Peephole optimization
• Loop invariant code motion
• Transform array element to simple variable
• Local Instruction scheduling
• Address calculation optimization
Program Reconstruction Optimizations
• Loop unrolling
• Loop interchange
Procedure Optimization
• Inlining mathematical functions
• Stack optimization
• SSE2 and SSE3 instructions
• Prefetch instructions
• Using fast input/output libraries
Link GNU C and assembly object files
LF64 supports static linking with GNU C or assembly. Combine your Fortran and C/C++ code into one executable. For the routines you don't want to develop yourself, you can also link with C/C++
routines from commercially available libraries.
Legacy Fortran Support
LF64 extends its language support in other directions adding many legacy Fortran features, including VAX structures and the various UNIX service routines. These features further facilitate your move
to cost/performance efficiency on the PC platform:
• Unlimited number of continuation lines in free or fixed source form
• DO UNTIL statement
• FIND statement
• STRUCTURE and END STRUCTURE statements
• UNION and END UNION statements
• MAP and END MAP statements
• RECORD statement
• Non-standard POINTER statement
• AUTOMATIC statement
• STATIC statement
• VALUE statement
• BYTE statement
• Hollerith constants
• Alternative forms of binary, octal, and hexadecimal constants
• Binary, octal, or hexadecimal constants in a DATA, declaration statement
• Period structure component separator
• IMPLICIT UNDEFINED statement
• Namelist input/output on internal file
• FORM = 'BINARY'
• TOTALREC specifier
• STATUS = 'SHR'
• Gw, $, \, and R edit descriptors
• LOC intrinsic function
• The following service subroutines: ABORT, BIC, BIS, CLOCK, CLOCKM, CLOCKV, DATE, ERROR, ERRSAV, ERRSET, ERRSTR, ERRTRA, EXIT, FDATE, FLUSH, FREE, GETARG, GETCL, GETDAT, GETENV, GETLOG, GETPARM,
GETTIM, GETTOD, GMTIME, IBTOD, IDATE, IETOM, IOSTAT_MSG, ITIME, IVALUE, LTIME, MTOIE, PERROR, PRECFILL, PRNSET, PROMPT, QSORT, REDLEN, SETBIT, SETRCD, SLEEP, SLITE, SLITET, TIMER
• The following service functions: ACCESS, ALARM, BIT, CHDIR, CHMOD, CTIME, DRAND, DTIME, ETIME, FGETC, FORK, FPUTC, FSEEK, FSEEKO64, FSTAT, FSTAT64, FTELL, FTELLO64, GETC, GETCWD, GETFD, GETGID,
GETPID, GETUID, HOSTNM, IARGC, IERRNO, INMAX, IOINIT, IRAND, ISATTY, JDATE, KILL, LINK, LNBLNK, LONG, LSTAT, LSTAT64, MALLOC, NARGS, PUTC, RAN, RAND, RENAME, RINDEX, RTC, SECNDS, SECOND, SH,
SHORT, SIGNAL, STAT, STAT64, SYMLNK, SYSTEM, TCLOSE, TIME, TIMEF, TTYNAM, UNLINK, WAIT
ANSI/ISO-Compliant Fortran 95
LF64 is a complete implementation of the ANSI/ISO Fortran 95 standard. Fortran 95 offers some small but important improvements over Fortran 90, including the ability to create your own elemental
procedures, default initialization for structure components, the NULL intrinsic for initializing pointers, the FORALL construct, and a standard CPU_TIME intrinsic procedure.
Free Technical Support
LF64 Linux Express includes e-mail technical support at no extra charge.
Automatic Parallelization
The LF64 compiler automatically parallelizes DO loops and array operations without you having to make modifications to the program. This makes it easy to migrate source programs to other platforms
(as long as the program conforms with the Fortran Standard). The effect is to save elapsed execution time by using two or more CPUs simultaneously. For instance, if a DO loop can be executed in
parallel by dividing it in half, then, theoretically, the execution time of this DO loop may be cut in half. In practice, improving performance requires some care and some work on the part of the
programmer. During compilation, the auto-parallel function will return information regarding which processes were (and which were not) parallelized and why. While certain loops can be analyzed
sufficiently to be parallelized by the compiler without input from the programmer, many loops have data dependencies that prevent automatic parallelization because of the potential for incorrect
results. For that reason, LF64 PRO also includes optimization control lines (OCLs) that provide information necessary for the compiler to parallelize these otherwise unparallelizable loops. The OCLs
are Fortran comments in a particular format, for example:
Note that programs with OCLs are standard-conforming and can be compiled with other compilers that do not support OCLs.
Four compiler switches control automatic parallelization: --parallel, --threads, --threadstack, and --ocl. Details of automatic parallelization (loop slicing, interchange, distribution, fusion, and
reduction, as well as OCL syntax and specifiers) are documented in the LF64 User's Guide and at www.lahey.com/doc.htm.
OpenMP v2.0 Compatibility
OpenMP specifies a set of compiler directives, library routines, and environment variables for shared-memory parallelism in Fortran and C/C++ programs. LF64 PRO v8.1 supports the OpenMP v2.0
specification for Fortran. Like automatic parallelization, OpenMP directives are used to parallelize a program that runs on a computer with more than one processor. With OpenMP you have more control
over how code is parallelized, but also more coding to do.
The LF64 Linux PRO v8.1 includes the OpenMP v2.0 Fortran specification in PDF. You can also view the specification at www.lahey.com/doc.htm. You can learn more about OpenMP at www.openmp.org.
Winteracter Starter Kit
Use the Winteracter Starter Kit - WiSK - for creating true X/Windows programs with Fortran. WiSK is a subset of the X/Winteracter Library created by Interactive Software Services, Ltd. (X/Winteracter
is available from Lahey.) X/Winteracter is a Fortran 95-callable, 64-bit, X/Windows, user-interface and graphics development kit. Derived from X/Winteracter, WiSK provides a library of subroutines
for window management, input handling, dialog management, and high resolution graphics. Designed for use with X11R6 and Open Motif 2.2.
X/Winteracter offers a wide range of powerful GUI capabilities to the Fortran 9x developer under X Windows, including:
• Multiple windows.
• Memory bitmap manipulation & bitmap viewer windows.
• Text editor windows, with optional command lines.
• Event handling.
• Text based menus, including floating menus (toolbars have still to be implemented).
• Dialog handling, including tabbed dialogs and nearly all control types (grid controls are the current exception).
• Common dialogs, e.g. file selector, message box, etc.
• Presentation graphics.
• and more!
Below are examples of WiSK's capabilities, visit the WiSK Examples Page for more examples.
BLAS and LAPACK
BLAS is a library for vector and matrix operations. The BLAS thread-safe version is based on BLAS provided on Netlib. BLAS includes 57 functions. The total number of routines for all precision types
amounts to approximately 170.
BLAS thread-safe version provides the following routines:
Level 1 BLAS : Vector operations
Level 2 BLAS : Matrix and vector operations
Level 3 BLAS : Matrix and matrix operations
Sparse-BLAS : Sparse vector operations
The thread-safe implementation of BLAS has exactly the same subroutine names and calling parameters as those of the Netlib baseline version.
Differences include:
• the thread-safe version can be used in the environment of SMP (Symmetric Multiple Processing)
• subroutines of the thread-safe version can be called from an OpenMP Fortran program
The purpose of using BLAS thread-safe version is to have a subroutine concurrently perform operations on different sets of data that are independent from each other, and thus reduce the time
necessary to finish all the operations.
LAPACK is a library of linear algebra routines. The LAPACK thread-safe version is based on LAPACK 3.0 provided on Netlib. LAPACK includes approximately 300 functions. The total number of routines for
all precision types amounts to approximately 1100.
LAPACK provides the following routines:
• Linear equations
• Linear least squares problems
• Eigenvalue problems
• Singular value decomposition
The LAPACK thread-safe version, like the BLAS version, can be called from an OpenMP program in the environment of SMP.
Fujitsu Scientific Subroutine Library 2
The Fujitsu Scientific Subroutine Library 2 (SSL2) has been in use for years in Japan on Fujitsu mainframe and workstation hardware. SSL2 offers over 250 optimized thread-safe routines in the
following areas:
Linear Algebra
Matrix Storage Mode Conversion
Matrix Manipulation
Linear Equations and Matrix Inversion (Direct Method)
Least Squares Solution
Eigenvalues and Eigenvectors
Eigenvalues and Eigenvectors of a Real Matrix
Eigenvalues and Eigenvectors of a Complex Matrix
Eigenvalues and Eigenvectors of a Real Symmetric Matrix
Eigenvalues and Eigenvectors of a Hermitian Matrix
Eigenvalues and Eigenvectors of a Real Symmetric Band Matrix
Eigenvalues and Eigenvectors of a Real Symmetric Generalized Eigenproblem
Eigenvalues and Eigenvectors of a Real Symmetric Band Generalized Eigenproblem
Nonlinear Equations
Polynomial Equations
Transcendental Equations
Nonlinear Simultaneous Equations
Minimization of Function with a Variable
Unconstrained Minimization of Multivariable Function
Unconstrained Minimization of Sum of Squares of Functions (Nonlinear Least Squares Solution)
Linear Programming
Nonlinear Programming (Constrained Minimization of Multivariable Function)
Interpolation and Approximation
Discrete Real Fourier Transforms
Discrete Cosine Transforms
Discrete Sine Transforms
Discrete Complex Fourier Transforms
Laplace Transform
Numerical Differentiation and Quadrature
Differential Equations
Special Functions
Elliptic Integrals
Exponential Integral
Sine and Cosine Integrals
Fresnel Integrals
Gamma Functions
Error Functions
Bessel Functions
Normal Distribution Functions
Pseudo Random Numbers
Pseudo Random Generation
Pseudo Random Testing
LF64 System Requirements
• Intelİ EM64T or AMDİ AMD64 64-bit processor.
• 32 MB of RAM.
• 70 MB of available hard disk space for LF64 Linux PRO; 40 MB for LF64 Linux Express.
• X-Windows to useWiSK and view the online PDF documentation.
• 64-bit version of as, the GNU assembler.
• 64-bit version of ld, the GNU linker.
• 64-bit versions of C startup and support object files crt1.o, crti.o, crtn.o, crtbegin. o, and crtend.o.
• 64-bit versions of C runtime and support libraries libc, libm, libpthread, librt, libgcc, libgcc_eh, libgcc_s, and libelf.
• A compatible version of the Linux operating system. Table 1 shows the versions of Linux that are known to be compatible with LF64. Other Linux variants might be compatible if they include kernel
version 2.6.9 or later and libc version 2.3.4 or later.
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Nontrivial tangent bundle that is diffeomorphic to the trivial bundle
up vote 13 down vote favorite
Is there an example of a smooth $n$-manifold $M$ whose tangent bundle is nontrivial as a bundle but is nonetheless (abstractly) diffeomorphic to the trivial bundle $M \times \mathbb{R}^n$?
(This question was inspired by Trivial fiber bundle.)
dg.differential-geometry differential-topology
A trivial observation: such an isomorphism would cover an automorphism of $M$. A possibly unhelpful observation: One could try to put a Riemannian metric on $M$ and see what constraints this gives
in the light of my first remark. – David Roberts Mar 18 '11 at 0:12
1 @David, is no reason for the diffeomorphism $TM\to M\times \mathbb R^n$ to cover a diffemorphism of $M$. @Faisal, the answer to your question is almost certainly yes, but I do not see an explicit
example at the moment, cf. my answer to mathoverflow.net/questions/58685/trivial-fiber-bundle. – Igor Belegradek Mar 18 '11 at 1:22
1 I think I can show that the unit tangent bundle over $S^5$ is diffeomorphic to the trivial one... but I'm having trouble otherwise, since most of the diffeomorphisms I can think of rely on
theorems about compact spaces. – Dylan Wilson Mar 18 '11 at 6:30
1 @Dylan - That was the example I was thinking of too! It is clear that $TS^5$ is not the trivial rank $5$ bundle over $S^5$. It only has $3$ linearly independent nowhere vanishing vector fields
(for lack of anything elementary one can use Adams' vector fields on spheres result). Although I don't know how to do it myself, if you can prove (extend your sphere bundle result to vector
bundles) your claim, we're done! – Somnath Basu Mar 18 '11 at 15:36
@Dylan and @Somnath: Tangent bundles over spheres never give such examples. Moreover, in general, if the homotopy-equivalence $M\to M$ induced by a diffeomorphism $TM\to M\times {\mathbb R}^n$ is
1 homotopic to a diffeomorphism $M\to M$, then $TM$ is trivial. See "Diffeomorphism of total spaces and equivalence of bundles" by De Sapio and Waldschap. This still leaves open the case when $M$ is
a homotopy-sphere though. – Misha Apr 2 '12 at 0:54
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W. Garrett Mitchener: My Web Pages
Here are electronic versions of some of my articles, posters, and slide shows. You may download them and read them for educational purposes, but please do not distribute them. Where possible, I have
included a link to the journal in which they were published.
Mathematica tools
Chat Room
Other Stuff from Duke
This site Copyright © 2002-2013 by W. Garrett Mitchener, all rights reserved except where otherwise noted.
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A characterization of lambda de in categorical models of implicit polymorphism
, 2002
"... Logical relations and their generalizations are a fundamental tool in proving properties of lambda-calculi, e.g., yielding sound principles for observational equivalence. We propose a natural
notion of logical relations able to deal with the monadic types of Moggi's computational lambda-calculus ..."
Cited by 19 (7 self)
Add to MetaCart
Logical relations and their generalizations are a fundamental tool in proving properties of lambda-calculi, e.g., yielding sound principles for observational equivalence. We propose a natural notion
of logical relations able to deal with the monadic types of Moggi's computational lambda-calculus. The treatment is categorical, and is based on notions of subsconing and distributivity laws for
monads. Our approach has a number of interesting applications, including cases for lambda-calculi with non-determinism (where being in logical relation means being bisimilar), dynamic name creation,
and probabilistic systems.
- In Typed Lambda Calculi and Applications (TLCA'99), Lecture Notes in Computer Science 1581 , 1999
"... We develop a notion of Kripke-like parameterized logical predicates for two fragments of intuitionistic linear logic (MILL and DILL) in terms of their category-theoretic models. Such logical
predicates are derived from the categorical glueing construction combined with the free symmetric monoidal co ..."
Cited by 11 (4 self)
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We develop a notion of Kripke-like parameterized logical predicates for two fragments of intuitionistic linear logic (MILL and DILL) in terms of their category-theoretic models. Such logical
predicates are derived from the categorical glueing construction combined with the free symmetric monoidal cocompletion. As applications, we obtain full completeness results of translations between
linear type theories.
- in Proc. International Conference on Theoretical Aspects of Computer Software, Springer LNCS 1281 , 1999
"... Flexibility of programming and efficiency of program execution are two important features of a programming language. Unfortunately, however, these two features conflict each other in design and
implementation of a modern statically typed programming language. Flexibility is model of computation, whi ..."
Cited by 6 (1 self)
Add to MetaCart
Flexibility of programming and efficiency of program execution are two important features of a programming language. Unfortunately, however, these two features conflict each other in design and
implementation of a modern statically typed programming language. Flexibility is model of computation, while efficiency requires optimal use of low-level primitives specialized to individual data
structures. The motivation of this work is to reconcile these two features by developing a mechanism for specializing polymorphic primitives based on static type information. We analyze the existing
methods for compiling a record calculus and an unboxed calculus, extract their common structure, and develop a framework for type-directed specialization of polymorphism. 1
- Homology, Homotopy and Applications , 2003
"... ..."
"... We develop a notion of equivalence between interpretations of the simply typed -calculus together with an equationally dened abstract data-type, and we show that two interpretations are
equivalent if and only if they are linked by a logical relation. We show that our construction generalises from th ..."
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We develop a notion of equivalence between interpretations of the simply typed -calculus together with an equationally dened abstract data-type, and we show that two interpretations are equivalent if
and only if they are linked by a logical relation. We show that our construction generalises from the simply typed -calculus to include the linear -calculus and calculi with additional type and term
constructors, such as those given by sum types or by a strong monad for modelling phenomena such as partiality or nondeterminism. This is all done in terms of category theoretic structure, using -
brations to model logical relations following Hermida, and adapting Jung and Tiuryn's logical relations of varying arity to provide the completeness results, which form the heart of the work.
, 2002
"... A variety of results which enable model checking of important classes of infinite-state systems are based on exploiting the property of data independence. The literature contains a number of
definitions of variants of data independence which are by syntactic restrictions in particular formalisms. Mo ..."
Add to MetaCart
A variety of results which enable model checking of important classes of infinite-state systems are based on exploiting the property of data independence. The literature contains a number of
definitions of variants of data independence which are by syntactic restrictions in particular formalisms. More recently, data independence was defined for labelled transition systems using logical
relations, enabling results about data independent systems to be proved without reference to a particular syntax. In this paper, we show that the semantic definition is suciently strong for this
purpose. More precisely, it was known that any syntactically data independent symbolic LTS denotes a semantically data independent family of LTSs, but here we show that the converse also holds.
- IN PROC. CSL/KGL'03, VOLUME 2803 OF LNCS , 2003
"... Pitts and Stark's nu-calculus is a typed lambda-calculus which forms a basis for the study of interaction between higher-order functions and dynamically created names. A similar approach has
received renewed attention recently through Sumii and Pierce's cryptographic lambda-calculus, which deals ..."
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Pitts and Stark's nu-calculus is a typed lambda-calculus which forms a basis for the study of interaction between higher-order functions and dynamically created names. A similar approach has received
renewed attention recently through Sumii and Pierce's cryptographic lambda-calculus, which deals with security protocols. Logical relations are a powerful tool to prove properties of such a calculus,
notably observational equivalence. While Pitts and Stark construct a logical relation for the nu-calculus, it rests heavily on operational aspects of the calculus and is hard to be extended. We
propose an alternative Kripke logical relation for the nu-calculus, which is derived naturally from the categorical model of the nu-calculus and the general notion of Kripke logical relation. This is
also related to the Kripke logical relation for the name creation monad by Goubault-Larrecq et al. (CSL'2002), which the authors claimed had similarities with Pitts and Stark's logical relation. We
show that their Kripke logical relation for names is strictly weaker than Pitts and Stark's. We also show that our Kripke logical relation, which extends the de nition of Goubault-Larrecq et al., is
equivalent to Pitts and Stark's up to rst-order types; our de nition rests on purely semantic constituents, and dispenses with the detours through operational semantics that Pitts and Stark use.
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20 August 2006
17 August 2006
With its time complexity of O(n log(n)) heapsort is optimal. It utilizes a special data structure called heap. This data structure is explained in the following.
29 July 2006
GTL, the Graph Template Library GTL can be seen as an extension of the Standard Template Library STL to graphs and fundamental graph algorithms.
1 (3 marks)
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probability & statistics
Number of results: 23,295
1. The probability that a student passes mathematics is 2/3 and the probability that he passes biology is 4/9. If the probability of passing one course is 4/5, what is the probability that he will
pass both courses?
Saturday, October 23, 2010 at 5:30pm by Anonymous
I will let a = 1/175,711,536= 5.591145*10^-9 , to save typing. The expected number, among 200,000,000, is L = 2*10^8 a = 1.13823 The probability of P(k) of k winners is given by Poisson statistics: P
(k) = L^k*e^-L/k! Probability of 0 winners = 0.32039 Probability of 1 winner...
Tuesday, March 18, 2008 at 4:38am by drwls
The below table shows the probabilities generated by rolling one die 50 times and noting the up face. What is the probability of getting an odd up face? roll 1 Probability 0.22 roll2 probability 0.10
roll 3 probability 0.18 roll 4 probability 0.12 roll 5 probability 0.18 roll ...
Thursday, May 3, 2012 at 8:05pm by sc
if n units are independent and the probability of each being up is 2/3 then the probability of all n being up is (2/3)^n HOWEVER I can not see your flow chart and therefore can not evaluate parallel
paths if there are two ways to get from A to B and the probability of failure ...
Monday, February 17, 2014 at 8:34pm by Damon
Probability and Statistics
Z = (score-mean)/SD Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion from the Z score. This is the probability for one
shark. The probability for all events is found by multiplying the probabilities ...
Sunday, April 29, 2012 at 3:06am by PsyDAG
statistics homework help A drinking game involves a 14 sided die. Six of the 14 faces are squares A, B, C, D, E,and F for short. The other eight are triangles which are called 1,2,3,4,5,6,7, and 8.
Each of the squares is equally likely. but he triangle probability differs from...
Tuesday, December 18, 2012 at 9:18pm by Anonymous
The probability that Luis will pass his statistics test is 0.37. Find the probability that he will fail his statistics test.
Tuesday, July 12, 2011 at 7:25pm by candy
With "or", it is unclear whether you want the probability between the scores, above them or below them. Assuming that you want the probability between the Z scores, find table in the back of your
statistics text labeled something like "areas under normal distribution" to find ...
Saturday, September 7, 2013 at 3:39pm by PsyDAG
What is an example of a research problem at your organization that would benefit from the use of either descriptive statistics or probability distribution statistics?
Monday, October 11, 2010 at 3:10pm by Kathy
Is that value the Z score or the probability? Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.095) and its
Z score.
Sunday, March 9, 2014 at 8:59pm by PsyDAG
Online "^" is used to indicate and exponent, e.g., x^2 = x squared. Z = (score-mean)/SD Find table in the back of your statistics text labeled something like "areas under normal distribution" to find
the proportion/probability related to the Z score. If the events are ...
Friday, April 12, 2013 at 5:56am by PsyDAG
Probability & Statistics
A basketball player shoots free throws and makes them with probability 1/3. What is the probability the player will miss three in a row?
Thursday, July 29, 2010 at 5:50pm by anonymous
A. Probability of 0 or 1 job offer. B. Probability of 2 or 3 job offers. To find either-or probability, add the individual probabilities.
Wednesday, May 19, 2010 at 9:48pm by PsyDAG
statistics help needed
Poisson statistics is not the only way to do these problems, but it can provide an approximate result for some. For B, the probability is that of no-green 24 times in a row. (36/38)^24 = (18/19)^24 =
0.273 That was easy For A, add the probabilities of getting green 4,5,6...24 ...
Sunday, February 22, 2009 at 2:20am by drwls
The probability of drawing an Ace is 4/52, while the probability of drawing some other card is 48/52. The probability of all events occurring is found by multiplying the probability of the individual
events. I hope this helps.
Tuesday, January 26, 2010 at 7:02pm by PsyDAG
Z = (score-mean)/SD Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score. Multiply that
probability by 2000 to get your second answer.
Monday, November 12, 2012 at 9:59am by PsyDAG
For any married couple who are members of a squash club, the probability that the husband has a degree is 2/4 and the probability that the wife has a degree is 1/4. The probability that the husband
has a degree given that his wife has a degree is 3/9. A married couple is ...
Saturday, December 3, 2011 at 8:23pm by Nabiha
Let a random variable be distributed as shown below X=x : 0,1,2,3,4,5,6 P(x): .1 .09 .2 .15 .16 .2 (a) Find the probability p(6) (b) Find the probability P(3< X < 5) (c) Find the probability P(X < 4)
(d) Find the probability P(X > 2)
Saturday, September 17, 2011 at 11:01am by kim wallace
In probability and statistics, mean is used to refer to one measure of the central tendency either of a probability distribution or of the random variable characterized by that distribution. In
statistics and probability theory, the standard deviation (represented by the Greek...
Thursday, October 17, 2013 at 6:05pm by John
The answer to part a) is 1/2 As for the others, I have never heard of "personal probability" There are two kinds of probability: 1. Empirical Probability which is based on the results of experiments
to collect data about an event 2. Mathematical Probability base on the ...
Sunday, March 9, 2008 at 8:54pm by Reiny
statistics math
The unemployment rate is 5.8% (Bureau of Labor Statistics, April 3, 2003). Suppose that 100 employable people are selected randomly. 1)What is the probability that exactly six people are unemployed
(to 4 decimals)? 2)What is the probability that exactly six people are ...
Sunday, October 18, 2009 at 10:30pm by aaron
Could anyone explaint to me what role the Binomial Theory plays in statistics and probability? Using statistics, we can make statements about a population based on sample data. Probability helps us
make those statements. The binomial theory can be used to determine ...
Tuesday, March 6, 2007 at 10:11pm by Jennie
Assume a binomial probability distribution has p = .60 and n = 200. c. What is the probability of 100 to 110 successes (to 4 decimals)? d. What is the probability of 130 or more successes (to 4
Sunday, April 6, 2014 at 9:01pm by Vanessa
Assume a binomial probability distribution has p = .60 and n = 200. c. What is the probability of 100 to 110 successes (to 4 decimals)? d. What is the probability of 130 or more successes (to 4
Monday, April 7, 2014 at 5:14pm by vanessa
The probability of a cherry on each of the two wheels = 1/20, while the middle wheel = 9/20. The probability of all events occurring is found by multiplying the probability of the individual events.
Tuesday, March 23, 2010 at 4:45pm by PsyDAG
Statistics - Probability
The probability of receiving a "false positive" from a mammogram is 7%. What is the probability that out of 10 mammograms, a patient receives at least 1 false positive?
