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Partial Fractions Question
Hi Need help to express the following in partial fractions: $\frac{x^2-2}{(x+1)(x-2)}$ P.S
What do you need to do with this expression? The order of the numerator is the same as the denominator, you may not need to make partial fractions. If you need to consider $\frac{x^2-2}{(x+1)(x-2)} =
\frac{A}{x+1}+\frac{B}{x-2}$ $x^2-2 = A(x-2)+B(x+1)$ choose values for x to solve for A and B.
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Derivative help using Square Root
If you know that a square root is the same as a power 1/2, then you can just apply the power/exponent rule.
wild_flowr69 I understand your problem as, $2\sqrt{x}+3\sqrt[3]{x}$ which you can express with exponents, $2x^{1/2}+3x^{1/3}$ then, use the exponent rule, $x^{-1/2}+x^{-2/3}$ $\left<\begin{array}
{ccc}\mathbb{Q}&\ & \ \\ \ & \mathbb{E}& \ \\ \ & \ & \mathbb{D}\right>$
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complex roots confusion
August 20th 2012, 09:41 AM #16
Re: complex roots confusion
Look at this solution.
The polynomial has degree four and has all real coefficients. That tells us that if there are any complex roots (i.e. the imaginary parts non-zero) they must occur in conjugate pairs. Thus there
could be zero, two, or four complex roots. In fact there are two. There are also two irrational roots. See the web link above.
Re: complex roots confusion
Ok thank you again plato it's a question on a web site and it say that it has four complex roots
thats what was confusing me.
Re: complex roots confusion
Any sites were I can go and test myself on this matter?
Re: complex roots confusion
Are you sure that you copied the equation correctly?
You can use WolframAlpha to solve such equations.
Re: complex roots confusion
Yeah your right that x^2 should be possitive
Re: complex roots confusion
Question state the number of complex roots,the possible number of real roots
of x^3 + 4x^2 + 5x -1 = 0
complex roots = 3 why(complex numbers include real numbers)
real roots = 1 or 3 why(1 in case two are complex,3 non may be complex)
are my statements correct?
Re: complex roots confusion
Yes, that is true.
real roots = 0,2,4 ( the reason for my answers are 0 the answer may all be complex) and this is where i'm confused
real numbers are on the number line -3,-2,-1,0,1,2,3ect
Well, not just integers, of course, numbers like " $\pi$", " $-\sqrt{3}$", etc. are real numbers. And you can say a little bit more. By "DeCarte's rule of signs", there can be either 0 or 2
positive real roots (because the sign changes at " $-3x^3$" and at "+3"- two sign changes. DesCarte's rule of signs says the number of positive real roots is at most the number of sign changes
but may be less by an even number. Now, replacing "x" by "-x" swaps positive and negative roots and $(-x)^4- 3(-x)^3+ (-x)^2- (-x)+ 3= x^4+ 3x^3+ x^2+ x+ 3$ which also has 0 or 2 sign changes.
But if a complex number is say 3(imaginary part=0) is 3 complex or
Both. The set of all real numbers is a subset of the set of all complex numbers.
For the question about rational roots, you can use the "rational roots theorem": if the rational number $\frac{m}{n}$, with m and n integers, of the polynomial equation $a_nx^n+ a_{n-1}x^{n-1}+ \
cdot\cdot\cdot+ a_1x+ a_0= 0$, all coefficients being integers, then n must evenly divide the "leading coefficient", $a_n$ and m must evenly divide the "constant term", $a_0$.
Here, $x^4 - 3x^3 - x^2 - x + 3 = 0$ so the "leading coefficient" is 1 which is evenly divisible only by 1 and -1. The "constant term" is 3 which is evenly divisible only by 1, -1, 3, and -3.
That is, the only possible rational roots are the integers 1, -1, 3, and -3. You can put each into the polynomial to see which, if any, actually satisfy the equation.
Re: complex roots confusion
Did you miss my last post.
Re: complex roots confusion
I think that you have misunderstood the whole point of this discussion.
Here is a direct quote: " the cubic polynomial $x^3 + x$ has two complex roots"
Now that polynomial has three roots, one real and two complex.
As an accident of history, the language of roots of a polynomial we use the words complex root to mean a complex number with a nonzero imaginary part. Thus $x^3 + 4x^2 + 5x -1 = 0$, has at least
one real root and at most two complex roots.
NOTE: above we are talking about complex roots, not complex numbers as such.
Re: complex roots confusion
Yes exactly plato two complex and one real thats what I got because complex
roots always come in pairs so that web page is wrong.
Re: complex roots confusion
Re: complex roots confusion
Yes, because non-real solutions to an equation with real coefficients must come in complex conjugate pairs, there must be either 2 or no non-real solutions. That leaves either 1 or 3 real roots.
And we can say if there only one real root it must be positive. If all three roots are real, one is positive and the other two negative.
Re: complex roots confusion
Just need my work checking(I'm sure it's all correct now)
For each equation state the number of complex roots,the possible number or real roots
and the possible rational roots.
1) -x^4 = 0 (complex roots 0 or2 or 4)
(real roots 2 or 4)
(rational roots 0)
2) 2x^5 - 4x^4 - 4x^2 + 5 = 0 (complex roots 0 or 2 or 4)
(real roots 1 or 3 or 5)
(rational roots + or - 5/2 or 1/2 or 1 or 5)
3)4x + 8 = 0 (complex roots 0)
(real roots 1)
(rational roots + or - 1/4 or 1/2 or 1 or 2 or or 8)
4)-2x^6 - x^2 + x - 7 = 0 (complex roots 0 or 2 or 4 or 6)
(real roots 2 or 4 or 6)
(rational roots + or - 1 or 1/2 or 7 or 7/2)
5)x^10 + x^8 - x^4 + 3x^2 - x + 1 = 0 (complex roots 0 or 2 or 4 or 6 or 8 or 10)
(real roots 0or 2 or 4 or 6 or 8 or 10)
(rational roots + or - 1)
Re: complex roots confusion
Please someone Please check my work above
Thank you.
Re: complex roots confusion
We don't see your answers in order to check them. Add your responses.
August 20th 2012, 09:48 AM #17
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August 27th 2012, 01:41 PM #30
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SHABNAM AKHTARI, Queen's
Quartic Thue Equations [PDF]
Applying the theory of linear forms in logarithms, we will give upper bounds upon the number of integral solutions to binary quartic Thue equations. We will treat a certain family of quartic
binary forms using a classical theorem of Thue based on Padé approximation to binomial functions. We shall also discuss the relation between the number of integral solutions to quartic Thue
equations and integral points on elliptic curves.
YURI BILU, Université Bordeaux 1
Uniformity in Galois Representations [PDF]
We prove that there exists an integer p[0] such that for any non-CM elliptic curve E over Q and any prime p > p[0] the image of the representation of Gal([`(Q)]/Q) induced by the Galois action on
the p-division points of E is not contained in the normalizer of a split Cartan subgroup. This gives a partial answer to an old question of Serre.
NILS BRUIN, Simon Fraser University, Burnaby, BC
Fake Selmer sets of curves [PDF]
We consider the problem of determining if a curve has any rational points. A first step in deciding this is to see if the curve has points everywhere locally. This is a necessary, but not
sufficient condition for having rational points.
A refined criterion is obtained by considering an unramified Galois cover of the curve. If the curve has a rational point, then one of finitely many twists of the cover has a rational point as
well, and hence there must be a twist that has point everywhere locally. This method is referred to as a finite descent on the curve, and the collection of everywhere locally solvable covers is
called a Selmer set of the curve.
We present a method that computes an object that is closely related to the Selmer set, but is much easier to compute. We also present some statistics on the effectiveness of this descent
obstruction compared to local obstructions.
BRYDEN CAIS, McGill
STEPHEN CHOI, Simon Fraser University
An extension to the Brun-Titchmarsh theorem [PDF]
The Siegel-Walfisz theorem states that for any B > 0, we have å[\substackp £ x, p º d mod v] 1 ~ x/j(v) log(x) for v £ log^B(x) and (v,d)=1. This only gives an asymptotic formula for the number
of primes in an arithmetic progression for quite a small modulus v compared to x. However, if we are concerned only with an upper bound, the Brun-Titchmarsh theorem says that for any 1 £ v £ x,
we have å[\substackp £ x, p º d mod v] 1 << x/j(v) log(x/v). In this talk, we will discuss an extension to the Brun-Titchmarsh theorem that concerns the number of integers with exactly s distinct
prime factors in an arithmetic progression.
This is joint work with Kai Man Tsang and Tsz Ho Chan.
ALINA C. COJOCARU, University of Illinois at Chicago
Serre curves in one parameter families [PDF]
In 1972, Serre proved that the Galois groups of the n-division fields of a non-CM elliptic curve over Q are as large as possible, provided that n is sufficiently large. I will discuss what
"sufficiently large" means when one looks at a one-parameter family of elliptic curves.
This is joint work with David Grant and Nathan Jones.
CHANTAL DAVID, Concordia University
Almost prime orders of elliptic curves over finite fields [PDF]
Let E be an elliptic curve over the rationals. A conjecture of Neal Koblitz predicts an exact asymptotic for the number of primes p such that the order of E over the finite field with p element
is prime. This conjecture is still open. Using sieve techniques, one can find a lot of primes p such that the order p+1-a[P](E) is almost prime. The best result that one may hope to achieve by
sieve techniques was obtained by Iwaniec and Jimenez Urroz for complex multiplication curves using Chen's sieve. They showed that there are infinitely many primes p such that p+1-a[p](E)=P[2],
where n=P[k] means that the integer n has at most k prime factors. For elliptic curves without complex multiplication, it is not known how to apply the switching principle of Chen's sieve to get
such a result.
For curves without complex multiplication, we show that there are many primes p such that p+1-a[p](E)=P[8] with an explicit lower bound (in terms of the constant C(E) of Koblitz's conjecture),
using Greaves' sieve and under the GRH. This improves previous work of Steuding and Weng. One can also show that there are many primes such that p+1-a[p](E) has at most 6 distinct prime factors,
but still cannot improve the number of (not necessarily distinct) primes from 8 to 6. This surprising result is related to the difficulty of sieving square-free numbers in the sequence p+1-a[p](E
This is joint work with Jie Wu (CNRS, Institut Elie-Cartan, Nancy).
ERNST KANI, Queen's University, Kingston, Ontario
Curves of genus 2 and a Conjecture of Gauss [PDF]
The purpose of this talk is to explain how an (extended) conjecture of Gauss plays a role in determining pairs of elliptic curves (E,E¢) which have the property that there is a curve of genus 2
on the product surface E×E¢.
STEPHEN KUDLA, University of Toronto, 40 St. George St., Toronto, ON, M5S 2E4
Arithmetic cycles for unitary groups [PDF]
In this talk, I will discuss some recent work with Rapoport on arithmetic cycles for Shimura varieties associated to U(n-1,1). In particular, we establish a relation between the arithmetic
degrees of certain 0-cycles and the nonsingular, nondegenerate Fourier coefficients of the derivatives of certain incoherent Eisenstein series on U(n,n). If time permits, I plan to discuss
various low dimensional examples and some explicit formulas for local height contributions.
RAM MURTY, Queen's University, Kingston, Ontario
Special values of L-series and transcendental numbers [PDF]
We discuss some recent progress in transcendental number theory. More precisely, we will examine special values of certain L-series and determine when these are transcendental numbers. In
particular, we study class group L-functions attached to imaginary quadratic fields.
This is a report of joint work with V. Kumar Murty as well as Sanoli Gun and Purusottam Rath.
MICHAEL RUBINSTEIN, University of Waterloo, 200 University Ave. W, Waterloo, ON, N2L 3G1
Computing lower terms for the moments of the zeta function [PDF]
A 100-year-old problem asks to determine the moments of the Riemann zeta function on Âs = 1/2. The second and fourth moments are well understood, but little has been proven about the higher
moments. These moments are needed to understand the distribution of the zeta function and its extreme behaviour.
In recent years, a detailed conjectural picture has emerged concerning the full asymptotics of the moments of the zeta function. I will describe these developments and describe methods to compute
the coefficients of the polynomials that appear in these asymptotics.
KENNETH WILLIAMS, Carleton University, Ottawa
The (p,k)-calculus for quadratic forms [PDF]
A simple method will be described for determining the number of representations of a positive integer n by a quadratic form a[1] x[1]^2 + ¼+ a[2k] x[2k]^2, where k is a positive integer and a
[1],...,a[2k] are positive integers whose prime factors belong to {2,3}.
This is joint work with A. Alaca and S. Alaca.
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Three Point Charges Are Located On A Circular Arc ... | Chegg.com
Please solve and explain part c. Thanks!
Image text transcribed for accessibility: Three point charges are located on a circular arc as shown in the figure below. (Take r = 4.36 cm. Let to the right be the +x direction and up along the
screen be the +y direction.) (a) What is the total electric field at P due only to the two positive charges? E rightarrow two positives only = ( N/C) + ( N/C) (b) What is the total electric field at
P, the center of the arc? E rightarrow = ( N/C)t + ( N/C) (c) Find the electric force that would be exerted on a -5.15-nC point charge placed at P. Frightarrow =( N) + ( N)
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Jumping to Conclusions
Though some rules can be overlooked without consequence, some are simply too important to not follow. Two rules, in particular, are at the fundamental analysis of all power systems: Ohm’s Law and
Kirchoff’s Law. A month rarely goes by when someone doesn’t send me data from a power quality monitor that, if taken at face value, would discredit these two long-standing and proven rules. Of
course, the flat-Earth theory was proven false, so these laws may someday follow suit. But it is best to assume that these rules still are true and that following them will lead to the correct
For example, I recently received data that showed a positive xxV DC bus going to negative zzV DC for a few milliseconds every yy milliseconds during certain times of the day (data was from a
classified site; hence, I’ve obscured the exact numbers). The change in voltage levels occurred in microseconds. The current did not change abruptly during this time. In addition, one would imagine
equipment powered off, such a voltage bus, would be “letting out the trapped smoke” that we often see during out-of-limits operation, particular with electronic components. Yet, none of this was
Ohm’s Law says there are relationships among voltage, current and resistance. This has been extended from pure resistance to the term “impedance,” which accounts for inductance and capacitance that
is found in most circuits of any size or capacity. The circuit’s current is directly proportional to the applied voltage and inversely proportional to the impedance of the circuit (I = V/Z). Or, the
voltage drop across two points is equal to the current times the impedance (V = I × Z). Voltage also is referred to as the difference in potential energy between those two points, while current is
the pressure pushing the electrons between those two points. Impedance, as the word sounds, is trying to impede or resist that pressure. For the same voltage potential, a higher impedance means less
current is flowing between those two points and vice versa.
Kirchoff’s Laws are twofold: One involves voltage, and the other involves currents. The voltage rule is that the sum of the voltage drops around a closed loop will equal zero. In the figure above,
tracing a route around the black portion of the circuit would say that Vs + Vload – Vg = 0. Note that since the generated voltage is the source rather than the sink or consumer of energy, the voltage
drop has a negative sign relative to the direction of the current, as compared to all of the impedances, which have a positive to negative relationship in the voltage relative to the current
direction. This is just a convention to also keep with the laws of physics, even in AC circuits where the current is really flowing in one direction during one half cycle and in the other direction
during the other half cycle of the sine wave.
A simple math operation would make it more clear, namely Vg – Vs = Vload, with Vg standing for the generated voltage, Vs representing the voltage drop across the source impedance (all the
transformers and wires leading up to the load) and Vload being the voltage remaining that appears across the load(s).
Since all three loads are in parallel in this example, the voltage across each is the same. Assuming the impedances (Z1, Z2 and Z3) are different, the current through each of the loads would be
different by nature of Ohm’s Law, where Vload/Z1 = Iz1, Vload/Z2 = Iz2 and so on. Since the voltage is the same and impedance changes, so should the current being different.
Kirchoff’s other law deals with the currents, where the sum of the currents at any node or junction should equal zero. The current of the source (Is) is the same for the generator and the source
impedance. However, when we get to node 1, where Z1, Z2 and Z3 connect to source impedance on the top branch and generator on the bottom branch, there are multiple paths for the current to flow. If
we consider just node 1, the current flowing in from Is should equal the current leaving through Z1, Z2 and Z3. Again, a positive and negative sign convention is used to be consistent in the math. In
this case, current entering the node is positive, and current leaving is negative. So Is – Iz1 – Iz2 – Iz3 = 0, or again with some quick math, Is = Iz1 – Iz2 – Iz3.
To put it in perspective, the source impedance is typically 10 to 50 times smaller than the load impedances. That way, most of the voltage from the generator ends up at the loads, doing work. In the
example mentioned above, it would take a tremendous current in the opposite direction of normal for the voltage at the loads to reverse polarity and increase the magnitude of the voltage.
In addition, the wiring inductance and capacitance don’t let these things happen instantaneously. Those facts alone make it nearly impossible for the data to be real. And, the absence of smoke
released (along with parts flying in the air) supports the idea that it is unlikely for this to be real. Instead, we look for other outside influences, such as very large electromagnetic fields, that
can be coupled into the measuring inputs and make the measured values appear as such but not represent the true voltage levels on the xxV DC bus.
Once again, the rules were not broken. The simple explanation is the measurement included things happening in the air, not just on the voltage bus.
BINGHAM, a contributing editor for power quality, can be reached at 732.287.3680.
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Linear second order, homogeneous, constant coefficients, unchanged by all rotations
January 22nd 2011, 12:09 PM #1
MHF Contributor
Mar 2010
Linear second order, homogeneous, constant coefficients, unchanged by all rotations
Show that the only second-order linear homogeneous equation with constant coefficients in x and y whose form is unchanged by all rotations of axes is $u_{xx}+u_{yy}+ku=0$.
$\omega_{\xi\xi}(\cos^2+\sin^2)+\omega_{\eta\eta}(\ cos^2+\sin^2)+k\omega=0\Rightarrow\omega_{\xi\xi}+ \omega_{\eta\eta}+k\omega=0\Rightarrow u_{xx}+u_{yy}+ku=0$
Showing this doesn't change is the easy part (see above), but how do I show it is the only second-order linear?
Show that the only second-order linear homogeneous equation with constant coefficients in x and y whose form is unchanged by all rotations of axes is $u_{xx}+u_{yy}+ku=0$.
$\omega_{\xi\xi}(\cos^2+\sin^2)+\omega_{\eta\eta}(\ cos^2+\sin^2)+k\omega=0\Rightarrow\omega_{\xi\xi}+ \omega_{\eta\eta}+k\omega=0\Rightarrow u_{xx}+u_{yy}+ku=0$
Showing this doesn't change is the easy part (see above), but how do I show it is the only second-order linear?
Start with
$a u_{xx} + b u_{xy} + c u_{yy} + du_x + e u_y + f u = 0$.
Introduce your change of variables and require that
$a \omega_{\xi \xi} + b \omega_{\xi \eta} + c \omega_{\eta \eta} + d \omega_{\xi} + e \omega_{\eta} + f u = 0$.
January 22nd 2011, 02:04 PM #2
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Geometric interpretation of Hida isomorphism
up vote 10 down vote favorite
[EDIT]: After getting a very nice answer by Kevin Buzzard I realize that my question was a little bit too vague and I try to restate it more precisely.
As the title says, I would like to understand an isomorphism of Hida from a more geometric perspective than what I normally read. What bothers me is that there are two construction of the universal
(ordinary) Hida-Hecke algebra and they turn out to give isomorphic objects: fix a prime $p\geq 5$, and a tame level $N$ prime to $p$.
1. Take the projective limit over the level $r$ of the Hecke algebra acting on $S_k(\Gamma_1(Np^r),\mathbb{Z}_p)$ where $k$ is any weight. By applying the usual idempotent, one gets the Hida-Hecke
ordinary algebra $h_k^0(Np^\infty;\mathbb{Z}_p)$, where I adopt notations as in Hida's paper in Inventiones, 1986, "Galois representations into $\mathrm{GL}(2,\mathbb{Z}_p[[X]])$...".
2. Consider now the injective limit over the weight of the spaces of cusp forms $S_k(Np;\mathbb{Z}_p)$. By taking a suitable completion of this injective limit, one sees that the projective limit
(over the weight, now) of Hecke algebras acts on the above completion. Applying again the idempotent, we get the Hida-Hecke ordinary algebra $h^0(N,\mathbb{Z}_p)$.
Theorem 1.1 in the quoted paper by Hida shows that these two algebras are isomorphic (in the most compatible way one can dream of, in particular inducing the same Hecke action on spaces of cusp
forms) but his proof is entirely algebraic.
My question is: is there a reasonable way to prove this isomorphism geometrically?As Kevin Buzzard suggested, several papers of Katz (and successive work by Coleman-Mazur, Buzzard himself et al.)
discuss geometric interpretation of $p$-adic modular forms and $p$-adic families of modular forms. Still, I do not understand how Hida's isomorphism comparing the Hecke algebra as acting on the
projective limit over the level (so ''at the top of the modular tower'') or on the inductive limit over the weight (so, ''over the first curve $X_1(Np)$'') can be given a geometric interpretation.
nt.number-theory arithmetic-geometry iwasawa-theory
Dear Filippo, Here is a more geometric formulation of what is happening: fix a tame level $N$. Then over $X_1(N)$ there is a line bundle $\omega$; weight $k$ modular forms are global sections of $\
omega^{\otimes k}$. Think of this in the old-fashioned way as a sheaf with structure group $\mathbb G_m$. The fact that we can raise it to integral powers reflets the fact that the character group
of $\mathbb G_m$ is precisely $\mathbb Z$. Now if we restrict to $X_1(N)_{ord}$ (the ordinary locus, an affinoid open subspace of $X_1(N)_{/ \mathbb Q_p}$, Katz shows that we may write ... –
Emerton Apr 21 '12 at 12:11
... $\omega = \mathcal O \otimes \mathcal L$, where $\mathcal L$ is an etale local system of rank one over $\mathbb Z_p$. (More precisely, the fibre of $\mathcal L$ over a point $x$ --- which
corresponds to an ordinary elliptic curve $E_x$ --- is the etale part of the $p$-divisible group of $E_x$, which is rank one because we are over the ordinary locus. [And maybe I should replace $\
mathcal L$ by $\mathcal L^{-1}$ here; I didn't think this through right now.]) This means that we have reduced the structure group from $\mathbb G_m$ to $\mathbb Z_p^{\times}$ --- which is why we
can define ... – Emerton Apr 21 '12 at 12:14
... $p$-adic modular forms of arbitrary weight $\kappa \in Hom(\mathbb Z_p^{\times}, \overline{\mathbb Q}_p^{\times})$: because we are allowed to raise $\mathcal L$ (and hence $\mathbb \omega$
restricted to the ordinary locus) to powers that are arbitrary characters of the structure group $\mathbb Z_p^{\times}$. Now if we pass from $X_1(N)_{ord}$ to $X_1(Np^r)_{ord}$, we reduce the
structure group even more, from $\mathbb Z_p^{\times}$ to $1 + p^r \mathbb Z_p$. In the limit over all levels $p^r$, we thus make the structure group trivial, and so $\omega$ as become the trivial
bundle. – Emerton Apr 21 '12 at 12:18
The conclusion is that if we add all $p$-power levels, the notion of weight has disappeared, because $\omega$ (and hence all its powers) have gone away. This is the source of what you call Hida's
isomorphism. To see this discussed in the literature you should look at Katz's LNM 350 article and the subsequent articles he wrote; a good summary is provided by his paper "$p$-adic $L$-functions
via moduli" in the Arcata proceedings. You can also look at Gross's "Tameness criterion" paper, where he works mod $p$, and shows that just by going to $X_1(Np)$ the distinction between weight and
... – Emerton Apr 21 '12 at 12:20
... disappears for mod $p$ forms. (This is because the structure group $1 + p\mathbb Z_p^{\times}$ is already trivial once we are looking at $\omega$ on the mod $p$ ordinary locus.) None of these
sources use the language of "reduction of the structure group" that I am using here, although I find it a convenient point-of-view. Instead, you will find discussion of the Hasse invariant, of a
$p-1$st root of the Hasse invariant (this is in Gross), and of lifts of these mod powers of $p$ (this is in Katz). These are just the particular objects which, in my phrasing, achieve the reduction
of ... – Emerton Apr 21 '12 at 12:24
show 2 more comments
2 Answers
active oldest votes
I am not sure what your criteria would be for a proof to be given a geometric interpretation, but the reason why weights "disappear" when we take the inverse limit on the level stems from
the contraction property of Hecke operators (at $p$), or informally from the fact that Hecke operators at $p$ diminish the level.
As you know, the proof of the isomorphism between the two different Hecke algebras requires the definition of a map between Hecke algebras acting on forms of weight 2 and forms of weight
$k$. Because theses two algebras are sub-algebras of endomorphisms generated by the same abstract elements but acting on different objects, this amounts to constructing a map between the
cohomology of (one of the level of) the modular tower with coefficients in the constant sheaf and (one of the level of) the modular tower with coefficients in a sheaf of weight $k$ (or the
same thing with the modular tower replaced by the Igusa tower, as in Kevin's answer). This last map is really no big deal: if memory serves, on the sheaves it is just projection on the
last component. The remarkable fact is that the map on cohomology then is surjective with a finite kernel (and is an isomorphism in the ordinary case); the proof of this assertion being
up vote 3 exactly the contraction property. Note that the proof necessarily requires the choice of a level at some point; how else would you even state the result?
down vote
accepted Note for instance that for a tower of more general Shimura varieties, it is not at all obvious how the contraction property will play out: group-theoretic properties of $\operatorname{GL}_
{2}(\mathbb Q_{p})$ really do play an important role in the proof. See the reference below though, for an answer to these questions.
So in the end, the isomorphism between the two Hecke algebras seems to me to come from the interplay between the cohomology of modular varieties and group-theoretic properties of the Hecke
algebra. A very general formulation of this fact can be found in D.Mauger Algèbres de Hecke quasi-ordinaires universelles. Ann. Sci. École Norm. Sup. (4) 37 (2004) (section 2.4 to be
Dear Olivier, this sounds good – and sad. I guess you are saying that you expect the isomorphism only to show up at level of (Betti?) realizations of our motives and not to be motivic
itself. May be that is what I would have called "geometric". Or, in more down-to-earth terms that I myself can understand, I would love the existence of a sheaf (?) $\omega_\infty(k)$ on
the pro-scheme $\varprojlim_r X_1(Np^r)$ that for ''some reason'' is isomorphic to a ind-sheaf $\varinjlim_k \omega^{\otimes k}$ over $X_1(N)$. Beside the nonsense, the ''some reason''
might spoil all the fun... Filippo – Filippo Alberto Edoardo Apr 11 '12 at 10:28
Dear Filippo, I am not sure that this is what I am saying honestly, because I am not sure I understand what you are looking for (through no fault of yours). My (very shallow)
understanding of the situation is that a motivic isomorphism would be an isomorphism between the towers (for the level) of Jacobians of modular curves and the "tower with respect to $k$"
(that is to say the simplicial scheme) of canonical desingularizations of the $k$-fold fibre product of the universal generalized elliptic curve over the compact modular curve. – Olivier
Apr 11 '12 at 11:55
It may be easy to cook up a proof by combining the classical proof I outlined and Scholl's result on motives for modular forms. In fact, I think Scholl does everything needed at the end
of his Invent. Math. 100 article (well, strictly speaking you would need to work with finite coefficients instead of $\mathbb Q$ coefficients but this does not look too serious). –
Olivier Apr 11 '12 at 12:02
Dear Olivier, this seems very nice and very close to what I was looking for. I'll look at School's paper, merci! Filippo – Filippo Alberto Edoardo Apr 11 '12 at 12:20
add comment
As you've spotted, there are two ways to do $p$-adic modular forms. The point is that at some point you take a limit of classical modular forms, and there are a whole host of modular forms
which may have e.g. very big weight but whose $q$-expansions are $p$-adically very close 1. Multiplication by such a gadget can completely change the "classical data" (weight, level) attached
to a modular form but might not change the $p$-adic properties of its $q$-expansion much at all. I'm being a little informal, but the upshot is that as long as you let something tend to
infinity you might well have enough space to get a decent theory. Oh by the way you speak of Hida theory but there is no ordinarity assumption built in at this point -- you may as well just
think of the entire huge space of $p$-adic modular forms when formulating your question.
A precise way of formulating an answer to your question can be found in work of Katz from the 1970s (several papers). I have paper copies of all the relevant papers in my office but I am not
in my office, and am too lazy to google for the exact references, so here is a cheat: find Fernando Gouvea's thesis, which was published in Springer Lecture Notes in Mathematics, where he
gives a summary of Katz' approach, and then follow up the references there. You'll find that in some sense you're "nearly" right: but you can't work with $X_1(Np^r)$ over $\mathbf{Q}_p$ or $\
mathbf{Z}_p$ immediately -- the trick is to work with a model of $X_1(Np^r)$ over $\mathbf{Z}/p^n\mathbf{Z}$, look at differentials or functions on such spaces, and then take a direct limit
over $r$ (corresponding to a projective limit of the schemes but with base $\mathbf{Z}/p^n\mathbf{Z}$), and then take a projective limit over $n$. The two limits don't commute so you have to
up be careful, but I think that Katz' approach is the approach that you're looking for. Note that these limits do not give you just Hida's ordinary modular forms, but all $p$-adic modular forms;
vote 9 you can then cut out the ordinary subspace using the $U_p$ operator.
vote But as you already know, there is more than one approach to the theory, and here is another one that I find quite geometric; the catch is that it relies on the theory of rigid spaces, i.e. (in
this setting) a $p$-adic version of Riemann surfaces. If you're happy with this theory then let me briefly explain how to use it to give another geometric interpretation. Take the modular
curve $X_1(Np)$ and consider it as a $p$-adic Riemann surface. Now remove all the points corresponding to elliptic curves with good supersingular reduction. The curve falls into two pieces --
take the piece containing infinity. This is an affinoid rigid space and it has a sheaf $\omega$ on it (the pushforward of the differentials on the universal generalised elliptic curve). We
could call this space the ordinary locus. Now because we are in a $p$-adic setting we can make sense of $\omega^\kappa$ for $\kappa$ in a much more general space than the integers -- basically
$\kappa$ can be any continuous group homomorphism $\mathbf{Z}_p^\times\to\mathbf{Z}_p^\times$ (and even more, in fact). It's a bit like the fact that for $p>2$ you can raise $(1+p)$ to the
power $k$ for $k$ in $\mathbf{Z}_p$ using the binomial theorem, but you can't raise 2 to the power $k$; the point is that $1+p$ is sufficiently close to 1. Similarly the ordinary locus is
sufficiently close to the cusp for things to work. You could define a $p$-adic modular form of weight $\kappa$ to be a section of $\omega^\kappa$ and you could define a family to be a
continuous, or analytic, family of such things, as $\kappa$ varies.
@Kevin. Thanks, I will try to look at Katz' paper(s) – I just read his Antwerp one and thought that the gadget I was looking for did not show up. By the way, I agree that I needed no
ordinarity assumption, but thought that if something geometric happens over the eigencurve it should also be clear on the $0$-slope subspace, that's why I restricted to that case. But I am
very happy with that theory, too. Anyhow, my point was to try to understand "informally" why we do not expect a weight anymore: your sheaves $\omega^\kappa$ over the ordinary locus do depend
on $\kappa$... – Filippo Alberto Edoardo Apr 9 '12 at 9:30
I don't really know what you mean by "we do not expect a weight anymore". These Katz space of generalised $p$-adic modular functions are basically a natural candidate for a space for which
the topology is governed by the $p$-adic properties of the $q$-expansion, and such that we can see classical forms of all weights. So you can take a form of weight $2$ and a form of weight
$4$ and add them up, and the resulting form will not have a weight. But many of the forms in the space have weights. Using the "let the level go to infinity" model, the forms with weights
are exactly the forms which are... – Kevin Buzzard Apr 11 '12 at 11:57
...eigenvalues for the diamond operators (which induce an action of $\mathbf{Z}_p^\times$ on the space). Is this a sufficient informal explanation of why some forms have weights and others
don't or are you looking for something else? – Kevin Buzzard Apr 11 '12 at 11:58
I think this explanation of ''do not having a weight anymore'' is nice and convincing - and pretty much what I was looking for: thanks! In the meanwhile, I realized (thanks to your answer)
that my question was a little bit too vague, and I edited it. I apologize for the first misleading formulation - although it turned out to be fruitful in understanding this weight issue ;-).
Filippo – Filippo Alberto Edoardo Apr 11 '12 at 12:23
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Not the answer you're looking for? Browse other questions tagged nt.number-theory arithmetic-geometry iwasawa-theory or ask your own question.
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Why do I find this result interesting- MOD 17 SAT
I find the following result to be really interesting but can't quite say why: (For this exposition ≤
means poly-m-reduction)
Let SAT[17] be the set of formulas such that the number of satisfying assignments is a multiple of 17.
If SAT[17] ≤[m] S, S sparse, then SAT[17] ∈ P.
This is one of those results where the proof in the literature is hard because they prove something far more powerful (btt reductions- and more). Hence I have my own exposition that I made for my
I presented it in class recently and the students questioned why it was interesting. I can usually answer questions like this (even about such things as the Polynomial VDW theorem) but this one is
harder to say. I DO find it interesting (not just the proof, but the result) but can't quite say why.
SO, here is my challenge: either tell me a reason the result is interesting OR tell me a result that YOU find interesting but can't quite say why.
14 comments:
1. For those of us who are not in the know, can you please say why mod 17, rather than some other prime? Is it suspected this result is true modulo other primes or is 17 really special?
2. 17 is not special.
See my writeup that I
point to. Works for any
bill g.
3. I think it's pretty intuitive. What do you mean by interesting?
4. From context, interesting == worthy of attention / spending time going over in class, perhaps? As for further reasoning, that's probably the question, yes? :)
5. How different is the result (or the proof) from Mahaney's theorem?
6. Is it also true that if SAT_p is not in P then it's NP-Complete?
7. I'm actually not entirely sure why I find P!=NP? interesting, since if the answer is "TRUE", it's only a proof for a very intuitive claim, and if the answer is "FALSE", I'm pretty sure the
polynom would be so big that the NP-hard problems would still remain infeasible.
I think the main reason I find it interesting is just that it sounds so intutively obvious, but still no one has been able to prove it. Okay, also P!=NP is a big assumption that the entire
complexity field is built on, so it's interesting to find out if its really true, but still, I'm not sure why it's interesting in the "if someone who doesn't know anything about computer science
asked me why it's interesting I could explain" kind of way.
8. hi,dear all,sorry to post an irrelevant question here.
I wonder who can kindly give me some links of the "theory/algo qualifying exams" from univerisitys such as MIT,Princeton,Berkeley,CMU,Stanfordand Harvard(not limited to these schools).
Many thanks!
9. All universities publish their qualifying examinations (if they have them) in a single online repository; the UQED (University Qualifying Exam Depository). I've embedded a direct link as follows:
Direct Link Here
10. dammit number 9! i completely fell for that.
11. The direct link from the previous post points to some song on YouTube.
This blog has officially hit rock bottom. Sigh.
12. Sorry,I can't find that UQED. Thanks anyway.
Any other students or professors in these universities can give me a hand on this?
I found that UIUC,Gatech,Wisconsin had put there qualifying exams on their theory group web,but seems others did not:-(
13. Not every school has qualifying exams.
14. They have finally posted the official FOCS 2007 trailer:
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an SAT prep class. I like to work one on one with students as I can adapt to their learning styles and figure out what teaching techniques work best for them.
19 Subjects: including algebra 1, reading, algebra 2, SAT math
...I have been completing needlework projects since I was a child. My recent projects have been having patterns created from photos. I have won 2 blue ribbons at the county fair for this work.
19 Subjects: including algebra 1, reading, English, anatomy
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Difference Between arsinh and arcsinh Functions
Date: 05/13/2004 at 13:18:17
From: Matt
Subject: The missing c in arsinh
Why isn't there a "c" in arsinh, arcosh, and artanh? These are the
equivalents of arcsin, arccos and arctan but they are hyperbolic. I
was just curious as to why there is no "c". Is there a reason?
- Matt (from England)
Date: 05/13/2004 at 23:04:56
From: Doctor Peterson
Subject: Re: The missing c in arsinh
Hi, Matt.
None of us seem to have heard of this; we all use arcsinh, arccosh,
and arctanh for the inverse functions. Can you tell me where you see
your versions--in a math text, a programming language, or what? Is it
the standard British usage?
Searching the web for the terms, I see indications that some
programming languages use your forms. Perhaps because they were
limited to 6 characters in a name. I also seem to see them used in
German and Finnish pages, and some in the UK.
I also find a few references to "arsinh" as meaning "area hyperbolic
sine"; this seems to be the reason for using "ar" rather than "arc" in
some languages. The idea evidently is that whereas the trig functions
take an angle (or arc) as their argument, so that the inverse function
returns an angle (or arc), the hyperbolic functions actually take an
area (which in the case of circular functions is proportional to the
arc length, so we don't see the difference). So it really does make
more sense to use "area" rather than "arc"; the inverse function
returns the "area" whose hyperbolic sine is such and such. I suppose
we use "arc" here just because we use it for trig functions without
thinking of its meaning, and therefore think of "arc" as if it just
meant "inverse". I never thought about this before!
Here is a reference that gives both names and the reason; it doesn't
indicate just how common "arsinh" is, or where it is used:
Here are some other interesting references:
The latter uses the more usual (at least in my part of the world)
"arc" names, though it earlier emphasized that the argument is not an
angle but an area. Your "ar" names turn out to be better. I'd love
to know the history of this, and how some parts of the world use one
set of names while the rest use one that is less meaningful.
If you have any further questions, feel free to write back.
- Doctor Peterson, The Math Forum
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Homework Help
Recent Homework Questions About Drawing
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What does your drawing show?
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Thursday, July 25, 2013 at 2:19pm
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Saturday, July 20, 2013 at 7:27pm
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Sunday, July 7, 2013 at 8:17pm
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Friday, July 5, 2013 at 8:27pm
Chemisty Please Help
I think drawing them the regular way will obey the octet rule.
Wednesday, July 3, 2013 at 7:16pm
Part (a) of the drawing shows a bucket of water suspended from the pulley of a well; the tension in the rope is 69.0 N. Part (b) shows the same bucket of water being pulled up from the well at a
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Sunday, June 30, 2013 at 5:30pm
A man is standing on a platform that is connected to a pulley arrangement, as the drawing shows. By pulling upward on the rope with a force the man can raise the platform and himself. The total mass
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Sunday, June 30, 2013 at 5:29pm
A man is standing on a platform that is connected to a pulley arrangement, as the drawing shows. By pulling upward on the rope with a force the man can raise the platform and himself. The total mass
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Sunday, June 30, 2013 at 1:19pm
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Sunday, June 30, 2013 at 1:18pm
Part (a) of the drawing shows a bucket of water suspended from the pulley of a well; the tension in the rope is 69.0 N. Part (b) shows the same bucket of water being pulled up from the well at a
constant velocity. What is the tension in the rope in part (b)?
Sunday, June 30, 2013 at 1:15pm
The drawing shows box 1 resting on a table, with box 2 resting on top of box 1. A massless rope passes over a massless, frictionless pulley. One end of the rope is connected to box 2, and the other
end is connected to box 3. The weights of the three boxes are W1 = 56.2 N, W2...
Sunday, June 30, 2013 at 1:04pm
A car is being pulled out of the mud by two forces that are applied by the two ropes shown in the drawing. The dashed line in the drawing bisects the 30.0° angle. The magnitude of the force applied
by each rope is 2900 newtons (N). (a) How much force would a single rope ...
Friday, June 28, 2013 at 12:40am
The drawing shows a person looking at a building on top of which an antenna is mounted. The horizontal distance between the person s eyes and the building is 87.8 m. In part a the person is looking
at the base of the antenna, and his line of sight makes an angle of 30.3o ...
Thursday, June 27, 2013 at 11:38pm
Use words from the book. Remember that a collage is made up of many different pictures. http://www.google.com/search?q=collage&tbm=isch&tbo=u&source=univ&sa=X&ei=SfrMUZejF8-QyQHm-4DYCQ&ved=0CEgQsAQ
&biw=711&bih=453 Instead of drawing the river, I suggest ...