Thursday, July 14, 2011 at 10:11am by Ana
Oops, I forgot. Here it is, an expected value distribution b. a discrete probability distribution c. none of these d. a continuous probability distribution e. a conditional probability distribution
Wednesday, October 10, 2012 at 12:42am by Tyson
odd numbers = 1, 3, 5, 7, 9 even numbers = 2, 4, 6, 8 you are looking for the probability of getting 1, 2, or 3 balls with odd numbers. 5/9*4/8*3/7 = probability of selecting one odd number 5/9*4/8*4
/7 = probability of selecting two odd numbers 5/9*4/8*3/7 = probability of ...
Monday, September 13, 2010 at 10:25pm by PsyDAG
odd numbers = 1, 3, 5, 7, 9 even numbers = 2, 4, 6, 8 you are looking for the probability of getting 1, 2, or 3 balls with odd numbers. 5/9*4/8*3/7 = probability of selecting one odd number 5/9*4/8*4
/7 = probability of selecting two odd numbers 5/9*4/8*3/7 = probability of ...
Monday, September 13, 2010 at 10:25pm by PsyDAG
probability and statistics
Four trials x 50% probability of a red each time = 2
Monday, November 24, 2008 at 5:39pm by drwls
statistics - probability
A total of 16 mice are sent down a maze, one by one. From previous experience, it is believed that the probability a mouse turns right is .38 a) What is the probability that 8 or fewer turn right? b)
What is the probability that 8 or more turn right? c) What is the probability...
Friday, October 12, 2012 at 2:01pm by andy
What is the probability of pat choosing a black marble
Saturday, October 23, 2010 at 5:30pm by Hailey
What is the probability of pat choosing a black marble
Saturday, October 23, 2010 at 5:30pm by Hailey
What is the probability of pat choosing a black marble
Saturday, October 23, 2010 at 5:30pm by Hailey
Probability of winning = 12/52 = 3/13 Probability of losing = 40/52 = 10/13 Probability of losing = 3 1/3 times the probability of winning. What would you consider fair? X * 3 1/3 = 5
Sunday, September 26, 2010 at 11:00pm by PsyDAG
And please disregard the probability of W question, It's .33 I'm pretty sure
Sunday, July 7, 2013 at 12:00pm by y912f
probability and statistics I NEED HELP
The MSU basketball team is playing OU. If the probability that MSU wins the game is 0.68 and the probability that OU wins is 0.27. 1. Find the probability that MSU or OU wins. 2. Find the Probability
that there is a tie.
Wednesday, May 11, 2011 at 9:17pm by Emily
Red has probability .30 and black has probability .70. What is the probability that patron will win all ten wagers if alternating between red and black on each wager? What would be the standard
Tuesday, February 21, 2012 at 8:04pm by Joe Dee.
Hint: You don't want the probability from mean to z. Remember the problem is asking "120 or less" or "170 or more" for the probability. You use those values. Once you have those values, add them
together for the total probability.
Tuesday, August 28, 2012 at 6:12pm by MathGuru
if Z is a standard normal variable, find the probability. the probability that z lies between -2.41 and 0
Tuesday, May 25, 2010 at 5:03pm by shasha
With replacement, the probability of selecting a good apple stays at 60/80=3/4. For a two-step experiment, the probability is the product of individual steps, hence, probability of select a good
apple in each pick is (3/4)^2=9/16
Monday, July 15, 2013 at 8:10am by MathMate
A mini-computer system contains two components, A and B. The system will function so long as either A or B functions. The probability that A functions is 0.95, the probability that B functions is
0.90, and the probability that both function is 0.88. What is the probability ...
Saturday, March 5, 2011 at 3:54am by Paula
Please note that your School Subject is NOT college. It's probably Math or Probability or Statistics.
Tuesday, October 6, 2009 at 1:02pm by Ms. Sue
probability and statistics
The MSU basketball team is playing OU. If the probability that MSU wins the game is 0.68 and the probability that OU wins is 0.27. 1. Find the probability that MSU or OU wins. 2. Find the Probability
that there is a tie.
Wednesday, May 11, 2011 at 7:16pm by Emily
2, 4 or 6, each with 1/6 probability. To find either-or probability, add the individual probabilities. Primes are 2, 3 or 5. Use same process.
Tuesday, May 17, 2011 at 5:22pm by PsyDAG
Probability and statistics
The probability that at most 3 tosses of a balanced dice are required to get a prime number on top is equal to a. 7/8, b.1/4, c. 1/2, d.3/4.
Thursday, March 1, 2012 at 12:32am by Arun
That would the probability of 1 - probability of no tails. 1 - .5^7 = ?
Wednesday, November 17, 2010 at 5:22am by PsyDAG
Suppose that each unit of a system is up with probability 2/3 and down with probability 1/3. Different units are independent. For each one of the systems shown below, calculate the probability that
the whole system is up (that is, that there exists a path from the left end to ...
Monday, February 17, 2014 at 8:34pm by sam
Suppose we want to determine the (binomial) probability (p) of getting 4 heads in 15 flips of a 2-sided coin. Using the Binomial Probabilities Table in Appendix B of the text, what values of n, x and
p would we use to look up this probability, and what would be the probability?
Thursday, October 13, 2011 at 8:43am by destiny
Try a binomial probability table or use a binomial probability function, which is this: P(x) = (nCx)(p^x)[q^(n-x)] For (A) x = 3 n = 15 p = .20 q = .80 (q is 1-p) For (B) Using a table might be
easier for this one. Find P(0), P(1), P(2), P(3), P(4), and P(5). Add together, ...
Sunday, October 7, 2012 at 9:46am by MathGuru
Use the binomial probability function: P(x) = (nCx)(p^x)[q^(n-x)] n = 10 x = 0,1 p = 0.10 q = 1 - p = 0.90 Find P(0), P(1). Add together for your probability. Note: You can also use a binomial
probability function table with the values listed to find the probability as well. ...
Wednesday, April 10, 2013 at 3:00pm by MathGuru
Find P(0), then take 1 - P(0) for your probability (since the problem says "what is the probability that at least one" is defective). Using the binomial probability function (you can use a binomial
probability table as well): P(x) = (nCx)(p^x)[q^(n-x)] n = 4 x = 0 p = 17/53...
Sunday, September 9, 2007 at 9:58pm by MathGuru
1)Let X have a normal distribution with mean 6 and standard deviation 3. a) What is the probability that X > 8.7 2)What is the probability that X<0? 3)What is the probability that |X-6| > 1.9 ? I
have got so far: P(Z > 0.9) but I do not know where to go from here! ...
Sunday, October 7, 2012 at 2:22pm by rachel
1. Assume that q and z are two random variables that are perfectly positively correlated. q takes the value of 20 with probability 0.5 and the value of zero with probability 0.5, while z takes the
value of 10 with probability 0.5 and the value of zero with probability 0.5. ...
Sunday, December 6, 2009 at 3:53am by Susan
a. The probability of being a senior citizen = .15. The probability of a senior citizen getting the flu = .12. The probability of both/all events occurring is found by multiplying the individual
probabilities. Use the same process for the remaining problems.
Tuesday, October 30, 2012 at 11:09am by PsyDAG
Z = (score-mean)/SD Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score. If the events
are independent, the probability of both/all events occurring is determined by ...
Tuesday, February 18, 2014 at 1:04am by PsyDAG
Statistics and Probability
In a game of skill, a player has probability 1/3,5/12,1/4 of scoring 0,1,2 respectively and the game ends when he scores a zero. Assuming independence of trials, find the probability of the total
score being 'n' when the game ends?
Friday, October 7, 2011 at 11:18pm by Saurabh
a) Yes, because of replacement, the probabilities do not change for the second card. b, c) probability of both events occurring is found by multiplying the probability of the individual events. d)
probability of either-or events occurring is found by adding the probability of ...
Friday, April 15, 2011 at 1:03pm by PsyDAG
a) Yes, because of replacement, the probabilities do not change for the second card. b, c) probability of both events occurring is found by multiplying the probability of the individual events. d)
probability of either-or events occurring is found by adding the probability of ...
Friday, April 15, 2011 at 1:03pm by PsyDAG
50 question multiple choices with A, B, C, D Probability of an unsual score? Probability of between 20 and 30 correct inclusive?
Saturday, November 10, 2007 at 9:45am by Anonymous
Here is the easiest way to do this problem. Use a binomial probability table. n = 5 x = 0, 1, 2 p = .75 Add P(0), P(1), and P(2). This will be your probability. I hope this helps.
Tuesday, October 18, 2011 at 8:02pm by MathGuru
Find the normal approximation for the binomial probability that x = 5, where n = 12 and p = 0.7. Compare this probability to the value of P(x=5) found in Table 2 of Appendix B in your textbook.
Thursday, October 20, 2011 at 7:50am by jane
Consider a binomial random variable X with parameters(4,1/2). Find the conditional probability mass function of X given that X is odd
Wednesday, January 25, 2012 at 3:28am by sand
i need to learn probability & statistics. completly.
Friday, February 8, 2008 at 4:02pm by sai
If I am interpreting you table correctly, the probability of being a man (M) is 528/991 (total men dividing by the grand total) and the probability of being a heavy smoker (HS) is total HS divided by
the grand total. The probability of either one event or another occurring is ...
Saturday, July 18, 2009 at 3:40pm by PsyDAG
Using formula P(x) = x/15 For each of the values x = 1, 2, 3, 4, and 5. a. probability distribution of x. b. Show that the probability distribution of x satisfies the properties of a discrete
probability distribution. c. mean of x. sketch the two specified normal curves on the...
Monday, October 26, 2009 at 2:35pm by Chinadoll
use the normal distribution to approximate the desired probability. find the probability that in 200 tosses we will obtain at least 40 fives.
Saturday, October 16, 2010 at 4:08pm by monique
Probability and statistics
6. The probability that at most 3 tosses of a balanced dice are required to get a prime number on top is equal to Choose one answer a. 7/8 b. ¼ c. ½ d. ¾
Friday, March 2, 2012 at 3:24am by Arun
Probability and statistics
8. A random variable X takes exactly the 5 values 1, 2,3,4,5, all with same probability. The mean of X is Choose one answer a. 2.5 b. 15 c. 7.5 d. 3
Friday, March 2, 2012 at 3:25am by Arun
Assume a binomial probability distribution has p = .60 and n = 200. a)What are the mean and standard deviation (to 2 decimals)? b)Why can the normal probability distribution be used to approximate
this binomial distribution? c)What is the probability of 100 to 110 successes (...
Saturday, April 5, 2014 at 11:42pm by Vanessa
statistics - conditional probability
Event A has probability 0.9 of occurring. Event B has 0.2 chance of occurring if, and only if, event A has occurred. What is the overall probability that event B will occur? I'm so confused. This
seems so simple, I'm not sure why I can't reason through it. They're not ...
Wednesday, September 25, 2013 at 10:19pm by Pam
statistics - conditional probability
Event A has probability 0.9 of occurring. Event B has 0.2 chance of occurring if, and only if, event A has occurred. What is the overall probability that event B will occur? I'm so confused. This
seems so simple, I'm not sure why I can't reason through it. They're not ...
Wednesday, September 25, 2013 at 10:20pm by Pam
Hi Shanice, Numbers in total : 1,2,3,4,5,6,7,8 In the first draw : 2 was selected, therefore the probability of drawing a 2 is 1/8. (since there is only one 2 in the list) In the second draw, 3 was
selected. The probability is again 1/8. Therefore the overall probability is: 1...
Thursday, October 10, 2013 at 10:41pm by Swaastikaa
In a particular suburb 30% of housholds have installed electronic security systems. If 2 households are chosen at random from this area, what is the probability that neither has installed a security
system? Using the binomial probability function for this problem: P(x) = (nCx...
Monday, February 26, 2007 at 3:30pm by Jason L
If df = 10 and P = .01, t = 3.169 From a table in my statistics text labeled "distribution of t probability."
Wednesday, June 22, 2011 at 10:54pm by PsyDAG
Z = (score-mean)/SD Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score. OR… what is the
probability of being higher than the mean in a normal distribution?
Sunday, May 12, 2013 at 11:25am by PsyDAG
Use the binomial probability function. If you use a binomial probability table, this problem will be easily solved. In most tables, you will need x, n, and p. x = 8, n = 10, and p = .5 (look up the
probability in the table using those values). I hope this will help.
Sunday, March 13, 2011 at 5:33pm by MathGuru
You can use a binomial probability table to determine the probability. Find P(18) through P(25), then add all the probabilities together for a total.
Monday, November 14, 2011 at 2:32pm by MathGuru
From a bag containing 6 black and 4 white balls, 2 balls are selected,one after the other,without replacement. Find (a) the probability that they are both black (b)the probability that they are both
the same colour, (c) probability that they are different colours.
Saturday, May 21, 2011 at 10:16am by Moses joseph
The probability of not hitting oil is .8 The probability of not hitting oil in 5 tries is .8^5 = .32768. Ergo, the probability of getting at least 1 hit is (1-.32768) = .67232
Wednesday, February 3, 2010 at 7:31pm by economyst
There is a probability of 1 for getting the first card. Assuming that you are the only one being dealt, once the first card is obtained, the probability of getting the second card for the flush is 12
/51. For the third card. it is 11/50. From this, you should be able to figure ...
Saturday, November 10, 2007 at 9:46am by PsyDAG
Event Probability Too much enamel 0.25 Too little enamel 0.2 Uneven application 0.35 No defects noted 0.42 (1) What is the probability of a paint defect? What is the probability of a paint defect
which includes an improper amount of paint?
Wednesday, September 29, 2010 at 6:57pm by Jordan
I need to post some thought on using either the multiplication rule or probability to answer a question on my online statistics.
Sunday, July 10, 2011 at 3:07pm by keri
Use of normal distribution to approximate the desired probability. Find the probability of getting at least 30 fives in 200 tosses of a fair 6 sided die.
Friday, October 8, 2010 at 10:58pm by Lori
Consider a binomial experiment with 20 trials and a probability of 0.45 on a single trial.Use the normal distribution to find the probability of exactly 10 successes.
Friday, March 21, 2014 at 10:18pm by john
You can use a binomial probability table, or calculate by hand using the following formula: P(x) = (nCx)(p^x)[q^(n-x)] p = .39 q = 1 - p n = 12 For (a): find P(3) For (b): find P(0),P(1),P(2),P(3).
Add for a total, then subtract from 1 for your probability. For (c): find P(0...
Saturday, March 30, 2013 at 1:18pm by MathGuru
Easy Statistics
The probability Joe is at work is 49%,. The probability Joe is at home is 25%. What is the probability Joe is neither at work nor at home? a) 39% b)74% c)52% d) 26% *I personally think it is a or c.
I am just needing some help thanks!
Friday, December 6, 2013 at 9:22pm by Dawn
Suppose that 50 identical batteries are being tested. After 8 hours of continuous use, assume that a given battery is still operating with a probability of .70 and has failed with a probability of
.30. What is the probability that between 25 and 30 batteries (inclusive) will ...
Wednesday, December 26, 2012 at 9:14pm by Camela
Statistics - Probability
Probability of choosing ONLY good funds from the US market = (5/6)*(4/5) = 2/3 Probability of choosing ONLY good funds from the foreign market = (3/4)*(2/3)=2/4=1/2 Probability of choosing ONLY good
funds from both markets = (2/3)*(1/2)=1/3 Probability of choosing at least one...
Sunday, April 8, 2012 at 11:04pm by MathMate
The experiment consists of 10 bernoulli (either true or false) experiments over one week. The probability of success p is 0.3 (so failure, q=0.7) The probability does not change throughout the week.
This is a binomial distribution, where the probability of r success out of n ...
Tuesday, June 12, 2012 at 5:00pm by MathMate
Ignoring twins and other mutiple births, assume babies born at a hospital are independent events with the probability that a baby is a boy and the probability that a baby is a girl both equal to 0.5.
Referring to the information above, the probability that at least one of the ...
Wednesday, September 19, 2007 at 4:53pm by Anonymous
Events and are mutually exclusive. Suppose event occurs with probability and event occurs with probability . Compute the probability that occurs or does not occur (or both). Compute the probability
that either occurs without occurring or and both occur.
Monday, April 8, 2013 at 1:07pm by Jack
Use a binomial probability table or a formula like the following: P(x) = (nCx)(p^x)[q^(n-x)] For a): Find P(2) For b): Find P(2), P(3), and P(4). Add together for your probability. For c): Find P(0)
and P(1). Add together for your probability.
Tuesday, February 12, 2013 at 5:18pm by MathGuru
Statistics -- Probability
I am confused too. If "the probability a woman is an actress is 0.550", how can the probability a woman is an actress, given that she’s a musical performer be .785. Do you mean .785 of the .550? The
same applies for the man as a ballet dancer. I hope this helps. Thanks for ...
Friday, April 17, 2009 at 3:19pm by PsyDAG
AP Statistics
if the probability of a swan drowning is .27, find the probability of exactly 4 out of the 7 swans drowning.
Wednesday, January 6, 2010 at 9:28pm by Belle
This is the probability of drawing a 5 and a 3. P(5) = 1/6 P(3) = 1/6 The probability of both/all events occurring is found by multiplying the individual events.
Wednesday, November 17, 2010 at 5:24am by PsyDAG
Probability of one tire NOT exploding = 1-80%=0.2 Probability of ALL FOUR tires NOT exploding = 0.2^4 = 0.0016
Monday, May 28, 2012 at 11:15am by MathMate
The probability that a house is burglarized in the city is 5%. If 50 houses are randomly selected, what is the probability that none will be burglarized?
Saturday, July 14, 2012 at 9:15pm by Sherry
A total of 16 mice are sent down a maze, one by one. From previous experience, it is believed that the probability a mouse turns right is .38 a) What is the probability that exactly 8 of these 16
mice turn right? b) What is the probability that 8 or fewer turn right? c) What ...
Friday, October 12, 2012 at 2:58pm by please help
If the probability of running out of gas is .03 and the probability the electronic starting system will not work is .01 a.) what is the probability there will be enough gas and that the starting
system will work? assume the two events are independent b.) when may independence ...
Friday, September 2, 2011 at 1:11am by kkg
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Countryside, IL Trigonometry Tutor
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How Many Cubic Meters Of Helium Are Required To ... | Chegg.com
How many cubic meters of helium are required to lift a balloon witha 428 kg payload to a height of7992 m? (Take ρ[He]= 0.180 kg/m^3.) Assume that the balloon maintains aconstant volume and that the
density of air decreases with altitudez according to the expression ρ[air]= ρ[0]e^-z/8000,where z is in meters and ρ[0] = 1.25kg/m^3 is the density of air at sea level.
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FYI: Lambda Calculus on Perl 6
Front page | perl.perl6.language | Postings from September 2005
FYI: Lambda Calculus on Perl 6
Thread Next
Dan Kogai
September 4, 2005 20:35
FYI: Lambda Calculus on Perl 6
Message ID:
I recently needed to write a series of codes on lambda calculus in
perl. As MJD has shown Perl 5 can handle lambda calculus but I am
beginning to get tired of whole bunch of 'my $x = shift' needed.
our $ZERO =
sub { my $f = shift;
sub { my $x = shift; $x }};
our $SUCC =
sub { my $n = shift;
sub { my $f = shift;
sub { my $x = shift;
$f->($n->($f)($x)) }}};
our $ADD =
sub{ my $m = shift;
sub { my $n = shift;
sub { my $f = shift;
sub { my $x = shift;
$m->($f)($n->($f)($x)) }}}};
our $MULT =
sub { my $m = shift;
sub { my $n = shift;
sub { my $f = shift;
$m->($n->($f)) }}};
our $POW =
sub { my $m = shift;
sub { my $n = shift;
$n->($m) }};
And I found that these can be made much, much simpler and more
intuitive with Perl 6, even more so than scheme!
our $ZERO = sub($f){ sub($x){ $x }};
our $SUCC = sub($n){ sub($f){ sub($x){ $f.($n.($f)($x)) }}};
our $ADD = sub($m){ sub($n){ sub($f){ sub($x){ $m.($f)($n.($f)
($x)) }}}};
our $MULT = sub($m){ sub($n){ sub($f){ $m.($n.($f)) }}};
our $POW = sub($m){ sub($n){ $n.($m) }};
You can even make it simpler by removing dots but I leave it that way
because it looks more like the original notation that way (i.e.
zero := λf.λx.x).
Runs perfectly fine on Pugs 6.2.8. Add the code below and see it for
my $one = $SUCC.($ZERO);
my $two = $SUCC.($one);
my $four = $ADD.($two)($two);
my $eight = $MULT.($two)($four);
my $sixteen = $POW.($four)($two);
for($one, $two, $four, $eight, $sixteen) -> $n {
$n.(sub($i){ 1 + $i})(0).say
Maybe we can use this for advocacy.
Dan the Perl 6 User Now
P.S. I am surprised to find Pugs does not include this kind of
sample scripts.
Thread Next
• FYI: Lambda Calculus on Perl 6 by Dan Kogai
nntp.perl.org: Perl Programming lists via nntp and http.
Comments to Ask Bjørn Hansen at ask@perl.org | Group listing | About
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The Institute For Figuring // Online Exhibit: Hyperbolic Space
In 1997 Cornell University mathematician Daina Taimina finally worked out how to make a physical model of hyperbolic space that allows us to feel, and to tactilely explore, the properties of this
unique geometry. The method she used was crochet.
Dr Taimina’s inspiration was based on a suggestion that had been put forward in the 1970’s by the geometer William Thurston (also now at Cornell). Noting that one of the qualities of hyperbolic space
is that as you move away from a point the space around it expands exponentially, Thurston designed a paper model made up of thin cresent-shaped annuli taped together.
But Thurston’s model is difficult to make, hard to handle, and inherently fragile. Taimina intuited that the essence of this construction could be implemented with knitting or crochet simply by
increasing the number of stitches in each row. As you increase, the surface naturally begins to ruffle and crenellate. Taimina, who grew up in Latvia with a childhood steeped in feminine handicrafts,
immediately set about making a model. At first she tried knitting - and you can indeed knit hyperbolic surfaces - but the large number of stitches on the needles quickly becomes unmanageable and
Taimina realized that crochet offered the better approach.
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A Unification Algorithm for Typed l-Calculus
, 1998
"... The paper proposes a theory relating syntax, semantics, and intonational prosody, and covering a wide range of English intonational tunes and their semantic interpretation in terms of focus and
information structure. The theory is based on a version of combinatory categorial grammar which directly p ..."
Cited by 111 (5 self)
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The paper proposes a theory relating syntax, semantics, and intonational prosody, and covering a wide range of English intonational tunes and their semantic interpretation in terms of focus and
information structure. The theory is based on a version of combinatory categorial grammar which directly pairs phonological and logical forms without intermediary representational levels.
- Journal of Automated Reasoning , 1996
"... This is a description of TPS, a theorem proving system for classical type theory (Church’s typed λ-calculus). TPS has been designed to be a general research tool for manipulating wffs of first-
and higher-order logic, and searching for proofs of such wffs interactively or automatically, or in a comb ..."
Cited by 71 (6 self)
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This is a description of TPS, a theorem proving system for classical type theory (Church’s typed λ-calculus). TPS has been designed to be a general research tool for manipulating wffs of first- and
higher-order logic, and searching for proofs of such wffs interactively or automatically, or in a combination of these modes. An important feature of TPS is the ability to translate between expansion
proofs and natural deduction proofs. Examples of theorems which TPS can prove completely automatically are given to illustrate certain aspects of TPS’s behavior and problems of theorem proving in
higher-order logic. 7
- Extensions of Logic Programming , 1992
"... The theory of partial inductive definitions is a mathematical formalism which has proved to be useful in a number of different applications. The fundamentals of the theory is shortly described.
Partial inductive definitions and their associated calculi are essentially infinitary. To implement them o ..."
Cited by 28 (1 self)
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The theory of partial inductive definitions is a mathematical formalism which has proved to be useful in a number of different applications. The fundamentals of the theory is shortly described.
Partial inductive definitions and their associated calculi are essentially infinitary. To implement them on a computer, they must be given a formal finitary representation. We present such a finitary
representation, and prove its soundness. The finitary representation is given in a form with and without variables. Without variables, derivations are unchanging entities. With variables, derivations
can contain logical variables that can become bound by a binding environment that is extended as the derivation is constructed. The variant with variables is essentially a generalization of the pure
GCLA programming language.
, 1998
"... Consistency Class) Let Ñ S be a class of sets of propositions, then Ñ S is called an abstract consistency class, iff each Ñ S is closed under subsets, and satisfies conditions (1) to (8) for all
sets F 2 Ñ S . If it also satisfies (9), then we call it extensional. 1. If A is atomic, then A = 2 F or ..."
Cited by 5 (1 self)
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Consistency Class) Let Ñ S be a class of sets of propositions, then Ñ S is called an abstract consistency class, iff each Ñ S is closed under subsets, and satisfies conditions (1) to (8) for all sets
F 2 Ñ S . If it also satisfies (9), then we call it extensional. 1. If A is atomic, then A = 2 F or :A = 2 F. 2. If A 2 F and if B is the bh-normal form of A, then B F 2 Ñ S 2 . 3. If ::A 2 F, then A
F 2 Ñ S . 4. If AB2F, then F A 2 Ñ S or F B 2 Ñ S . 5. If :(AB) 2 F, then F :A :B2 Ñ S . 6. If P a A 2 F, then F AB 2 Ñ S for each closed formula B 2 wff a (S). 7. If :P a A 2 F, then F :(Aw a ) 2 Ñ
S for any witness constant w a 2 W that does not occur in F. 8. If :(A = a!b B) 2 F, then F :(Aw a = Bw) 2 Ñ S for any witness constant w a 2 W that does not occur in F. 9. If :(A = o B) 2 F, then F
[fA;:Bg 2 Ñ S or F[f:A;Bg 2 Ñ S . Here, we treat equality as an abbreviation for Leibniz definition. We call an abstract consistency class saturated, iff for all F 2 Ñ S and all...