Thursday, June 27, 2013 at 10:52pm
how do Huck and Jim feel about the sanctuary of the raft and how do they feel about the violence and shame of the shore? I have to make a collage on this and I don't really know what to put. I was
thinking of drawing a river and in the inside of the river would be a ...
Thursday, June 27, 2013 at 10:32pm
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Mode, Median, and Mean
Date: 08/23/97 at 09:49:48
From: Jennifer
Subject: Mode, median, and mean
Find a set of five data values with modes 0 and 2, median 2, and
mean 2. Explain how you found your answer.
I just don't understand how I'm supposed to do it!
I'm really confused. Please help.
Date: 08/29/97 at 09:38:25
From: Doctor Rob
Subject: Re: Mode, median, and mean
Mode = values appearing most often,
Median = value with as many other values above as below,
Mean = average (sum of the values divided by the number of
There are five values. If all were different, there would be five
modes, but there are only two. The two modes must appear at least
twice. They cannot appear three times each, because then you would
have at least six values, not five. Thus four of the values must be
0, 0, 2, and 2.
For 2 to be the median, the remaining value, call it x, must be
greater than 2. If 0 < x < 2, then x would be the median, and if
x < 0, 0 would be the median.
Then the mean is (0+0+2+2+x)/5 = 2, which you can solve for x.
-Doctor Rob, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
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Fast SSE2 pow: tables or polynomials?
We found that for many applications a substantial part of the time spent in software vertex processing was being spend in the powf function. So quite a few of us in Tungsten Graphics have been
looking into a faster powf.
The basic way to compute powf(x, y) efficiently is by computing the equivalent exp2(log2(x)*y)) expression, and then fiddle with IEEE 754 floating point exponent to quickly estimate the log2/exp2.
This by itself only gives a very coarse approximation. To improve this approximation one has to also look into the mantissa, and then take one of two alternatives: use a lookup table or fit a
function like a polynomial.
Lookup table
See also:
union f4 {
int32_t i[4];
uint32_t u[4];
float f[4];
__m128 m;
__m128i mi;
#define EXP2_TABLE_SIZE_LOG2 9
#define EXP2_TABLE_SIZE (1 << EXP2_TABLE_SIZE_LOG2)
#define EXP2_TABLE_OFFSET (EXP2_TABLE_SIZE/2)
#define EXP2_TABLE_SCALE ((float) ((EXP2_TABLE_SIZE/2)-1))
/* 2 ^ x, for x in [-1.0, 1.0[ */
static float exp2_table[2*EXP2_TABLE_SIZE];
void exp2_init(void)
int i;
for (i = 0; i < EXP2_TABLE_SIZE; i++)
exp2_table[i] = (float) pow(2.0, (i - EXP2_TABLE_OFFSET) / EXP2_TABLE_SCALE);
* Fast approximation to exp2(x).
* Let ipart = int(x)
* Let fpart = x - ipart;
* So, exp2(x) = exp2(ipart) * exp2(fpart)
* Compute exp2(ipart) with i << ipart
* Compute exp2(fpart) with lookup table.
exp2f4(__m128 x)
__m128i ipart;
__m128 fpart, expipart;
union f4 index, expfpart;
x = _mm_min_ps(x, _mm_set1_ps( 129.00000f));
x = _mm_max_ps(x, _mm_set1_ps(-126.99999f));
/* ipart = int(x) */
ipart = _mm_cvtps_epi32(x);
/* fpart = x - ipart */
fpart = _mm_sub_ps(x, _mm_cvtepi32_ps(ipart));
/* expipart = (float) (1 << ipart) */
expipart = _mm_castsi128_ps(_mm_slli_epi32(_mm_add_epi32(ipart, _mm_set1_epi32(127)), 23));
/* index = EXP2_TABLE_OFFSET + (int)(fpart * EXP2_TABLE_SCALE) */
index.mi = _mm_add_epi32(_mm_cvtps_epi32(_mm_mul_ps(fpart, _mm_set1_ps(EXP2_TABLE_SCALE))), _mm_set1_epi32(EXP2_TABLE_OFFSET));
expfpart.f[0] = exp2_table[index.u[0]];
expfpart.f[1] = exp2_table[index.u[1]];
expfpart.f[2] = exp2_table[index.u[2]];
expfpart.f[3] = exp2_table[index.u[3]];
return _mm_mul_ps(expipart, expfpart.m);
#define LOG2_TABLE_SIZE_LOG2 8
#define LOG2_TABLE_SIZE (1 << LOG2_TABLE_SIZE_LOG2)
#define LOG2_TABLE_SCALE ((float) ((LOG2_TABLE_SIZE)-1))
/* log2(x), for x in [1.0, 2.0[ */
static float log2_table[2*LOG2_TABLE_SIZE];
void log2_init(void)
unsigned i;
for (i = 0; i < LOG2_TABLE_SIZE; i++)
log2_table[i] = (float) log2(1.0 + i * (1.0 / (LOG2_TABLE_SIZE-1)));
log2f4(__m128 x)
union f4 index, p;
__m128i exp = _mm_set1_epi32(0x7F800000);
__m128i mant = _mm_set1_epi32(0x007FFFFF);
__m128i i = _mm_castps_si128(x);
__m128 e = _mm_cvtepi32_ps(_mm_sub_epi32(_mm_srli_epi32(_mm_and_si128(i, exp), 23), _mm_set1_epi32(127)));
index.mi = _mm_srli_epi32(_mm_and_si128(i, mant), 23 - LOG2_TABLE_SIZE_LOG2);
p.f[0] = log2_table[index.u[0]];
p.f[1] = log2_table[index.u[1]];
p.f[2] = log2_table[index.u[2]];
p.f[3] = log2_table[index.u[3]];
return _mm_add_ps(p.m, e);
static inline __m128
powf4(__m128 x, __m128 y)
return exp2f4(_mm_mul_ps(log2f4(x), y));
For more details see:
#define EXP_POLY_DEGREE 3
#define POLY0(x, c0) _mm_set1_ps(c0)
#define POLY1(x, c0, c1) _mm_add_ps(_mm_mul_ps(POLY0(x, c1), x), _mm_set1_ps(c0))
#define POLY2(x, c0, c1, c2) _mm_add_ps(_mm_mul_ps(POLY1(x, c1, c2), x), _mm_set1_ps(c0))
#define POLY3(x, c0, c1, c2, c3) _mm_add_ps(_mm_mul_ps(POLY2(x, c1, c2, c3), x), _mm_set1_ps(c0))
#define POLY4(x, c0, c1, c2, c3, c4) _mm_add_ps(_mm_mul_ps(POLY3(x, c1, c2, c3, c4), x), _mm_set1_ps(c0))
#define POLY5(x, c0, c1, c2, c3, c4, c5) _mm_add_ps(_mm_mul_ps(POLY4(x, c1, c2, c3, c4, c5), x), _mm_set1_ps(c0))
__m128 exp2f4(__m128 x)
__m128i ipart;
__m128 fpart, expipart, expfpart;
x = _mm_min_ps(x, _mm_set1_ps( 129.00000f));
x = _mm_max_ps(x, _mm_set1_ps(-126.99999f));
/* ipart = int(x - 0.5) */
ipart = _mm_cvtps_epi32(_mm_sub_ps(x, _mm_set1_ps(0.5f)));
/* fpart = x - ipart */
fpart = _mm_sub_ps(x, _mm_cvtepi32_ps(ipart));
/* expipart = (float) (1 << ipart) */
expipart = _mm_castsi128_ps(_mm_slli_epi32(_mm_add_epi32(ipart, _mm_set1_epi32(127)), 23));
/* minimax polynomial fit of 2**x, in range [-0.5, 0.5[ */
#if EXP_POLY_DEGREE == 5
expfpart = POLY5(fpart, 9.9999994e-1f, 6.9315308e-1f, 2.4015361e-1f, 5.5826318e-2f, 8.9893397e-3f, 1.8775767e-3f);
#elif EXP_POLY_DEGREE == 4
expfpart = POLY4(fpart, 1.0000026f, 6.9300383e-1f, 2.4144275e-1f, 5.2011464e-2f, 1.3534167e-2f);
#elif EXP_POLY_DEGREE == 3
expfpart = POLY3(fpart, 9.9992520e-1f, 6.9583356e-1f, 2.2606716e-1f, 7.8024521e-2f);
#elif EXP_POLY_DEGREE == 2
expfpart = POLY2(fpart, 1.0017247f, 6.5763628e-1f, 3.3718944e-1f);
return _mm_mul_ps(expipart, expfpart);
#define LOG_POLY_DEGREE 5
__m128 log2f4(__m128 x)
__m128i exp = _mm_set1_epi32(0x7F800000);
__m128i mant = _mm_set1_epi32(0x007FFFFF);
__m128 one = _mm_set1_ps( 1.0f);
__m128i i = _mm_castps_si128(x);
__m128 e = _mm_cvtepi32_ps(_mm_sub_epi32(_mm_srli_epi32(_mm_and_si128(i, exp), 23), _mm_set1_epi32(127)));
__m128 m = _mm_or_ps(_mm_castsi128_ps(_mm_and_si128(i, mant)), one);
__m128 p;
/* Minimax polynomial fit of log2(x)/(x - 1), for x in range [1, 2[ */
#if LOG_POLY_DEGREE == 6
p = POLY5( m, 3.1157899f, -3.3241990f, 2.5988452f, -1.2315303f, 3.1821337e-1f, -3.4436006e-2f);
#elif LOG_POLY_DEGREE == 5
p = POLY4(m, 2.8882704548164776201f, -2.52074962577807006663f, 1.48116647521213171641f, -0.465725644288844778798f, 0.0596515482674574969533f);
#elif LOG_POLY_DEGREE == 4
p = POLY3(m, 2.61761038894603480148f, -1.75647175389045657003f, 0.688243882994381274313f, -0.107254423828329604454f);
#elif LOG_POLY_DEGREE == 3
p = POLY2(m, 2.28330284476918490682f, -1.04913055217340124191f, 0.204446009836232697516f);
/* This effectively increases the polynomial degree by one, but ensures that log2(1) == 0*/
p = _mm_mul_ps(p, _mm_sub_ps(m, one));
return _mm_add_ps(p, e);
The accuracy vs speed for several table sizes and polynomial degrees can be seen in the chart below.
The difference is not much, but the polynomial approach outperforms the table approach for any desired precision. This was for 32bit generated code in a Core 2. If generating 64bit code, the
difference between the two is bigger. The performance of the table approach will also tend to degrade when other computation is going on at the same time, as the likelihood the lookup tables get
trashed out of the cache is higher. So by all accounts, the polynomial approach seems a safer bet.
14 comments:
Jon Smirl said...
Do the tables need to be CONST in a shared library? Then multiple users could share the same copy in cache.
Do the tables need to be CONST in a shared library? Then multiple users could share the same copy in cache.
If the tables are const in a shared
library and used by other applications, then the likelihood of the table is in the cache is indeed higher. But that does not help if, for example, precedent/subsequent computations need to lookup
a lot of texture data, which evicts the table from the cache.
The polynomial approach not only does not suffer from that problem, as it has faster / more accurate than table lookups, at least for SSE.
Did you check if most of the pow's were in fact e^x?
That was the case when I looked at this with idr before.
Maybe optimising for the e^x case might make things better also.
As a general rule, on modern systems, lookup tables rarely help unless they replace a substantial amount of calculation. The overhead of memory access, and the high speed of modern processors,
makes it reasonable to expend many instructions to avoid a single memory access.
Did you check if most of the pow's were in fact e^x?
It might be the case in some examples, but not the case of a particular application we were interested in.
Maybe optimising for the e^x case might make things better also.
Optimize the e^x case when compiling the shaders would be the ideal, but I'm not sure if it is implemented or possible at all. Detecting in runtime might not compensate overall, but still worth
giving it a try.
first of all, thanks for the code.
But the title 'Fast sse2..' is a bit misleading.
There's not a single sse2 intrinsic in your code, rather it's just sse (float) that you're using.
Would be nice to have a true sse2 variant (double) of the code posted here.
Can I use this code like under a BSD license or something? I would use it in the context of our research project software, and that is currently not public, and we definitely cannot introduce GPL
code there.
I am an occasional contributor to eigen (http://eigen.tuxfamily.org/index.php?title=Main_Page), a wonderful linear algebra library. I am looking to adapt the pow function presented here. For
this, I need to be clear on the licensing of the code. What is it's license?
Eigen's licensing FAQis here. (http://eigen.tuxfamily.org/index.php?title=FAQ#Licensing).
Thanks for the nice example.
BTW, did you use asciidoc for formatting your code?
@Anonymous, many months later...
Actually, there ARE SSE2 instructions in there: the int32 arithmetic stuff cannot be done on 4 values in a shot with pure SSE.
I guess the version processing doubles was left as an exercise to the reader.
congratulations on your great work, its fantastic to see more Portuguese on OSS :)
What's the license of this code? Is it public domain/GPL(2/3)/BSD? I'd like to include it in my synthesizer (haruhi.mulabs.org), which is GPL-3 licensed.
Anonymous> SSE2 does not mean double precision support. this code is indeed SSE2, solely because of the integer intrinsics that are used.
_mm_cvtepi32_ps, _mm_srli_epi32, all these do not exist in the original SSE instruction set, and you need SSE2 support to execute them.
> What's the license of this code? Is it public domain/GPL(2/3)/BSD?
Consider the code in my post to be MIT license
> I'd like to include it in my synthesizer (haruhi.mulabs.org), which is GPL-3 licensed.
This code has been incorporated and lives on in Mesa3D source code:
- SSE2 instrinsics
- LLVM IR
The SSE2 intrinsic version has now been abandoned. We only use the LLVM IR JIT version which has much more improvements (more accuracy, NaN handling, etc.) The principle is still the same though.
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scipy.stats.binned_statistic_2d(x, y, values, statistic='mean', bins=10, range=None)[source]¶
Compute a bidimensional binned statistic for a set of data.
This is a generalization of a histogram2d function. A histogram divides the space into bins, and returns the count of the number of points in each bin. This function allows the computation of the
sum, mean, median, or other statistic of the values within each bin.
New in version 0.11.0.
x : (N,) array_like
A sequence of values to be binned along the first dimension.
y : (M,) array_like
A sequence of values to be binned along the second dimension.
values : (N,) array_like
The values on which the statistic will be computed. This must be the same shape as x.
statistic : string or callable, optional
The statistic to compute (default is ‘mean’). The following statistics are available:
☆ ‘mean’ : compute the mean of values for points within each bin. Empty bins will be represented by NaN.
Parameters ☆ ‘median’ : compute the median of values for points within each bin. Empty bins will be represented by NaN.
: ☆ ‘count’ : compute the count of points within each bin. This is identical to an unweighted histogram. values array is not referenced.
☆ ‘sum’ : compute the sum of values for points within each bin. This is identical to a weighted histogram.
☆ function : a user-defined function which takes a 1D array of values, and outputs a single numerical statistic. This function will be called on the values in each bin. Empty
bins will be represented by function([]), or NaN if this returns an error.
bins : int or [int, int] or array-like or [array, array], optional
The bin specification:
☆ the number of bins for the two dimensions (nx=ny=bins),
☆ the number of bins in each dimension (nx, ny = bins),
☆ the bin edges for the two dimensions (x_edges = y_edges = bins),
☆ the bin edges in each dimension (x_edges, y_edges = bins).
range : (2,2) array_like, optional
The leftmost and rightmost edges of the bins along each dimension (if not specified explicitly in the bins parameters): [[xmin, xmax], [ymin, ymax]]. All values outside of this
range will be considered outliers and not tallied in the histogram.
statistic : (nx, ny) ndarray
The values of the selected statistic in each two-dimensional bin
xedges : (nx + 1) ndarray
The bin edges along the first dimension.
Returns :
yedges : (ny + 1) ndarray
The bin edges along the second dimension.
binnumber : 1-D ndarray of ints
This assigns to each observation an integer that represents the bin in which this observation falls. Array has the same length as values.
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A calculator for exact real number computation. B.Sc. project report, Univ. of Edinburgh; available from ftp://ftp.tardis.ed.ac.uk/ users/dbp/report.ps.gz
, 2000
"... . A new treatment of data refinement in typed lambda calculus is proposed, based on pre-logical relations [HS99] rather than logical relations as in [Ten94], and incorporating a constructive
element. Constructive data refinement is shown to have desirable properties, and a substantial example of ..."
Cited by 12 (7 self)
Add to MetaCart
. A new treatment of data refinement in typed lambda calculus is proposed, based on pre-logical relations [HS99] rather than logical relations as in [Ten94], and incorporating a constructive element.
Constructive data refinement is shown to have desirable properties, and a substantial example of refinement is presented. 1 Introduction Various treatments of data refinement in the context of typed
lambda calculus, beginning with Tennent's in [Ten94], have used logical relations to formalize the intuitive notion of refinement. This work has its roots in [Hoa72], which proposes that the
correctness of a concrete version of an abstract program be verified using an invariant on the domain of concrete values together with a function mapping concrete values (that satisfy the invariant)
to abstract values. In algebraic terms, what is required is a homomorphism from a subalgebra of the concrete algebra to the abstract algebra. A strictly more general method is to take a homomorphic
, 1999
"... : The usual approach to real arithmetic on computers consists of using oating point approximations. Unfortunately, oating point arithmetic can sometimes produce wildly erroneous results. One
alternative approach is to use exact real arithmetic. Exact real arithmetic allows exact real number computat ..."
Add to MetaCart
: The usual approach to real arithmetic on computers consists of using oating point approximations. Unfortunately, oating point arithmetic can sometimes produce wildly erroneous results. One
alternative approach is to use exact real arithmetic. Exact real arithmetic allows exact real number computation to be performed without the roundo errors characteristic of other methods.
Conventional representations such as decimal and binary notation are inadequate for this purpose. We consider an alternative representation of reals, using the golden ratio. Firstly we look at the
golden ratio and its relation to the Fibonacci series, nding some interesting identities. Then we implement algorithms for basic arithmetic operations, trigonometric and logarithmic functions,
conversion and integration. These include new algorithms for addition, multiplication, multiplication by 2, division by 2 and manipulating nite and innite streams. Acknowledgements I would especially
like to than my supe...
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Macmerry, Scotland
Macmerry, Scotland
Macmerry, Scotland
Macmerry, the place in East Lothian, Scotland, the United Kingdom
village, geographical area, settlement
the United Kingdom, East Lothian, Lothian and Borders, East Lothian, Scotland
Wikipedia Page
"NT435723", "NT4372"
Postcode District
55°56'25.48"N, 55°56'32"N, 2°54'21.92"W, 2°54'16"W
Nearest Settlement of
▶ the area of uk postcode eh33 1qa, the area of uk postcode eh33 1ra, the area of uk postcode eh33 1pa, the area of uk postcode eh33 1pb, the area of uk postcode eh33 1qb, …
the area of uk postcode eh33 1qd, the area of uk postcode eh33 1qe, the area of uk postcode eh33 1qf, the area of uk postcode eh33 1pf, the area of uk postcode eh33 1qg, the area of uk postcode
eh33 1pg, the area of uk postcode eh33 1ph, the area of uk postcode eh33 1qh, the area of uk postcode eh33 1qj, the area of uk postcode eh33 1pj, the area of uk postcode eh33 1pl, the area of uk
postcode eh33 1ql, the area of uk postcode eh33 1pn, the area of uk postcode eh33 1pp, the area of uk postcode eh33 1pr, the area of uk postcode eh33 1ps, the area of uk postcode eh33 1qs, the
area of uk postcode eh33 1pt, the area of uk postcode eh33 1pu, the area of uk postcode eh33 1pw, the area of uk postcode eh33 1rw, the area of uk postcode eh33 1px, the area of uk postcode eh33
1ry, the area of uk postcode eh33 1py, the area of uk postcode eh33 1pz, the area of uk postcode eh33 1nz
village, settlement, geographical area
Related Websites
http://www.railscot.co.uk/Macmerry_Branch/frame.htm, http://www.eastlothian.gov.uk/documents/contentmanage/2001%20Census%20-%20Macmerry-2794.PDF
Freebase ID
GeoNames URL
DBPedia URI
Freebase Primary MID
Edit this profile
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25 Subjects: including statistics, writing, calculus, geometry
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Klein bottle
A mathematician named Klein,
Thought the Möbius band was divine,
He said, "If you glue,
The edges of two,
You'll get a weird bottle like mine."