- In ACM SIGPLAN International Conference on Functional Programming , 2002
"... We propose an extension of Haskell's type class system with lambda abstractions in the type language. Type inference for our extension relies on a novel constrained unification procedure called
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Cited by 5 (1 self)
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We propose an extension of Haskell's type class system with lambda abstractions in the type language. Type inference for our extension relies on a novel constrained unification procedure called
guided higher-order unification. This unification procedure is more general than Haskell's kind-preserving unification but less powerful than full higher-order unification.
- Higher Order Logic Theorem Proving and Its Applications: 6th International Workshop, HUG '93, volume 780 of Lecture Notes in Computer Science , 1994
"... This is a demonstration of TPS, a theorem proving system for classical type theory (Church's typed l-calculus). TPS can be used interactively or automatically, or in a combination of these
modes. An important feature of TPS is the ability to translate between expansion proofs and natural deduction ..."
Cited by 1 (1 self)
Add to MetaCart
This is a demonstration of TPS, a theorem proving system for classical type theory (Church's typed l-calculus). TPS can be used interactively or automatically, or in a combination of these modes. An
important feature of TPS is the ability to translate between expansion proofs and natural deduction proofs. CATEGORY: Demonstration 1. Introduction This presentation is a demonstration of TPS, a
theorem proving system for classical type theory (Church's typed 3 l-calculus [14]) which has been under development at Carnegie Mellon University for a number of years. TPS is based on an approach
to automated theorem proving called the mating method [2], which is essentially the same as the connection method developed independently by Bibel [13]. The mating method does not require reduction
to clausal form. TPS handles two sorts of proofs, natural deduction proofs and expansion proofs. Natural deduction proofs are human-readable formal proofs. An example of such a proof which was
produced aut...
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Math Help
December 12th 2011, 08:53 PM #1
Dec 2011
Eek, so I'm not sure if this is the right forum, but it is technically statistics
Psychology student trying to work out a t-score, but I'm not really good at math, so I'm really struggling with this equation!
please help!!
T = (5.65-511)√(29+77-2)22x77
√(29x16.5² + 77x18.57²)(29=77)
I think that's typed out right ):
If someone could help me, I would very grateful!!!!
Last edited by mr fantastic; December 13th 2011 at 01:28 AM. Reason: Title.
Re: Help please on T-scores!! URGENT!!
looks like a two sample t statistic
where your n's are off
you need to pool correctly and thats looks like a 29+77-2 below
and 29 should be a 28 and 77 a 76 in the pooled sample variance
BUT do explain where this is coming from
that would make my job easier
Re: Help please on T-scores!! URGENT!!
It was basically a re-creation of Brugger et al's sheep/goat effect
people were asked to iagine a dice rolled 66 times and to read out the numbers they imagined, then were asked to rate their belief in ESP on a scale of 1-6
1-2 are the 'sheep' 5-6 are the 'goats'
We then took the amount of time the participents said a number twice in a row
e.g 6, 2, 3,[COLOR=rgb(0, 0, 0)] 4, 4[/COLOR], 1
number of participants in group 1 = 29
number of participants in group 2 = 77
group 1 mean = 5.65
group 2 mean = 5.11
group 1 SD = 4.23
group 2 SD = 4.3
group 1 varience = 17.94
group 2 varience = 18.57
Thank-you for replying by the way, I've been at this for over 6 hours now ):
Re: Help please on T-scores!! URGENT!!
It's the same as the other post I replied to tonight
It's a s-pooled t situation
December 12th 2011, 10:10 PM #2
December 12th 2011, 11:39 PM #3
Dec 2011
December 12th 2011, 11:46 PM #4
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polar coordinates
August 30th 2008, 12:06 PM
polar coordinates
Determine the main set of polar coordinates of points which are rectangular coordinates:
How do I find the angle?
The answer is:
August 30th 2008, 12:25 PM
Hi Apprentice123,
Given the coordinates $(a,b)$ the angle $(\theta)$ is given by $\arctan\left(\frac{b}{a}\right)$ Thus in this case $\theta=\arctan\left(\frac{-2}{0}\right)$ Note two things. The first is that $\
frac{-2}{0}$ is undefined so we want $\theta$ such that $\tan(\theta)$ is undefined. The second is that the coordinates $(0,-2)$ are in the fourth quadrant and thus so is your angle.
The answer follows.
August 30th 2008, 12:30 PM
* (0, -2)
We see that the angle is $\frac{3\pi}{2}$ away from the polar axis.
August 30th 2008, 03:33 PM
Thanks to all
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Piscataway Prealgebra Tutor
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Finding max clique in perfect graphs
up vote 5 down vote favorite
A fast algorithm to find the size of the largest clique in a perfect graph(this one having odd cycles with at least 1 chord) with about 100 vertices ??
And is there any simpler method than brute force as this is a perfect graph and there should be a polynomial time solution to it. But I am not able to find the algorithm.
Does greedy coloring give optimal coloring in all perfect graphs??
c++ algorithm graph clique
Have you made any attempt? – mvid Jun 11 '10 at 5:11
I hv attempted few approaches but all of them were too slow. – copperhead Jun 11 '10 at 5:20
Just found this in wikipedia: in all perfect graphs, the graph coloring problem, maximum clique problem, and maximum independent set problem can all be solved in polynomial time (Grötschel, Lovász
1 & Schrijver 1988) Grötschel, Martin; Lovász, László; Schrijver, Alexander (1988). Geometric Algorithms and Combinatorial Optimization. Springer-Verlag. See especially chapter 9, "Stable Sets in
Graphs", pp. 273–303. – yogsototh Jun 11 '10 at 5:30
add comment
2 Answers
active oldest votes
See page 296, with some work you should write the right linear programming constraint to solve this problem.
up vote 0 down vote http://www.scribd.com/doc/5710463/Geometric-Algorithms-And-Combinatorial-Optimization
+1: Only answer which addresses perfect graphs, for which the clique problem is in P. – Aryabhatta Jun 11 '10 at 12:22
add comment
100 vertices? Pffft. Brute force it in a few seconds (perhaps fraction of a second) with Cliquer. http://users.tkk.fi/pat/cliquer.html
up vote 3
down vote
Can u explain the algorithm(I have seen the documentation) but in simpler terms – copperhead Jun 11 '10 at 6:56
Sure. First Cliquer defines a permutation of the vertices. I think by default it is in whatever order you used in the input. Secondly, cliquer iterates finding the largest clique in the
1 set [i....n], from i = n-1 to i=1. Along the way it remembers the largest clique it has found so far and when testing for new cliques it prunes the search when it becomes apparent that
from the previously calculated clique sizes it will be impossible for that path of the search to yield a larger clique. – Chad Brewbaker Jun 14 '10 at 7:23
add comment
Not the answer you're looking for? Browse other questions tagged c++ algorithm graph clique or ask your own question.
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Equipotential Contour Theory
1. The problem statement, all variables and given/known data
The lines show equipotential contours in the plane of three point charges, Q1, Q2, and Q3. The positions of the charges are marked by dots. The values of the potentials are in kilovolts as indicated,
e.g., +5 kV, −5 kV; the contour interval is 1 kV. The letters denote locations on the contours. For distances use the scale below the picture, to an appropriate accuracy. Select all the correct
statements, e.g., AB.
A) The electric field at g is zero.
B) Charge Q2 is the largest negative charge.
C) Q1 is a negative charge.
D) The electric field at d is stronger than at b.
E) Charge Q3 is the largest positive charge.
F) The force on an electron at g points to the top of the page.
2. Relevant equations
No equations necessary, theory only.
3. The attempt at a solution
Last Answer: CE
Last Answer: BCE
Last Answer: BCEF
Last Answer: ABCEF
Last Answer: ABCE
Last Answer: ACE
Last Answer: ACEF
I just don't know the theory behind this, I've been tried, failed, and I don't understand. If anyone can explain the concepts and what the right answer is, I'd really appreciate it.
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Patterns Across Cultures: The Fibonacci Sequence in Visual Art
Exploring where the Fibonacci Sequence and Golden Ratio appear in nature and the visual arts.
Key Staff
Primary instructor. If primary instructor is not a visual art teacher, it would be a good idea to enlist the assistance of a visual art expert.
Key Skills
Developing Arts Literacies: Analyzing and Evaluating - Critique
Making Art: Producing, Executing and Performing
The Fibonacci Sequence manifests in nature and visual arts. Through videos, lectures and images, students will learn what the Fibonacci Sequence is, where and how it appears in nature, and its role
in the visual arts in various cultures. Students will also learn about the Golden Mean/Ratio and Golden Spiral, an important concept in art history. Students will identify works of art in which the
Golden Spiral or Golden Ratio appear.
Learning Objectives
Students will:
• Learn about the origin of the Fibonacci Sequence
• Learn about the Golden Mean/Ratio and Golden Triangle, important concepts in art history
• Be able to identify the Fibonacci Sequence in nature and visual arts
Teaching Approach
• Thematic
• Project-Based Learning
• Arts Integration
Teaching Methods
• Discovery Learning
• Discussion
• Experiential Learning
• Reflection
Assessment Type
What You'll Need
Required Technology
• Television
• DVD Player
• Projector
Lesson Setup
Teacher Background
Teachers should familiarize themselves with the definition and examples of the Fibonacci Sequence as it applies to art and the Golden Ratio.
Prior Student Knowledge
Students should have some background in geometry and visual arts.
Physical Space
Since images to be shown to students are on the computer, make sure your computer is connected to an LCD projector and that there is a screen in the room.
Accessibility Notes
Students who are deaf or hard of hearing will need handouts/reading materials to explain the Fibonacci Sequence and Golden Ratio.
1. Write the following sequence of numbers on the board and ask students if they know what it is. Ask them if they can identify the pattern and ask what the next number in the sequence would be.
2. Explain that this sequence is called the Fibonacci Sequence. The first two numbers of the sequence are always 0 and 1. Each following number is the sum of the previous two (so, the next number in
the sequence after 144 would be 233 because it is the sum of 89 and 144).
You could also explain that in mathematical terms, in order to determine a given number F[n] of the sequence, you could use the recurrence relation (F[n]= F[n]-1 + F[n]-2) when F[0]= 0 and F[1]= 1.
3. Describe the history of the Fibonacci sequence. It was named after a Medieval Italian monk (and mathematician) named Leonardo de Pisa (known as Fibonacci --- a combination of filius Bonaccio,
which means “son of Bonaccio” in Italian). The Fibonacci sequence was introduced to Western Europe in his 1202 book, Liber Abaci. Though the Fibonacci sequence is attributed to Leonardo de Pisa, it
also appeared in other cultures in India and Northern Africa before the publication of his book.
4. Explain that the Fibonacci sequence appears in nature and, as a result, has highly influenced visual arts. Using your computer and an LCD projector, show students this video.
5. To reinforce the notion of the Fibonacci sequence in nature, you may want to pull some examples from this site to share with students.
Build Knowledge
1. Explain that artists have always been influenced by what they see in nature. The Fibonacci Sequence is related to the Golden Ratio, a concept that appears in both nature and visual arts.
2. Have students calculate the Golden Ratio with calculators: 1/1, 2/1, 3/2, 5/3, 8/5, 13/8, 21/13, 34/21, 55/34, 89/55 ...They should get 1, 2, 1.5, 1.6666..., 1.6, 1.625, 1.615384615...,
1.619047619..., 1.617647059..., 1.618181818...
Explain that these ratios will eventually reach the Golden Ratio, (√5 + 1)/2, a number approximately equal to 1.6180339887498948482. The Greek letter Phi is used to refer to this ratio. In the
Fibonacci Sequence (0, 1, 1, 2, 3, 5, 8, 13 ...), each term is the sum of the two previous terms (for instance, 2+3=5, 3+5=8 ...). As you go farther and farther to the right in this sequence, the
ratio of a term to the one before it will get closer and closer to the Golden Ratio.
3. Show students the graph showing the relationship between the Fibonacci sequence and golden ratio.
4. A similar concept is the Golden Spiral, a logarithmic spiral whose growth factor is related to the golden ratio.
5. Show students this video of how to construct a Fibonacci Spiral: http://library.thinkquest.org/27890/theSeries6a.html
6. Explain that the Golden Ratio and Golden Spiral have informed the work of artists throughout history. Talk about the ways in which the Fibonacci Sequence informed the work of M.C. Escher. M. C.
Escher went to Alhambra, Spain and was inspired by the tile patterns he saw there to develop his signature style of images repeating and creating themselves (Islamic design). These tile patterns are
also examples of geometric repetition and the Fibonacci Sequence in visual art. Tiles at Alhambra. These designs are based on the Golden Spiral.
For repetition in Escher’s work: http://www.mcescher.com/ (go to “picture gallery” and then “symmetry”)
Recommended Resources
• One calculator for each student
1. Ask students to think about the different representations of the Fibonacci Sequence and Golden Ratio they’ve seen. Ask them to identify pieces of art that use these themes or formulas. Assign them
to find a piece online that corresponds to these ideas.
You may want to assign students specific pieces. Examples of pieces of art that use the Fibonacci Sequence and Golden Ratio can be found at:
http://www.geom.uiuc.edu/~demo5337/s97b/art.htm or http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibInArt.html#daVinci
2. Ask students to prepare a short presentation on the piece they have been asked to research. The presentation should include the history of the piece, how the Fibonacci/Golden Ratio is used, and
information about the artist.
Recommended Resources
1. Have students present their findings about the Fibonacci sequence in an existing work of art. They should have an image of their piece to show the class as well as the information listed above
(history of the piece, relation to Fibonacci Sequence/Golden Ratio, information about the artist).
2. End the lesson with a discussion about now the Fibonacci sequence has influences art across cultures and time periods (use student projects to exemplify this).
Assess the students' final projects. Do they reflect an understanding of the Fibonacci Sequence, Golden Mean and Golden Spiral?
Key Vocabulary
• Fibonacci Sequence
• Nature
• Golden Mean/Ratio
• Golden Spiral
• Visual Art
• Pattern
Extending the Learning
Arrange a field trip to a local art museum. Assign students to research different pieces found in the museum and identify those that use the Fibonacci Sequence or Golden Mean/Ratio. Have students
serve as “guides” and talk about these pieces while you are at the museum.
Engage students in a lesson about engineering and architecture. Identify the presence of the Fibonacci Sequence or Golden Mean/Ratio in processes related to these disciplines.
The Common Core State Standards Initiative seeks to bring diverse state curricula into alignment through a set of common learning goals and assessments. In 2010, Standards were released for English
language arts and mathematics. Common standards have not yet been released for science, social studies, and other subject areas, including the arts. In addition, some states have yet to, or have
chosen not to, adopt the Common Core standards.
During this transitional period, ArtsEdge will present all relevant state and nationals standards as they apply to our lessons.
National Standards for Arts Education
For the full text of the content and achievement standards in Arts Education, visit our Standards section.
Lessons connect to the National Standards for Arts Education, the Common Core Standards, and a range of other subject area standards.
Common Core/State Standards
Select state and grade(s) below, then click "Find" to display Common Core and state standards.
National Standards For Arts Education
Visual Arts
Grade 9-12 Visual Arts Standard 2: Using knowledge of structures and functions
Grade 9-12 Visual Arts Standard 3: Choosing and evaluating a range of subject matter, symbols, and ideas
Grade 9-12 Visual Arts Standard 4: Understanding the visual arts in relation to history and cultures
Grade 9-12 Visual Arts Standard 5: Reflecting upon and assessing the characteristics and merits of their work and the work of others
Grade 9-12 Visual Arts Standard 6: Making connections between visual arts and other disciplines
National Standards in Other Subjects
Science Standard 11: Understands the nature of scientific knowledge
Math Standard 5: Understands and applies basic and advanced properties of the concepts of geometry
Language Arts
Language Arts Standard 1: Uses the general skills and strategies of the writing process
Jen Westmoreland Bouchard
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How to solve a 4x4x4 Rubik's Cube (Part One)
In this Tutorial you'll learn how to solve the 4 by 4 Rubik's Cube.
to go to part two.
1) L'*, 2U, L*
2) R*, 2U, R'*
3) L', R, U, F, U', L
4) D*, R, F', U, R', F, D'*
1) 2U*, 2L*, 2U, 2l, 2U, 2L*, 2U*, R, U', L, 2U, R', U, R, L', U', L, 2U, R', U, L', U
2) 2R*, 2B, 2U, L*, 2U, R'*, 2U, R*, 2U, 2F, R*, 2F, L'*, 2B, 2R*
3) 2r, 2U, 2r, 2U*, 2r, 2u
4) 2U*, 2L*, 2U, 2l, 2U, 2L*, 2U* F', U', F, U, F, R', 2F, U, F, U, F', U', F, R
5) 2r, 2U, 2r, 2U*, 2r, 2u
3 comments:
Thanks for the great tutorial on the 4 by 4 Rubix cube. On to part 2 (:
Thanks a lot for your enthusiastic introduction! I'm also going directly to part 2....
Rubiks kub said...
Thanks for that great tutorial on how to solve the 4x4x4, moving on to part two :)
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Weekly Problem 43 - 2007
Copyright © University of Cambridge. All rights reserved.
'Weekly Problem 43 - 2007' printed from http://nrich.maths.org/
The diagram shows 10 identical coins which fit exactly inside a wooden frame. As a result each coin is prevented from sliding. What is the largest number of coins that may be removed so that each
remaining coin is still unable to slide?
If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.
This problem is taken from the UKMT Mathematical Challenges.
View the previous week's solutionView the current weekly problem
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Proof involving hyberbola
March 31st 2012, 08:41 AM #1
Proof involving hyberbola
Prove that the line $x cos \alpha + y sin \alpha = p$ touches the hyperbola $x^2/a^2 - y^2/b^2 = 1$ if $p^{2} = a^2 cos^2 \alpha - b^2 sin^2 \alpha$.
How do I start?
Edit: Sorry for the error in the question.
Last edited by Don; March 31st 2012 at 07:17 PM.
Re: Proof involving hyberbola
Check the wording of the question. What you've written isn't a line.
Re: Proof involving hyberbola
Corrected the question.
Re: Proof involving hyberbola
Again have you copied the question correctly. That is an ellipse not a hyperbola.
Re: Proof involving hyberbola
biffboy, thanks for your patience. Error again and corrected again.
Re: Proof involving hyberbola
Make x the subject of the line equation and substitute this into the hyperbola equation. Rearrange this as a quadratic in y. For the line to be a tangent we require this quadratic to have
repeated roots. Applying this condition should produce the result stated.
Re: Proof involving hyberbola
So, do I substitute the value of y in the line equation with the value of y with respect to x?
In that case, the equation becomes,
So the line equation will become
I tried to simplify the equation and it is becoming messy. Am I doing things right?
(Sorry for the small image, laTEX is pain and I do not know how to put the special characters here. Tried to use an online latex editor but there seems to be some problem)
Re: Proof involving hyberbola
Replace y in the hyperbola equation with (p-xcosa)/sina
Re: Proof involving hyberbola
$(p - ysin\alpha/cos\alpha )^2/a^2 - y^2/b^2 = 1$
$(p^2 + y^2sin^2\alpha - 2pysin\alpha)/a^2 cos^2 \alpha - y^2/b^2 = 1$
$(p^2 + y^2sin^2\alpha - 2pysin\alpha)/a^2 cos^2 \alpha = 1 + y^2/b^2$
$(p^2 + y^2sin^2\alpha - 2pysin\alpha)b^2 = (b^2 + y^2)(a^2 cos^2 \alpha)$
This doesn't seem to be leading anywhere. Is this what I am supposed to do?
Re: Proof involving hyberbola
Rearrange your last line to get a quadratic in y. You then want this quadratic to have a repeated root.
Re: Proof involving hyberbola
What you have substituted for x should be p/cosa-(sina/cosa)y
Re: Proof involving hyberbola
So I got the equation
$\left (p - y sin\alpha \right )^2 = (a^2+b^2)(a^2cos^2\alpha )/b^2$
Is this correct?
Re: Proof involving hyberbola
I can't see what's happened there. Looking back your last line of post 9 was correct so look at my post 10 again
Re: Proof involving hyberbola
If you want one last fresh start try this. I'll write A instead of alpha. Line can be written y=-(cosA/sinA)x+p/sinA
This is y=mx+c with m=-cosA/sinA and c=p/sinA
Substitute y=mx+c into the hyperbola. So x^2/a^2-(mx+c)^2/b^2=1
Multiply throughout by a^2b^2 and remove the brackets and then rearrange as a quadratic in x
Apply the condition for this to have repeated roots
After some symplifying this gives b^2+c^2-(a^2m^2)=0
Put back in what m and c stood for to get the required result
Re: Proof involving hyberbola
OK, here's my attempt:
$x^2/a^2 - (mx + c)^2/b^2 = 1$
Multiplying both the terms by a^2b^2,
$x^2b^2 - (m^2x^2 + c^2 + 2mxc)a^2 = 1$
Rearranging, I get
$x^2(b^2 - m^2a^2) + x(-2mca^2) + (-a^2c^2-1) = 0$
This is of the form ax^2+bx+c=0
What do I do now? I don't know how to make it have repeated roots.
I hope I am on the right track.
March 31st 2012, 09:16 AM #2
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March 31st 2012, 09:22 AM #3
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March 31st 2012, 07:18 PM #5
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April 1st 2012, 08:01 AM #7
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April 1st 2012, 09:23 AM #9
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April 4th 2012, 08:46 AM #15
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Overview Package Class Tree Deprecated Index Help
PREV CLASS NEXT CLASS FRAMES NO FRAMES
SUMMARY: NESTED | FIELD | CONSTR | METHOD DETAIL: FIELD | CONSTR | METHOD
Class Dtrsen
public class Dtrsen
extends java.lang.Object
Following is the description from the original
Fortran source. For each array argument, the Java
version will include an integer offset parameter, so
the arguments may not match the description exactly.
Contact seymour@cs.utk.edu with any questions.
* ..
* Purpose
* =======
* DTRSEN reorders the real Schur factorization of a real matrix
* A = Q*T*Q**T, so that a selected cluster of eigenvalues appears in
* the leading diagonal blocks of the upper quasi-triangular matrix T,
* and the leading columns of Q form an orthonormal basis of the
* corresponding right invariant subspace.
* Optionally the routine computes the reciprocal condition numbers of
* the cluster of eigenvalues and/or the invariant subspace.
* T must be in Schur canonical form (as returned by DHSEQR), that is,
* block upper triangular with 1-by-1 and 2-by-2 diagonal blocks; each
* 2-by-2 diagonal block has its diagonal elemnts equal and its
* off-diagonal elements of opposite sign.
* Arguments
* =========
* JOB (input) CHARACTER*1
* Specifies whether condition numbers are required for the
* cluster of eigenvalues (S) or the invariant subspace (SEP):
* = 'N': none;
* = 'E': for eigenvalues only (S);
* = 'V': for invariant subspace only (SEP);
* = 'B': for both eigenvalues and invariant subspace (S and
* SEP).
* COMPQ (input) CHARACTER*1
* = 'V': update the matrix Q of Schur vectors;
* = 'N': do not update Q.
* SELECT (input) LOGICAL array, dimension (N)
* SELECT specifies the eigenvalues in the selected cluster. To
* select a real eigenvalue w(j), SELECT(j) must be set to
* .TRUE.. To select a complex conjugate pair of eigenvalues
* w(j) and w(j+1), corresponding to a 2-by-2 diagonal block,
* either SELECT(j) or SELECT(j+1) or both must be set to
* .TRUE.; a complex conjugate pair of eigenvalues must be
* either both included in the cluster or both excluded.
* N (input) INTEGER
* The order of the matrix T. N >= 0.
* T (input/output) DOUBLE PRECISION array, dimension (LDT,N)
* On entry, the upper quasi-triangular matrix T, in Schur
* canonical form.
* On exit, T is overwritten by the reordered matrix T, again in
* Schur canonical form, with the selected eigenvalues in the
* leading diagonal blocks.
* LDT (input) INTEGER
* The leading dimension of the array T. LDT >= max(1,N).
* Q (input/output) DOUBLE PRECISION array, dimension (LDQ,N)
* On entry, if COMPQ = 'V', the matrix Q of Schur vectors.
* On exit, if COMPQ = 'V', Q has been postmultiplied by the
* orthogonal transformation matrix which reorders T; the
* leading M columns of Q form an orthonormal basis for the
* specified invariant subspace.
* If COMPQ = 'N', Q is not referenced.
* LDQ (input) INTEGER
* The leading dimension of the array Q.
* LDQ >= 1; and if COMPQ = 'V', LDQ >= N.
* WR (output) DOUBLE PRECISION array, dimension (N)
* WI (output) DOUBLE PRECISION array, dimension (N)
* The real and imaginary parts, respectively, of the reordered
* eigenvalues of T. The eigenvalues are stored in the same
* order as on the diagonal of T, with WR(i) = T(i,i) and, if
* T(i:i+1,i:i+1) is a 2-by-2 diagonal block, WI(i) > 0 and
* WI(i+1) = -WI(i). Note that if a complex eigenvalue is
* sufficiently ill-conditioned, then its value may differ
* significantly from its value before reordering.
* M (output) INTEGER
* The dimension of the specified invariant subspace.
* 0 < = M <= N.
* S (output) DOUBLE PRECISION
* If JOB = 'E' or 'B', S is a lower bound on the reciprocal
* condition number for the selected cluster of eigenvalues.
* S cannot underestimate the true reciprocal condition number
* by more than a factor of sqrt(N). If M = 0 or N, S = 1.
* If JOB = 'N' or 'V', S is not referenced.
* SEP (output) DOUBLE PRECISION
* If JOB = 'V' or 'B', SEP is the estimated reciprocal
* condition number of the specified invariant subspace. If
* M = 0 or N, SEP = norm(T).
* If JOB = 'N' or 'E', SEP is not referenced.
* WORK (workspace/output) DOUBLE PRECISION array, dimension (LWORK)
* On exit, if INFO = 0, WORK(1) returns the optimal LWORK.
* LWORK (input) INTEGER
* The dimension of the array WORK.
* If JOB = 'N', LWORK >= max(1,N);
* if JOB = 'E', LWORK >= M*(N-M);
* if JOB = 'V' or 'B', LWORK >= 2*M*(N-M).
* If LWORK = -1, then a workspace query is assumed; the routine
* only calculates the optimal size of the WORK array, returns
* this value as the first entry of the WORK array, and no error
* message related to LWORK is issued by XERBLA.
* IWORK (workspace) INTEGER array, dimension (LIWORK)
* IF JOB = 'N' or 'E', IWORK is not referenced.
* LIWORK (input) INTEGER
* The dimension of the array IWORK.
* If JOB = 'N' or 'E', LIWORK >= 1;
* if JOB = 'V' or 'B', LIWORK >= M*(N-M).
* If LIWORK = -1, then a workspace query is assumed; the
* routine only calculates the optimal size of the IWORK array,
* returns this value as the first entry of the IWORK array, and
* no error message related to LIWORK is issued by XERBLA.
* INFO (output) INTEGER
* = 0: successful exit
* < 0: if INFO = -i, the i-th argument had an illegal value
* = 1: reordering of T failed because some eigenvalues are too
* close to separate (the problem is very ill-conditioned);
* T may have been partially reordered, and WR and WI
* contain the eigenvalues in the same order as in T; S and
* SEP (if requested) are set to zero.
* Further Details
* ===============
* DTRSEN first collects the selected eigenvalues by computing an
* orthogonal transformation Z to move them to the top left corner of T.
* In other words, the selected eigenvalues are the eigenvalues of T11
* in:
* Z'*T*Z = ( T11 T12 ) n1
* ( 0 T22 ) n2
* n1 n2
* where N = n1+n2 and Z' means the transpose of Z. The first n1 columns
* of Z span the specified invariant subspace of T.
* If T has been obtained from the real Schur factorization of a matrix
* A = Q*T*Q', then the reordered real Schur factorization of A is given
* by A = (Q*Z)*(Z'*T*Z)*(Q*Z)', and the first n1 columns of Q*Z span
* the corresponding invariant subspace of A.
* The reciprocal condition number of the average of the eigenvalues of
* T11 may be returned in S. S lies between 0 (very badly conditioned)
* and 1 (very well conditioned). It is computed as follows. First we
* compute R so that
* P = ( I R ) n1
* ( 0 0 ) n2
* n1 n2
* is the projector on the invariant subspace associated with T11.
* R is the solution of the Sylvester equation:
* T11*R - R*T22 = T12.
* Let F-norm(M) denote the Frobenius-norm of M and 2-norm(M) denote
* the two-norm of M. Then S is computed as the lower bound
* (1 + F-norm(R)**2)**(-1/2)
* on the reciprocal of 2-norm(P), the true reciprocal condition number.
* S cannot underestimate 1 / 2-norm(P) by more than a factor of
* sqrt(N).
* An approximate error bound for the computed average of the
* eigenvalues of T11 is
* EPS * norm(T) / S
* where EPS is the machine precision.
* The reciprocal condition number of the right invariant subspace
* spanned by the first n1 columns of Z (or of Q*Z) is returned in SEP.
* SEP is defined as the separation of T11 and T22:
* sep( T11, T22 ) = sigma-min( C )
* where sigma-min(C) is the smallest singular value of the
* n1*n2-by-n1*n2 matrix
* C = kprod( I(n2), T11 ) - kprod( transpose(T22), I(n1) )
* I(m) is an m by m identity matrix, and kprod denotes the Kronecker
* product. We estimate sigma-min(C) by the reciprocal of an estimate of
* the 1-norm of inverse(C). The true reciprocal 1-norm of inverse(C)
* cannot differ from sigma-min(C) by more than a factor of sqrt(n1*n2).
* When SEP is small, small changes in T can cause large changes in
* the invariant subspace. An approximate bound on the maximum angular
* error in the computed right invariant subspace is
* EPS * norm(T) / SEP
* =====================================================================
* .. Parameters ..
│ Method Summary │
│ static │ dtrsen(java.lang.String job, java.lang.String compq, boolean[] select, int _select_offset, int n, double[] t, int _t_offset, int ldt, double[] q, int _q_offset, int ldq, double[] wr, int │
│ void │ _wr_offset, double[] wi, int _wi_offset, intW m, doubleW s, doubleW sep, double[] work, int _work_offset, int lwork, int[] iwork, int _iwork_offset, int liwork, intW info) │
│ Methods inherited from class java.lang.Object │
│ clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait │
public Dtrsen()
public static void dtrsen(java.lang.String job,
java.lang.String compq,
boolean[] select,
int _select_offset,
int n,
double[] t,
int _t_offset,
int ldt,
double[] q,
int _q_offset,
int ldq,
double[] wr,
int _wr_offset,
double[] wi,
int _wi_offset,
intW m,
doubleW s,
doubleW sep,
double[] work,
int _work_offset,
int lwork,
int[] iwork,
int _iwork_offset,
int liwork,
intW info)
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Topic: Some where equidistant
Replies: 5 Last Post: Jul 16, 2013 4:29 AM
Messages: [ Previous | Next ]
Re: Some where equidistant
Posted: Jul 13, 2013 12:10 PM
On Fri, 12 Jul 2013 21:05:15 -0700, William Elliot <marsh@panix.com>
>Let (S,d) be a compact connected metric space and
>f:S -> S a homeomorphism. How would one show
>there's some distinct x,y with d(f(x),f(y)) = d(x,y)?
You still haven't given a correct statement of the
problem! One of the counterexamples quasi
mentioned is a counterexample to this version
as well.
Date Subject Author
7/13/13 William Elliot
7/13/13 Re: Some where equidistant David C. Ullrich
7/13/13 Re: Some where equidistant William Elliot
7/14/13 Re: Some where equidistant David C. Ullrich
7/13/13 Re: Some where equidistant quasi
7/16/13 Re: Some where equidistant William Elliot
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Graph Theory - To Euler or not to Euler? With attempted working.
February 20th 2010, 08:26 AM
Graph Theory - To Euler or not to Euler? With attempted working.
I'm having a bit of difficulty with this question. I have posted part of my working (though it may be completely wrong) below. I'm assuming I need to use Euler's formula.
Find an upper bound on the number of edges of a connected simple plane graph with number of vertices - n (greater than or equal to) 3 and girth - g (greater than or equal to) 3. Use this to show
that the Petersen graph is non-planar.
Attempted Answer
m= no. of edges
n= no. of vertices
L= no. of faces
From the question we know that each face is bounded by at least 3 edges and each edge seperates 2 faces.
Therefore: m (is greater than or equal to) 3L/2
Combining with Euler's formula: n - m + L = 2
We get m (is less than or equal to) 3n - 6.
This is what I believe is the upper bound. By plugging in the values for the Petersen graph (n=10, m=15) I was hoping to get a false inequality to show that the graph is in fact non-planar.
However the petersen graph passes this 'test'.
So I was wondering if my initial upperbound is wrong, or if that is not the correct way to prove the petersen graph is non-planar.
Any help would be appreciated. Thanks
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
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Graphing Quadratic Functions (vertex formula)
I am having a really hard time with this section. My teacher will be absent tomorrow and I have a test on Monday. I don't understand this. So, for practice could anyone help me out and show me step
by step how to do some problems?
In my book, it says:
"use the Vertex Form: y - k = a (x - h)^2
Write an equation of the parabola having V as the vertex and containing P."
V(3, 5) ; P(4,7)
V(-1 ,1) ; P(1, 3)
I don't understand what to do. Do I just plug these in to the above formula and that's the answer or do I solve to get a = something?
You are given the formula $y\, =\, a(x\, -\,h)^2\, +\, k,$ and are given values for x, y, h, and k. Plug these values in, and solve for "a".
Then rewrite the formula, plugging in the values for h, k, and a.
Re: Graphing Quadratic Functions (vertex formula)
Thanks stapel for the simple explanation.
Here's the work I did for the 1st example.
y = a(x - h)^2 + k
7 = a(4 - 3)^2 + 5
7 = a(1)^2 +5
7= a + 5
2 = a
And the final answer is: y = 2(x-3)^2 + 5
For the second one I got 1/2 = a
and the final answer/equation is: y = 1/2(x - (-1))^2 + 1
Are these correct ?
The only adjustment I would recommend is simplifying inside the parentheses in the second one, to get:
$y\, =\, \frac{1}{2}\left(x\, +\, 1)^2\, + \,1$
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the first resource for mathematics
New classes of exact solutions to general nonlinear equations and systems of equations in mathematical physics.
(English) Zbl 1217.35166
From the text: New classes of exact solutions to some general nonlinear equations and systems of equations of mathematical physics are described that involve arbitrary functions. Special attention is
given to equations and systems of equations encountered in the theory of mass and heat transfer, mathematical biology, and wave theory.
In this paper, an exact solution is regarded in the sense of the definition given in [A. D. Polyanin, V. F. Zaitsev and A. I. Zhurov, Methods for solving nonlinear equations in mathematical physics
and mechanics. Fizmatlit, Mowcow, (2005)], page 10.
35Q53 KdV-like (Korteweg-de Vries) equations
35Q79 PDEs in connection with classical thermodynamics and heat transfer
35Q92 PDEs in connection with biology and other natural sciences
35C05 Solutions of PDE in closed form
80A20 Heat and mass transfer, heat flow
35A24 Methods of ordinary differential equations for PDE
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To begin, we load a graph.
load ../graphs/clr-25-2.mat
Next, let's check the documentation to see which functions to implement for the visitor
help dijkstra_sp
DIJKSTRA_SP Compute the weighted single source shortest path problem.
Dijkstra's algorithm for the single source shortest path problem only
works on graphs without negative edge weights.
This method works on weighted directed graphs without negative edge
The runtime is O(V log (V)).
See the shortest_paths function for calling information. This function
just calls shortest_paths(...,struct('algname','dijkstra'));
The options structure can contain a visitor for the Dijkstra algorithm.
See http://www.boost.org/libs/graph/doc/DijkstraVisitor.html for a
description of the events.
visitor is a struct with the following optional fields
Each visitor parameter should be a function pointer, which returns 0
if the shortest path search should stop. (If the function does not
return anything, the algorithm continues.)
load graphs/clr-25-2.mat
See also SHORTEST_PATHS, BELLMAN_FORD_SP.
The help states that dijkstra_sp allows visitors functions for initialize_vertex, discover_vertex, examine_vertex, examine_edge, edge_relaxed, edge_not_relaxed, and finish_vertex.
Rather than implementing 7 functions ourselves, we define two helper functions. These helper functions return functions themselves. There is one helper that returns a vertex visitor function and one
helper than returns an edge visitor function.
vertex_vis_print_func = @(str) @(u) ...
fprintf('%s called on %s\n', str, char(labels{u}));
edge_vis_print_func = @(str) @(ei,u,v) ...
fprintf('%s called on (%s,%s)\n', str, char(labels{u}), char(labels{v}));
These anonymous functions return functions themselves.
ev_func = vertex_vis_print_func('examine_vertex');
examine_vertex called on s
I hope you see how these functions are useful in saving quite a bit of typing.
Calling dijkstra_sp
We are almost done. Now, we just have to setup the visitor structure to pass to the dijkstra_sp call.
vis = struct();
vis.initialize_vertex = vertex_vis_print_func('initialize_vertex');
vis.discover_vertex = vertex_vis_print_func('discover_vertex');
vis.examine_vertex = vertex_vis_print_func('examine_vertex');
vis.finish_vertex = vertex_vis_print_func('finish_vertex');
vis.examine_edge = edge_vis_print_func('examine_edge');
vis.edge_relaxed = edge_vis_print_func('edge_relaxed');
vis.edge_not_relaxed = edge_vis_print_func('edge_not_relaxed');
With the visitor setup, there is hardly any work left.
dijkstra_sp(A,1,struct('visitor', vis));
initialize_vertex called on s
initialize_vertex called on u
initialize_vertex called on x
initialize_vertex called on v
initialize_vertex called on y
discover_vertex called on s
examine_vertex called on s
examine_edge called on (s,u)
edge_relaxed called on (s,u)
discover_vertex called on u
examine_edge called on (s,x)
edge_relaxed called on (s,x)
discover_vertex called on x
finish_vertex called on s
examine_vertex called on u
examine_edge called on (u,x)
edge_not_relaxed called on (u,x)
examine_edge called on (u,v)
edge_relaxed called on (u,v)
discover_vertex called on v
finish_vertex called on u
examine_vertex called on x
examine_edge called on (x,u)
examine_edge called on (x,v)
edge_not_relaxed called on (x,v)
examine_edge called on (x,y)
edge_relaxed called on (x,y)
discover_vertex called on y
finish_vertex called on x
examine_vertex called on v
examine_edge called on (v,y)
edge_not_relaxed called on (v,y)
finish_vertex called on v
examine_vertex called on y
examine_edge called on (y,s)
examine_edge called on (y,v)
finish_vertex called on y
Understanding the output
To understand the output, we find it helpful to have a copy of Introduction to Algorithms by Cormen, Leiserson, and Rivest. The source for the graph is Figure 25-2 in that book and the authors use
the graph to illustrate how Dijkstra's algorithm runs. In particular, Figure 25-5 shows a sample run of Dijkstra's algorithm.
Perhaps the first thing to notice is that the initialize vertex visitor is never called. This results from an error in the MatlabBGL and Boost documentation. Once it is resolved, we will update the
MatlabBGL documentation to match the Boost graph library.
The results: discover_vertex is called before examine_vertex. For the edges, examine_edge is always called before either edge_relaxed or edge_not_relaxed. The edges that are relaxed are the shaded
edges in Figure 25-5.
Finally, finish vertex is called on a vertex after all of its edges have been examined and possibly relaxed.
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Probability problem.
Re: Probability problem.
N numbers are randomly chosen from U(0,1). What's the value of
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
Re: Probability problem.
Hi gAr;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Probability problem.
Hi bobbym,
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
Re: Probability problem.
Hi gAr;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Probability problem.
Hi bobbym,
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
Re: Probability problem.
Hi gAr;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Probability problem.
Hi bobbym,
Logging out, see you later..
Last edited by gAr (2014-01-06 05:31:17)
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
Re: Probability problem.
Hi gAr;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Probability problem.
Hi bobbym,
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
Re: Probability problem.
Hi gAr;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Probability problem.
Hi bobbym,
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
Re: Probability problem.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Probability problem.
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
Re: Probability problem.
Hi gAr;
Until you mentioned it in the post #276, I did not even think about looking for a formula.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Probability problem.
Okay, now we know!
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
Re: Probability problem.
Are we done with this topic?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Probability problem.
I think so, completed most of the things I had in mind.
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
Re: Probability problem.
I had never done any problems of this type before and did not have any idea of how to solve one, even with a CAS. I have a little idea of how to do them now so I say thanks!
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Probability problem.
Me neither, thanks to wikipedia and mathworld for the formulas.
But was surprised that I could find no numerical examples anywhere. What will anyone do with the formula if the usage documentation is not there?!
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
Re: Probability problem.
I could not find any book even that would give a decent explanation or example.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Probability problem.
New problem:
A certain pattern with a probability of occurrence 'p' has to occur 'k' consecutive times. What is the average number of trials required?
E.g. When 2 dice are thrown, the average number of throws required to get '66' three successive times.
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
Re: Probability problem.
Hi gAr;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Probability problem.
Hi bobbym,
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
Re: Probability problem.
Hi gAr;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Probability problem.
Hi bobbym,
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
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|
The rank of a class elliptic curves
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MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.
For elliptic curve $y^{2}=x(x+a^{2})(x+(1-a)^{2})$,($a$ is a rational number and does not equal 0,1,1/2),is its rank always 0?
Without further explanation or motivation, this looks to me like a homework exercise. Perhaps math.stackexchange.com would be a better venue for this question.
David Roberts May 3 '11 at 2:25
The elliptic curves in my question,is related to a number theory problem I considered.If for some elliptic curves like these,ranks are not zero,the infomation given by the rational points on the
elliptic curves of this kind is not so useful as to the problem I considered. I have the impression that to determine the ranks of elliptic curves is very hard,even for some specific elliptic
curve.I wonder if someone has considered similar questions for certain class of elliptic curves. Thanks to comments made by David Roberts.
leiweixi May 3 '11 at 6:12
For $a=7$, the rank is nonzero, as it is for $a=31$. A third example is $a=751$.
Junkie May 3 '11 at 13:02
Thanks a lot.Junkie have just give me a negative answer.
leiweixi May 3 '11 at 13:41
Junkie, your comment is definitely good enough to go in the answer box.
S. Carnahan♦ May 3 '11 at 14:58
show 1 more comment
Your family of elliptic curves are exactly those elliptic curves with torsion subgroup containing $\Bbb Z / 2 \Bbb Z \times \Bbb Z / 4 \Bbb Z$. To see this make the linear change of
variables $t = a/(1-a)$ and look at the parameterizations on page 101 of Husemöller's book "Elliptic Curves".
up vote 16 So in particular there are values of $a$ for which your elliptic curve has Mordell-Weil rank 8, the first of which was discovered by Elkies in 2005. See Dujella's website. Moreover,
down vote Eroshkin found that there are infinitely many such curves with rank at least 5.
These happen to be my favorite elliptic curves. I'd be very interested to know why you were interested in them.
add comment
Although Junkie has answered the question, I'd like to point out that in the case of parametrized families of elliptic curves (such as this) it is often easy to find an explicit subfamily
with positive rank. In the present case, let us take $x=2a^2$ and see what condition on $a$ forces this to give a point on the elliptic curve. Making this substitution we obtain. $$ y^2=6 a^4
up vote (3 a^2-2a+1). $$ We can simplify by defining $b=y/6 a^2$. Thus $$ 6b^2=3 a^2-2a+1. $$ This is a conic with the point $(a,b)=(-1,1)$ and so we can parametrize all the solutions: $$ a=(-t^2 +
15 down 4t - 10/3)/(t^2 - 2), \qquad b=(t^2 - 8t/3 + 2)/(t^2 - 2) $$ where $t$ is rational. The argument gives that $$ (x,y)=(2a^2,6a^2 b)$$ is a rational point on the elliptic curve provided $a$,
vote $b$ have the above shape. It should be possible through a slightly tedious computer algebra calculation to determine all rational numbers $t$ where the point above is torsion (using Mazur's
Theorem), and for all other values of $t$ the rank is positive.
How did it occur to you to take x = 2a^2?
KConrad May 4 '11 at 7:17
Actually, we can set $x = ka^2$ for more or less any $k$. This will give $$y^2 = k(k+1)a^4 ((k+1)a^2 - 2a + 1)$$. After setting $y = k(k+1)b$ you get $$ k(k+1)b^2 = (k+1)a^2 - 2a +1. $$
Multiply by $k+1$ and set $b = c/(k+1)$ to get $$ k c^2 = (k+1)^2 a^2 - 2(k+1)a + (k+1) = ((k+1)a - 1)^2 + k. $$ and setting $a = 1 + kd/(k+1)$ you get $$ c^2 = kd^2 + 1 $$ which has one,
and therefore infinitely many rational points. Most of them will give non-torsion points on the original elliptic curve. I can't read Samir's mind, but setting x = ka^2 seems like an easy
way to get rid of the $x(x+a^2)$.
Abhinav Kumar May 5 '11 at 4:01
add comment
I'm probably going out on a limb here, but I think that the following conjecture is reasonable.
Conjecture. Let $E:y^2=x^3+a(T)x+b(T)$ be an elliptic curve defined over $\mathbb{Q}(T)$ with the property that its $j$-invariant is not in $\mathbb{Q}$. Then there are infinitely many
rational numbers $t$ such that $E_t(\mathbb{Q})$ has positive rank.
up vote
15 down Of course, I have no idea how one might prove this. As Samir pointed out, in some cases one can find a covering $C\to \mathbb{P}^1$ such that the pullback of $E$ has positive rank over $C$
vote and such that $C$ has genus 0 (or 1) with infinitely many rational points. But in general this won't be possible.
One final note. It's possible to create families for which the sign of the functional equation is always even, in which case the fibers have even rank (if we believe Birch-Swinnerton-Dyer).
But even in this case, I'd expect infinitely many fibers to have rank 2 (or greater).
I wonder if your conjecture follows from the 'rank distribution conjecture' (due to Goldfeld, Katz-Sarnak?) that (on average) half of all elliptic curves have rank 0, and half have rank 1.
I've just taken a look at a preprint of Bhargava, and one of his theorems is that, assuming finiteness of sha, a positive proportion of all elliptic curves have rank 1. I would think that
such a result could be used to prove that a positive proportion of curves in a family (like in your conjecture) have positive rank.
Barinder Banwait May 9 '11 at 9:07
@Joe Silverman: What is the expected rank distribution in a family with constant parity? I guess the minimalist philosophy would say 100% of specializations $t$ would have either rank $r$
or rank $r+1$, depending on which parity is obtained where $r$ is the Mordell-Weil rank of $E$ over $\Bbb Q(T)$. Are there examples of these constant parity families over $\Bbb Q$? I know
of twist families over number fields, the so called "lawful curves" of Dokchitser & Dokchitser, but not of any over $\Bbb Q$. If so could you point me to an example?
Jamie Weigandt May 10 '11 at 6:01
@Jamie Weigandt: I agree that the minimalist philosophy is reasonable. I don't know a reference for constant parity families over $\mathbb{Q}$, but the following paper says that for any
open interval $I$ in $[-1,1]$, there is a twist families such that the average rank lies in $I$. MR1641093 Rizzo, Ottavio G., Average root numbers in families of elliptic curves. Proc.
Amer. Math. Soc. 127 (1999), no. 6, 1597–1603.
Joe Silverman May 10 '11 at 13:15
Rohrlich's paper for twist families is: archive.numdam.org/article/CM_1993__87_2_119_0.pdf He quotes (0.5) Cassels and Schinzel for $7(1+t^4)y^2=x^3-x$, the idea resting on Stephens, that
the 2-selmer rank is odd for twists by 6,7 (mod 8) of the congruent number curve. blms.oxfordjournals.org/content/14/4/345.full.pdf
Junkie May 11 '11 at 22:31
add comment
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Ph.D. Candidate
My research area is applied probability. Currently I am designing and investigating new simulation optimization algorithms for continuous parameter spaces. In general I am interested in various
problems involving probability theory, such as Monte-Carlo methods, risk management and information engineering. In addition I am interested in high-frequency trading algorithms and market
microstructure theory.
My Ph.D. advisors are Professor Shane G. Henderson and Professor Peter I. Frazier.
I received a Bachelor and Master degree in Mathematics from ETH Zurich. My master thesis was on multivariate extreme value theory and was advised by Professor Paul Embrechts.
Publications and Working Papers
Journals and Refereed Conference Proceedings
• Stoikov, S., and Waeber, R. (2012)
Optimal Asset Liquidation Using Limit Order Book Information
submitted. [available at SSRN]
• Waeber, R., Frazier, P. I, and Henderson, S. G. (2012)
Bisection Search with Noisy Responses
• Waeber, R., Frazier, P. I, and Henderson, S. G. (2011)
A Bayesian Approach to Stochastic Root-Finding
Proceedings of the 2011 Winter Simulation Conference. [pdf]
*1st Place, WSC '11 Best Student Paper Competition.
• Waeber, R., Frazier, P. I, and Henderson, S. G. (2012)
A Framework for Selecting a Selection Procedure
ACM Transactions on Modeling and Computer Simulation (TOMACS) 22(3):16. [pdf]
• Waeber, R., Frazier, P. I, and Henderson, S. G. (2010)
Performance Measures for Ranking and Selection Procedures
Proceedings of the 2010 Winter Simulation Conference. [pdf]
• Lysenko, N., Roy, P., and Waeber R., (2009)
Multivariate Extremes of Generalized Skew-Normal Distributions
Statistics and Probability Letters, 79/4, 525-533. [pdf]
Working Papers
• Waeber, R., Frazier, P. I., and Henderson, S. G.
Probabilistic Bisection and Stochastic Root-Finding
• Cheridito, P., Frazier, P. I., Stadje, M., and Waeber R.
Dynamic Programming with Risk Constraint
• Waeber, R., and Weber, S.