—Leo Moser
Take a rectangle and join one pair of opposite sides to make a cylinder. Now join the other pair with a half-twist. The result is a Klein bottle. Sound easy? It is if you have access to a fourth
dimension because that's what is needed to carry out the second step and to allow the surface to pass through itself without a hole. A true Klein bottle is a four-dimensional object. It was
discovered in 1882 by Felix Klein when he imagined, as in the limerick, joining two Möbius bands together to create a single-sided bottle with no boundary.
An ordinary (three-dimensional) bottle has a crease or fold around the opening where the inside and outside of the bottle meet. A sphere doesn't have this crease or fold, but it has no opening. A
Klein bottle has an opening but no crease: like a Möbius band, it is a continuous one-sided structure. Because it has no crease or fold, it has no verifiable definition of where it's inside and
outside begin. Therefore, the volume of a Klein bottle is considered to be zero, and the bottle has no real contents – except itself! As the joke goes: "In topological hell the beer is packed in
Klein bottles." Take a coin, slide it across the surface of a Klein bottle until it returns to its starting point, and the coin, as if by magic, will be flipped over. This is because, unlike a sphere
or a regular bottle, a Klein bottle is non-orientable.
Although a Klein bottle can't be "embedded" (that is, fully realized) in three dimensions, it can be "immersed" in three dimensions. Immersion is what happens when a higher dimensional object cuts
through a lower dimensional one, producing a cross section. When a sphere is immersed in a plane, for example, it produces a circle. A three-dimensional glass model of a Klein bottle can be made by
stretching the neck of a bottle through its side and joining its end to a hole in the base. Except at the side-connection (the nexus), this properly gives the shape of a four-dimensional Klein
bottle. Just as a photo of such a bottle is of a three-dimensional Klein bottle immersion, so the immersion in real life is like a photo of the true four-dimensional bottle.
Related categories
• SPACE AND TIME
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Physics Forums - View Single Post - Work done by a spring & its potential energy
According to work - mechanical energy theorem ,
W = K(final) - K(initial) + U(final) - U(initial) . . . . (1)
as we define Potential energy as negative of work done by conservative force and assuming that the only force in this situation is Spring force then ,
W(spring) = K(final) - K(initial)
As work done is calculated by finding component of spring force in direction of displacement. How can we say that U(final) - U(initial) applies for all possible conditions of extension of spring as
displacement may not be in direction of force ?
Spring force = 0.5kx^2
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Westport, MA Math Tutor
Find a Westport, MA Math Tutor
I am currently a researcher in international relations with extensive academic preparation in history, political science, writing and research. I also have taken extensive schooling in theology,
philosophy and French. I have worked as a teaching assistant in a Political Science Department, as well as a tutor and teacher at the high school and university level.
16 Subjects: including prealgebra, English, writing, ESL/ESOL
...I look forward to meeting you.I strive to develop a student's Algebra skills from the basics of Middle School math and Pre - Algebra. I like to use models that help a student visualize the
concepts of Algebra for greater true understanding (and less memorization). I have succeeded at helping Alg...
11 Subjects: including algebra 1, algebra 2, calculus, geometry
...I have also helped students with psychology, US History and World history. I read all the texts and literature that my students read. At the elementary level, I worked with students on basic
phonics instruction, reading and math.
30 Subjects: including prealgebra, geometry, reading, English
...Please contact me if this sounds like a fit. I look forward to working with you!I have certification in Rhode Island in elementary education, K-6. I have worked with preschool age students
using a variety of phonics programs.
16 Subjects: including prealgebra, SAT math, reading, algebra 1
...I have tutored students of all ages from elementary school students to adults who have decided to go back to school. My philosophy is to instill confidence which is a key to a student
obtaining success. I encourage students to think outside the box when tackling a mathematical problem.I have taught Algebra 2 for many years in a public school setting.
11 Subjects: including algebra 1, algebra 2, geometry, prealgebra
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Types of Loans
Loan Descriptions
Below, you will see a brief description of the many types of mortgage loans. Some of them are not used as much as others.
This just a small sample of the available loan programs that we have to offer. We understand that each borrower is unique and that is why we strongly encourage our clients to meet with us for a free
mortgage consultation to make sure the loan program that you choose is the one that best fits your family and financial situation.
Conforming Loan
A loan in which the amount borrowed is less than or equal to $417,000 (this number could be different depending on the bank) .
Jumbo Loan
A loan in which the amount borrowed is greater than $417,000 (this number could be different depending on the bank) .
Fixed Rate Loans
This type of loan has monthly payments that remain the same for the entire loan term after which time the loan is paid in full. The monthly payment is based on an interest rate which does not change
over the term of the loan (hence the term "fixed rate"). Fixed rate loans are typically 30, 15 & 20 year terms.
Adjustable Rate Mortgage (ARM)
The interest rate on an ARM may vary up or down at fixed intervals. The changes are tied to a financial index such as one year Treasury notes. The ARM often offers a low beginning interest rate as a
"teaser". However, this rate will go up after a certain time. If interest rates are low, an arm may be a good option. This is especially true if its cap (the highest interest you may be charged) is
not more than a few points higher than the current interest rate. ARMs are of special interest to buyers who know their income will rise in the future or who don't plan to own the home for many
years. Arm Loan Programs are available for a variety of terms. Typical terms are 1, 3, 5 and 7 years.
Interest Only Loan
A mortgage is "interest only" if the monthly mortgage payment does not include any repayment of principal for some period. The payment consists of interest only. During that period, the loan balance
remains unchanged.For example, if a 30-year fixed-rate loan of $100,000 at 8.5% is interest only, the payment is .085/12 times $100,000, or $708.34. Otherwise, the payment would be $768.92. This is
the "fully amortizing payment" - the payment that, if maintained over the term of the loan, will pay it off completely. The interest only loan thus reduces the monthly payment by 7.9%. A loan that is
interest-only for the full term would not amortize. The loan balance would be the same at term as it was at the outset. Back in the twenties, loans of this type were the norm. Borrowers typically
refinanced at term, which worked fine so long as the house didn't lose value and the borrower didn't lose his job. But the depression of the thirties caused a large proportion of these loans to go
into foreclosure. Lenders stopped writing them and have never brought them back. They want loans that eventually amortize. Hence, the interest only loans of today are interest only for a specified
period, such as 5 years. At the end of that period, the payment is raised to the fully amortizing level. In such case, the new payment will be larger than it would have been if it had been fully
amortizing at the outset. Suppose, for example, the interest only period on the loan described above is 5 years. Then the payment starting in month 61 would be $805.23. To reduce the payment by
$60.58 for the first 5 years, the borrower would pay an additional $36.31 for the next 25. The longer the interest only period, the larger the new payment will be when the interest only period ends.
If the same loan is interest only for 10 years, for example, the fully amortizing payment beginning in month 121 is $867.83. To reduce the payment by $60.58 for the first 10 years, the borrower would
pay an additional $98.91 for the next 20. Interest only mortgages are for borrowers who want a lower initial payment, and have some confidence that they will be able to deal with a payment increase
in the future.
Balloon Loans
This type of loan has fixed monthly payments for the balloon term of the loan that are based on a 30 year repayment schedules. At the end of the balloon term, the outstanding principal balance of the
loan is due plus any unpaid interest. This loan program generally has a refinance option at the end of the balloon period that gives the borrower the option to extend the loan at a fixed rate for the
remaining term. Balloon loans are typically 5 and 7 year terms.
3-2-1- Buy down Loan
This type of loan program is based on an interest rate (actual rate) that does not change over the term of the loan and has fixed monthly payments that are based on a 30 year repayment schedule.
However, the monthly payments that are made during the first 36 months (three years) are calculated based on an interest rate that is less than the actual rate. The first 12 monthly payments of the
loan are calculated based on an interest rate that is 3% less than the actual rate. For the second year of the loan, payments 13 through 24 are based on an interest rate that is 2% less than the
actual rate of the loan. For the third year of the loan, payments 25 through 36 are based on an interest rate that is 1% less than the actual rate. After the third year, the monthly payments to be
made over the remaining 27 years of the loan are based on the actual rate.
This type of loan is typically used to help borrowers who are unable to qualify for a loan at current interest rates. By "buying down" the interest rate, the borrower decreases the initial monthly
payments that are required to be made which increases the borrower's ability to qualify for the loan. The cost of "buying down" an interest rate for a period of time is generally determined by
calculating the difference between (a) the total monthly payments that would have been made during the buy down period if the loan did not have a buy down feature and (b) the total monthly payments
to be made during this same period with the buy down feature in place. This amount is generally paid for at time of closing by the Lender or the Seller, depending on how it is structured.
B/C Credit Loan
These types of loans are available to borrowers who have or have had credit problems such as being late on or defaulting on the repayment of loans or credit cards. Although such loans are available
as fixed rate or adjustable rate mortgage loans, the interest rate and/or costs associated with such loans are generally higher than loans available to borrowers who do not have a history of credit
issues to reflect the fact that the risk associated with such loans is generally higher. Borrowers who do not have a history of credit issues are said to have "A" credit. Those with a history of
credit issues are said to have "B", "C" or "D" credit depending on the severity of the credit issues.
No Income/No Asset Verification Loan
This type of loan is a No Income Verification Loan and a No Asset Verification Loan. It is used by borrowers who do not wish to or are unable to verify their income and their assets. Once again, the
interest rate and/or costs for such loans may be slightly higher than normal to reflect the higher degree of risk involved in loaning to borrowers without verifying their income or assets. Such risk
is offset by borrowers who have an excellent credit history.
Construction Loan
This type of loan is used to finance the construction of a home. It may or may not also include the purchase of the land upon which the home is to be built. Unlike a mortgage loan where the entire
amount of the loan is disbursed to the borrower at the time the loan transaction is consummated, a construction loan involves a series of disbursements which are linked to a construction schedule.
Some construction loans have fixed interest rates, others have variable interest rates. In addition, some construction loans automatically convert to a regular mortgage (referred to as "permanent"
financing) once construction has been completed, while others require another loan transaction to take place so the borrower can payoff the construction loan and obtain permanent financing.
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We prsent a multiscale method for computing the effective behavior of a class of stiff and highly oscillatory ordinary differential equations. The oscillations may be in resonance with one another
and thereby generate hidden slow dynamics. The proposed method relies on correctly tracking a set of slow variables whose dynamics is closed up to $\epsilon$ perturbation, and is sufficient to
approximate any variable and functional that are slow under the dynamics of the ODE. This set of variables is detected numerically as a preprocessing step in the numerical methods. Error and
complexity estimates are obtained. The advantages of the method is demonstrated with a few examples, including a commonly studied problem of Fermi, Pasta, and Ulam.
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Rule of division by 23.
Oh! You want practical!
Actually, not really. I just thought the lack of practicality was worth mentioning.
For example, one borderline-practical test for divisibility by 7 would be to add up a number by groups of 6:
111,222,333,444,555,666,777,888,999 is divisible by 7 iff 111+222333+444555+666777+888999 = 2222775 is, and 2,222,775 is divisible by 7 iff 2+222775 = 222,777 is. Of course to go further you need
trial division (or the standard 'shorten by 1 digit' rule).
But the analogue for 17 (summing digits in groups of 16) is seldom, if ever, worthwhile, and I can't imagine anyone working their way through that rule for 59.
I'm more interested in seeing where the rules come from. So far they come from the factorization of
* n^a, by looking at the last a digits
* n^a - 1, by summing in groups of a digits
* n^a +/- n^b, by (alternate) summing
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Word Algebras and Strange Grammars
Ars Mathematica
the issue of
Algebraic Combinatorics on Words
, by the anonymous group of authors called 'M. Lothaire'. This is a sequel to the original
Combinatorics on Words
It may come as a surprise to some (at least it did to me) that you can define interesting algebraic structures on words (sequences of letters from a (finite) alphabet). Combinatorics on Words is a
great book that builds up the theory of free algebras over words. It was in this book that I first came across a beautiful proof of the precursor of the Robertson-Seymour program:
Consider any infinite sequence of finite-lengthwords w[1], w[2], ..., over a finite alphabet. There always exists an i and a j > i such that w[i] is a subsequence of w[j]
Viewed another way, what this says is that if we define a partial ordering on strings such that one string is greater than another if it contains the second as a substring, then
there is no infinite antichain
Why is this a precursor of the Robertson-Seymour program ? Well, the above fact shows that the set of strings with the given order is a well-quasi order, which in turn means that any filter (i.e any
set that is "upward" closed, or for each element x contains all elements y greater than x) has a finite basis, which means that there is a finite set of strings whose upward closure defines the
filter). Conversely, this means that any ideal (a set of strings closed under the substring operation) has as complement a filter, and thus membership in the ideal can be described in terms of
"forbidden" strings (the elements of the finite basis)
The Robertson-Seymour program can be stated as: graphs with the order defined by minors (G is greater than G' if G' is a minor of G) form a well-quasi order, which thus means that any upward closed
family has a finite basis, and thus any idea (any minor-closed family) has a finite set of excluded graphs that defines it.
The result on strings is not hard to prove: I used it in
this paper
to show that membership for a certain grammar was decidable. Unfortunately, we were never able to prove a stronger result, and the grammar seemed so simple that we felt there must be a more efficient
membership result.
Here's the grammar, for your entertainment.
All production rules are of the form
a -> ba
cd -> d
a, b, c, d
are elements of the alphabet. There is a special symbol
, and the language of such a grammar is defined as all strings
such that
wx ->* x.
Notice that in the productions, a symbol either inserts or deletes a new symbol to its left. We called these grammers (somewhat clunkily) "leftist grammars".
Update (12/23/05):
Leftist grammars are
context-free. In FCT 2005,
Tomasz Jurdzinski and Krzysztof Lorys
present a context-sensitive language that can be expressed as a leftist grammer.
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Pot Odds
Pot Odds Stuff: Pot Odds : The Rule of 4 and 2 : Pot Odds Examples
Learning how to use pot odds puts an incredibly useful weapon in your poker arsenal. Knowledge of this basic concept is fundamental in determining whether or not you will become a winning or losing
poker player.
This guide aims to explain how pot odds work and how to effectively incorporate them into your game. It shouldn't take more than 10 minutes to read this guide from start to finish, which is pretty
good considering it could be saving (and winning) you more money for the rest of your poker career.
What are pot odds?
Pot odds simply involves using the odds or likelihood of winning when on a drawing hand to decide whether or not to call a bet or a raise.
Therefore when you are on a flush or straight draw, you will be able to work out whether or not to call or fold depending on the size of the bet you are facing by making use of pot odds. Pretty handy
Check out my rankings of the top Texas Hold'em poker rooms for US players. You may be surprised.
A familiar situation you will find yourself in Texas Hold'em is holding 2 cards of the same suit with another 2 cards of that suit on the flop. In poker this is called a flush draw or sometimes
referred to as a ‘four flush’. We will use this as an example in learning the use of pot odds.
Working out pot odds.
There are two ways that you can work out pot odds in Texas Hold'em.
1. Ratio method
2. Percentage method.
Both of these methods provide the same results, so the one you decide to use is simply a matter of preference.
The ratio method is the most commonly used method for working out pot odds, but I personally found the percentage method the easiest to get to grips with when I was calculating pot odds for the first
1) Ratio Method.
The majority of books and forums will put pot odds in the ratio format, so it's definitely worth while getting used to this method of calculating and working with pot odds.
You Hold:
Now say there are two people left in the pot, you and your opponent. There is $80 in the pot and your opponent bets $20. What should you do?
1] Calculating the "card odds".
First of all we need to find out how likely we are to catch another heart on the turn. This can be done in many ways, but the most popular way is to find the ratio of cards in the deck that we don’t
want against cards that we do want.
• There are 5 cards in this hand that we know, our 2 holecards and the 3 cards on the flop.
• This leaves us with 47 cards in the deck that we do not know.
• Out of those 47, there are 9 cards that will make our flush and 38 that will not.
• If we put this into a ratio it gives us 38:9, or roughly 4:1.
2] Compare with pot odds.
Now we know that the odds of hitting a heart on the next card are 4:1 (our card odds). This means for every 4 times we don’t catch a heart, 1 time we will.
Next we calculate the same ratio of odds using the size of the pot and the size of the bet.
• Our opponent has bet $20 into an $80 pot making it $100.
• This means we have to call $20 to stand a chance of winning $100.
• This makes our odds $100:$20 which works out to equal 5:1 pot odds.
Card Odds: 4:1
Pot Odds: 5:1
This means that we should call as the odds we are getting from the pot are bigger than the odds that we will hit our flush on the next card. In the long run we will be winning more money than we are
Remember! You should only call if the pot odds are greater than the "card odds" (odds of completing your draw).
If finding the card equity by working them out in your head is too time consuming (which most beginners will) . You can find them more quickly by using odds charts. These are handy if you print them
out and stick them next to your computer and refer to them the next time you end up with a draw.
Try SPOC if you're just starting out. It's a very handy tool for helping you work out pot odds during play.
2) Percentage Method.
The percentage method was easier for me to get to grips with when I first starting learning pot odds. Unfortunately, it is not as widely used as the ratio method.
For the percentage method I will use an example with a straight draw.
You Hold:
This time your opponent bets $30, making the pot $90 in total (so the pot was originally $60, but that doesn't matter). We want find out whether or not to call by finding out the pot odds using
1] Finding the "card equity" (same as "card odds", but just using %s).
To find the chance of making the straight on the next card we again need to find the number of outs (‘outs’ are cards that will complete the hand we are trying to make, in this example we are trying
to make a straight.). There are 4 fives and 4 tens that will complete our straight giving, us a total of 8 outs.
To find the percentage chance of making the straight on the next card we simply need to double the outs and add one.
• Finding the percentage "card equity".
• Double the outs: 8 * 2 = 16
• Add one: 16 + 1 = 17%
• = 17% chance of making the straight
2] Compare with pot odds.
Our opponent has bet $30 making the pot $90. This means we have to call $30 to stand a chance of winning $120.
As you can see we have to add our own bet that we will call onto the size of the pot to find the total pot size. This part is very important, as finding the percentage of $30 in a $90 pot will give a
very different result that the percentage of $30 in a $120 pot. Using basic mathematics we know that $30 is 25% of the $120.
Card Equity: 17%
Pot Odds: 25%
As we have already found out we have 17% chance of making the straight on the next card, which means that we should only call 17% of what is in the pot. Because we are being forced to pay 25% to play
on, we should fold. We would be losing money in the long run if we called.
Remember! You should only call if the percentage chance of making your hand is greater than the percentage of the pot you have to call.
The percentage card equity can also be found in odds charts if you find it easier to use them instead of work them out. These are useful as a guide as you start incorporating pot odds into your game,
or if you have trouble working out the odds in the short space of time you are given to make decisions whilst playing online.
Try playing flush and straight draws for an alternative explanation of using pot odds in poker.
Question: Why are we working out the odds for the next card only if there are two cards to come?
Good question. If we are on the flop with a flush draw, our odds of making the best hand on the turn are roughly 4 to 1 or 20%. However, seeing as we are on the flop there are indeed 2 more cards to
come (and not just the 1), shouldn't the "card equity" be more like 2 to 1 or 40%?
The answer.
Generally, no. This is one of the biggest mistakes players make when using pot odds.
The explanation.
When you work out your pot odds, you are comparing the pot odds for the current size of the pot (and bet) to the chances of making your draw on the next card. If you work using the odds of making
your draw over the next two cards, you need to factor in any extra money that you will have to pay on the turn also.
Seeing as it's incredibly unlikely that we're going to accurately guess how much more money we might have to pay on the turn, it's far easier and infinitely more reliable to take it one card at a
time. This way, you won't end up paying more money than you should for your drawing hands when on the flop.
The only time that you should ever use the odds for making the best hand over the next two cards combined (e.g. using 2 to 1 odds instead of 4 to 1 odds for a flush draw) is when your only opponent
in all-in on the flop. In this instance, you can guarantee that you won't face another bet on the turn, as your opponent has no more money to bet.
I briefly mention this stuff on my percentage odds chart and my ratio odds chart. There is also an explanation in my article on the rule of 4 and 2 for pot odds. It's obviously a very common mistake!