Estimation of OCE Risk Measures with Liquidity Adjustments
• Master Thesis, ETH Zurich, Switzerland, 2008
Advisor: Professor Paul Embrechts
Multivariate Skew-Normal Distributions and their Extremal Properties [pdf]
• Bachelor Thesis, ETH Zurich, Switzerland, 2006
Advisor: Professor Philipp J. Schönbucher
Bewerten von Rainbow-Optionen: Ein Dualitätsansatz
(Valuation of Rainbow Options: A Duality Approach) [pdf] (available only in German)
Curriculum Vitae
CV is available upon request.
Contact Information
294 Rhodes Hall
Cornell University
Ithaca, NY, 14853
rw339 at cornell.edu
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Summary: MONODROMY GROUPS OF HURWITZ-TYPE
Abstract. We solve the Hurwitz monodromy problem for degree
4 covers. That is, the Hurwitz space H4,g of all simply branched
covers of P1
of degree 4 and genus g is an unramified cover of
the space P2g+6 of (2g + 6)-tuples of distinct points in P1
. We
determine the monodromy of 1(P2g+6) on the points of the fiber.
This turns out to be the same problem as the action of 1(P2g+6)
on a certain local system of Z/2-vector spaces. We generalize our
result by treating the analogous local system with Z/N coefficients,
3 N, in place of Z/2. This in turn allows us to answer a question
of Ellenberg concerning families of Galois covers of P1
with deck
group (Z/N)2
A ramified cover C of P1
of degree d is said to have simple branching if
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"Zapiski Nauchnyh Seminarov POMI"
VOL. 279
This issue is entitled "Geometry and Topology. Part 6"
editors S. V. Buyalo and N. Yu. Netsvetaev
• Aleksandr Aleksandrovich Ivanov (to the eightieth anniversary) .......7 (.ps.gz)
• Akhmet'ev P. M. A higher-order analog of the helicity number for a pair of divergent-free vector fields .......15 (.ps.gz)
• Berestovskii V. N., Gichev V. M. Metrized semigroups .......24 (.ps.gz)
• Duzhin S. V. Infinitesimal classification of systems of two PDE of first order .......61 (.ps.gz)
• Zhubr A. V. On a paper by Barden .......70 (.ps.gz)
• Ivanov A. A. Metric axioms of Euclidean space .......89 (.ps.gz)
• Ivanov A. O., Hong Van Le, Tuzhilin A. A. Planar Manhattan local minimal and critical networks .......111 (.ps.gz)
• Ivanov D. V. Totally geodesic subsets in the manifold of directions of physical space .......141 (.ps.gz)
• Itenberg V. S., Itenberg I. V. Symmetric sextics and auxiliary conics .......154 (.ps.gz)
• Karpunin G. A. Universal boundary sets in the generalized Steiner problem .......168 (.ps.gz)
• Makeev V. V. On inscribing a regular octahedron in a three-dimensional convex body with smooth boundary .......183 (.ps.gz)
• Makeev V. V. Equipartition of a mass continuously distributed on a sphere or in space.......187 (.ps.gz)
• Malyutin A. V. Fast algorithms for identification and comparison of braids.......197 (.ps.gz)
• Svetlov P. V. Geometrization of the mapping tori of Dehn twists on an infinite genus surface.......218 (.ps.gz)
• Fel'shtyn A. L. The Reidemeister number of any automorphism of a Gromov hyperbolic group is infinite .......229 (.ps.gz)
• Finashin S. M. An integral formula for the complex intersection number of real cycles in a real algebraic variety with topologically rational singularities .......241 (.ps.gz)
• Reviews .......246 (.ps.gz)
• Paging: 251 pp.
• Language: Russian Back to the Petersburg Department of Steklov Institute of Mathematics
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vector projection
January 15th 2008, 11:04 AM #1
Nov 2005
vector projection
[quote]Let T be the triangle with vertices at (4,10), (−8,8),(3,8). The area of T is = ? [quote]
i used the projection forumla and got the length of 1 side as: 4.2426 (not sure if that is right). should i do the same thing to find the other sides?
[QUOTE=viet;98972][quote]Let T be the triangle with vertices at (4,10), (−8,8),(3,8). The area of T is = ?
i used the projection forumla and got the length of 1 side as: 4.2426 (not sure if that is right). should i do the same thing to find the other sides?
You can get the area of the triangle directly using Heron's formula.
It is really easy we recall a basic formula:
$A = \frac{1}{2}a \cdot b\sin \left( \phi \right) = \frac{1}{2}\left\| {a \times b} \right\|.$
That is one-half the dot product of the two vectors forming two sides of the triangle times the sine of the angle between. But sine is related to the cross product.
January 15th 2008, 11:08 AM #2
January 15th 2008, 12:11 PM #3
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Comment on ``On Discriminative vs. Generative Classifiers: A Comparison of Logistic Regression and Naive Bayes"
Jinghao Xue and Mike Titterington
Neural Processing Letters 2007.
Comparison of generative and discriminative classifiers is an ever-lasting topic. Based on their theoretical and empirical comparisons between the na\"{i}ve Bayes classifier and linear logistic
regression, Ref.~\citeauthor{Ng:2001} claimed that there existed two distinct regimes of performance between the generative and discriminative classifiers with regard to the training-set size.
However, our empirical and simulation studies, as presented in this paper, suggest that it is not so reliable to claim such an existence of the two distinct regimes. In addition, for real world
datasets, so far there is no theoretically correct, general criterion for choosing between the discriminative and the generative approaches to classification of an observation $\mathbf{x}$ into a
class $y$; the choice depends on the relative confidence we have in the correctness of the specification of either $p(y|\mathbf{x})$ or $p(\mathbf{x}, y)$. This can be to some extent a demonstration
of why Ref.~\citeauthor{Efron:1975} and~\citeauthor{ONeill:1980} prefer LDA but other empirical studies may prefer linear logistic regression instead. Furthermore, we suggest that pairing of either
LDA assuming a common diagonal covariance matrix (LDA-$\Lambda$) or the na\"{i}ve Bayes classifier and linear logistic regression may not be perfect, and hence it may not be reliable for any claim
that was derived from the comparison between LDA-$\Lambda$ or the na\"{i}ve Bayes classifier and linear logistic regression to be generalised to all the generative and discriminative classifiers.
PDF - Requires Adobe Acrobat Reader or other PDF viewer.
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Example: Canonical
This example is based on a fictitious data file (Factor.sta) describing a study of life satisfaction. This data file is also analyzed in the Example section of Factor Analysis.
Suppose that a questionnaire is administered to a random sample of 100 adults. The questionnaire contains 10 items that are designed to measure satisfaction at work, satisfaction with hobbies
(leisure time satisfaction), satisfaction at home, and general satisfaction in other areas of life. Responses to all questions are recorded via computer, and scaled so that the mean for all items is
approximately 100.
The results are entered into the Factor.sta data file (see the partial listing shown below). Open this data file via the File - Open Examples menu; it is in the Datasets folder.
Purpose of Analysis. Suppose that you want to learn about the relationship of work satisfaction to satisfaction in other domains. Conceptually, you will treat the work satisfaction items as the
explanatory or independent variables, and the other satisfaction items as the dependent variables.
Initial Computations. From the Statistics - Multivariate Exploratory Techniques menu, select Canonical Analysis to display the Canonical Analysis Startup Panel. Click the Variables button on the
Startup Panel to display the Select variables for canonical analysis dialog. Here, select all variables in this file and click the OK button. The Startup Panel will now look like this:
Canonical correlation analysis is based on the correlation matrix of variables. Therefore, the first step of the analysis is to compute that correlation matrix (unless a Correlation Matrix input file
is specified via the Input File drop-down box, in which case the input needs to be a correlation matrix). Note that you can later select variables for the analysis (for the two sets) from among those
that are specified at this point.
Select the Review descriptive statistics and correlation matrix check box in order to compute the detailed descriptive statistics (i.e., means, correlations, covariances) for the variables in the
current analysis. Now, click the OK button to display the Review Descriptive Statistics dialog.
Plots. In order to visualize the distribution of the variables, two types of plots are available from the Review Descriptive Statistics - Advanced tab: Box & whisker plot of vars and Matrix plot of
Click the Box & whisker plot of vars button, select all variables in the data set in the resulting variable selection dialog, and click the OK button. The Box-Whisker Type dialog will then be
displayed in which you can select from four types of box and whisker plots (for this example, select the Median/Quart./Range option button and then click the OK button). The central tendency (e.g.,
median) and variability (e.g., quartile and range) of the selected variables are displayed in these plots (note that the specific method for computing the quartiles and the median can be specified on
the Options - General tab).
Box & whisker plots are useful to determine if the distribution of a variable is symmetrical. If the distribution is not symmetrical, then you may want to view the histogram for the respective
Now, click the Matrix plot of correlations button (and select all of the variables in the variable selection dialog) to display a matrix of scatterplots. These plots should be examined for outliers,
which may greatly bias the computation of the correlation coefficients, and thus the canonical analysis (see the discussion in the Introductory Overview).
Next, click the Means & standard deviations button on the Review Descriptive Statistics - Advanced tab to display a spreadsheet containing the means and standard deviations of the selected variables.
As in most other modules, the default plot for the spreadsheet of means and standard deviations is the histogram of the distribution of the respective variable. This histogram will show the normal
curve superimposed over the observed distribution, to provide a visual check for any violations of the normality assumption. For example, to produce the histogram for variable Work_1, right-click the
Mean for variable Work_1 (97.0296) and then select Graphs of Input Data - Histogram Work_1 - Normal Fit in the resulting shortcut menu to produce the graph shown below.
The distribution of this variable (of subjects' responses to the first item) follows the normal distribution. Thus, there is little reason to suspect that this variable violates the normality
Specifying the Canonical Analysis. To proceed with the canonical correlation analysis, click the OK button to exit the Review Descriptive Statistics dialog and display the Model Definition dialog in
which you can select the left and right set variables.
Specifying variables. To specify variables for the two sets, click the Variables for canonical analysis button on the Model Definition - Quick tab to open the standard variable selection window.
Select the work satisfaction items (i.e., variables Work_1, Work_2, and Work_3) in the First variable list, and the remaining satisfaction items (i.e., variables Hobby_1, Hobby_2, Home_1, Home_2,
Home_3, Miscel_1, Miscel_2) in the Second variable list.
Note that the designation of first and second list is arbitrary here; that is, you could also select the work satisfaction items into the second list and the remaining satisfaction items into the
first list. In a sense, canonical analysis is completely "symmetrical," that is, it will compute the same statistics (loadings, weights, etc.) for the variables in both lists.
You may also use some other options, such as the Means & standard deviations or Correlations buttons, as well as create some useful descriptive graphs via the Model Definition - Descriptives tab.
Reviewing Results. After specifying the two lists of variables, you are now ready to begin the analysis. Click the OK button on the Model Definition dialog. After a few moments, the Canonical
Analysis Results dialog will be displayed.
The nature of most of the statistics reported here in the results summary are reviewed in the Introductory Overview section of this module. Therefore, the focus in this example will be on the
interpretation of the results.
Click the Summary: Canonical results button on the Quick tab or the Canonical factors tab to display the Canonical Analysis Summary spreadsheet.
Canonical R. The overall canonical R is fairly substantial (.88), and highly significant (p< .001). Remember that the canonical R reported here pertains to the first and most significant canonical
root. Thus, this value can be interpreted as the simple correlation between the weighted sum scores in each set, with the weights pertaining to the first (and most significant) canonical root.
Overall Redundancy. The values in the rows labeled Variance extracted and Total redundancy give an indication of the magnitude of the overall correlations between the two sets of variables, relative
to the variance of the variables. This is different from the canonical R-square, because the latter statistic expresses the proportion of variance accounted for in the canonical variates (see the
Introductory Overview for further details).
Variance extracted. The values in the Variance extracted row indicate the average amount of variance extracted from the variables in the respective set by all canonical roots. Thus, all three roots
extract 100% of the variance from the left set, that is, the three work satisfaction items, and 54% of the variance in the right set. Note that one of these values will always be 100% because
STATISTICA extracts as many roots as the minimum number of variables in either set. Thus, for one set of variables, there are as many independent sums (canonical variates) as there are variables.
Intuitively, it should be clear that, for example, three independent sum scores derived from three variables will explain 100% of all variability.
Redundancy. The computation of the Total redundancy is explained in the Introductory Overview section. These values can be interpreted such that, based on all canonical roots, given the right set of
variables (the seven non-work related satisfaction items), you can account for, on the average, 61.6% of the variance in the variables in the left set (the satisfaction items). Likewise, you can
account for 33.3% of the variance in the non-work related satisfaction items, given the work related satisfaction items. These results suggest a fairly strong overall relationship between the items
in the two sets.
Testing the Significance of Canonical Roots. Now, check whether all three canonical roots are significant. Keep in mind that the canonical R reported in this spreadsheet represents only the first
root, that is, the strongest and most significant canonical correlation. To test the significance of all canonical roots, click the Chi square tests button on the Canonical factors tab of the
Canonical Analysis Results dialog. These results will be displayed in Chi-Square with Successive Roots Removed spreadsheet.
The maximum number of roots that can be extracted is equal to the smallest number of variables in either set. Since three work satisfaction items were selected into the first set, STATISTICA would be
expected to extract three canonical roots.
The sequential significance test works as follows. First, look at all three canonical variables together, that is, without any roots removed. That test is highly significant. Next, the first (and, as
you know, most significant) root is "removed" and the statistical significance of the remaining two roots is determined. That test (in the second row of the spreadsheet) is not significant. You can
stop at this point and conclude that only the first canonical root is statistically significant, and it should be examine further. If the second test were also statistically significant, you would
then proceed to the third line of the spreadsheet to see whether the remaining third canonical root is also significant.
Factor Structure and Redundancy. You now know that you should consider further only the first canonical root. How can this root be interpreted, that is, how is it correlated with the variables in the
two sets? As discussed in the Introductory Overview, the interpretation of canonical "factors" follows a similar logic to that in Factor Analysis. Specifically, you can compute the correlations
between the items in each set with the respective canonical root or variable (remember that the canonical variable in each set is "created" as the weighted sum of the variables). Those correlations
are also called canonical factor loadings or structure coefficients.
You can compute those values (as well as the variance extracted for each set) via the Factor structures & redundancies button of the Factor structures tab on the Canonical Analysis Results dialog.
Click this button to generate four results spreadsheets; namely (1) Factor Structure, left set, (2) Variance Extracted (Proportions), left set, (3) Factor Structure, right set, and (4) Variance
Extracted (Proportions), right set.
Factor structure in the left set. First, examine the loadings for the left set.
Remember that only the first canonical root is statistically significant, and it is the only one that should be interpreted. As you can see, the three work satisfaction items show substantial
loadings on the first canonical factor, that is, they correlate highly with that factor.
As a measure of redundancy, the average amount of variance accounted for in each item by the first root could be computed. To do so, you could sum up the squared canonical factor loadings and divide
them by 3 (the number of variables in this set). Click the Factor structures & redundancies button on the Canonical Analysis Results - Factor structures tab to obtain various spreadsheets. First,
look at the Variance Extracted (Proportions), left set results spreadsheet.
As you can see in this spreadsheet, the first canonical root extracts an average of about 77% of the variance from the work satisfaction items. If you multiply that value with the proportion of
shared variance between the canonical variates in the two sets (i.e., with R-square), then the number in the Reddncy column of the spreadsheet (i.e., redundancy) is obtained. Thus, given the
variables in the right set (the non-work related satisfaction items), you can account for about 60% of the variance in the work related satisfaction items, based on the first canonical root.
Factor structure in right set. In the Factor Structure, right set spreadsheet, the first canonical root or factor is marked by high loadings on the leisure satisfaction items (Hobby_1 and Hobby_2).
The loadings are much lower for the home-related satisfaction items. Therefore, you can conclude that the significant canonical correlation between the variables in the two sets (based on the first
root) is probably the result of a relationship between work satisfaction, and leisure time and general satisfaction. If you consider work satisfaction as the explanatory variable, you could say that
work satisfaction affects leisure time and general satisfaction, but not (or much less so) satisfaction with home life.
The Variance Extracted (Proportions), right set spreadsheet shows the redundancies for the right set of variables.
As you can see, the first canonical root accounts for an average of roughly 42% of variance in the variables in the right set; given the work satisfaction items, you can account for about 33% of the
variance in the other satisfaction items, based on the first canonical root. Note that these numbers are "pulled down" by the relative lack of correlations between this canonical variate and the home
satisfaction items.
Canonical Scores. Remember that the canonical variates represent weighted sums of the variables in each set. You can review those weights by using the Canonical Analysis Results - Canonical Scores
tab on the dialog.
Click the Left & right set canonical weights button to display the following separate spreadsheets:
The weights shown in the Canonical Weights spreadsheets pertain to the standardized (z-transformed) variables in the two sets. You can use those weights to compute scores for the canonical variates.
The scores computed from the data in the current data file may be saved via the Save canonical scores button on the Canonical Analysis Results - Canonical Scores tab.
Plotting canonical scores. Now, plot the canonical scores for the variables in the left set against the scores for the variables in the right set. Click the Scatterplot of canonical correlations
button on the Canonical Analysis Results - Factor structures tab to display the Scatterplot of Canonical Correlations dialog. To produce a scatterplot for the first (and only significant) canonical
variate, select Root 1 in the Left set box and Root 1 in the Right set box.
Now, click the OK button to produce the desired scatterplot. Note that in the illustration shown below a linear regression line was added to the plot via the Plot: Fitting options pane of Graph
Options dialog.
There are no outliers apparent in this plot, nor do the residuals around the regression line indicate any non-linear trend (e.g., by forming a U or S around the regression line). Therefore, you can
be satisfied that no major violations of a main assumption of canonical correlation analysis are evident.
Clusters of cases. Another interesting aspect of this plot is whether or not there is any evidence of clustering of cases. Such clustering may happen if the sample is somehow heterogeneous in nature.
For example, suppose respondents from two very different industries who are working under very different conditions were included in the sample. It is conceivable that the canonical correlation
represented by the plot above could then be the result of the fact that one group of respondents is generally more satisfied with their work and leisure time (and life in general). If so, this would
be reflected in this plot by two distinct clusters of points: one at the low ends of both axes and one at the high ends. However, in this example, there is no evidence of any natural grouping of this
kind, and therefore you do not have to be concerned.
Conclusion. It can be concluded from the analysis of this (fictitious) dataset that satisfaction at work affects leisure time satisfaction and general satisfaction. Satisfaction with home life did
not seem to be affected. In practice, before generalizing these conclusions, you should replicate the study. Specifically, you should ensure that the canonical factor structure that led to the
interpretation of the first canonical root is reliable (i.e., replicable).
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Student Solutions Guide, Volume 2 for Larson's Calculus, 7th
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Designed for the three-semester engineering calculus sequence,
Calculus: Early Transcendental Functions
offers fully integrated coverage of exponential, logarithmic, and trigonometric functions throughout the first semester, within the hallmark balanced approach of the Larson team. A rich variety of
applications encountered earlier in the course prepares students for concurrent physics, chemistry, and engineering courses. This edition features nearly 10,000 diverse and flexible exercises,
carefully graded in sets progressing from skill-development problems to more rigorous application and proof problems.
• New! P.S. Problem Solving sections thought-provoking and challenging exercises at the end of each chapter require students to use a variety of problem-solving skills as they work with calculus
• New! Getting at the Concept exercises, boxed and titled for easy reference, check a student's understanding of the basic concepts of each section.
• Think About It conceptual exercises require students to use their critical-thinking skills and help them develop an intuitive understanding of the underlying theory of the calculus.
• Multi-part Modeling Data questions ask students to find and interpret mathematical models to fit real-life data, often through the use of a graphing utility.
• Section Projects, extended applications at the end of selected exercise sets, can be assigned to individual students or used in a collaborative or peer-assisted learning environment.
• Writing Exercises help students develop reasoning skills and discuss mathematical concepts.
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• Revised! Review Exercises, now grouped and correlated by text section, provide students with a more effective study tool, enabling them to target the concepts they need to review.
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• Interactive Calculus Early Transcendental Functions 3.0 CD-ROM and Internet Calculus Early Transcendental Functions 2.0 web site, referred to in the text by an IC icon, offer enhanced
opportunities for exploration and visualization using the program itself and/or a computer algebra system.
• CAS Examples, also identified in the text by an icon, offer opportunities for interactive exploration using Maple, Mathcad, Mathematica, or Derive.
• Explorations, optional boxed projects, enable students to discover selected concepts on their own, before being exposed to them in the text, making them more likely to remember the results.
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Product Details
• ISBN-13: 9780618149230
• Publisher: Cengage Learning
• Publication date: 8/20/2001
• Edition description: Older Edition
• Edition number: 7
• Pages: 240
• Product dimensions: 8.50 (w) x 10.80 (h) x 0.40 (d)
Table of Contents
1. PREPARATION FOR CALCULUS. Graphs and Models. Linear Models and Rates of Change. Functions and Their Graphs. Fitting Models to Data. Inverse Functions. Exponential and Logarithmic Functions. 2.
LIMITS AND THEIR PROPERTIES. A Preview of Calculus. Finding Limits Graphically and Numerically. Evaluating Limits Analytically. Continuity and One-Sided Limits. Infinite Limits. Section Project:
Graphs and Limits of Trigonometric Functions. 3. DIFFERENTIATION. The Derivative and the Tangent Line Problem. Basic Differentiation Rules and Rates of Change. Product and Quotient Rules and
Higher-Order Derivatives. The Chain Rule. Implicit Differentiation. Section Project: Optical Illusions. Derivatives of Inverse Functions. Related Rates. Newton's Method. 4. APPLICATIONS OF
DIFFERENTIATION. Extrema on an Interval. Rolle's Theorem and the Mean Value Theorem. Increasing and Decreasing Functions and the First Derivative Test. Section Project: Rainbows. Concavity and the
Second Derivative Test. Limits at Infinity. A Summary of Curve Sketching. Optimization Problems. Section Project: Connecticut River. Differentials. 5. INTEGRATION. Antiderivatives and Indefinite
Integration. Area. Riemann Sums and Definite Integrals. The Fundamental Theorem of Calculus. Section Project: Demonstrating the Fundamental Theorem. Integration by Substitution. Numerical
Integration. The Natural Logarithmic Function: Integration. Inverse Trigonometric Functions: Integration. Hyperbolic Functions. Section Project: St. Louis Arch. 6. DIFFERENTIAL EQUATIONS. Slope
Fields and Euler's Method. Differential Equations: Growth and Decay. Differential Equations: Separation of Variables. The Logistic Equation. First-Order Linear Differential Equations. Section
Project: Weight Loss. Predator-Prey Differential Equations. 7. APPLICATIONS OF INTEGRATION. Area of a Region Between Two Curves. Volume: The Disk Method. Volume: The Shell Method. Section Project:
Saturn. Arc Length and Surfaces of Revolution. Work. Section Project: Tidal Energy. Moments, Centers of Mass, and Centroids. Fluid Pressure and Fluid Force. 8. Integration Techniques, L'Hôpital's
Rule, and Improper Integrals. Basic Integration Rules. Integration by Parts. Trigonometric Integrals. Section Project: Power Lines. Trigonometric Substitution. Partial Fractions. Integration by
Tables and Other Integration Techniques. Indeterminate Forms and L'Hôpital's Rule. Improper Integrals. 9. INFINITE SERIES. Sequences. Series and Convergence. Section Project: Cantor's Disappearing
Table. The Integral Test and p-Series. Section Project: The Harmonic Series. Comparisons of Series. Section Project: Solera Method. Alternating Series. The Ratio and Root Tests. Taylor Polynomials
and Approximations. Power Series. Representation of Functions by Power Series. Taylor and Maclaurin Series. 10. CONICS, PARAMETRIC EQUATIONS, AND POLAR COORDINATES. Conics and Calculus. Plane Curves
and Parametric Equations. Section Projects: Cycloids. Parametric Equations and Calculus. Polar Coordinates and Polar Graphs. Section Project: Anamorphic Art. Area and Arc Length in Polar Coordinates.
Polar Equations of Conics and Kepler's Laws. 11. VECTORS AND THE GEOMETRY OF SPACE. Vectors in the Plane. Space Coordinates and Vectors in Space. The Dot Product of Two Vectors. The Cross Product of
Two Vectors in Space. Lines and Planes in Space. Section Project: Distances in Space. Surfaces in Space. Cylindrical and Spherical Coordinates. 12. VECTOR-VALUED FUNCTIONS. Vector-Valued Functions.
Section Project: Witch of Agnesi. Differentiation and Integration of Vector-Valued Functions. Velocity and Acceleration. Tangent Vectors and Normal Vectors. Arc Length and Curvature. 13. FUNCTIONS OF
SEVERAL VARIABLES. Introduction to Functions of Several Variables. Limits and Continuity. Partial Derivatives. Section Project: Moire Fringes. Differentials. Chain Rules for Functions of Several
Variables. Directional Derivatives and Gradients. Tangent Planes and Normal Lines. Section Project: Wildflowers. Extrema of Functions of Two Variables. Applications of Extrema of Functions of Two
Variables. Section Project: Building a Pipeline. Lagrange Multipliers. 14. MULTIPLE INTEGRATION. Iterated Integrals and Area in the Plane. Double Integrals and Volume. Change of Variables: Polar
Coordinates. Center of Mass and Moments of Inertia. Section Project: Center of Pressure on a Sail. Surface Area. Section Project: Capillary Action. Triple Integrals and Applications. Triple Integrals
in Cylindrical and Spherical Coordinates. Section Project: Wrinkled and Bumpy Spheres. Change of Variables: Jacobians. 15. VECTOR ANALYSIS. Vector Fields. Line Integrals. Conservative Vector Fields
and Independence of Path. Green's Theorem. Section Project: Hyperbolic and Trigonometric Functions. Parametric Surfaces. Surface Integrals. Section Project: Hyperboloid of One Sheet. Divergence
Theorem. Stoke's Theorem.
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Donald A. Martin
Professor of Mathematics and Philosophy
• Offices:
□ 7935 Mathematical Sciences Building
□ 355 Dodd Hall
• Phone:
□ (310) 825-4701 (messages)
□ (310) 206-6673 (FAX)
• E-mail : < dam@math.ucla.edu>
• Mailing Address Dept. of Mathematics UCLA Los Angeles, California, 90095-1555 Two version of Borel determinacy proof. From time to time people ask me for a copy of my 1985 "A Purely inductive
proof of Borel determinacy." I recently realized that I have a hard copy, and so I have scanned and posted it. I have also posted what I regard as a better version of the proof of Borel
determinacy, plus a discussion of the two versions.
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Crystal Growth Behaviors of Silicon during Melt Growth Processes
International Journal of Photoenergy
Volume 2012 (2012), Article ID 169829, 16 pages
Review Article
Crystal Growth Behaviors of Silicon during Melt Growth Processes
Institute for Materials Research (IMR), Tohoku University, Katahira 2-1-1, Aoba-ku, Sendai 980-8577, Japan
Received 31 August 2011; Accepted 29 December 2011
Academic Editor: Teh Tan
Copyright © 2012 Kozo Fujiwara. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any
medium, provided the original work is properly cited.
It is imperative to improve the crystal quality of Si multicrystal ingots grown by casting because they are widely used for solar cells in the present and will probably expand their use in the
future. Fine control of macro- and microstructures, grain size, grain orientation, grain boundaries, dislocation/subgrain boundaries, and impurities, in a Si multicrystal ingot, is therefore
necessary. Understanding crystal growth mechanisms in melt growth processes is thus crucial for developing a good technology for producing high-quality Si multicrystal ingots for solar cells. In this
review, crystal growth mechanisms involving the morphological transformation of the crystal-melt interface, grain boundary formation, parallel-twin formation, and faceted dendrite growth are
discussed on the basis of the experimental results of in situ observations.
1. Introduction
The expectations for solar cells have been increasing yearly toward solving energy and environmental problems worldwide. The Si multicrystal (mc-Si) is one of the most important materials along with
the Si single crystal (sc-Si) for the substrate of solar cells in the future although other materials are being developed. The crystal structure of an mc-Si ingot obtained by casting based on a
unidirectional growth technique is markedly different from that of sc-Si, as illustrated by the formation of grain boundaries and the distribution of crystallographic orientations, which prevent the
realization of high-efficiency solar cells. Various types of defect included in an mc-Si ingot, such as grain boundaries, dislocations, sub-grain boundaries, and metallic impurities, affect the
properties of solar cells. Therefore, there is increasing importance to control the macro- and microstructures of mc-Si ingots. In recent years, various new methods for controlling the macro- and
microstructures have been presented, which focus on controlling the nucleation or crystal growth mechanism in the earlier stage of casting. Dendritic casting is a method in which seed crystals are
created from the melt during crystallization by inducing dendrite growth at the bottom of the crucible [1–3]. Seeded casting is a method in which prearranged seed crystals are set at the bottom of a
crucible before crystallization, which initiates from seed crystals [4–6]. Other methods have been proposed, in which the nucleation site is controlled by controlling the cooling area in the initial
stage of casting [7–9]. Many studies to establish such technologies for obtaining high-quality mc-Si ingots have been continuously performed on all these methods. In this review, such technologies
for mc-Si ingot growth will not be discussed; only the melt growth mechanisms of Si will be discussed because fundamental understanding of crystal growth mechanisms is crucial to developing all such
technologies. The crystal growth phenomena during melt growth processes including the morphological transformation of crystal-melt interfaces, grain boundary formation, parallel-twin formation, and
faceted dendrite growth will be discussed by providing a review of our recent studies.
2. Morphological Transformation of Crystal-Melt Interface
2.1. Crystal-Melt Interface during Unidirectional Growth
To control the morphology of the crystal-melt interface during unidirectional growth processes is crucial to obtaining high-quality crystals because it affects the macro- and microstructures and
eventually the mechanical, optical, and electrical properties of materials. It has been suggested that the generation of crystal defects, such as dislocations and twin boundaries, is related to the
morphology of the crystal-melt interface [12–14]. The segregation of impurities is also dependent on the interfacial morphology [15]. According to Jackson’s theory [16, 17], only the crystal-melt
interfaces of Si planes, called facet planes, are atomically smooth; other crystallographic planes are atomically rough. Because the planes have the lowest surface energy [18], a zigzag-faceted
interface bounded by planes is expected to form at the crystal-melt interface of rough planes. The crystal-melt interface was observed in recrystallization processes in Si thin films [19–28], and
micron-sized zigzag facets at the Si (100) crystal-melt interface were observed. Molecular dynamics simulation showed the existence of atomic-scale zigzag facets at the Si (100) crystal-melt
interface [29, 30]. Such a zigzag-faceted interface at the Si (100) crystal-melt interface was also observed in bulk samples [31, 32]. Therefore, as one possible formation mechanism of a
zigzag-faceted interface, it was considered that atomic-scale facets are initially formed, which gradually enlarge to macroscopic facets during crystallization [26]. On the other hand, recently, the
zigzag-facet formation at the Si (100), (112), (110), and (111) crystal-melt interfaces has been investigated by in situ observation [10, 11].
Figure 1 shows the Si (100) crystal-melt interface whose growth velocity was 162μm/s [10]. In this experiment, a piece of Si (100) wafer was set between quartz plates inside the furnace to keep the
surface of the Si melt flat during crystal growth. The morphology of the interface transformed from planar to zigzag facets during the growth. It was shown that a wavelike perturbation is introduced
into a planar interface, the perturbation is amplified, and the zigzag facets are formed finally. A similar morphological transformation from planar to zigzag facets of the moving interface was
observed at the Si (112) and (110) crystal-melt interfaces at higher growth velocities, as shown in Figure 2 [11]. On the other hand, when the growth velocity was lower, the planar interfaces were
maintained throughout the crystallization, as shown in Figure 3. The critical growth velocity for the morphological transformation, , was found to be 123μm/s < < 147μm/s, 107μm/s < < 124μm/s, and
102μm/s < < 129μm/s at the Si (100), Si (110), and Si (112) crystal-melt interfaces, respectively [10, 11]. In those experiments, the temperature gradient along the growth direction in the furnace
was about 8K/mm. It was found that the critical growth velocity was dependent on the temperature gradient in the furnace. When the temperature gradient was about 4K/mm, the critical growth velocity
of the Si (100) interface was 36μm/s < < 100μm/s [33]. Figure 4 shows isochrones of the interface of Si (100) and Si (110) when the planar interface transformed to a zigzag-faceted interface. It
was shown that the wavelength of the wavy interface completely agreed with that of the zigzag-faceted interface. The perturbation was amplified with time, and one perturbation peak formed one zigzag
facet peak. Here, it should be determined why the perturbation is amplified when the growth velocity is high, using stability arguments [34]. Tokairin et al. calculated the thermal field in the Si
crystal and melt during crystallization along the growth direction at a constant growth velocity, before the zigzag facet formation [10]. Generally, a crystal-melt interface becomes unstable, leading
to the amplification of the perturbation, when the temperature gradient at the interface is negative along the growth direction [35]. However, the temperature gradient in the furnace was positive in
Figures 1–4. Therefore, it was considered that the latent heat of crystallization increases the temperature at the crystal-melt interface, and that the temperature gradient in the Si melt at the
interface becomes negative when growth velocity is high, because the amount of generated latent heat per unit time increases with growth velocity. The thermal field of the Si crystal and melt, , is
governed by the partial differential equation [36] where , , , , and are the heat capacity of the Si crystal and melt, the thermal conductivity of the Si crystal and melt, the thermal conductivity of
the quartz plate, the thickness of the quartz plate, and the thickness of the Si, respectively. The origin of the coordinate is the crystal-melt interface, which moves with the growth velocity . is
the furnace temperature, where is the temperature gradient in the furnace, and is the furnace temperature at the interface. The first and second terms on the right-hand side of (1) come from the heat
diffusion along the growth direction and the heat conduction between the Si crystal or melt and the furnace through the quartz plates, respectively. From the solution of (1), the thermal fields of
the Si crystal and melt during crystal growth for various growth velocities were obtained, as shown in Figure 5 [10]. The physical properties of Si used were based on those indicated in [37], and , ,
and were based on experimental values. The temperature gradient in the Si melt at the interface changes from positive to negative as growth velocity increases. When the growth velocity is low, the
temperature gradient is positive, and this means that the interface is stable and the planar interface is maintained. On the other hand, when the growth velocity exceeds its critical value, the
temperature gradient in the Si melt at the interface changes from positive to negative, and thus the perturbation introduced into the planar interface is amplified, forming zigzag facets at high
growth velocities.
Figure 6 shows the crystal-melt interface of the Si (111) plane growing at a high growth velocity of 200μm/s. The Si (111) interface maintained a planar shape even at a high growth velocity although
a negative temperature gradient should have been formed at the interface at such a high growth velocity. The difference in interface morphology between the (111) plane and the other planes could be
explained by considering the anisotropy of growth velocity. The growth velocity along the direction is the lowest. Moreover, the growth velocity on the Si (111) plane was more than 2 orders lower
than that on the Si (100) plane at a given undercooling [38]. Thus, even though perturbation is introduced into the (111) planar interface, the morphology of the interface remains planar owing to the
faster growth in the lateral direction than in the direction. For added consideration regarding the growth of the Si (111) interface, it was reported that the step-flow mode is stable up to an
undercooling of 40K by molecular dynamics simulation [39]. The step velocity at the (111) interface was found to be quite high because of the large step kinetic coefficients, that is, 0.29–0.79m/
sK. Such a high step velocity makes the interface planar in this growth mode.
It has been shown that the interface morphology is controlled by temperature gradient and growth velocity. When the zigzag facets are formed at the crystal-melt interface of rough planes during a
unidirectional growth process of an mc-Si ingot by casting, the growth velocity difference between the rough planes and facet planes becomes larger [32, 40]. When two crystal grains, the one has
(100) interface and the other one has (111) interface, exist side by side, the (100) grain with zigzag facets interface grows much faster than (111) grain with planar interface, and the (100) grain
gradually occupies the space and becomes larger than (111) grain during the unidirectional growth. In such competitive growth process between the two crystal grains, the shape of the grain boundary
between two crystal grains is also changing during the growth [32, 40]. The stress given to the grain boundary maybe increased with the shape change, which leads to increasing the defect generation,
such as dislocations or twin boundaries, at the grain boundary. Moreover, when the zigzag facets are formed at the crystal-melt interface of rough planes during a unidirectional growth process, the
impurities segregate at the valleys of zigzag facets, which lead to in-plane inhomogeneity in an ingot. Those phenomena seem to affect the quality of an mc-Si ingot in an adverse way. On the other
hand, Si-faceted dendrites easily grow from a zigzag facets interface [12]. In the dendritic casting, faceted dendrites are promoted to grow along the bottom of the crucible in the earlier stage of
casting [1]. In this method, the growth condition to form the zigzag facets interface should be selected to initiate the dendrite growth in the initial stage of casting, which leads to the formation
of large-size grains at the bottom of the ingot. Thus, we should carefully control the growth mechanism during casting for obtaining high-quality mc-Si ingots.
2.2. Growth Kinetics in Melt Growth of Silicon
In the previous section, the morphological transformation of crystal-melt interfaces for different orientations was shown. Those phenomena observed at crystal-melt interface depend on the anisotropy
of crystal-melt interfacial energy, which governs the growth kinetics and growth modes. It is well known that Si (111) is a facet plane, which means that the surface of crystal-melt interface is
atomically smooth, and other planes are atomically rough. Therefore, the growth kinetics in melt growth processes seems to be different between the (111) plane and other planes. In general, the
growth on the (111) facet plane is discussed by a 2D nucleation mode, and that of rough planes are discussed by a normal growth mode [38]. The normal growth velocity, , on Si rough plane is expressed
as , where are the kinetic coefficient for rough plane and undercooling at the crystal-melt interface, respectively. The value of the kinetic coefficient for Si (100) rough plane has been derived by
computations as approximately 0.1m/(sK) [38, 39]. Thus, the growth velocity on Si (100) rough plane linearly increases with increasing the amount of undercooling. On the other hand, the growth
velocity on a Si (111) facet plane increases exponentially with undercooling in a 2D nucleation growth mode [38, 42]. Therefore, the growth velocity on a facet plane seems to be much smaller than
that of the rough plane at a low undercooling. However, direct evidences by experiments for the growth kinetics in melt growth processes have been very limited comparing with those in the solution
growth or the vapor growth. Thus, a more detailed discussion of the growth mode should be reserved for future works. The growth mode on a Si (111) facet plane has been well summarized in [42].
2.3. Crystal-Melt Interface at the Grain Boundary Formation
In mc-Si grown by a casting, grain boundaries are one of the main factors governing the mechanical, optical, and electrical properties. For instance, some of the grain boundaries inhibit the increase
in the energy conversion efficiency of solar cells because they act as recombination centers of photocarriers. Therefore, the relationships between grain boundary properties and electrical activity
have been extensively investigated [44–51]. Moreover, a grain boundary, which is usually straight, is electrically inactive, but other CSL (coincidence site lattice) boundaries or random grain
boundaries, which are straight, wavy or curved, are electrically active. Grain boundary structures are also investigated by transmission electron microscopy [52–54] or computer simulation [55–57].
Almost all studies of grain boundaries in Si have been performed after crystallization. Here, the crystal growth behaviors in grain boundary formation during crystallization from Si melt are shown.
Figure 7(a) shows the crystal growth behavior when two crystals converge during the crystallization process [41]. Most of the crystal-melt interface between the two crystals is linear, which suggests
that the growth direction vertical to those interfaces is . In the underlying crystal, a sharp triangular corner is observed at the right edge of the interface when there is sufficient distance
between the two growing crystals. This difference in interfacial shape is due to the difference in the growth direction owing to the existence of a grain boundary at the location marked by a red
arrow in Figure 7(a). The sharp corner gradually flattens as the two crystals become approach each other. Figure 7(b) shows the result of crystallographic orientation analysis after crystallization
determined by the electron backscattering diffraction pattern (EBSP) method [41]. The directions vertical to the grain boundary are shown in color using the inverse pole figure triangle. A linear
grain boundary was formed corresponding to the shape of crystal-melt interfaces just before grain boundary formation. It is found that the growth directions vertical to the linear growing interfaces
were (shown in blue), and that the growth direction with a sharp-cornered interface in the underlying crystal was (shown in red). Grain boundary characteristics are also indicated by colored lines in
Figure 7(b). The grain boundary characteristic differs depending on location. A (111) grain boundary, shown by a red line, is formed where facet planes impinge on each other. On the other hand, a
random grain boundary, shown by a black line, is formed at a location where the shape transition of the growing interface occurs just before grain boundary formation. Figure 8(a) shows isochrones of
the position of the growing interfaces at 1/3 s intervals for the sample shown in Figure 7, and Figure 8(b) shows the t-x plot of the growing interface at the two parts marked by the red and blue
lines on the isochrones [41]. The color of dots in Figure 8(b) corresponds to the line color in Figure 8(a). The initial point of the shape transition of the interface is indicated in Figure 8(b). It
is shown that the growth velocity of the growing interfaces, the gradient of the t-x curve, decreases from this point. This means that the amount of undercooling in front of the growing interface
decreased and the melt temperature in front of the growing interface increased. We explained this phenomenon by considering the thermal fields in front of the growing interface. When a crystal is
growing from the melt, the heat of crystallization is released. The heat of crystallization can be taken away through the melt and/or crystal [58] when the two crystals are sufficiently separated.
However, as the distance between two crystals becomes smaller, thermal fields in front of the two growing crystals overlap and the heat of crystallization cannot be taken away; then the temperature
between the two crystals increases. This leads to the reduction in the growth velocity at the tip of the corner on the growing interface, and, as a result, the shape of the interface becomes flat.
Figure 9(a) shows the growth behavior of the two crystals with zigzag-faceted interfaces [41]. Although the distance between the two crystals is large, well-developed facet planes appear on the
growth surfaces, and, thus, sharp corners are formed. As the distance between the two crystals becomes smaller, the shape transition of the growing interface is observed. The sharp corners on the
interfaces gradually disappear, and wavy interfaces are formed. Just at impingement, the shape of the interfaces becomes more linear. This behavior of the growing interface is similar to that
observed in Figure 7. Figure 9(b) shows the result of orientation analysis. The grain boundary was wavy owing to the shape transition of the growing interface, and a random grain boundary was formed
[41]. The results in Figure 7 to Figure 9 show that the grain boundary shape and grain boundary characteristics are dependent on the behavior of the growing interface before impingement, which, in
turn, strongly depends on the shape of the faceted interface. Figure 10 shows the growth behavior of the dendrite at the grain boundary formation. The tip of the dendrite is narrow and sharp when the
distance between the two crystals is large. As the distance between the two crystals becomes smaller, the tip of the dendrite becomes wider and flatter. To continue the dendrite growth, sufficient
undercooling, that is, more than 10K, is necessary [12]. When the distance between the two crystals becomes smaller, melt temperature seems to increase owing to the overlapping thermal fields in
front of the crystal-melt interfaces, as explained before, and then the dendrite growth cannot continue; thus, the tip of the dendrite is flattened.
3. Parallel-Twin Formation
In mc-Si ingots, twin boundaries are often observed at a high density [59]. The twin boundary formation energy is very low [60, 61], and some origins of twin boundary formation have been reported [16
, 62–64]. Figure 11 shows an mc-Si wafer (left) and a faceted dendrite crystal examined by EBSP analysis (right). Most straight lines observed in the mc-Si wafer and the red lines at the center of
the dendrite by EBSP analysis are twin boundaries. It is observed that at least two twin boundaries often exist parallel to each other, with a narrow spacing. Here, the parallel-twin formation, which
is related to the faceted dendrite growth, is discussed.
The fundamental study of faceted dendrite growth has had a long history since the first report by Billig [65]. It has been shown that more than two parallel twins exist at the center of a faceted
dendrite [65–69]. There have been some discussions concerning when and/or where a twin boundary is formed, that is, whether a nucleus originally contains a twin boundary [70] or whether a twin is
formed after nucleation [71]. There was a report that twin boundaries are generated from a normal grain boundary during directional growth [72]. It has also been reported that parallel twins can form
at (111) microfacets at a crystal-melt interface [43]. Pohl et al. studied this issue by classical molecular dynamics simulation [13]. In their simulations, parallel twin formation at the normal
grain boundary was observed, but that at microfacets on a crystal-melt interface was not observed.
Figure 12 shows the growth behavior of a faceted dendrite grown from a crystal-melt interface. In this experiment, the crystal growth was initiated from a Si single-crystal seed. A faceted dendrite
always appears following the zigzag facet formation on a crystal-melt interface during unidirectional growth. Figure 13 shows the result of the EBSP analysis of the faceted dendrite after
crystallization. The EBSP analysis was performed at the origin of the dendrite growth. The direction along the dendrite growth was colored using the inverse pole figure triangle. grain boundaries
were indicated by red lines, and other grain boundaries were labeled by black lines if they existed in the crystal. It was found that no grain boundaries without twin boundaries related to the
dendrite growth existed in the crystal. This suggests that the grain boundary is not always necessary for twin boundary formation. To explain the experimental result, it should be considered that the
twin boundary is formed on microfacets at the crystal-melt interface. Figure 14 schematically shows a model for generating parallel twins at the crystal-melt interface [43]. The formation of a
zigzag-faceted interface bounded by planes was discussed in Section 2.1. The shape of the faceted interface is dependent on crystal growth orientation, and crystal growth is always promoted on facet
planes at the zigzag-faceted interface. When an atom attaches to a facet plane with a twin relationship, a layer that maintains the twin relationship is formed on the facet plane after lateral
growth, and then one twin boundary is generated on the layer, as shown in Figure 14(b). The crystal continuously grows in the lateral-growth mode on the plane (Figure 14(c)), and then another twin
boundary forms parallel to the previous twin boundary, as shown in Figure 14(d). In this model, the formation of parallel twin boundaries is ensured when one twin boundary is formed on a facet plane
at the zigzag-faceted interface. When the undercooling of the melt is sufficient, a faceted dendrite grows from the parallel twin boundaries.
4. Faceted Dendrite Growth
Dendritic growth is a widespread phenomenon that appears during crystallization from a liquid or vapor phase in almost all types of materials containing metals, semiconductors, oxides, and organic
materials. Dendrites of faceted materials, so-called “faceted dendrites,” which are distinguished from those of nonfaceted materials in metals and alloys, were discovered in the 1950s [65]. It is
known that the surface of a dendrite is bounded by habit planes, and at least two parallel twin boundaries exist at the center of the dendrite [65–69]. Moreover, the preferential growth directions of
Si-faceted dendrites are and [69], and the growth rate of faceted dendrites is higher than that of equiaxed grains. Such features can be applied in technologies for growing thin Si ribbon crystals [
76, 77] and mc-Si ingots [1–3] for solar cells. The growth model of faceted dendrites preferentially grown in the direction, hereafter referred to as dendrites, was first proposed in 1960 by Hamilton
and Seidensticker [67]. Recently, the growth model has been modified to apply to an actual growth based on experimental evidence by in situ observation of the growth behavior of dendrites [74]. It
was also shown that the modified model is adequate to explain the growth behavior of dendrites [73].
Figure 15 shows a dendrite and a dendrite growing from a portion of a faceted crystal-melt interface. In both dendrites, parallel twins exist parallel to the surface. The shape of the tip of the
growing dendrite is markedly different between the and dendrites. Although the tip of the dendrite becomes wider during growth, that of the dendrite remains narrow during growth. Figure 16 shows the
growth process of the dendrite. It was found that triangular corners with an angle of 60° were formed at the tip of the dendrite and that the direction of the corners alternately changed from outward
to forward in the direction of growth [74].
The modified growth model of the dendrite based on experimental evidence is shown in Figure 17 [74]. Figure 17(a) shows the equilibrium form of the Si crystal with two parallel twins. The crystal is
bounded by habit planes, and parallel twin planes exist parallel to the upper surface. In the explanation, the twins are distinguished by labeling them twin[1] and twin[2]. A reentrant corner with an
external angle of 141° (type I) appears at the growth surface only at twin[1] when the dendrite is growing in one direction. Nucleation more readily occurs at the reentrant corner than at flat
surfaces [69, 73, 74, 78–80], and thus the rapid growth occurs there. In this mechanism, it is considered that a triangular corner with an angle of 60° is formed at the growth tip of a dendrite owing
to the rapid growth at the reentrant type I corner, as shown in Figure 17(b). This is the major difference from the previous growth mechanism of the dendrite presented by Hamilton and Seidensticker [
67]. In their explanation, it was considered that the formation of a triangular corner disturbs the continuous growth of a dendrite because the reentrant type I corner disappears. However, the
formation of a triangular corner at the tip of the dendrite is observed in Figure 16 by in situ observation [74]. Crystal growth can continue on the flat surface although the rapid growth is
inhibited owing to the disappearance of the type I corner (Figures 17(b)-17(c)). After the propagation of the crystal, two new type I corners are formed on the growth surface at twin[2] (Figure 17(c)
). Thus, rapid growth occurs there again, and triangular corners with an angle of 60° are formed in the same manner as before (Figure 17(d)). Crystal growth is promoted on the flat surface, leading
to the formation of a new reentrant type I corner at twin[1] (Figure 17(e)), and the rapid growth occurs again (Figure 17(f)). A faceted dendrite continues to grow by the repetition of the same
processes. In this model, the tip of the dendrite becomes wider during the growth.
Figure 18 shows the growth mechanism of a dendrite [73]. Two parallel twins are also contained in this dendrite, and thus the initial shape of the crystal should be the same as that shown in Figure
17(a). This means that the elemental process for the growth of the dendrite is the same as that of the dendrite. It is found that type I corners exist at both twin[1] and twin[2] on the growth
surface (Figure 18(a)). This is the major difference from the case of the dendrite, in which type I corners alternately appear at each twin. Rapid growth occurs at these type I corners (Figure 18(b)
), leading to the formation of triangular corners. Although the rapid growth is inhibited after the formation of triangular corners, the crystal continuously grows on the flat surface (Figures 18(b)-
18(c)). When triangular crystals propagate across the other twin, two type I corners appear, and rapid growth occurs there again (Figures 18(c)-18(d)). When triangular crystals propagate across the
other twin, two type I corners appear (Figure 18(e)). The tip of the dendrite remains narrow during crystal growth. Thus, the shapes of the and dendrites during growth are explained by the models
shown in Figures 17 and 18.
The equation for the growth velocity of a dendrite and a dendrite is derived on the basis of the growth model shown in Figures 17 and 18 [75]. The growth velocities of both dendrites are,
respectively, described as where is the height of a triangular corner, is the twin spacing, is the growth velocity at a type I corner, and is the growth velocity on the plane. It is found that the
growth velocity of a dendrite is inversely proportional to twin spacing. Figure 19(a) shows the growth of three dendrites (d1, d2, and d3) from the same crystal-melt interface at the same time [75].