Pot odds evaluation.
Although upon first glance pot odds may appear difficult, it is one of the most basic applications of mathematics in the game of poker. If you base your drawing decisions on pot odds, then you will
mathematically be a winner in the long run, regardless of whether or not you win the hand or not.
Now you've got some strategy under your belt, use it against the terrible players at Bodog Poker and win even more money than before.
In addition to deciding whether or not to call, pot odds can be used to influence how much you should bet to "protect" your hand. If you believe your opponent is drawing to a flush then you should
bet a large enough sum into the pot to give your opponents the wrong odds to call if you think you have the best hand. Once again, regardless of whether or not your opponent wins the particular hand,
they will be losing and you will be winning in the long run.
For another take on explaining pot odds, try this pot odds guide from FirstTimePokerPlayer.com. There are some very handy tables and examples in this Texas Hold'em strategy section that should help
to broaden your understanding of the basics of pot odds in poker.
Note: The pot odds examples used in this guide have been in the situation where you have seen the flop and are waiting to see the turn. The same mathematics can be applied for when you are on the
turn waiting to see the river, as both odds are almost exactly the same. However, you should remember that there will be one less unknown card left in the deck when working out the odds because you
now know what the turn card is.
Go back to poker mathematics.
Go back to the awesome Texas Hold'em Strategy.
How Much More Money Could
You Be Winning?
“I played break-even online poker for 4 years before finding
DeucesCracked, for the last 5 months I've made more money playing
poker than at my full-time job.”
- liquid_quik, DC Member
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Families of number fields of prime discriminant
up vote 12 down vote favorite
When I am testing conjectures I have about number fields, I usually want to control the ramification, especially minimize to a single prime with tame ramification. Hence, I usually look for fields of
prime discriminant (sometimes positive, sometimes negative).
I get the feeling that I cannot be the only one who does this...
And so, are there families of number fields of prime discriminant for each degree? Or at least degree 3 and 4? (They are the coolest. Except quadratics. Of course.) What about: given a prime - can I
find a polynomial of degree d with the prime as its discriminant?
nt.number-theory number-fields prime-numbers
hum...but it seems we only have good formulas of discriminant for good (simple) field extensions. If we want to good (simple) discriminants, then the field extension itself would be quite
complicated...no? – natura Jan 4 '10 at 21:59
I might be missing something, but doesn't Stickleberger's theorem imply that your final question is false for all primes of the form 4k+3? – Ben Linowitz Jan 4 '10 at 22:10
@basic: No. For instance $\mathbb{Q}(\zeta_p)/\mathbb{Q}$ (here $\zeta_p$ is a primitive $p$th root of unity) is a very simple extension for which a very precise formula for the discriminant is
available. In particular, it is only ramified at $p$ and is tamely ramified there. It doesn't quite answer Dror's question, because there is at most one such extension of any given degree. – Pete
L. Clark Jan 4 '10 at 22:11
@Ben Linowitz: sure, as written. Maybe he means $\pm$... – Pete L. Clark Jan 4 '10 at 22:11
1 @Pete, Yes that's what I mean. Good simple extensions like cyclotomic extensions have good formula of Discriminants, but if we want to have a discriminant as simple as a prime number, then the
extension itself probably won't be very simple. – natura Jan 4 '10 at 22:22
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5 Answers
active oldest votes
Klueners Malle online might be just the thing you're looking for. Make your own lists! And here's some they made themselves, if you run out of ideas.
up vote 7 down
so e.g. reh.math.uni-duesseldorf.de/cgi-klueners/… gives you a list of number fields of degree 5 with prime power discriminant, and many of the later entries have prime
discriminant. – Kevin Buzzard Jan 4 '10 at 22:26
add comment
A recent paper of Bhargava and Ghate discusses the enumeration of quartic fields of prime discriminant (see section 7).
up vote 5 down
I was not familiar with Bhargava or his thesis work, but I must say, as I am reading his series of papers on composition laws, that it is awesome! So glad I posted this question...
– Dror Speiser Jan 5 '10 at 20:04
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As for your second question: if $f$ is an irreducible polynomial woth degree $n$ and prime discriminant (actually, squarefree discriminant is sufficient), then the roots of $f$ generate an
extension with Galois group $S_n$ (thus these examples are not necessarily the best candidates for testing conjectures since you miss out on all the more interesting Galois groups). Since
$S_n$ is not solvable for $n \ge 5$, class field theory is your friend only for $n = 2, 3, 4$.
• For $n = 2$, the situation is trivial.
• For $n = 3$, there is a number field with prime discriminant $p$ if and only if the
quadratic field with discriminant $p$ has class number divisible by $3$.
• For $n = 4$, there is an $S_4$-extension with prime discriminant if and only if there is a quadratic number field with discriminant $p$ and class number divisible by $3$ such that one
up vote 5 of its unramified cubic extensions has class number divisible by $2$ (and thus necessarily by $4$).
down vote
There's a very nice article by Shanks (A survey of quadratic, cubic and quartic algebraic number fields (from a computational point of view), Proc. 7th southeast. Conf. Comb., Graph
Theory, Comput.; Baton Rouge 1976, 15-40 (1976)) where you will find more.
Scholz, during the 1930s, showed how to construct (using class field theory, which means you will not get generators, just the existence) of Galois groups with small solvable groups; in
his construction, the number of ramified primes can be controlled.
Then, who is my friend at $n>=5$ and $G=S_n$? Also, for 3 or 4, is there a fast computational method to find a class of order 3 in the quadratic field? Would be interesting only if it
didn't involve computing the whole class group. – Dror Speiser Feb 16 '10 at 20:49
"if $f$ is an irreducible polynomial woth degree $n$ and prime discriminant (actually, squarefree discriminant is sufficient), then the roots of $f$ generate an extension with Galois
group $S_n$" How does one prove this? – Timo Keller Apr 14 '10 at 18:37
This goes back to Arnold Scholz (1934); a recent reference is Nakagawa, "On the Galois group of a number field with square free discriminant", Comm. Math. Univ. St. Pauli 37 (1988),
95-99 – Franz Lemmermeyer Apr 26 '10 at 12:58
@Franz: Just reading the review on Nakagawa's paper on Mathscinet, I see he did the case when the discriminant of the field is a prime. However it is not clear to me that when the
polynomial has prime discriminant, its splitting field would have prime discriminant. (while I am waiting for my interlibrary requested article) Would you be a little more precise as did
Nakagawa/Scholz do that? – Ying Zhang May 2 '10 at 18:01
add comment
John Jones' Tables http://hobbes.la.asu.edu/NFDB/ are my favorite. I think his data tables are the most complete online. Check for example in Klueners-Male tables for cubic fields of prime
discriminant -3299, and you'll see that there are no results shown. However Jones' tables contains the 4 cubic fields with such discriminant.
Now about your question on the primes, as Ben mentioned p must be 1 mod 4 so the question has some hope. Even for p that are "allow" to be discriminants there might not exist a field of fix
up vote degree d of discriminant p. For example, class field theory tells you that there is a cubic field of discriminant p if and only if the 3-Sylow part of $Cl(\mathbb{Q}(\sqrt{p}))$ is
4 down non-trivial. In fact you can even tell how many of them there are! As an example of this we can conclude that there are no cubic fields of discriminant p=-3, even though p is a fundamental
vote discriminant. A similar analysis can be done for quartic fields, and I think for them the behavior is related to the 2-Sylow part of $Cl(\mathbb{Q}(\sqrt{p}))$.
Also you might want to look at this paper of Jone's which I think is very close to your question http://hobbes.la.asu.edu/papers/OnePrimeJR.pdf
very interesting links! do you mind explaining this statement? thank you! "For example, class field theory tells you that there is a cubic field of discriminant p if and only if the
3-Sylow part of Cl(Q(\sqrt{p})) is non-trivial." – natura Jan 5 '10 at 4:25
2 @basic: I have written a paper that uses this, so instead of copy and paste, I better give you the link to it. math.wisc.edu/~mantilla/PagProb/pruebas.pdf What you are wondering can be
found at the beginning of section 4. – Guillermo Mantilla Jan 5 '10 at 4:59
this sounds real cool, I didn't know of the relation to the 3 part... Btw, I mention in my question that positive and negative is fine, so negative 3 mod 4 is also fine. – Dror Speiser
Jan 5 '10 at 5:45
yes -3 mod 4 is the same as 1 mod 4. Now, the impossibility I was referring to is p=2,3 mod 4. The point that Ben was trying to make in his comment is that the discriminant of a number
field is always congruent to 0 or 1 mod 4(this is known as Stickleberger's theorem). In particular if a prime p is a discriminant, so the question has some hope to be answered, then p has
to be 1 mod 4 (regardless if p is positive or negative). Hence, p = 2,3 mod 4 is never a possibility. – Guillermo Mantilla Jan 5 '10 at 17:19
Usually p=2,3 mod 4 refers to a positive prime, and it really does seem that that is what Ben meant, in which case, negative of a prime 3 mod 4 is fine. Hence my comment, as well as
Pete's. – Dror Speiser Jan 5 '10 at 19:11
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I am going to annoy all the people who answered above, but I am pretty sure the answer to Dror's question is basically no. In particular, is it known that there are infinitely many cubic
fields of prime discriminant? I have not heard of such a result -- if one is out there then I would be extremely grateful if someone would share the appropriate references with me.
As is pointed out above, there is a classical correspondence between such fields and subgroups of $Cl(\mathbb{Q}(\sqrt{p}))$ of index 3. However, I'm not aware that this makes the question
easier to answer.
There is also the work of Bhargava and Ghate, Delone-Faddeev, Davenport-Heilbronn, etc. which says that cubic (and quartic and quintic) fields are parameterized by integral orbits on nice
prehomogeneous vector spaces which meet certain local conditions. For example, in the cubic case, cubic rings are parameterized by integral binary cubic forms up to $GL_2(\mathbb{Z})$
equivalence, and maximal cubic orders are those cubic rings which meet a certain local condition at each prime.
This allows you to prove formulas for the number of cubic fields with $Disc(K) < X$ with good error terms, and this works if you ask for the condition $d | Disc(K)$. This allows you to run
up vote 2 a sieve.
down vote
However sieves are notoriously bad at finding primes! The information above is also essentially available in the twin prime problem, but all we can prove there is that there are infinitely
many primes $p$ so that $p + 2$ has at most two prime factors.
You can use this argument to find cubic fields with three (I think) prime factors -- there is a paper of Belabas and Fouvry that does this. Maybe you could push their arguments a little
bit better. But one cannot hope to find primes this way.
Of course there are excellent computational results, and I don't want to take anything away from these. But I feel like the question is asking if there are infinite families, and I'm
pretty sure this is widely expected but not at all known.
You can find infinitely many cubic fields of prime discriminant if you assume a conjecture of Hardy and Littlewood on primes in quadratic progressions -- just take the "simplest cubic
fields" where the quadratic function giving the discriminant represents a prime value. – Cam McLeman Nov 17 '10 at 2:07
Simplest cubic fields have a square discriminant. – Dror Speiser Nov 17 '10 at 7:00
Oops. . – Cam McLeman Nov 17 '10 at 12:43
1 @Frank: I wonder how far Heath-Brown's results on primes represented by binary cubic forms can be extended to proving infinitely many primes represented by the discriminant form (of a
general cubic). Sure - the degree goes up by one, but the number of variables goes up by two! – Dror Speiser Nov 17 '10 at 20:02
Dror - interesting suggestion! I'm not familiar enough with Heath-Brown's results to hazard a guess. – Frank Thorne Nov 18 '10 at 1:28
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Not the answer you're looking for? Browse other questions tagged nt.number-theory number-fields prime-numbers or ask your own question.
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o Sparse Graph Representation.
Can anyone explain how to get from an adjacency matrix to a sparse graph representation.
I have provided a picture of the example I am trying to understand. I have looked everywhere for information about the subject (books, youtube, scholar articles, google).
I understand that a sparse graph representation has non-zero values entities but other than that I have no idea how get from the adjacency matrix to sparse graph.
If anyone can provide a reference or explain it will be greatly appreciated.
Sorry but I could not rotate image. You can just download image and rotate it in your image viewer.
This post has been edited by eiarzate: 04 June 2012 - 05:16 PM
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Michal Migurski's notebook, listening post, and soapbox. Subscribe to . Check out the rest of my site as well.
One of the core gestures in a multi-touch interface is the two-finger deforming drag, a descendent of the traditional mouse-driven drag and drop. The difference is that with two points of contact,
interface elements such as windows can be moved, stretched, and turned. See what this would look like in a real interface five seconds into the big-ass table video. Implementing two-finger drag turns
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Drag the fingers and pretend they're your own:
(Please install the Flash Player plugin)
There are two main difficulties: figuring out how precisely the two contacts should act on an object, and then translating those into the appropriate placement, sizing, and rotation of the object. We
start with two rules: the object can be moved and turned, but not skewed, squashed, or otherwise deformed, and the fingers should stay in contact with the same points on the object throughout their
Both troubles can be solved with the use of affine transformation matrices, the closest thing computer science has to a true, working hammer. I've described before how to derive a complete
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Since version 8 or so, Flash has exposed proper matrix transformations on all clips in addition to the usual x, y, rotation, and scale. Unfortunately, the documentation leaves something to be
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Look out for two important functions in the source code:
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2. deformBox() applies the three fingers to repeatedly transform the photograph so that the fingers appear to be dragging it around the screen.
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Bond graph based simulation of nonlinear inverse systems using physical performance specificat
Gawthrop, P.J., and Ngwompo, R.F. (1999) Bond graph based simulation of nonlinear inverse systems using physical performance specificat. Journal of the Franklin Institute: Engineering and Applied
Mathematics, 336 (8). pp. 1225-1247. ISSN 0016-0032 (doi: 10.1016/S0016-0032(99)00032-0)
Full text not currently available from Enlighten.
Publisher's URL: http://dx.doi.org/10.1016/S0016-0032(99)00032-0
Analysis and simulation of non-linear inverse systems are sometimes necessary in the design of control systems particularly when trying to determine an input control required to achieve some
predefined output specifications. But unlike physical systems which are proper, the inverse systems are very often improper leading to numerical problems in simulation as their models sometimes have
a high index when written in the form of differential-algebraic equations (DAE). This paper provides an alternative approach whereby performance specifications and the physical system are combined
within a single bond graph leading to a greatly simplified simulation problem.
University Staff: Request a correction | Enlighten Editors: Update this record
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Sir Francis Galton - 100 years on
Monday 17th January 2011 marked the 100th anniversary of the death of Sir Francis Galton FRS (Fellow of the Royal Society). A man who was undoubtedly one of the most prominent scientific figures of
his time, his work across a plethora of fields – including anthropology, dermatoglyphics, geography, meteorology, psychometrics and statistics – has shaped and influenced much subsequent research via
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For most, Galton is probably characterized by one, if not both of these descriptions: 1) a forefather of eugenics (he invented the term in 1883), and 2) the half-cousin of Charles Darwin. Despite his
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Galton’s explorations in the fields of heredity and ancestry led him to writing about “regression toward the mean” - the statistical phenomenon that if an initial measurement of some variable is an
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exciting research.
Galton studied the Normal distribution during the 1870s and 1880s, and subsequently invented the Quincunx, or Galton Board to demonstrate the Central Limit Theorem – that the average of a large
collection of independent measurements of some variable (e.g. height, weight) will be Normally distributed (for an illustration of this, watch the following video). He has also been credited with the
“discovery” of the standard deviation, which enabled him to understand the variability of numerous characteristics within a population.
Perhaps his greatest statistical legacy was far more indirect, via the supervision of a doctoral student at University College London called Karl Pearson. Galton’s initial work on correlation was
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scientific journal on biometrics and the analysis of biological phenomenon using statistics.
Galton worked throughout his life in almost every academic discipline – and often provided fundamental contributions that are still evident today. On top of his work in statistics, he is also
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science and even founded the field of psychometrics. A true polymath of his time, Galton’s contributions to the sciences should not be underestimated, nor overshadowed by some of their applications.
Sir Francis Galton. Born 16th February 1822. Died 17th January 1911.
M. Eileen Magnello
Francis Galton never held a university post and, therefore, never supervised any doctoral students at UCL. Moreover, Karl Pearson did not do a PhD in statistics at UCL, especially as UCL was not even
offering any courses or degrees in mathematical statistics when he was a student. Pearson not only created the foundations to modern mathematical statistics at UCL beginning in 1892, but he set up
the first known university undergraduate degree in statistics in the world in 1917.
By the time Weldon introduced Galton to Pearson, he had already set up the infrastructure to his statistical methodology. Additionally, it was W.F.R. Weldon who first suggested to Pearson that they
establish Biometrika in November 1899: Galton was asked later to ask in consultation on the journal, but virtually all of the editing from 1901 to 1933 was done by Pearson.
Finally, Pearson was not Galton’s protégé; the inpetus to Pearson’s change of careers from an elastician to that of a mathematical statistician came from Weldon in the early 1890s. It was Pearson’s
later exploration of the mathematical properties of Galtonian correlation that enabled him to devise a battery of correlational techniques: Galton’s influence does not extend any further than this.
Pearson developed a body of methods in mathematical statistics that lay outside Galton’s conceptual framework and beyond his technical scope. He was never Galton’s student and he was his intellectual
heir in a limited sense only. Three men may be viewed as having inherited Galton’s statistics to a more notable extent include W F R Weldon, Francis Ysidro Edgeworth and George Darwin.
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Showing WithOUT A Dollar Sign
I am a new user both to Excel and these forums, but I am very glad they both exist! What I have is this (it is C13 for those playing at home):
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Duzgun, H.S.B. and Usul, N.
In many GIS application, it is essential to perform an accuracy assessment analysis which basically relies on sampling method and number of selected samples. This paper outlines the process of
employing simple and stratified random sampling and finding the optimum sample size in GIS. Two sampling methods such as simple and stratified random sampling are implemented on Güvenc Creek Basin,
Ankara, Turkey in order to draw samples for the elevation variable in GIS environment. For various sample sizes the optimum sample size is established for each sampling scheme as well as assessing
the efficiency of the sampling methods.
1. Introduction
Determination of required sample size is a very important task in many spatial problems, since the accuracy of estimations about the population is basically dependent on the sample size. Although
there are some rules of thumb on the required sample size in conventional data analysis (Walfond, 1995), when sampling frame is spatial the optimum sample size relies on the area of concern.
In this study, a GIS-based methodology is proposed for implementing the simple and stratified random sampling techniques. The optimum sample sizes from both sampling methods are determined by using
ArcGIS for the elevation variable. The two sampling methods are implemented on Güvenc Creek Basin, Turkey, in order to draw elevation samples in GIS environment. For various sample sizes, summary
statistics such as mean, mode, variance standard deviation are obtained as well as computing the confidence intervals. The optimum sample size for each sampling scheme is found based on the length of
confidence interval and standard error of the mean. The performance of two sampling methods is evaluated by comparing the sample statistics with the population parameters.
Moreover, the Güvenc Creek Basin s mean elevation found from hypsometric curve is compared with the ones obtained from both sampling methods implemented in GIS framework. In addition to that, the
elevations of existing five rain gauge stations are evaluated in terms of their representative ability of the Basin.
2. Sampling Methods
The sampling methods are basically divided into two categories such as, probability and non-probability sampling. The non-probability sampling is based on subjective judgement, while the probability
sampling uses random chance as determining factor for an observation to be involved in the sample (Walford, 1995). In this respect, probabilistic sampling has advantages over non-probabilistic
sampling since it ensures that all the population members have equal chance of being included in the sample. It also minimizes the bias introduced into the sample by subjective judgement of the
There are mainly five sampling tyres in probabilistic sampling scheme, namely:
- Simple
- Stratified
- Nested
- Cluster
- Systematic
Among these methods of sampling, stratified and simple random sampling techniques are the two of most widely used ones. The simple random sampling is usually used for the situations where the
population size is known. In this method each member of the population is identified and assigned a unique reference number. Then, based on randomly generated numbers, samples are drawn from the
population. Principally, simple random sampling forms the basis of other probabilistic sampling methods. Stratified random sampling is composed of grouping the members of the population into strata.
By using simple random sampling scheme, samples are drawn from each stratum and than the selected observations are pooled to form a single sample set.