The preferential growth direction of these three dendrites is . The growth velocities are clearly different among the three dendrites. Figure 19(b) shows the parallel twins observed at the center of
d1, d2, and d3 by EBSP measurement. The average twin spacings in d1, d2, and d3 are 4.5μm, 10μm, and 16.5μm, respectively. Dendrite growth velocity as a function of twin spacing is plotted in
Figure 19(c). As twin spacing decreases, dendrite growth velocity nonlinearly increases. The equation above was used to fit the experimental results, as shown by the red line in Figure 19(c). It is
found that the growth velocity of a dendrite eventually approaches a certain value as twin spacing increases to infinity. The dependence of the growth velocity of faceted dendrites on undercooling
was also investigated [81]. Dendrite growth velocity increased linearly with increasing undercooling. It was also found that the relationship between dendrite growth velocity and undercooling is most
sensitive to twin spacing [81].
5. Summary
Some crystal growth behaviors observed in melt growth processes were reviewed in this paper. Other significant issues not treated in this review include the dislocation/sub-grain-boundary formation [
82–87] and impurity behavior [88–93] in crystal growth processes. For the complete understanding of crystal growth mechanisms and the control of the macro- and microstructures of mc-Si ingots,
further data accumulation is required. It is expected that such ongoing studies will lead to the establishment of a technology for producing high-quality mc-Si ingots in the near future.
The author would like to acknowledge S. Uda and K. Nakajima for their fruitful discussions. This work was partially funded by the JST PRESTO Program and the Cabinet Office, Government of Japan
through its “Funding Program for Next Generation World-Leading Researchers.”
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92. P. Gundel, M. C. Schubert, W. Kwapil et al., “Micro-photoluminescence spectroscopy on metal precipitates in silicon,” Physica Status Solidi—Rapid Research Letters, vol. 3, no. 7-8, pp. 230–232,
2009. View at Publisher · View at Google Scholar · View at Scopus
93. T. Buonassisi, A. A. Istratov, M. D. Pickett et al., “Micro-photoluminescence spectroscopy on metal precipitates in silicon,” Progress in Photovoltaics, vol. 14, no. 6, pp. 513–531, 2006.
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Why is it important that partial derivatives commute?
up vote 17 down vote favorite
I am asking this in the context of differential geometry (specifically Riemannian).
When the Levi-Civita Connection is defined, we require that the torsion tensor is 0, which in local coordinates translates to the requirement that $\Gamma_{ij}^{k} = \Gamma_{ji}^{k}$; which is the
covariant derivative version of saying partial derivatives commute: $\nabla_{\partial_i}(\partial_j)=\nabla_{\partial_j}(\partial_i)$.
This is obviously true in the Euclidian settings, and I understand all the details of the proofs. But why is this such an essential property? Why does this capture our intuitive sense of derivatives?
6 A related question might be: What uses, if any, are there for connections that aren't torsion free? – Harald Hanche-Olsen Feb 23 '13 at 15:42
1 There are nice interpretations of torsion in mathoverflow.net/questions/20493/… Especially, check out Tom Boardman's answer. – Claudio Gorodski Feb 23 '13 at 20:24
4 So that exterior differentiation makes a chain complex! – David Corwin Feb 24 '13 at 16:11
1 Isn't this a repeat of the thread Claudio links to above? – Ryan Budney Feb 24 '13 at 19:44
1 "Which is the covariant derivative version of saying partial derivatives commute" not quite. It only says that for scalar functions. – Willie Wong Feb 25 '13 at 9:51
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4 Answers
active oldest votes
To me, a Riemannian metric and the Levi-Civita connection associated with the metric represent the intrinsic geometric properties of a submanifold in Euclidean space induced by the inner
product and natural flat connection on Euclidean space. Since they are intrinsic, their definitions can be extended from submanifolds of Euclidean space to abstract manifolds.
up vote
8 down If you don't assume the connection is torsion-free, then there are an infinite number of connections that are compatible with the metric (instead of exactly one), so the link between the
vote geometric properties of the metric and that of the connection is much weaker.
Thanks. This makes sense, although in a way it kind of cancels the entire point of differential geometry as I see it so far (which is not a lot): I thought the motivation is to define
calculus again in an intrinsic, coordinate-free way on general smooth manifolds; specifically, since we believe our universe is modeled well by one. Reducing it all back to the Euclidian
setting when we stumble along some complication does not seem to follow the same spirit. Perhaps this means we should think our universe is some complicated embedding in a larger Euclidian
space, unreachable to us? – R S Feb 23 '13 at 15:59
Maybe it's circular in this setting, but there is Nash's embedding theorem. – horse with no name Feb 23 '13 at 16:09
I have two responses: Designing an intrinsic geometric structure modeled on submanifolds of Euclidean space does not mean we are restricting our attention to submanifolds of Euclidean
space, notwithstanding the Nash isometric embedding theorem. Nor does it mean that the geometric structure is not co-ordinate-free. The study of Riemannian geometry does not depend on
co-ordinates. Of course, even the study of submanifolds in Euclidean space does not, too. – Deane Yang Feb 23 '13 at 16:52
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Here is another way of obtaining the Christoffel symbols with the symetry imposed by the torsion free condition
$$ \Gamma^i_{k\ell}=\Gamma^i_{\ell k}. $$
This goes back to Riemann's Habillitation.
Suppose that $(M,g)$ is a Riemann manifold of dimension $N$, $p\in M$. By fixing an orthonormal frame of $T_pM$ we can find local coordinates $(x^1,\dotsc, x^N)$ near $p$ such that, $\
newcommand{\pa}{\partial} $
$$ x^i(p)=0, \;\; g=\sum_{i,j} g_{ij}(x) dx^i dx^j, $$
$$g_{ij}(x)= \delta_{ij} +\sum_{i,j}\left(\sum_k\pa_{x^k}g_{ij}(0) x^k\right) dx^i dx^j + O(|x|^2). $$
In other words, in these coordinates,
$$ g_{ij}(x)=\delta_{ij} +O(|x|). $$
Riemann was asking whether one can find new coordinates near $p$ such that in these coordinates the metric $g$ satisfies $g_{ij}=\delta_{ij}$.
As a first step, we can ask whether we can find a new system of coordinates such that, in these coordinates the metric $g$ is described by
$$ g=\sum_{ij}\hat{g}_{ij} dy^idy^j, $$
$$\hat{g}(y)=\delta_{ij}+ O(|y|^2). \tag{1} $$
up vote 6 The new coordinates $(y^j)$ are described in terms of the old coordinates $(x^i)$ by a family of Taylor approximations
down vote
$$y^j= x^j + \frac{1}{2}\sum_{ij}\gamma^j_{\ell k} x^\ell x^k + O(|x|^3),\;\; \gamma^j_{\ell k}=\gamma^j_{k\ell}. $$
The constraint (1) implies
$$ \gamma^j_{\ell k}=\frac{1}{2}\left(\pa_{x^\ell}g_{jk}+\pa_{x^\ell}g_{jk}-\pa_{x^j}g_{\ell k}\right)_{x=0}. $$
We see that, in the $x$ coordinates
$$ \Gamma^i_{k\ell}(p)=\gamma^i_{k\ell}, $$
because $g^{ij}(p)=\delta^{ij}$.
It took people several decades after Riemann's work to realize that the coefficients $\Gamma^i_{k\ell}$ are related to parallel transport, and ultimately, to a concept of connection.
Ultimately, to my mind, the best explanation for the torsion-free requirement comes from Cartan's moving frame technique. The clincher is the following technical fact: given a
connection $\nabla$ on $TM$ and a $1$-form $\alpha\in \Omega^1(M)$ then for any vector fields $X,Y$ on $M$ we have
$$d\alpha(X,Y)= X\alpha(Y)-Y\alpha(X)-\alpha([X,Y]) $$
$$= (\nabla_X\alpha)(Y)-(\nabla_Y\alpha)(X)+\alpha(\nabla_XY-\nabla_YX)-\alpha([X,Y]) $$
$$= (\nabla_X\alpha)(Y)-(\nabla_Y\alpha)(X)+\alpha\bigl(\;T_\nabla(X,Y)\;\bigr). $$
If the torsion is zero, the above equality looses a term, and one obtains rather easily Cartan's structural equations of a Riemann manifold.
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The covariant derivative version of trying to commute partial derivatives is: $\nabla_{\partial_i}\nabla_{\partial_j}-\nabla_{\partial_j}\nabla_{\partial_i} - 0 = R(\partial_i,\
up vote 4 Torsion is measuring something different: It is the covariant derivative of the soldering form $\sigma\in\Omega^1(M,E)$ which you use to identify the vector bundle $E$ with $TM$, where
down vote $E$ is the bundle you are considering your covariant derivative on.
Thanks. It is possible to explain this in simpler terms ("soldering form" seems to be advanced from the basic place in diff. geometry in which I now stand)? Or do you think a true
understanding of the torsion tensor requires more advanced concepts? – R S Feb 23 '13 at 16:05
There are lots of nice explanations of torsion here mathoverflow.net/questions/20493/… although you may consider them to be advanced also. – Paul Reynolds Feb 23 '13 at 19:12
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The Levi-Civita connection is just a very special one - torsion free. It is interesting that the same geometry may be described by switching to another, non-torsion-free connection. E.g.
there is such a version of general relativity which is called teleparallel formulation. While the curvature tensor (based on the new connection) vanishes, all the deviation from flatness has
up vote been shifted to the torsion tensor (better: vector-valued 2-form). Einstein exchanged his ideas with Cartan in the 1920 about that. Torison has also an equivalent in physics as dislocation
0 down density (disclocations are defects in crystals). The theory has been developed in the 1950's by Kondo, Bilby and Kröner. See also the book Ricci calculus by J.A. Schouten. To summarize, it is
vote not that important that the connection is symmetric, it is merely a matter of choice. The metric, for instance, is independent of the connection.
You put the cart before the horse. First, you fix a metric. Then you fix a connection compatible with the metric. If you are interested only in geodesic then, as E. Cartan observed, there
are several connections compatible with the metric that give the same geodesics. If you are interested in more than geodesics, then curvature matters, and the curvature does depend on the
choice of connection. – Liviu Nicolaescu Feb 25 '13 at 13:14
I don't think that contradicts what I said, but it is surely more precise to start the discussion with the metric, yes. Thanks for clarifying. – ClassicalPhysicist Feb 27 '13 at 10:45
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Not the answer you're looking for? Browse other questions tagged dg.differential-geometry connections riemannian-geometry torsion or ask your own question.
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Topic: How do you create a complex proof?
Replies: 1 Last Post: Nov 5, 2007 8:19 PM
Messages: [ Previous | Next ]
Stephen How do you create a complex proof?
Posted: Nov 5, 2007 4:02 PM
Posts: 1
From: Stratford, NJ with Hello,
classes at Regis Univ.
Registered: 11/5/07 I need help with Discrete Mathematics having been away from all Mathematics for over 20 years. I have been asked to provide proofs for the following problems and don't
have any idea where or how to start. Can someone please help or point out some resources for learning the basics of Discrete Mathematics?
In our text we use X' for not X also A -> B means "A implies B" or if A, then B.
1. (A' -> B') ^ B ^ (A -> C) -> C
2. (A -> C) ^ (C -> B) ^ B -> A'
3. [A -> (B v C)] ^ C' -> (A -> B)
I could use any assistance regarding learning how to prove problems like those above.
Date Subject Author
11/5/07 How do you create a complex proof? Stephen
11/5/07 RE: How do you create a complex proof? Ben Brink
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angular acceleration problem
1. The problem statement, all variables and given/known data
A flywheel turns through 22 rev as it slows from an angular speed of 6.5 rad/s to a stop. (a) Assuming a constant angular acceleration, find the time for it to come to rest. (b) What is its angular
acceleration? (c) How much time is required for it to complete the first 11 of the 22 revolutions?
2. Relevant equations
n revolutions = 2npi radians
2npi= 1/2 alpha(t^2)
omega final^2 = omega initial^2 + 2alpha (2npi)
3. The attempt at a solution
i already got parts a and b, but i am confused about part c
the problem states that the angular acceleration is constant, so why can't i just use the kinematic analogue that i used to find the time in part a to find part c. please help!
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Direct sum of a range and T-invariant subspace problem
March 30th 2010, 03:17 PM #1
Nov 2009
Direct sum of a range and T-invariant subspace problem
I've been working for this problem for quite a while, but am seriously stuck now, so I would appreciate any help!
Here's the problem:
Suppose $V=R(T) \oplus W$ and W is T-invariant. Prove that $W \subseteq N(T)$
where R(T) is the range of linear transformation $T:V \rightarrow V$ and N(T) is a null space of T, and W is a subspace of V.
Here's what I got:
I need to show that for all $w \in W, w \in N(T)$ and w is in N(T) if and only if T(w)={0}, so what i need to show is for all $w \in W, T(w) = 0$.
Since V is a direct sum of R(T) and W, then $V=R(T)+W$ and $R(T) \cap W = \{0\}$ . Since V=R(T)+W, i know that for all $v \in V$, there exists $y \in R(T)$ and there exists $w \in W$ such that v=
y+w. Since there exists y in R(T), then there exists an $x \in V$ such that T(x)=y. And from W being T-invariant, i know $T(w) \in W$.
Ok, knowing it is good, but i cannot relate that to N(T). What do I do next??
Thank you for any help!!
Last edited by vanishingpoint; March 30th 2010 at 03:49 PM.
If we take $w \in W$, then from W being T-invariant, we get that $T(w) = w'$ for some $w' \in W$; that is, $T(w) \in W$.
But we also know that $T(w) \in Range(T)$ by definition, therefore we get that $T(w) \in W \cap Range(T)$ but this is possible only if $T(w) = 0 \Rightarrow w \in Null(T) \Rightarrow W \subset
Oh, thank you so much! it is so easy now that you explained it! makes perfect sense. Thanks a lot!
March 30th 2010, 04:34 PM #2
Super Member
Aug 2009
March 30th 2010, 05:25 PM #3
Nov 2009
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Neglect Level Control at Your Peril
Neglect Level Control at Your Peril -- This first article in a four-part series examines reset mode tuning issues.
Figure 2 illustrates the response in level to an upset to the material balance. When the material balance is closed (imbalance is zero), vessel level is constant. In Figure 2, this is the case prior
to time 0. At that point the discharge valve opening is reduced by 10%, which decreases discharge flow and causes level to increase.
All examples we'll discuss pertain to a straight-walled vessel containing a constant density liquid, hence the ramp has a constant slope as in Figure 2. We'll express the level as a percentage of the
level measurement span. The response in Figure 2 is for a 12,000-L vessel. The average flow through the vessel is 200 L/min, giving a residence time of 60 min or 1 hr.
A simple characterization of a level process relies on two parameters whose value can be readily obtained from the response in Figure 2:
Process gain, K[F]. This is the effect of a 10% change in the controller output on the slope of the ramp. From Figure 2, a 10% reduction in the controller output causes the slope of the ramp to
change from zero to 0.49%/min. So:
K[F] = (0.49 %/min)/10% = 0.049 (%/min)/%
A decrease leads to an increase in level, so the process is reverse acting.
Process lag, θ. The material balance suggests the ramp should commence immediately, as indicated by the dashed line in Figure 2. Instead, the slope changes gradually from zero to 0.49 %/min. By the
time the slope reaches 0.49 %/min, the actual response lags by 0.4 min.
The process lag shown by the ramp in Figure 2 includes the following:
Control valve lag. A digital system can change its output by 10% very quickly but all final control elements exhibit some lag in responding to a change in their input signal. Rarely are the response
characteristics of the final control element well known.
Measurement device lag. This depends on the measurement technology employed and, sometimes, on how the device is installed. Rarely is this lag quantified.
Smoothing of the process variable. When smoothing is applied either within the measurement device or the control system, quantitative values are available. However, with some level measurement
technologies, smoothing can be applied externally.
The lag observed in Figure 2 is roughly the sum of these lags. The combined effect often is approximated by a transportation lag or dead time. In the simple approximations of the dynamics of a level
process, the process lag θ is considered to be entirely transportation lag.
Unfortunately, for many level loops, conducting a test such as in Figure 2 is impractical due to the presence of noise on the measured level value and variability in the feed flow to the vessel.
Testing procedures are available to determine K[F] and θ in face of both measurement noise and flow upsets. One approach is to use a pseudo-random binary signal (PRBS) for the output to the control
element. Model predictive control technology relies on such tests to find process characteristics. However, such tests are long-duration (days) and difficult to justify for level control
When a proportional-integral (PI) controller is used, the relationships between the tuning coefficients and the process characteristics are:
Controller gain, K[C]: inversely proportional to the process gain K[F]; inversely proportional to the lag θ.
Reset time, T[I]: not affected by K[F]; directly proportional to θ.
Especially for large tanks with long residence times, tuning equations often suggest unreasonably large values for K[C]. Most tuning relationships link the product K[F]K[C] (the loop gain) to the
process dynamics. The large value for K[C] results from two factors:
1. For responsive processes, the tuning equations suggest a large value for K[F]K[C]. For a vessel with a residence time of 1 hr, a lag of 0.4 min is trivial.
2. For large vessels, the process sensitivity K[F] is small.
Using the Ziegler-Nichols tuning equations, the suggested values for the tuning coefficients are:
K[C]= 1/(K[F] θ) = 0.9/{[0.049 (%/min)/%] × (0.4 min)} = 46 %/%
T[I]= 3.33 θ = 3.33 × (0.4 min) = 1.33 min
The performance objective for the Ziegler-Nichols tuning equations is a response with a quarter decay ratio, which usually provides a rapid response to a disturbance. Figure 3 presents the response
to a 10-min 50-L/min increase in one feed for the level process in Figure 1. These tuning coefficients maintain the vessel level very close to its set point — the maximum level deviation is
approximately 0.2%. The response period, P, is 2.7 min. Also note the feed flow change is translated quickly into a discharge flow change.
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Running engines for electric power - SailNet Community
Running engines for electric power
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Running engines for electric power
I hope this is the right area to ask this...
How can one determine how long an engine has to run (to recharge batteries) in order to provide enough power to run an electric appliance for a certain length of time?
For example, how many hours a day would a typical, modern, 30HP engine have to run (per day) to keep a set of batteries topped off if you were running a chart plotter and radar unit 24 hrs, nav
lights 10 hrs and recharging a laptop once (per day).
And how would you figure out how many/what type of solar panels could reduce that by X hours (assuming decent sunlight).
Is there an "easy" way to figure it out? I know there's math involved, but hopefully nothing too complex.
To top it off, I'm not all that clear on the difference between watts, volts and amps.
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Originally Posted by
I hope this is the right area to ask this...
How can one determine how long an engine has to run (to recharge batteries) in order to provide enough power to run an electric appliance for a certain length of time?
For example, how many hours a day would a typical, modern, 30HP engine have to run (per day) to keep a set of batteries topped off if you were running a chart plotter and radar unit 24 hrs, nav
lights 10 hrs and recharging a laptop once (per day).
And how would you figure out how many/what type of solar panels could reduce that by X hours (assuming decent sunlight).
Is there an "easy" way to figure it out? I know there's math involved, but hopefully nothing too complex.
To top it off, I'm not all that clear on the difference between watts, volts and amps.
Hello Greg,
Watts = volts x amps.
The boat's electrical system is usually a mixture of A/C and DC. The DC is most often 12 volts, or 12vdc (12 volts dc). If you have an inverter then you can convert that 12vdc into 120vac (120 volts
a/c) which is what is in your house. Watts = volts x amps, so if you have something using 12vdc boat power at 1 amp then it is using 12 watts. If you have something that needs 240 watts and it
requires 12 volts then it uses 20 amps. If you have a solar panel that is 100 watt and it is producing that amount then it is making approximately 100 divided by 12 or about 8 amps.
So if you aren't worried about losses, etc, then just add up what everything is using - let's say a fan uses 20 watts, a television 30 watts, etc, etc, and you end up with lets say 300 watts, then
you can figure out how much power you need per day in amp hours by dividing 300 by 12 equals 25 amps for 24 hours or 600 amp hours. One of those big deep cycle batteries is often 100 amp hours at 12
volts, so you can start to see about what sized batteries you would need as a "buffer".
Why a "buffer" ? Because you don't want to run the engine all the time, you want to charge up some batteries, let the load drain them to some safe level where they aren't overly discharged, then
charge them back up. The bigger the buffer the less often you have to run the motor. Solar panels are great because they are like running the motor at a low charge level all the time that the sun is
up, so they charge during the day and then don't do much of anything at night. Still, no matter how big the battery bank you still have to generate as much power as you use, so if your load is 300
watts constant then you need to be generating and storing enough power to replenish that 300 watts per hour into your battery bank.
There are tons of "but, if's ..." involved, there are losses that mean that you don't get all the power that you generate, there are concerns about how deeply you discharge the batteries before you
charge them back up, there are concerns about wearing out the diesel engine by running it for short periods of time which some say doesn't allow it to warm up and be properly lubricated, etc, but the
basics are pretty basic and it doesn't take much learning to get the basic idea. One concern you would have is that no matter how big your generator is you can't charge the battery bank any faster
than what it will accept, let's say 10% of its capacity for easy math, so if you have 10 deep cycle batteries which is 1000 damp hours then you can only charge it at 100 amps (10% of 1000 amp
capacity equals 100 amps), so 1200 watts (12 volts times 100 amps) is the most power the bank could even accept, hooking the bank up to a 10000 watt generator would just waste fuel.
There's more to it ...
What are you pretending not to know ?
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Thanks wind_magic! I think I understand most of what you're saying. Except I am confused about...
Originally Posted by
So if you aren't worried about losses, etc, then just add up what everything is using - let's say a fan uses 20 watts, a television 30 watts, etc, etc, and you end up with lets say 300 watts, then
you can figure out how much power you need per day in amp hours by dividing 300 by 12 equals 25 amps for 24 hours or 600 amp hours.
You say "if a TV uses 30 watts"... how do you know how many watts it uses? And is that "per hour"? Or is that "while it's on". I guess I'm confused about if watts is a measure of "flow" or a measure
of "amount". (Like "gallons per hour", or just "gallons")
So how much "power" does the TV suck out of the batteries if on for 3 hours? And what unit of measure do you use to express the absolute amount of power still in the battery, volts? or watts?
And about the engines... how many watts does a modern, 30HP marine engine produce? (I assume it's actually an alternator making the electricity, correct? Or do marine engines use something
I'm learning... little by little...
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Originally Posted by
Thanks wind_magic! I think I understand most of what you're saying. Except I am confused about...
You say "if a TV uses 30 watts"... how do you know how many watts it uses? And is that "per hour"? Or is that "while it's on". I guess I'm confused about if watts is a measure of "flow" or a measure
of "amount". (Like "gallons per hour", or just "gallons")
So how much "power" does the TV suck out of the batteries if on for 3 hours? And what unit of measure do you use to express the absolute amount of power still in the battery, volts? or watts?
And about the engines... how many watts does a modern, 30HP marine engine produce? (I assume it's actually an alternator making the electricity, correct? Or do marine engines use something
I'm learning... little by little...
Watt is a measure of work, you can think of it like flow. In terms of the water analogy volts is the speed the water moves at and amps is the size of the river, and multiplying flow rate times size
gives you volume [edit, really it is more like flow, or total inertia, not volume], which is sort of like what watts are.
How many watts does the motor produce, that depends on the alternator which is the electrical generator on the motor. Your alternator will be rated to produce a certain number of amps, you can look
up the model of your alternator and see what it will produce. You can also change the alternator to get a bigger one, usually, that will produce even more power. The alternator typically uses very
little of the power generated by your motor so a lot of power gets wasted unless you are also using the motor to actually move the boat somewhere, in which case the electrical power is kind of "free"
since you're using the motor anyway.
How much power do things use ? Turn your television around and look at the back and it will have a little plate saying how many watts it uses. Or you can look in the manual that came with it. Or if
it is something small you can look at the little wall wart (the power adapter) that came with it and see how many amps that generates. Or you can just get a table of common values and guess. Or you
can get something like a "watts up" meter and actually measure how much power the device is using.
Power usage is while it is on, typically, so if it uses 30 watts for 1/2 hour then that is the same as using 15 watts for a full hour, or 60 watts for 15 minutes. Does that make sense ? 30 watts at
12 volts is about 2.5 amps, so if you want to run it for an hour you need 2.5 amp hours of battery capacity, so a 100 amp hour battery (in theory) would run a 2.5 amp load for 40 hours before being
completely discharged.
Power usage isn't always constant - there are usually two kinds of power usage, one when you are using it, and one when it sits idle. So for example your old VCR player might use 20 watts while you
are using it but still use 0.2 watts when it is turned off (to keep the clock set, etc), so even if you aren't using something it may be using small amounts of power. Usually this is called "standby"
power and you can find it in the documentation that came with whatever device it is.
Keep asking questions and I or someone else will answer them ...
What are you pretending not to know ?
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Let me try using your water analogy again, I wasn't very clear ...