3. Description of the Basin and the Methodology
Güvenc Basin is located at Güvenc Village of Yenimahalle, Ankara,in Middle Anatolia region of Turkey (Kupcu, 1996). The basin has an area and perimeter of 15.9 km^2 and 19.5 km, respectively.The
location of the basin in Turkey and its shape is given in Figure 1. There are five rain gauge stations installed in the basin (Figure 2).
Figure 1. Location of the Guvenc Basin
Figure 2. Distribution of the rain gauge stations in the basin>
The available data for the basin were in the form of digital contour line information (Figure 3). In order to apply both sampling methods, the area of the basin is converted into set of points with
attributes of elevation. For this purpose, first a TIN layer from the contours is created (Figure 4). Then the raster TIN layer is transformed into a vector layer of points (Figure 5). Later the
points in Figure 4 are associated with the elevation attributes by using the Join operation in ArcGIS. In joining the elevation attributes to the points, each point is given a mean elevation value
from the contour lines (Esri, 2000) The newly obtained layer in Figure 5 is composed of 89728 points with elevation values. Point density in the basin is 177.7 points/m^2. The mean elevation for
Güvenc Basin is found to be 1230.06 m with a standard deviation of 73.92 m, which corresponds to coefficient of variation (c.o.v.) of 6%. The minimum and maximum elevation values in the basin are
1050 and 1440 m, respectively. In the basin, 17% of the area has elevation less than 1150 m, 42 % has elevation range between 1150 m and 1250 m, 38 % has elevation between 1250 m and 1350 m and
finally, 3 %, has elevation greater than 1350 m. In other words, 80 % of the elevation values in the basin change between 1150 and 1350 m range.
Figure 3. Digital contour lines of the basin
Figure 4. TIN layer created from the contour lines
Figure 5. Point layer created from the transformation of the TIN
Having obtained basin area in the form of points, the next step is to implement simple and stratified random sampling methods. For this purpose, Visual Basic scripts are coded within ArcMap. The
following algorithm is used for simple random sampling:
1. Generate random numbers between the coordinate ranges of the points
2. Search the total population and find the corresponding point which has coordinates generated randomly in step 1
3. Record the elevation of the selected point
4. Repeat steps 1-3 until desired number of samples are selected.
An example for simple random sampling with sample size of 10% of 89728 points is given in Figure 6.
Figure 6. Distribution of sample points from simple random sampling
For stratified random sampling, the basin is classified into four strata based on the elevation. The first stratum is composed of elevation value less than 1150 m, the second and third ones contain
points with elevation values changing between 1150-1250 m and 1250-1350 m, respectively. Finally the forth stratum consists of elevation values greater than 1350 m. In this sampling scheme, 17 % of
the total sample size is selected from the first stratum, 42 % and 38 % of the total sample size are obtained from the second and third strata, respectively. The rest (3 %) is drawn from the forth
The algorithm developed for stratified random sampling is similar to the simple random sampling and as follows:
1. Generate random numbers between the coordinate ranges of the points
2. Search the total population and find the corresponding point which has coordinates generated randomly in step 1.
3. Read the elevation of the selected point.
4. Locate the point into one of the four of the strata by checking the elevation value.
5. Repeat steps 1-4 until each stratum contains required number of samples.
4. Determination of Optimum Sample Size
In order to determine optimum sample size, several number of samples are drawn based on each sampling method. Statistics such as, mean (s), standard deviation (s), variance (s^2), mode, minimum
(min), maximum (max) and standard error of the mean (s[s]) are computed as well as establishing the confidence intervals. The results are tabulated in Tables 1 and 2 for simple and stratified random
sampling methods,respectively.
Table 1. Sample statsitics for simple random sampling
Table 2. Sample statsitics for stratified random sampling
Tables 1 and 2 indicate that when the sample size increases, the length of confidence internal and the standard error decrease, as expected. The graphs of sample size versus standard error (Figures 7
and 8) and sample size versus confidence interval length (Figures 9 and 10) for both sampling methods show that sample size of 5 % of total population (4486) can be considered as optimum sample size,
since the curves start levelling out at this sample size. This fact can also be seen from Figures 11 and 12, where the means from both sampling schemes become constant at the sample size of 4486.
Figure 7. Sample size versus standard error of the mean for simple random sampling
Figure 8. Sample size versus standard error of the mean for stratified random sampling
Figure 9. Sample size versus confidence interval length for simple random sampling
Figure 10. Sample size versus confidence interval length for stratified random sampling
Figure 11. Sample size versus sample mean for simple random sampling
Figure 12. Sample size versus sample mean for stratified random sampling
Moreover when the standard error of the mean values of the two sampling techniques are compared, it is seen that they give almost the same values. Hence this indicates that, both sampling schemes can
be effectively used in such analyses. However simple random sampling is computationally more efficient. Based on the proposed methodology, in this study the mean elevation is found to be 1230 m. On
the other hand, the mean elevation computed by Denli et al. (1996) is 1256m, where the median elevation of 1235 m is obtained from hypsometric curve (Kupcu, 1996), which is closer to the one computed
in this study (1230 m).
As it is mentioned before, the 80 % of the basin area has elevation ranging between 1150 and 1350 m. When the elevation of rain gauge stations are examined (Figure 13) it can easily be seen that
elevation values of four, out of five stations are in this range.
Figure 13. Elevation of the five rain gauge stations in the basin
5. Conclusion
This study presents a GIS-based methodology for estimating the mean elevation in a river basin. In hydrology, there are various methods for the determination of mean basin elevation and other
parameters related to the basin and these methods are usually cumbersome, especially the basin is large. If better and easier methods can be found by using GIS techniques, it will help hydrologist a
lot. In this respect, the proposed methodology is in fact a general and a straight forward one and can also be used for estimating other properties of basins such as total basin slope and aspect
The proposed methodology is implemented on a small basin in Turkey. The mean elevation is estimated based on simple and stratified random sampling methods. However, it can be applied by using other
sampling methods by coding appropriate scripts.
The optimum sample size for both sampling schemes is determined based on the standard error of the mean and the length of the confidence intervals. Since Güvenc Creek Basin is relatively homogeneous
in terms of elevation with c.o.v. of 6 % the optimum sample size is obtained at 5 % of total population. However note that for basins having elevation more variable than Güvenc, the optimum number of
samples would subject to change. On the other hand, since the methodology is a general one it can easily be implemented on greater basins.
6. References
Denli, O., Tekeli, I., Demirkiran, O. and Sevinc, N., 1996. Research on Rainfall and Flow Characteristics of Ankara-Yemimahalle-Guvenc Basin, Annual Final Report of General Directorate of Rural
Services (in Turkish).
Esri, 2000. ArcGIS User Manual, Esri Publications, Redland, USA
Kupcu, O., 1996. Application of GIS Techniques to Derive SCS Synthetic Unit Hydrograph, Master of Science Thesis, Middle East Technical University, Ankara, Turkey.
Walford, N.,1995. Geographical Data Analysis, John Wiley and Sons Inc., New York, USA.
H. Sebnem Düzgün
Geodetic and Geographic Information Technologies
Middle East Technical University
Civil Eng. Water Resources Lab.
K4-123 06531 Ankara Turkey
Tel: +90-312-2105415
Fax: +90-312-2101002
Nurünnisa Usul
Assoc. Prof. Dr.
METU, Civil Engineering Dept.
Ankara, 06531, Turkey
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Documented Problem Solving: Price Elasticity of Demand
The concept of price elasticity of demand was introduced in class. Elastic and inelastic goods were discussed. The impact that a change in price will have on total revenue was also presented.
Learning Goals
Students will:
• describe the relationship between the price of a good and the demand for the good;
• interpret the impact a change in price will have on demand based on the price elasticity;
• compare the impact that a change in price will have on total revenue based on elasticity.
Context for Use
This activity is appropriate to use during class or as a homework assignment. It will work in any size class or setting. It can be used with a Principles of Economics class or with an Intermediate
Microeconomics class.
Description and Teaching Materials
For this exercise, the instructor will need a MC, T/F, or short answer question that involves critical thinking. One is given below.
If the price elasticity of demand for a product is equal to 1, then a decrease in price will cause:
a) An increase in total revenue.
b) A decrease in total revenue.
c) No change in total revenue.
d) An increase in total revenue if the cost is low and a decrease in total revenue if the price is high.
Answer: c
Teaching Notes and Tips
Students sometimes struggle with the concept of price elasticity of demand, but if they focus on the idea that elasticity measures how much something will stretch or change, it will be easier for
them. If demand for a product is elastic, meaning it changes, then a change in price will cause an even bigger change in demand. The opposite is also true.
Below is a sample of a student solution that includes the thought processes used to solve the question.
1. First, I thought about the definition for price elasticity of demand. It measures how much demand will change in response to a change in price.
2. This reminded me of the law of demand - the quantity demanded of a good increases as the price of the good decreases.
3. I looked back in my notes for the formula for elasticity. It is the (percentage change in quantity demanded) รท (percentage change in price).
4. Based on the formula in order for elasticity to equal 1, the percentage change in quantity demanded must be the same as the percentage change in price.
5. I looked in the textbook for the formula for total revenue. It is equal to price X quantity.
6. Then I looked for the relationship between elasticity and a change in total revenue.
7. If elasticity is 1, that means that the numerator and the denominator are the same, so the percentage change in quantity demanded is equal to the percentage change in price.
8. If the percentage change on the bottom (quantity demanded) is equal to the percentage change on the top (price), then a price decrease will cause an increase in quantity demanded.
9. I know that total revenue is price X quantity, so if both change by the same percentage, then total revenue stays the same.
References and Resources
Angelo, T.A. and Cross, K.P. (1993). Classroom Assessment Techniques: A Handbook for College Teachers. San Francisco: Jossey-Bass.
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, of a Fraction, is the number which shews how many of those parts, which the integer is supposed to be divided into, are denoted by the fraction. And, in the notation the Numerator is set over the
denominator, or number that shews into how many parts the integer is divided, in the fraction. So, ex. gr. 3/4 denotes three-fourths, or 3 parts out of 4; where 3 is the numerator, and 4 the
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Dynamic Vehicle Routing Using an Improved Variable Neighborhood Search Algorithm
Journal of Applied Mathematics
Volume 2013 (2013), Article ID 672078, 12 pages
Research Article
Dynamic Vehicle Routing Using an Improved Variable Neighborhood Search Algorithm
^1Branch of Quality Management, China National Institute of Standardization, Beijing 100088, China
^2Department of Industrial Engineering, Tsinghua University, Beijing 100084, China
^3School of Economics and Management, Beihang University, Beijing 100191, China
Received 14 September 2012; Revised 5 December 2012; Accepted 25 December 2012
Academic Editor: Qiuhong Zhao
Copyright © 2013 Yingcheng Xu et al. 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.
In order to effectively solve the dynamic vehicle routing problem with time windows, the mathematical model is established and an improved variable neighborhood search algorithm is proposed. In the
algorithm, allocation customers and planning routes for the initial solution are completed by the clustering method. Hybrid operators of insert and exchange are used to achieve the shaking process,
the later optimization process is presented to improve the solution space, and the best-improvement strategy is adopted, which make the algorithm can achieve a better balance in the solution quality
and running time. The idea of simulated annealing is introduced to take control of the acceptance of new solutions, and the influences of arrival time, distribution of geographical location, and time
window range on route selection are analyzed. In the experiment, the proposed algorithm is applied to solve the different sizes' problems of DVRP. Comparing to other algorithms on the results shows
that the algorithm is effective and feasible.
1. Introduction
Dynamic Vehicle Routing Problem (DVRP) is a variant of the Vehicle Routing Problems (VRPs) that has arisen due to recent advances in real-time communication and information technologies. The VRP is a
nondeterministic polynomial hard (NP-hard) problem that calls for the determination of the optimal set of routes to be performed by a fleet of vehicles to serve a given set of customers. However, a
majority of the information is unpredictable before path optimization, such as the customer's geographic position, customer demand, and vehicle travel and service time. This information is dynamic
and new information may appear or existing information changes, and so forth. Many different factors must be considered when a decision about the allocation and scheduling of a new request is taken:
the current location of each vehicle, their current planned route and schedule, characteristics of the new request, travel times between the service points, characteristics of the underlying road
network, service policy of the company, and other related constraints. The DVRP is a complex problem compared to the classic VRP, and variable neighborhood search (VNS) is proposed as a means to
effectively and efficiently tackle the dynamic problem and optimize the planned routes between the occurrences of new events. VNS was initially proposed by Hansen and Mladenović [1, 2] for solving
combinatorial and global optimization problems. The main reasoning of this metaheuristic is based on the idea of a systematic change of neighborhoods within a local search method.
This paper has two main contributions. First, according to the characteristic of DVRP, we gave the graph expression and formulated the mathematical model. Second, the proposed algorithm is improved
in initial solution, shaking and local search based on classic VNS. The later optimization process is presented, and the results show that the algorithm is feasible and competitive with other
existing algorithms. The remainder of the paper is organized as follows: literature reviews are illustrated in Section 2 and the mathematical formulation problem is discussed in Section 3. Section 4
introduces the main ideas of the improved variable neighborhood search, while computational results are presented and discussed in Section 5. Section 6 concludes the paper.
2. Literature Reviews
The DVRP problem is closely related to the actual production and life. In recent years, DVRP gradually becomes a hot topic. Both domestic and foreign scholars mainly focus on the construction of
different scheduling models and the designing of simple and efficient heuristic algorithms. A survey of results achieved on the different types of DVRPs can be found in Gendreau and Potvin [3]. The
dynamic full-truckload pickup and delivery problem has been studied by Fleischmann et al. [4] and Yang et al. [5]. Montemanni et al. [6] proposed the vehicle scheduling model with random dynamic
demand and solved it using an ant colony algorithm. In Gendreau et al. [7] neighborhood search heuristics for dial-a-ride problems are finally presented. Goel and Gruhn [8] studied the mathematical
model of dynamic vehicle routing problems under the conditions of the real-life information. Schönberger and Kopfer [9] studied the real-time decision-making and autonomous decision-making of DVRP.
Novoa and Storer [10] introduced the approximate solution algorithm with stochastic demands. Branchini et al. [11] presented local search algorithm for a dynamic vehicle routing problem. Müller [12]
studied DVRP with time window and analyzed the algorithm for the model suboptimal solution.
In the past ten decades, a tremendous amount of work in the field of vehicle routing problems has been published, especially literature based on VNS. The Bräysy [13] gave the internal design of the
VND and RVNS algorithm in detail, analyzed the VRPTW problem, and indicated the VND algorithm as one of the most effective ways to solve VRPTW problems. Polacek et al. [14] designed VNS to solve the
multidepot vehicle routing problem with time windows MDVRPTW. His algorithm used the neighborhood structure of swap and cross to do shaking operation for the current solution, to do local search with
a constrained 3-opt operator to accept the part of the poor solution to avoid getting into a local optimum for the algorithm by Threshold Accepting. Kytöjoki et al. [15] designed the guided VNS
algorithm to handle the 32 existing large-scale VRP problems and compared it to the TS algorithm. The result showed that the VNS algorithm was more effective than TS algorithm in solving time. Goel
and Gruhn [16] introduced the RVNS to solve the general VRP problem including time windows, vehicle constraints, path constraints, travel departure time constraints, capacity constraints, the order
models compatibility constraints, multisupplier point of the orders, and transport and service position constraints. Hemmelmayr et al. [17] proposed the VNS algorithm for periodical VRP problem,
adopted the saving algorithm for the construction of the initial solution, designed the move and cross neighborhood, used 3-opt operator as local search strategies, and contrasted it with other
research results. Fleszar et al. [18] adopted VNS algorithm to solve the open-loop VRP problem and tested 16 benchmark problems.
Due to the complexity of the problem, the current solving quality and efficiency for the DVRP are far from the practical requirements. So, there are many problems need to make in-depth research such
as how to seek feasible solutions, how to prevent falling into local optimum, and how to control the solution within the acceptable range. This paper presents an improved variable neighborhood search
algorithm to solve DVRP; it integrates local search operator, optimization process, and the simulated annealing algorithm into the VNS algorithm framework. Through the comparison with other
algorithms, it shows that the proposed algorithm can get the better solution.
3. Problem Descriptions
3.1. Problem Definition
Larsen [19] defined DVRP with two aspects: not all information relevant to the planning of the routes is known by the planner when the routing process begins; information can change after the initial
routes have been constructed. And he illustrated a dynamic vehicle routing situation. The simple example is shown in Figure 1. As seen from it, two uncapacitated vehicles must service both advance
and immediate request customers without time windows. The advance request customers are represented by yellow nodes, while those that are immediate requests are depicted by black nodes. The solid
blue lines represent the two routes the dispatcher has planned prior to the vehicles leaving the depot. The two thick arcs indicate the vehicle positions at the time the dynamic requests are
received. Ideally, the new customers should be inserted into the already planned routes without the order of the nonvisited customers being changed and with minimal delay. This is the case depicted
on the right-hand side route. However, in practice, the insertion of new customers will usually be a much more complicated task and will imply a replanning of the nonvisited part of the route system.
This is illustrated by the left-hand side route where servicing the new customer creates a large detour.
We can also describe the DVRP with a mathematic approach. The problem is defined on a complete graph , where is the vertex set and is the arc set. The vertices’ set corresponds to the depots. Each
vertex has several nonnegative weights associated with it, namely, a demand , a arrival time , the waiting time , service time , and an earliest and latest possible start time for the service, which
define a time window [, ]. is the transportation cost from customer to customer in the period . Each vehicle has associated with a nonnegative capacity . refers to the number of time periods, is the
fixed cost for a vehicle, is the traveling cost per unit time, and is a very large number.
The variables are defined as follows:
3.2. Mathematical Formulation
3.2.1. Objective Function
The dynamic vehicle routing problem with time windows is formulated in this section as a mixed integer linear programming problem. The objective of the formulation is to minimize the total cost that
consists of the fixed costs of used vehicles and the routing costs:
3.2.2. Problem Constraints
The constraints of the problem consist of vehicle constraints, demand constraints, routing constraints, and other constraints.
Assignment of Nodes to Vehicles. Equation (3) ensure that each customer has a vehicle service for only one time and the vehicle does not return to the yard:
Relationship between the Vehicle and Depot. Constraint (4) ensures that the number of vehicles departed from the yard does not exceed the maximum number of vehicles belonging to the distribution
Relationship between the Vehicle and Route. Constraint (5) ensures that customers on the same route are delivered by the same vehicle:
Assignment of Nodes to Vehicles. Equation (6) states that every customer node must be serviced by a single vehicle:
Capacity Constraints. Constraint (7) states that the overall load to deliver to customer sites serviced by a used vehicle should never exceed its capacity:
Subcircuit Constraints. Equation (8) ensures to eliminate subcircuit:
Time Constraint Violations due to Early/Late Services at Customer Sites. One has
Other Constraints. One has
4. An Improved Variable Neighborhood Search Algorithm
VNS is a metaheuristic for solving combinatorial and global optimization problems proposed by Hansen and Mladenović [1, 2]. Starting from any initial solution, a so-called shaking step is performed
by randomly selecting a solution from the first neighborhood. This is followed by applying an iterative improvement algorithm. This procedure is repeated as long as a new incumbent solution is found.
If not, one switches to the next neighborhood (which is typically larger) and performs a shaking step followed by the iterative improvement. If a new incumbent solution is found, one starts with the
first neighborhood; otherwise one proceeds with the next neighborhood, and so forth. The description consists of the building of an initial solution, the shaking phase, the local search method, and
the acceptance decision. The flow of VNS is shown in Figure 2.
4.1. Initial Solution
Using variable neighborhood search algorithm, first, it needs to build one or more initial feasible solution; a clustering algorithm for an initial feasible solution mainly completes two tasks:
customer allocation and path planning. For obtaining an initial solution, each customer is assigned a visit day combination randomly. Routes are constructed by solving a vehicle routing problem for
each day using the Clarke and Wright savings’ algorithm [20]. The Clarke and Wright savings’ algorithm terminates when no two routes can feasibly be merged; that is, no two routes can be merged
without violating the route duration or capacity constraints shown in Algorithm 1. As a result, the number of routes may exceed the number of available vehicles. In that case, a route with the fewest
customers is identified and the customers in this route are moved to other routes (minimizing the increase in costs). Note that this may result in routes that no longer satisfy the duration or
capacity constraints. This step is repeated until the number of routes is equal to the number of vehicles. Since the initial solution may not be feasible, the VNS needs to incorporate techniques that
drive the search to a feasible solution.
The initial solution obtained by the above method can basically meet the needs of the follow-up work, building the foundation to achieve optimal feasible solution in the following algorithm.
4.2. Shaking
Shaking is a key process in the variable neighborhood search algorithm design. The main purpose of the shaking process is to extend the current solution search space, to reduce the possibility that
the algorithm falls into the local optimal solution in the follow-solving process, and to get the better solution. The set of neighborhood structures used for shaking is the core of the VNS. The
primary difficulty is to find a balance between effectiveness and the chance to get out of local optimal. In the shaking execution, it first selects a neighborhood structure from the set of
neighborhood structures of current solution ; then according to the definition of , corresponds to change and generate a new solution .
There are two neighborhood structures to achieve the shaking: insert and exchange. Insert operator denotes a certain period of consecutive nodes moving from the current path to another path; exchange
operator refers to interchange the two-stage continuous nodes belonging to different paths. The insert and exchange operators are shown in Figure 3. The cross-exchange operator was developed by
Taillard et al. [21]. The main idea of this exchange is to take two segments of different routes and exchange them. Compared with the VNS by Polacek et al. [14], the selection criterion is slightly
changed. Now it is possible to select the same route twice. This allows exploring more customer visit combinations within one route. An extension to the CROSS exchange operator is introduced by
Bräysy [13]; this operator is called improved cross-exchange—Cross exchange for short. Both operators are used to define a set of neighborhood structures for the improved VNS.
In each neighborhood, the insert operator is applied with a probability to both routes to further increase the extent of the perturbation; then the probability of the exchange operator is . IVNS
selects randomly an exchange operator to change path for each shaking execution. The shaking process is somewhat similar to the crossover operation of the genetic algorithm. When the process is
finished, the only two paths have the exchange of information; most of the features of the current solution will be preserved, to speed the convergence of the algorithm.
4.3. Local Search
In a VNS algorithm, local search procedures will search the neighborhood of a new solution space obtained through shaking in order to achieve a locally optimal solution. Local search is the most
time-consuming part in the entire VNS algorithm framework and decides the final solution quality, so computational efficiency must be considered in the design process of local search algorithm. Two
main aspects are considered in the design of local search algorithms: local search operator and the search strategy. Based on the previous studies, this paper selects and as a local search operator
in order to obtain the good quality local optimal solution in a short period; they are shown in Figure 4. According to the probability, one of the two operators is selected in each local search
process. The parameter represents the probability of selection for ; similarly, the probability of selection for can be expressed as . This mixed operator can develop optimization ability for and and
expand the solution space of the algorithm.
There are mainly two search strategies: first-improvement and best-improvement in local search algorithm. The former refers to access the neighborhood solution of the current solution successively in
the solution process, if the current access neighborhood solution is better than , to make and update neighborhood solution. We repeat these steps until all the neighborhood solutions of are
accessed. Finally, will be obtained as a local optimal solution. The latter refers to traverse all of the neighborhood solution of current solution in the solution process, to select the optimum
neighborhood solution as a local optimal solution. In this paper, we adopt the best-improvement strategy; it enables the algorithm to achieve a better balance in the solution quality and run time.
4.4. Later Optimization Process
In order to accelerate the convergence speed and improve the solution quality, the later optimization process is proposed in the IVNS algorithm. After the local search is completed, if the local
optimal solution is better than the global optimal solution , that is, , the later optimization process will be continued to be implemented in order to seek a better global optimal solution [22]. The
algorithm of later optimization process which was proposed by Gendreau et al. is suitable for solving the traveling salesman problem and the vehicle routing problem with time windows. The algorithm
processes can be simply described as follows.
Step 1. There are some assumptions. The path that will be optimized is , its length is , and its value of the evaluation function is . The final optimized path is , the value of the evaluation
function is , and , , and .
Step 2. The Unstring and a String processes [23] are, respectively, performed for the th customer in the , the optimized path can be obtained, and its value of the evaluation function value is .
Step 3. If , some processes are carried out; they are , , , , and , jump to Step 2; otherwise, .
Step 4. If , the algorithm will be terminated; otherwise, jump to Step 2.
4.5. Acceptance Decision
The last part of the heuristic concerns the acceptance criterion. Here we have to decide whether the solution produced by VNS will be accepted as a starting solution for the next iteration. To avoid
that the VNS becomes too easily trapped in local optima, due to the cost function guiding towards feasible solutions and most likely complicating the escape of basins surrounded by infeasible
solutions, we also allow to accept worse solutions under certain conditions. This is accomplished by utilizing a Metropolis criterion like in simulated annealing Kirkpatrick [24] for inferior
solutions and accepting them with a probability of (11), depending on the cost difference to the actual solution of the VNS process and the temperature . We update every iterations by an amount of ,
where denotes a random number on the interval , where denotes the maximal VNS iterations, and an initial temperature value is :
5. Numerical Experiments
In order to assess the performance of the improved variable neighborhood search algorithm to solve DVRP, three test problems with respect to different sizes (small, medium, and large) have been done.
We analyze the solution quality and efficiency of our proposed algorithm. IVNS algorithm is implemented by the C # language, and the main configuration of the computer is an Intel Core i3 1.8GHz,
2GB RAM running Windows XP.
5.1. Case 1
5.1.1. Experimental Data and Setting
In order to assess the performance of the improved variable neighborhood search algorithm to solve DVRP, the data sets from the literature [25] are used. Firstly, the VNS algorithm is applied to
solve the DVRP, and then the results are analyzed and compared with other existing algorithms.
The dynamic distribution network is randomly generated by the computer. The distribution area is a square of 50km 50km; 30 static demand customers and 10 the dynamic demand customers are randomly
generated. Each customer's demand is a random number of , the vehicle's capacity of distribution centers is 8t, and the maximum driving distance of vehicle once is 100km. These information
including the coordinates and demand of 30 static customers and 10 dynamic customers are randomly generated by the computer, the location of the distribution center is (25km, 25km), and the number
of vehicles is 3. The objective is to arrange the delivery route of the vehicle reasonably, so that the distribution mileage is the shortest. For simplicity, the distance between customer and
distribution center uses the straight-line distance. The coordinates and demands of static customers and dynamic customers are shown in Tables 1 and 2, respectively, the specific position
relationship for the customer and the customer, and the customer and distribution center are shown in Figure 5.
The initial values of the various parameters for IVNS algorithm are set as follows.(1)The parameter settings for simulated annealing accepted criteria are initial temperature , every generation to
update temperature , to end the algorithm.(2)The parameters value of the Shaking operation are as follows: , , and .(3)The value is 0.5 in local search.
5.1.2. Numerical Results
First, distribution of fixed-demand customer is optimized, solved, and generated initial distribution route, as shown in Table 3.
The customers’ demand information is updated at period 1; at this moment, the service requests are put forward by six dynamic customers of A, D, E, F, G, and I. According to the known information on
real-time optimization stage, the improved variable neighborhood search algorithm is used to solve the distribution network at period 1 and output scheduling plan, as shown in Table 4, and the
specific route is shown in Figure 6.
At period 2, the four dynamic demand customers of B, C, H, and J have the service request, and now according to customer requests, we update the relevant information. Based on known real-time
information on the dynamic distribution network, the improved variable neighborhood search algorithm is applied to solve the distribution network at period 2, and then the scheduling plan is output,
as shown in Table 5, while specific vehicle route is shown in Figure 7.
These results show that the improved variable neighborhood search algorithm can solve the real-time requirements of the dynamic vehicle routing problem; the best-improvement search strategy and the
insert and exchange operators speed the convergence of the algorithm and obtain better solution.
In order to verify that the improved variable neighborhood search for solving the dynamic vehicle routing problem, we compare it with the genetic algorithms (GAs), tabu search (TS), and a two-stage
algorithm (TPA) based on the above example. The parameters setting of GA, TS, and TPA can be seen in the relevant literature [22, 25]. A comprehensive comparison is made from the optimal value, the
worst value, the average value, search success rate, and the number of iterations. The results of the experiment are shown in Table 6.
As can be seen from Table 6, the four algorithms consistently find an optimal solution, but the worst and average values have obvious differences; there are differences in search success rate and
the number of iterations. The order of the four algorithms of search success rate from the smallest to largest is tabu search algorithm, two-stage algorithm, genetic algorithm, and IVNS; the order
for the worst value is as follows: the average value from small to big is IVNS, two-stage algorithm, genetic algorithm, and tabu search algorithm; it reflects the IVNS has the better global search
capability. The number of iterations for four algorithms from the smallest to the largest is IVNS, tabu search algorithm, two-stage algorithm, and genetic algorithm, this also shows that convergence
rate of IVNS is faster than the other algorithms, and it can be more appropriate to solve dynamic vehicle routing problem to some extent.
5.2. Case 2
The experimental test uses the benchmark data which was 100-node VRPTW calculation example and compiled by Solomon in 1987. Every sample contains 100 nodes and distributes into Euclidean plane. The
sample is divided into six categories: R1, R2, C1, C2, RC1, and RC2. DVRPTW adopts the Lackner dynamic test data set which is designed in 2004 based on the Solomon example. For each question in the
Solomon example, there are five data sets corresponding to it. They are 90%, 70%, 50%, 30%, and 10% five dynamic degree.
The dynamic degree is described as follows: is the number of dynamic customer demand, and is the number of static customer demand.
Table 7 gives the comparison results of the Solomon problem to IVNS and Lackner under different dynamic degrees. From the number of vehicles, the average driving distance, and the average calculation
time, we compared the calculation results and the relative error. For five dynamic degrees, the average values of the number of vehicles, respectively, are 8.637 and 8.81 for IVNS and Lackner, the
relative error is −2.01; the average values of driving distance respectively are 1118.63 and 1183.72 for IVNS and Lackner, and the relative error is −5.50; the average values of calculation time,
respectively, are 32.46 and 33.03 for IVNS and Lackner, and the relative error is −1.72. According to Table 7, the following conclusions can be drawn.(1)In terms of the total driving distance,
relative to the best results which are obtained by Lackner’s various algorithms, except for the RC1, R1, and C2, the proposed algorithm can obtain better calculation results for other type’s dynamic
degrees.(2)The present algorithm, in which the average computation time is less than 90 seconds, fully meets the requirements of real-time scheduling. The average calculation time of IVNS is lower
than Lackner; it demonstrates that the algorithm has better search capabilities.
5.3. Case 3
In order to verify the proposed variable neighborhood search algorithm to solve the large-scale dynamic vehicle routing problem effectively, this paper extends to large-scale test instances with 1000
customer which are presented by Gehring and Homberger. Table 8 gives the comparison results of the first path adjustment for the part instance based on Gehring and our proposed algorithm.
As shown in Table 8, the results of the objective value, the number of vehicles, and the calculation time are compared based on Gehring and IVNS. 107.33 average new demands are occurred, and 692.33
total demands are handled in the path adjustment instance; only 1.17 customer demands failed to be satisfied, while the service level is close to 100%. In dealing with a large-scale problem, the
algorithm running time is an important evaluation index; the average running time of our proposed algorithm is 274.10s; it is smaller than Gehring’s algorithm. The number of vehicle is closely
related to the cost, and it is decreased from 62.17 to 60.67. The objective value is reduced from 40044.58 to 39926.00; it demonstrates that the results have been improved. According to the analysis,
to some extent, the proposed algorithm can solve the large-scale dynamic VRP and improve the results.
6. Conclusions
In this research, we proposed a formulation for a dynamic vehicle routing problem. We also presented an improved variable neighborhood search (IVNS) metaheuristic for DVRP. In the initial solution,
the routes are constructed by solving a vehicle routing problem for each day using the Clarke and Wright savings algorithm Clarke and Wright [20]. In the shaking execution, there are two neighborhood
structures to achieve the shaking: insert and exchange. This paper selects and as a local search operator in order to obtain the good quality local optimal solution in a short period. In order to
accelerate the convergence speed and improve the solution quality, the later optimization process is proposed in the IVNS algorithm. To test the performance of the proposed improved variable
neighborhood search algorithm, we test three problems with respect to different sizes (small, medium, and large) and compare the results with other algorithms. The results show that the proposed
model and the algorithm are effective and can solve the DVRP within a very short time and improve the quality of solution to some extent.
Funding for this research is supported by the National Science Foundation of China under Grants nos. 90924020, 70971005, and 71271013, the Societal Science Foundation of China under Grant no.
11AZD096, China Postdoctoral Science Foundation funded project no. 2012M520008, and Quality Inspection Projects no. 200910088 and no. 201010268.
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Re: st: Doing something an observation-specific number of times
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Re: st: Doing something an observation-specific number of times
From Maarten Buis <maartenlbuis@gmail.com>
To statalist@hsphsun2.harvard.edu
Subject Re: st: Doing something an observation-specific number of times
Date Tue, 28 Aug 2012 19:39:29 +0200
On Tue, Aug 28, 2012 at 6:45 PM, robert hartman wrote:
> Imagine that observation 1 has v1 and v2 values of .41 and 78,
> respectively. <snip> For example, for observation 1, the new obs 1 v3
> value=((1+(.41^1))/2) + ((1+(.41^2))/2) ...((1+(.41^77))/2) +
> ((1+(.41^78))/2).
> I have begun to think of some klugy ways of doing this via looping or
> even the expand command.
Depending on the number of observations in your original dataset the
-expand- route may be the easiest. If the number of observations is
large than this strategy may be infeasible due to memory limitations.
When it comes to efficiency, you need to make the tradeoff between the
amount of time you need to write the more fancy code (and the effort
you will need to understand it again after some time...) against the
time you safe because it runs quicker. Often the balance will be
against the more fancy solutions(*).
*---------------- begin example ---------------
// create some example data
input v1 v2
.41 78
.23 50
// we need to keep track on who is who before
// expanding
gen id = _n
// create v2 rows per observation
expand v2
// create the appropriate exponent
bys id : gen expo = _n
// create the basic component of the computation
gen double value = (1+v1^expo)/2
// sum() returns a running sum
by id : replace value = sum(value)
// the final sum is the last of the running sum
bys id (expo) : replace value = value[_N]
//get rid of things that are no longer needed
drop expo
by id : keep if _n == 1
drop id
// see the result
*----------------- end example ----------------
(For more on examples I sent to the Statalist see:
http://www.maartenbuis.nl/example_faq )
Hope this helps,
(*) This of course ignores the pure joy you will get from figuring out
the fancy solution, but we are not payed to enjoy ourselves!
Maarten L. Buis
Reichpietschufer 50
10785 Berlin
* 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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Simple Proof Question
September 23rd 2010, 05:28 PM #1
Junior Member
Sep 2010
Simple Proof Question
Given gcd(a,b)=1 and a|bc prove a|c?
This is what I did, I just want someone to confirm it as I don't have any solutions.
as+bt=1 (coprime) (1) and bc=ak (2) for some integer s,t and k.
Multiply (1) by c --> acs+bct=c and use (2) to replace bc with ak
Then we get acs+akt=c --> factorize a and then we get a|c.
Is this simple proof enough, or have I missed out on something?
Given gcd(a,b)=1 and a|bc prove a|c?
This is what I did, I just want someone to confirm it as I don't have any solutions.
as+bt=1 (coprime) (1) and bc=ak (2) for some integer s,t and k.
Multiply (1) by c --> acs+bct=c and use (2) to replace bc with ak
Then we get acs+akt=c --> factorize a and then we get a|c.
Is this simple proof enough, or have I missed out on something?
Looks fine to me.
September 23rd 2010, 10:23 PM #2
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4 point interpolation
May 4th 2011, 11:17 PM #1
May 2011
4 point interpolation
Hello, I hope this post belongs to the right topic.
I have a problem where i want to calculate a value at a normalized (x,y) position on a square field where values for the four corner points are known. I know this sounds like a problem found
anyware so there should be a simple way of doing this.
Given a square field with corners (0,0)=A (1,0)=B (0,1)=C (1,1)=D, calculate interpolated value at point (x,y) where x={0..1} and y={0..1}
Solution (brutal force):
result = (x1+(x2-x1)x + y1+(y2-y1)y)/2
Is there a simple way?
Hello, I hope this post belongs to the right topic.
I have a problem where i want to calculate a value at a normalized (x,y) position on a square field where values for the four corner points are known. I know this sounds like a problem found
anyware so there should be a simple way of doing this.
Given a square field with corners (0,0)=A (1,0)=B (0,1)=C (1,1)=D, calculate interpolated value at point (x,y) where x={0..1} and y={0..1}
Solution (brutal force):
result = (x1+(x2-x1)x + y1+(y2-y1)y)/2
Is there a simple way?
Google and then usually wikipedia are your friends:
Bilinear interpolation - Wikipedia, the free encyclopedia
Exactly what I where looking for, just didn't know hot to name it. Thanks.
May 5th 2011, 02:09 AM #2
Grand Panjandrum
Nov 2005
May 5th 2011, 05:17 AM #3
May 2011
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Aircraft Response to Atmospheric Turbulence at Various Types of the Input Excitation
The effects of atmospheric turbulence on many of the modern sophisticated transport systems have become an important design parameter from both structural and performance aspects. Aircraft encounters
with turbulence represent a serious safety threat for airlines. The problem of gusty winds proved itself to be a major obstacle to successful flight. The history of aviation abounds with incidents
and accidents in which the variability of the wind in space or time played a decisive role. Loss of control of the altitude or the flight path and even the crashes of jet aircraft were not uncommon.
Aside from the human catastrophes, annual injuries to passengers and flight crew cost airlines millions in lost work time and medical expenses (Prince and Robinson, 2001). Turbulence refers to an
irregular or disturbed flow in atmosphere that produces gusts and eddies. The most economic and practical method to explore innovative concepts and to investigate configuration options at an early
stage is to first conduct an analytical and/or simulation study using an appropriate engineering mathematical model of the relevant physics (Buck and Newman, 2005). Often in an aircraft model
simulation development, the gust effects of the atmosphere are neglected for various reasons and removed in the final form of the equations. Here, gust effects are the key excitation of interest. The
models of the wind have to accommodate both events that are perceived as discrete (usually described as gusts), as well as the phenomenon described as continuous turbulence. Discrete events are
isolated encounters with steep gradients (horizontal or vertical) in horizontal or vertical speed of air. The discrete gust has evolved over the years from the isolated sharp-edged step function used
in the airworthiness requirements to the currently favored one-minus-cosine. Static gust loadings are still determined by one-minus-cosine vertical gust velocity shape with the aircraft motion
constrained to the plunge mode only. Haddadpour and Shams (2005) showed that the linear model analysis technique and linear qusi-steady aerodynamic are still used for structure modeling and
aerodynamic modeling, respectively. Random turbulence is a chaotic motion of air that is described by its statistical properties (Kim et al., 1999). The main statistical features that need to be
considered are: stationary, homogeneity, probability distribution and correlations and spectra. The power spectral approach offers a more realistic representation of the continuous nature of
atmospheric turbulent and it allows more rational consideration of design and operational variations such as configuration changes, mission changes and airplane degrees of freedom. The main object of
this paper is to analyze the response of aircraft under excitation of various types of turbulence atmosphere, based on statistical and spectral technique. Five new linear dynamics models are
developed to describe the normal acceleration throughout an aircraft due to vertical gust effect. To the best of authors knowledge, no attempts have been made to investigate the effect of atmospheric
turbulence on aircraft response with these various models for the base line Convair CV-880 jet transport aircraft model.
Aerodynamic and Stability Derivatives Model: The Aircraft selected as a model in this research work is the Convair CV-M880 jet transport operating at Mach = 0.86 and altitudes of 7005 m (23000 ft)
and 10661 m (35000 ft).The airframe fixed coordinate system, dimensional aerodynamic and stability derivatives influence coefficients of aircraft are access to flight data test from the model
original in accord with NASA convention in USA, 1973 (Schmidt, 1998). The Convair CV-M880 jet transport layout can be shown in Fig. 1. Flight conditions and stability derivatives of jet transport
were illustrated in Table 1.