Pretend you have "X" amount of water going past a bridge in a given period of time, that is kind of like watts, it is the rate times something like the area of the cross-section of the river. Now, if
you reduce the size of that river then the speed of the river has to increase to get the same amount of water to flow past the bridge in a given period of time, that is kind of like increasing the
voltage which reduces the amount of amps to get the same number of watts. Watts is kind of like the ability of the river to do work, its rate times its size, and amp hours at a given voltage is kind
of like the volume of the water. So if you have something that uses 12 volts at 10 amps that's 120 watts. But you could also have something that uses 24 volts at 5 amps and that is still 120 watts,
they both use the same amount of "power", but the 24 volt electricity is "moving faster" (24 volts), but the river is half as wide (5 amps instead of 10 amps). The lake holding the water is kind of
like a battery and that is amp hours at a given voltage, so if you have a 12 volt battery bank and it holds 100 amp hours then it can supply 12 watts (12 volts at 1 amp) for 100 hours which equals
100 amp hours, or it could supply 6 watts (12 volts at 1/2 amp) for 200 hours, etc (in theory).
What are you pretending not to know ?
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Buy Nigel Calder's books on boat maintenance. Tons of excellent info all spelled out for you on alternator power, battery banks, battery meters, inverters, solar & wind, load calculations, etc, etc.
Big book full of info that could not be translated in a few posts here.
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Thanks Xort, I'll look it up.
Edit: It now on my Amazon wishlist.
Last edited by GregX999; 12-20-2009 at 07:37 AM.
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Mass of block given μK and Force?
1. The problem statement, all variables and given/known data
A student pushes a block with 240 N of force across a horizontal surface with a coefficient of kinetic friction of 0.4. The block accelerates at a rate of 0.88 m/s
. Find the mass of the block.
2. Relevant equations
= F
/ F
3. The attempt at a solution
My first attempt was to figure out mass using F=ma
m = f/a
= 240 N / 0.88 m/s^2
= 272.7 kg
but this is not taking into account the kinetic friction...
So what I did was
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x F
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Bernhard Riemann had the touch of gold. Everything he worked with, he revolutionized. He was born in 1826 to a Lutheran pastor, the second of six children. He was born in the little town of Hanover,
Germany. He was a shy boy and young man when he was around people, but he was very bold in his scientific thought.
His dad was his first teacher and he began his first arithmetic at the age of six. He took right off and never stopped. At age ten, he took his instruction from a professional teacher, but the pupil
soon began to be the teacher. When he was fourteen years old, Riemann lived with his grandmother, and he sent gifts to his family. One gift to his parents was an original perpetual calendar. He
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true for very small portions of space.”^1
Five years later, Riemann wrote a hypothesis on prime numbers that is still one of the outstanding challenges to mathematicians. His hypothesis on prime numbers was to determine how many prime
numbers are less than any given number n. His equation is as follows:
where s is a complex number, say s = u + iv , i = s.^2
Bernhard Riemann had poor health throughout much of his life. He contracted pleurisy at the age of 35 and passed away four years later. During his life, he held closely to his Christian faith and
considered it to be the most important aspect of his life. At the time of his death, he was reciting the Lord’s Prayer with his wife and passed away before they finished saying the prayer. He left
her with one child. Sometimes people’s lives are short, but they make great contributions while they are here. Bernhard Riemann’s life was of this type.
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Lecture 4
G22.2590 - Natural Language Processing - Spring 2003 Prof. Grishman
Lecture 4 Outline
February 13, 2003
Part-of-Speech Tagging Using Hidden Markov Models
Corpus-Based Methods
Natural language is very complex
- we don't know how to model it fully,
so we build simplified models which provide some approximation to natural language
How can we measure 'how good' these models are?
- we build a corpus,
annotate it by hand with respect to the phenomenon we are interested in,
and then compare it with the predictions of our model
- for example, how well the model predicts part-of-speech or syntactic structure
How to train the model?
- need a goodness metric
- train by hand, by adjusting rules and analyzing errors
- train automatically
* develop new rules
* build probabilistic model (generally very hard to do by hand)
Statistical Part-of-Speech Tagging (J&M sec 8.5)
Looking at words in isolation:
Given a word w, what tag t should we assign to it?
We want to assign the tag which maximizes the number of correct assignments.
The probability of getting the assignment correct is P ( t | w )
So we want to assign the tag = argmax(t) P ( t | w )
We can estimate P ( t | w ) as
(number of times w is tagged as t in corpus) / (number of times w appears in corpus)
(this is the 'maximum likelihood estimator', J&M p. 200)
Using Bayes' rule, we can rewrite
P ( t | w ) = P ( t ) * P ( w | t ) / P ( w )
This will be helpful when we take context into account.
Looking at the tag of the previous word
Call the tag of the previous word 'u'
Want to find argmax(t) P ( t | u, w)
Under independence assumptions, = argmax(t) P ( t | u ) * P ( w | t )
P ( t | u, w ) = P ( t ) P ( u, w | t ) / P (u, w)
= P ( t ) P ( u | t ) P ( w | t ) / ( P ( u ) P ( w ) ) [ independence assumption ]
= P ( t | u ) P ( w | t ) / P ( w )
Taking into account both the previous and following tags in deciding the tag of the current word.
We want to find the most likely sequence of tags T = t[1], t[2], ... , t[n]
given the sequence of words W = w[1], w[2], ... w[n]
I.e., find argmax(T) P ( T | W ) = argmax(T) P ( T ) P ( W | T ) / P ( W )
= argmax(T) P ( T ) P ( W | T )
= argmax(T) product(i) P( t[i] | t[i-1], ... t[1] ) P ( w[i] | t[i] w[i-1] t[i-1] ... w[1] t[1] )
which we will approximate as
argmax(T) product(i) P ( t[i] | t[i-1] ) P ( w[i] | t[i] )
Markov Model
Finite state networks with transition probabilities:
probability of next state depends only on current state ('Markov assumption')
By associating a particular word with each state, we get a probabilistic model of word sequences ...
a probabilistic generative model of sentences.
Hidden Markov Model (HMM)
Suppose we associate a part-of-speech with each state in a Markov Model.
We then associate an 'emission probability' P ( w | t )
of emitting a particular word when in a particular state.
This is a hidden Markov Model ...
the sequence of words generated do not uniquely determine the sequence of states.
Training an HMM
Training an HMM is simple if we have a completely labeled corpus:
we have marked the POS of each word.
We can then directly estimate both P ( t[i] | t[i-1] ) and P ( w[i] | t[i] ) from corpus counts
using the Maximum Likelihood Estimator.
Using an HMM ('decoding')
The argmax(T) given above corresponds to finding the most likely path through the model, given W.
The Viterbi algorithm provides a fast (linear in number of tokens) algorithm for this task. (J&M p. 176)
It consists of a forward pass which computes probabilities,
and a backward pass which traces the most likely path.
In the forward pass,
it builds a probability matrix viterbi [number of states+2, number of tokens + 2], and
a back pointer matrix of the same size.
viterbi [ s, t ] = max (over all paths to [s,t]) of the probability of reaching state s at token t
(the algorithm is given on p. 179)
JET HMM POS Tagger
Uses the general HMM model classes within Jet.
File pos_hmm.txt: trained on 96% of Penn Tree Bank corpus (960,000 words).
Weak on handling of unknown words (words not in training corpus)
- determining which tags (open classes) are likely for unknown words
- using morphology, esp. suffixes (even more important for more inflected languages)
Limitations of bigram tagging (limited context)
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The Pot of Gold for Public School Teachers
By JOHN SULLIVAN • February 22, 2012
It’s always nice to see a fresh-faced 22-year-old begin a teaching career with eagerness, but is this happiness actually due to the fact that they have just entered an economic utopia where there is
no concern about inflation, retirement, job loss, long hours, vacation time, job stress, health care and raises in income for doing exactly the same thing? Let’s look at what a teacher starting at
$40,000 per year has to look forward to.
• Beginning salary of $40,000 per year.
• Benefits of fifty cents per dollar of salary, including 15 paid sick days, 4 paid personal days, 6-10 paid.
• holidays, medical, dental, prescription, eyeglasses, disability insurance, life insurance, etc.
• Eight months of work per year. Three months off in the summer, two weeks off in the winter, two weeks off in the spring, five hours per day of class time and three hours per day of work outside
of class.
• Healthcare rising at 6%/year – about twice the rate of inflation.
• Teacher salaries, including cost-of-living, rising at 6% per year.
• No access to Social Security or Medicare, nor any funding for these programs.
• Non-teachers receive Social Security and Medicare, pay for Social Security and Medicare, and pay for98% of any shortfalls in these programs. It is substantially a self-funded, pay-as-you-go
• Teacher pensions based upon 75% of ending salary and increasing at 3% per year.
• No raises for additional education.
• Life expectancy of 84 years.
• Cost of money of 5%.
Finally, the sense of entitlement on the part of teachers that, having worked one day, none of the conditions that are in effect on that day can ever be changed as long as they live.
$40,000 Starting salary
$60,000 Total for eight months of work
$90,000 Annualized at 1.5 $47,130 Pension expense (Calculations follow)
(1) $15,631
Expense for free healthcare in retirement. (Calculations follow) $152,761 Total compensation on an annualized basis (and 10% less than the King of Siam).
How does this compare to a college graduate in the private sector who typically receives 25 cents per dollar of salary in benefits? Well, if his starting salary is $30,552 with benefits, his total
compensation is $38,190 – exactly one-fourth of what a teacher starts at, and right around the median income($33,000) in this country. Guess what? He can actually be fired if he does a poor job!
Guess what? He has to pay the teacher salaries out of that amount when he makes a fourth of what they do!
Guess what? It gets worse! At age 40, when total compensation for the teacher above is $329,651, the private sector counterpart gets $65,016 – a fifth of the teacher compensation. At age 60, when
the teacher is retiring, total compensation for the teacher is $886,643 – 7.6 times what his counterpart in the private sector gets ($117,425). You see, in the private sector, people don’t get paid
more in real wages for doing the same thing. They are lucky if they can keep up with 3% inflation. Guess what? It gets worse! Productivity improves every year in the private sector while student
test scores have been declining for decades.
(1) In fairness, it is true that some teachers pay up to 9.4% of their salary into their pension plan. In this example, this would be $3760 of the $47,130. In many cases, the School Boards have
kindly agreed to let the taxpayers reimburse the teachers who have their contribution deducted by increasing their gross pay. In any case, the overall teacher contributions to their pension plans are
not significant.
Calculations follow.
Age 22 $152,761 – calculations above.
Age 40 $329,651 – $90,000 increasing at 6% per year compounded for 18 years plus a $47,130 pension.
Payment and a $15,631 payment for retirement healthcare.
Age 60 $886,643- $90,000 increasing at 6% per year compounded for 38 years plus a $47,130 pension.
Payment and a $15,631 payment for retirement healthcare.
Age 22 $40,000
Age 40 $114,173 – $40,000 increasing at 6% compounded for 18 years.
Age 60 $366,170 – $40,000 increasing at 6% compounded for 38 years.
$274, 627 – beginning pension at 75% of final salary.
$542,000 – pension payment at age 84 with a 3% increase every year compounded.
$9,454,438 – total pension payments with a 3% increase every year for 24 years.
$5,076,422 – lump sum equivalent (income stream discounted back to age 60 at 5% cost of money).
$47,130 – annual payment needed getting a 5% rate of return for 38 years to have $5,384,993 at age 60 to fund the pension while still receiving a 5% rate of return throughout retirement.
Retirement healthcare
$65,911 – cost of health insurance at age 60 for a single person in 38 years. ($600/month today = $7200 increasing at 6% compounded for 38 years.) This is seriously understated since the cost of
health insurance increases with age as well as time.
$266,869 – cost of health insurance at age 84.
$3,349,305 – total payments with 6% increase every year for 24 years.
$1,683,662 – lump sum equivalent (healthcare payments discounted back to age 60 at 5% cost of money).
$15,631 – annual payment needed getting a 5% rate of return for 38 years to have $1,683,662 at age 60 to fund the health insurance while still receiving a 5% rate of return throughout retirement.
Wow! No wonder why this 22-year-old teacher is so happy! He stands to receive $5,436,168 in salary and $2,718,084 in benefits over the next 38 years. Then, over the next 24 years, he will receive
$9,454,438 in pension payments and $3,349,305 in health insurance for a total of $18,242,629 from various taxpayers. This is equivalent to an annuity of $294,236 per year for 62 years simply for
working at the same part-time job. Even the King of Siam would be jealous!
But what of the School Board? Now, many of them are former teachers, but some of the others must be able to use a calculator. One would think that there is not a single taxpayer (or Jury) who would
conclude that the School Board members, who have a fiduciary duty to act in the interests of the taxpayers and students, have done anything but act in the sole interests of the teachers.
Compensating teachers somewhere near their counterparts in the private sector would allow the Board to cut its $150,000,000 annual budget in half and lower property taxes at least 30%. Is there some
reason why they should not?
John Sullivan is a financial analyst and adviser who lives in Palatine, Illinois.
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Surprise from Martin Schmaltz
We are done with April Fools
for this year. It is high time to resume boring theory blogging. I have a few delayed reports in store...
Martin Schmaltz flashed through CERN two weeks ago. Whenever he's around, there is action. The other day, when he talked about Little Higgs at CERN, somebody from the audience called him a little
taliban. No wonder that, this time, nobody at CERN TH gave him a seminar slot. Luckily enough, he was adopted by unsuspecting experimentalists from the EP seminar.
Martin came with a talk
MSSM scalar masses and hidden sector interactions. Although this title seems well-behaved,
a look at the abstract
promises a lot. Martin wrote ...
[concerning soft scalar masses]
from commonly used spectrum codes (such as ISAJET, SPHENO, SOFTSUSY, SUSPECT) may be completely wrong
... He bothered to enumerate all the popular codes in order to annoy a maximum number of people.
If the Minimal Supersymmetric Standard Model (MSSM) is discovered at the LHC and precisely measured at the ILC, then we will learn about the superparticle masses. We expect that those masses are
determined by a high energy theory which is more fundametal than the MSSM. One possibility is that the high energy theory is a string theory. Thus, by measuring the MSSM parameters we could get a
glimpse of the fundamental theory. If we are extremely lucky, the LHC and ILC measurements could point to some specific realization of string theory.
The relation between masses measured in low-energy experiments and the fundamental parameters is not straightforward. In colliders, we would measure the MSSM parameters at the 1 TeV scale. On the
other hand, the fundamental theory directly predicts the mass parameters at the high scale, for example at the GUT scale 10^16 GeV. The two sets of parameters are related by
renormalization group equations
. Relating the low- and high-energy parameters is theoretically so important, that the task has been automatized in the publicly available codes mentioned before. The nice thing is that, within the
standard MSSM paradigm, the renormalization group equations depend only on the quantities that can be measured at low energies. Therefore unambiguous predictions of high-energy parameters from
low-energy inputs can be made. Or so it seemed.
In his recent
, Martin pointed out that the relation between the high- and low-energy parameters can be obscured by the effects related to supersymmetry breaking. He claims that the renormalization group equations
for the MSSM scalar masses contain a potentially large contribution that has been overlooked so far (the gaugino masses are not affected). This contribution comes from self-interactions in the hidden
sector that is responsible for spontaneous supersymmetry. It has been assumed so far that this effect can change only the overall normalization of the scalar masses. Martin showed that this
assumption is in general wrong and gave examples where the new effects dominate the standard MSSM contributions. The masses of the hidden sector particles are typically of order 10^11 GeV, so we will
have no experimental access to the hidden sector parameters. Therefore collider measurements of scalar superparticle masses may reveal nothing about the fundamental theory.
According to Martin, a string theory fan after listening to his talk would look like this:
This is an artistic vision. However, there is still a large class of string models where the hidden sector renormalization is not dangerous. This is the case when supersymmetry breaking is mediated
by light moduli, which includes the beloved KKLT. The moduli interact weakly, with gravitational strength, therefore they cannot mess up the MSSM renormalization group equations. On the other hand,
in the case of dynamical supersymmetry breaking a-la Seiberg, the hidden sector interacts strongly and large effects are difficult to avoid.
For a comic book version of the story, have a look at the
. For technical details, you should consult the
1 comment:
In old fashioned string theory GUTs, the
different matter particles which form a
16 of SO(10) or 10 + 5 bar of SU(5) do not come from the same "points" in the extra dimensions anyway; unification occurs at the Kaluza-Klein scale and there is Wilson line breaking to 3-2-1
right there. So
Martin's very interesting work doesn't make those models look much different. It affects purely field theoretic models where over some range of scales one would have had e.g. SO(10) with the
pieces of a 16 all coming from the "same" place. In Martin's story, the gauge couplings still unify, so one still sees a hint of unification, in any case, if it occurs.
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ideal completion
ideal completion
The ideal completion of a category $C$ generalises the notion of ideal completion of a partially ordered set. It is formed by taking the category of ideals? of $C$. It is the completion of $C$ under
filtered colimits of monomorphisms.
See page 24 of An Outline of Algebraic Set Theory.
“The actual definition requires either some care in specifying choices of monomorphisms, as is done in [4], or a sheaf-theoretic approach as in [5].”
• [4] S. Awodey, C. Butz, A. Simpson and T. Streicher, Relating first-order set theories, toposes and categories of classes. In preparation, 2007. Preliminary version available here.
• [5] S. Awodey and H. Forssell, Algebraic models of intuitionistic theories of sets and classes, Theory and Applications of Categories 15(1): 147-163, 2005.
Revised on June 9, 2011 07:36:22 by
Andrew Stacey
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Oswego, IL Math Tutor
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Norm on a field
From Encyclopedia of Mathematics
A mapping
The norm of [4]. See also Valuation.
Examples of norms. If absolute value or modulus of the number
Finite fields and their algebraic extensions only have the trivial norm.
Examples of norms of another type are provided by logarithmic valuations of a field
Norms satisfying condition 4) are known as ultra-metric norms or non-Archimedean norms (as distinct from Archimedean norms which do not satisfy this condition (but do satisfy 3)). They are
distinguished by the fact that
A norm metric on
The structure of all Archimedean norms is given by Ostrowski's theorem: If
Any non-trivial norm of the field
A similar formula is also valid for algebraic number fields [2], [3].
If Complete topological space). One of the principal modern methods in the study of fields is the imbedding of a field Adèle). If
A norm on a field
[1] N. Bourbaki, "Elements of mathematics. Commutative algebra" , Addison-Wesley (1972) (Translated from French)
[2] Z.I. Borevich, I.R. Shafarevich, "Number theory" , Acad. Press (1966) (Translated from Russian) (German translation: Birkhäuser, 1966)
[3] S. Lang, "Algebra" , Addison-Wesley (1984)
[4] A.G. Kurosh, "Lectures on general algebra" , Chelsea (1963) (Translated from Russian)
[5] J.W.S. Cassels (ed.) A. Fröhlich (ed.) , Algebraic number theory , Acad. Press (1986)
Non-Archimedean norms satisfy Archimedean axiom, whence the appellation.
How to Cite This Entry:
Norm on a field. V.I. Danilov (originator), Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Norm_on_a_field&oldid=16545
This text originally appeared in Encyclopedia of Mathematics - ISBN 1402006098
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Reactance help please! (electrical engineering)
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Wikipedia and all about circuits give the following formula for reactance of an inductor: \(\large X_L=2\pi fL\). This is given in ohms. My question is about the concept of reactance.
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More specifically, I am wondering about its relation to resistance and why it is imaginary.
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Unlike resistance, reactance " impedes" the flow of AC current but does not consume energy. It effectively stores and returns energy during a cycle. Reactance is real but but is used as the
imaginary part in the complex number formalism that is useful in AC circuit analysis. Reactance comes naturally out of the differential equation for describing AC circuits with capacitors and
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How does the usage of complex numbers to represent impedance get incorporated into describing the phenomenon of impedance? I mean, when do they use j^2=-1?
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Unlike DC circuits with resistive loads AC voltages or currents do not rise and fall at the same time in fact voltages or currents at different parts of the circuit do not rise and fall at the
same time. In capacitors current is rising before voltage and in inductors voltage is rising before current. They are said to be out of phase. This phase difference has significant consequenses
in the behavior of the circuit. In AC circuits the voltages and currents have a sinusoidal form e.g. \[V=V _{m}\sin(\omega t+\phi) \]where w=2pi*f and f is the frequency and phi is a phase shift
relative to a reference. Since we need to multiply and divide currents and voltages in circuit analysis doing so with sines and cosines with different arguments can be exceedingly tedious. There
is a representation of cosine and sines in and exponential form given by Eulers Theorem. \[e ^{jx}=\cos(x) + jsin(x)\] so \[\cos(x)= real(e ^{jx}) \] and \[\sin(x)= Imag(e ^{jx)}\] In actual
circuits the voltages and currents are complex but after the arithmetic is finished you only use the real part of the result. With this approach the math becomes simple complex arithmetic. When
the final results are obtained the complex notation for current and voltage are converted back to sinusoidal notation.The validity of this approach is verified by comparing the results to those
using the fundamental differential equation to solve the circuit. This brief discussion may not be very clear until you work through some examples To show how j gets into the discussion. Take a
simple circuit of a capacitor driven by a voltage source of frequency f. For a capacitor\[ I = C(dV/dt)\] where \[V=\Re e(V _{0}e ^{j2\pi ft})\] so \[I=\Re e(V _{0}e ^{j2\pi f}/(-j/2\pi fC) )\]
Ohms law is valid for reactance. So \[X _{c}=-j/(2\pi fC)\] You can do the same for inductance or a combination of a R, C and L. where you find that Impedance Z can be written as\[Z = R +j(X _{L}
-X _{C})\].
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Individual investors might be able to properly analyze a stock or even a few stocks. But while an exchange traded fund (one based on an equity index) is essentially just a basket of stocks, it might
not be feasible for an individual investor to analyze an ETF as extensively as he or she would a stock, as the investor would need to value every single stock that the ETF represents. Fortunately,
some valuation methods and principles that apply to stocks can also be applied efficiently to an ETF.
In this article, I will be talking about two Benjamin Graham’s investing principles that I use to help me value ETFs: Avoiding stocks (or in our case, ETFs) that have a Price/Earnings ratio that’s
higher than the earnings yield of the stock + the stock’s growth rate (found this golden nugget at
www.Joshua Kennon.com
), look for stocks with price to book below 1.5, and investing with a margin of safety.
Look for ETFs with P/E below its earnings yield + growth rate
Say, an ETF you want to value has a P/E ratio of 15 (Each ETF provider might have its own way of calculating P/E, but we will get to that a little later). We start our analysis by finding the
earnings yield of the ETF. You can get the earnings yield by dividing earnings by the price or dividing 1 by the P/E ratio. For our example, 1/15 = 6.6% earnings yield. The earnings yield is the
profits that the company earned in the most recent 12-month period, expressed as a percentage of its market cap or share price, depending on how you look at it.
Let’s also say that the provider of the ETF estimates that the fund will grow earnings at 15% over the next 3-5 years. The 15% growth rate + the 6.6% earnings yield would equal to 21.6. The ETF has a
P/E ratio of 15, so it passes the test of having a P/E ratio below its earnings yield + growth rate. We still need to look at the other 2 principles before we make any kind of decision, though.
Identify ETFs with price to book below 1.5
Benjamin Graham said that, for defensive investors, the price to book should generally not be above 1.5, but exceptions can be made if the P/E ratio is below 15. As a rule of thumb, he said that the
P/E ratio times the price to book ratio shouldn’t be higher than 22.5. For simplicity’s sake, our example ETF will have a price to book of 1.5. So, 1.5 times 15 will equal to 22.5. This principle is
meant for defensive investors, but if you don’t mind taking a little bit more risk, and you know how to properly research an ETF, then it can be ok for you to buy that ETF at a higher valuation.
Margin of safety
The margin of safety principle states that investors should only invest in a stock if the stock’s price is significantly below its intrinsic value, so as to reduce the risk of losses. By applying the
first principle to our example ETF, we’ve found that the ETF’s rough intrinsic value can be said to be around a P/E of 21.6 (earnings yield 6.6% + 15% growth rate). And the current price of the ETF
is a P/E of 15 (of course in real life the ETF will have a share price, but we will just use the P/E ratio as a proxy for price here). So, there is a margin of safety in our example, whether the
margin of safety is significant enough is, of course, up to the individual investor.
With valuation methods (at least the ones that I talked about in this article) only giving us a very rough estimate of intrinsic value, forecasted growth rates that might not materialize or cover a
long enough period, and as I mentioned earlier, different ETF providers potentially calculating the P/E ratio differently, which can result in inaccurate P/E ratios, investing with a margin of safety
becomes especially important in the case of ETFs.
If you have any questions, or have anything that you would like to share, please don’t hesitate to comment. Thank you for reading, and may you always sustain good returns on your portfolio. Take
(Captain obvious side note: While I believe that the Benjamin Graham’s principles I talked about in this article can help us tremendously in valuing ETFs, investors also need to take into account
other things like fees and etc.)
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MATH M118 2806-2810 FINITE MATHEMATICS
Mathematics | FINITE MATHEMATICS
M118 | 2806-2810 | --
P: Two years of high school algebra or M014. Sets, counting, basic
probability, including random variables and expected values. Linear
systems, matrices, linear programming, and applications. Credit
given for only one of M118, A118, or the sequence D116-D117.
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Hi i'm new to this forum. I have a problem with a question on number theory which requires proving. The question is "Let a, b, c be integers. Prove that if a^2|b and b^3|c then a^4b^5|c^3." I do not
know how to go around to solve this question please help thanks. I tried expanding a^2|b an...
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