Fig. 1: Convair CV-880M jet transport layout
Fig. 2: One-minus-cosine discrete gust
Table 1: Flight conditions and stability derivatives of Convair CV-M880 jet transport flying at Mach = 0.86 in both conditions
Aircraft Response Model to Discrete Gust: The idealized sharp-edged gust is a very severe type of a velocity profile that seldom occurs in nature. Instead, a discrete gust may be modeled more
practically by a ramp input that reaches a peak value in a distance known as the gradient distance. The one- minus-cosine model (Fig. 2) is more frequently used in the determining gust-induced load
factors rather than a ramp rising to a steady peak gust. The aircraft response when interring one-minus-cosine gust is in the vertical (plunging) degree of freedom mode only. The load factor for
aircraft constrained to the plunging mode can be obtained from Eq. 1. The full derivation can be found in (Schmidt, 1998):
Δn[z] (t) = Local load factor,
λ = Time constant (in seconds)
d = Gradient distance,
W = Aircraft weight,
S = Wing span,
W[g] = Vertical gust velocity and
C[lα] = Lift curve slope.
The maximum load factor will occur near to the time for peak gust value.
Modified Aircraft Equations of Motion to Reflect the Gust Input: The use of the short period dynamic model will provide an insight as to import of increasing the airframe degrees of freedom when
representing the airframe dynamics. The simplified set of short period equations of motion can be expressed as (Schmidt, 1998):
= Normal force due to angle of attack rate,
Z[α] = Normal force due to angle of attack,
V = Aircraft velocity,
q = Pitch rate,
Z[q] = Normal force due to pitch rate,
Z[δ] = Normal force due to elevator,
= Pitching moment due to angle of attack rate,
M[u] = Longitudinal stability derivative,
M[q] = Pitch damping,
δ = Control input and
M[δ] = Pitching moment due to elevator.
The two coupled linear Eq. 2 will be restated as functions solely of α and q by the use of algebraic substitutions, i.e.,
A further simplification can be made by recognizing that both [q] are nearly zero in magnitude and most assuredly are negligible when compared to the free stream velocity in the preceding equations.
The short period approximation in a commonly used becomes
In present study the longitudinal model (short period response) modified to reflect gust inputs of α[g ](t) and q[g ](t) in place of control inputs. The longitudinal equations of motion can be
written in state space form as follow^ :
The normal acceleration output is given by
Power Spectral Technique: The power spectrum represents a frequency viewpoint for describing the square of random variable that is originally considered in time domain. The original time-varying
random signal or function x (t), shown in Fig. 3a, is processed (or filtered) through a unit rectangular filter, shown in Fig. 3b, to yield a truncated signal x[T ](t) that is zero when Fig. 3c and
this signal is absolutely integrable because is finite and the function is assumed to be bounded variation (McLean, 1990).
Fig. 3: Truncation of a random signal
Consequently, that a Fourier transformation of X[T](t) exists may be expressed as:
Since X[T](ω) in Eq. 6 is a complex quantity whereas X[T](t) is a real quantity.
From Parseval`s theorem, which can, which can be described in the preceding notation as:
The mean square expectation can be defined as:
The development of Power Spectral Density (PSD) follows from applying Parseval`s theorem to Eq. 8 to obtain an alternate form for the mean square that involves frequency -dependent function, i.e.,
The limiting action on the integr and in the preceding expression leads to the definition of the power spectral density,
Fig. 4: Load factor response to a vertical gust input
Therefore the expectation for mean square may be described statistically in term of frequency content by
Aircraft Response Models to Random Gust: The aircraft normal load factor, in response to a turbulent vertical gust may be found by the series application by Dryden vertical gust model`s transfer
function (squared) to the aircraft transfer function (squared) of normal load factor to vertical gust input. This statement can be shown in Fig. 4. The expectation of the normal load factor response
is obtained by integrating the power spectral density. The Dryden vertical gust model may be expressed in a transfer function format as (Schmidt, 1998)
σ[w] = Root mean square (rms) of stochastic gust.
The transfer function Gw[g] (S) can be expressed in terms of the Laplace transform variable as follows,
L[2 ] is the scale of vertical turbulence gust.
The aircraft longitudinal response is based on short period approximation where it is noted that α = w/V and α[g ]=[ ]w[g]/V. The state variable form,
α[n] = Normal acceleration, x = State variables
Where [A], [B], [C] and [D] are in accord with equation 4 whereas {X} = [wq]^T.
The transfer function of G[nwg(S)] becomes
Which leads to PSD as:
Table 2: Model assumption
PS: Power Spectral, θ[o]: Pitch angle, (rad), q[g]: Pitch rate gust, body axis, (rad sec^-1)
and the output PSD as:
Finally, the normal load factor is obtained by the integration of the output power spectral density
In an attempt to understand the nature of atmospheric turbulent better, to provide data through which mathematical modeling of turbulence may be made and an improve means for treating the response of
aircraft in turbulent air, many experimental studies have been made to predict and measure the vertical gust velocity in various circumstances, using aircraft probing (such as NASA probe used in
flight measurement of turbulence (Houbolt, 1973)^ and NASA B-757-200 research aircraft (Buck and Newman, 2005). In current study, the rms values of turbulent vertical gust (α[w]) are detected
experimentally from (Etkin, 1981).
Five new models with different gust excitation complexities are used in present work. The assumptions used in each models are presented in Table 2. Models 1 and 5 are the simplest models for short
period and Lyapunov approach (Farrell, 1994). As shown, the value of [g ] are zero. Models 3 and 4 are considered the effects of simulation model was built by using the MATLAB software.
The maximum load factor determination is the primary purpose of the current work to predict the aircraft response resulting from flight within a different turbulence atmospheric environment in
degrees of severity (moderate to severe turbulence, usually the latter is storm related, such as thunderstorm). The longitudinal equations of motion are modified to include the gust effect. New
models are developed to estimate the mentioned purpose. The base line aircraft was taken into consideration in this analysis, Convair CV-880 jet transport, when operating at Mach 0.86 at altitude
7005 m (23000 ft) and 10661 m (35000 ft). The response of aircraft tested under two categories of turbulence excitation, including discrete gusts (usually 1-cos gust) as well as the phenomenon
described as continuous turbulence. The maximum load factor (Δn[z])[max] = 1.79 g at time = 0.285 sec is found from the time history response to one-minus-gust (Fig. 5). The effects on gust response
of degree of freedom in present method can appear, with maximum load factor = (1.72 g) occurring at time = 0.252 sec. The addition of pitch angle rotation to response model results in maximum load
factor decreasing by
Fig. 5: Effects on gust response of degree of freedom for Convair CV-880 Jet transport at H = 7005 m (23000 ft): Mach = 0.86: W[g] = 21 m sec^-1
about 0.07 g after the startup transient has occurred. The other type of turbulence under consideration in this study is a random turbulence, which was modeled by an appropriate power spectral
density. The transfer function approach was applied here to determine the aircraft gust response based on short period approximation. The area under the power spectrum curve represented the mean
square of load factor. The numerical estimation for σ[w] (normalized input) and σ[w] (output) were determined by using trapezoidal integral approximation for finite frequency range Eq. 11 yielded to
σ[w] = 0.382 m sec^-1 (1 ft sec^-1) if ω[max] were infinite; however, the frequency truncation results in σ[W] estimation, which corresponds to 0.6% error. Figure 6a is a spectral representation of
Dryden vertical gust model when normalized to unit area. The aircraft normal load factor transfer function due to vertical gust input shown in Fig. 6b with speak response value occuring near the
short period frequency. Figure 6c represents the frequency distribution of aircraft
normal acceleration (product of |G[wg] (ω)|^2 |G[wg] (ω)|^2). The three sigma value for aircraft normal load factor estimation of 1.1 g with the probability of 99.7% does not exceed this value when
encountering a turbulent vertical gust at variance σ[w] = 6.1 m sec^-1 (Fig. 7).
To validate the numerical results, a comparison between the present work and data in reference (Schmidt, 1998) was made to determine the load factor for Lockheed jet transport when operating at Mach
0.75 and altitude 6092 m (20000 ft). The results show a good agreement with 1.6% error. These verification results are shown in Fig. 8. Five new models (previously discussed) for the aircraft
acceleration response are excited by vertical gust with different values of σ[w]. The effects of σ[w] (rms) values of stochastic vertical gust upon load factor (model 3 taken as an example for
calculation) is illustrated in Fig. 9. The results show that the load factor and σ[w] are directly related at different probabilities. Peak values of normal acceleration for all models are presented
in Fig. 10. An alternate approach adopted in this study was the application of the Lyapunov equation, which directly yielded
the mean square of the load factor (Farrell, 1994; Ogata, 1990), resulting in small error when the variance is estimated. This is noted when the values of load factor for models 1, 5 and 3, 4 is
compared, respectively. The most energetic responses are models 1 and 5. As expected, these models exhibit higher frequency content due to the non-causal transfer function structure resulting from
noted assumption. Models 1 and 5 predict the load factor with (2-3.5%) error compared with model 2 which considered all gust penetration effects (Fig. 10).
Fig. Aircraft spectral response resulting from a vertical gust: H = 7005 m: Mach = 0.86, a) Dryden vertical gust input, b) Aircraft transfer function, c) Aircraft normal load factor φ[n] (w) for σ
6: [w] = 0.328 m sec^-1 (1 ft sec^-1)
Fig. 7: Normal load factor estimation values for Convair 880M transport (Model 3): H = 7005 m: Mach = 0.86: at different probabilities not exceeding these values (load factor): σ[w ]= 6.1 m sec^-1
Fig. Lockheed jet transport comparison results of peak normal load factor at different probabilities of not exceeding these values (model 3): H = 6092 m, Mach = 0.75:1.65% error in the estimation
8: of load factor response
Fig. 9: Effects of σ[w] values of stochastic vertical gust upon load factor (model 3)
Fig. 10: Peak normal load factor for all suggestion models. Convair CV-880M, At H = 7005 m: Mach = 0.86: σ[w] = 6.1 m sec^-1
Fig. 11: Effects of σ[w] on peak normal load factor for all models for CV-880M transport with 99.7% probability not exceeding the value of peak normal load
Fig. 12: The load factor predication (model 3) at different values of σ[w] ( rms) of vertical gust at high altitude H = 10661, m: Mach 0.86
Figure 11 introduces the effect of σ[w] values on peak normal load factor. Other test conditions of aircraft at altitude 10661 m (35000 ft) was made at various σ[w] (rms) according to probability not
exceeding the predicted values of aircraft acceleration. These results are presented in Fig. 12.
Finally, this study provides increased motivation to improve airplane response in gust turbulence atmospheric by using modern optimal control methods.
Methodology to estimate aircraft transient response resulting from flight within turbulent atmosphere based on statistical and power spectral technique is presented. Modified longitudinal equations
of motion which includes the effects of atmospheric gust are solved. The responses of aircraft are tested under two categories of turbulence excitation input (discrete and continuous random
turbulence). Five linear dynamics models are developed to describe the normal acceleration throughout an aircraft when it encounters a vertical gust. The following conclusions have been obtained in
the present research:
• The numerical results show dependencies on which gust excitation type and evaluation criteria are considered.
• Models 1 and 5 exhibit higher frequency contents and give a rapid estimation of normal load factor in case complete data are not readily available. These models predict the load factor with
(2-3.5%) error compared with model 2 which considered all gust penetration effects.
• It can be concluded that the agreement between the finite frequency limit on integration of the spectral distribution and Lyapunov`s results obtaining an estimate for the output deviation of load
factor is well within the accuracy.
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How to tell your homeschool math program is working
Denise just posted an article
How to Recognize a Successful Homeschool Math Program
on her blog. I enjoyed that a lot and recommend you read it too, no matter what math curriculum you are using!
She summarizes it this way:
If you are wondering how well your homeschool math program is working, pay attention to your children.
□ Do they understand that common sense applies to math?
□ Can they give logical reasons for their answers?
□ Even when they get confused, do they know that math is nothing to fear?
If so, then be assured: your children are already miles ahead of most of their peers. Their foundations are solid, and the details will eventually fall into place as you continue to play with
mathematical ideas together.
She also notes her 'yardstick' for measuring
math anxiety
: if your child does not fear
word problems
, he/she is not suffering from math anxiety.
Photo by wecometolearn
There was a time when my second daughter actually
word problems and thought they were the BEST part of her math work (it was about 2nd- 3rd grade). Now she said she still enjoys them, but likes mental math problems best (she just started 5th).
When it comes to mental math, I sometimes give myself a little challenge (such as when making an answer key to my books): can I do this problem
instead of a calculator? It's not anything I fear - it's enjoyable in a sense.
I feel this is similar to when people do crossword puzzles, solve Sudoku, or even play Freecell: you actually
the mental challenge, right? The same can happen with mental math, or with math in general:- it doesn't have to be something fearful, disgusting, or repulsive -- far from that! : )
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define your variable and solve
April 24th 2012, 01:01 PM #1
Apr 2012
New York
define your variable and solve
A rock group plans three concert tours over a period of 38 weeks. The tour in Britian will be 4 weeks longer than in the tour in france and the tour in Germany will be 2 weeks shorter the tour in
France. How many weeks will they be in France?
Re: define your variable and solve
Let's denote :
$x$ - number of weeks in Britain
$y$ - number of weeks in Germany
$z$ - number of weeks in France
Assuming that time isn't relative one can formulate following system of equations :
$\begin{cases}x+y+z=38 \\x=4+z \\y=z-2\end{cases}$
April 24th 2012, 09:11 PM #2
Senior Member
Nov 2011
Crna Gora
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Metrically homogeneous subsets of the plane
up vote 10 down vote favorite
A metric space $M$ is metrically homogeneous if for every pair of points $x, y \in M$ there is an isometry $f$ of $M$ onto $M$ such that $f(x)=y$. What is known about metrically homogeneous spaces?
Any references? In particular, is there an explicit description of all metrically homogeneous subspaces (under the Euclidean metric) of the plane $\mathbb{R}^2$?
Remark. The same question in one dimension has an easy answer: A subset $M$ of $\mathbb{R}^1$ is metrically homogeneous if and only if $M$ is the union of two cosets $G+a\ $ and $G+b\ $ of an
additive subgroup $G$ of $\mathbb{R}^1$.
mg.metric-geometry plane-geometry
2 I do not know about subsets of $R^2$, but many results on metrically homogeneous spaces follow from the solution of Hilbert's 5th problem, see e.g. arxiv.org/pdf/0908.4205v2.pdf . However, these
results typically assume much more regularity about the metric space than you would have for arbitrary subsets of the plane. – Misha Jul 11 '13 at 22:40
1 No, sorry, older! – Wlodzimierz Holsztynski Jul 12 '13 at 1:18
1 @WK, Leonard M. Blumenthal (or someone younger?) consider homogeneous spaces in one of a chapter. The title had something like "G-spaces", like geodesic spaces. (Also Borsuk considered this
problem; S.Spież considered a related paper, where he proved that S×S cannot be homogeneous). – Wlodzimierz Holsztynski Aug 24 '13 at 17:38
add comment
2 Answers
active oldest votes
Here is an description of homogeneous subsets of ${\mathbb R}^2$ using group theory, similar to the one you have in dimension 1. Let $M\subset {\mathbb R}^2$ be homogeneous and let $G< I=
Isom({\mathbb R}^2)$ be its group of isometries. Then $G$ fits into short exact sequence $$ 1\to T\to G\to S \to 1 $$ where $T$ is a group acting by translations on the plane and $S$ is a
subgroup of $O(2)$. Unlike in 1-dimensional case, this sequence need not split, but you can still find a subset $C\subset G$ of coset representatives of $S$ so that $G=TC$ (I am using
action on the left). Then $M=G\cdot m$ for some $m\in M$ and, hence, $$ M= T\cdot (Cm), $$ i.e., $M$ is a disjoint union of $T$-orbits, where $T$ is a subgroup of ${\mathbb R}^2$. This is
an analogue of the description that you liked in the 1-dimensional case, except now $S$ is typically infinite. Thus, you have a (typically) infinite disjoint union, indexed by the set $S/
G_m$, where $G_m$ is the projection of the stabilizer of $m$ in $G$ to the group $S$. I think, this is the best one can do without introducing further restrictions on the set $M$. Even in
the case when $S$ is a finite dihedral group, there will be infinitely many possibilities for choosing a subgroup $T\subset {\mathbb R}^2$ normalized by $S$.
In order to get something more geometrically appealing, let's assume that $M$ is a closed subset of ${\mathbb R}^2$. This immediately implies that the group $G$ is a closed subgroup of the
Lie group $I=Isom({\mathbb R}^2)$. Basic Lie theory tells you that $G$ is a Lie subgroup of $I$. Now, the problem essentially reduces to classification of Lie subgroups of $I$. Here it is:
1. $dim(T)=2$, then $T$ acts transitively on ${\mathbb R}^2$ and, hence, $M={\mathbb R}^2$.
2. $dim(T)=1$. Then:
up vote 8 a. Either $T$ is either isomorphic to ${\mathbb R}$, acting via translations along a line $L\subset {\mathbb R}^2$, or
down vote
accepted b. $T\cong {\mathbb R}\times {\mathbb Z}$, where the first factor acts by translations along a line $L$ and the second acts by translations in the orthogonal direction.
In either case, clearly, $M$ is a product ${\mathbb R}\times M_1$, where $M_1$ is a discrete homogeneous subset of the real line (there are four possibilities for $M_1$ which you already
know: single point, two points, orbit of the discrete group of translations or union of two such orbits).
3. $dim(T)=0$. Thus, $G$ is a discrete group of Euclidean isometries.
There are again 3 subcases here, depending on the rank of the free abelian group $T$ ($0, 1$, or $2$). If $T=0$ then $G< O(2)$ and, hence, $G$ is either 1-dimensional (in which case $M$ is
a round circle) or finite cyclic or dihedral group; in the finite case everything reduces to vertex sets of regular or semiregular planar convex polygons. If $rank =1$ then $S$ is a
subgroup of $Z_2\times Z_2$ and $M$ is either contained in one line or in two parallel lines.
The most interesting case is when $T$ has rank 2 and, hence, $G$ is a Euclidean crystallographic group (Russians call them "Fedorov groups"). If you so desire, you can now go through the
well-known list of such groups and identify $G$-orbits in ${\mathbb R}^2$. Some will give you esthetically pleasing vertex sets of regular and semiregular "floor-tiling" patterns on ${\
mathbb R}^2$. If I were Joseph O'Rourke, I would add some nice pictures here, but I am not.
Nice, Micha. But something seems missing: the group of rotations (that would be case 2c), producing circles. – Wlodek Kuperberg Jul 12 '13 at 17:55
@WlodzimierzKuperberg: You are right, I forgot one more case when $T=0$. – Misha Jul 12 '13 at 18:01
1 "But I am not". – Andres Caicedo Jul 12 '13 at 18:11
3 @AndresCaicedo: Yes, to the best of my knowledge, I am not Joseph O'Rourke! – Misha Jul 12 '13 at 18:15
add comment
There's a complete classification that has been known since 1967: look past the Springer paywall here in the paper
Branko Grünbaum, L. M. Kelly. Metrically homogeneous sets, Israel Journal of Mathematics April 1968, Volume 6, Issue 2, pp 183-197.
If I recall correctly, there's a flaw in one of the lemmas as written, but a correction was published. I'll edit this answer once I find that correction.
Thanks to Francois Ziegler, we now have a link to the correction (see comment below). As pointed out in the other comments, the classification holds only for compact subsets of the plane.
Also, as pointed out in the comments, the definition of metric homogeneity used in this paper differs from the OP's as follows: in their definition, a metric space $(M,d)$ is homogeneous if
for each pair $x,y$ of points, the set of distances $d(x,z)$ equals the set of distances $d(y,z)$ when $z$ ranges over points in $M$. This is weaker than the requirement that there be an
isometry taking $x$ to $y$, so the objects of interest to the OP are certainly a subset of the classification and I hope that the paper is at least a starting point: the literature seems
up vote 4 rather thin aside from this.
down vote
On the other hand, let's take the paper's definition of metric homogeneity. Fix $x,y$ in a space $(M,d)$ satisfying this definition and note that there exists a set-valued map $F:M \to 2^M$
defined as follows: for each $z \in M$, define $F(z) \subset M$ by
$$F(z) = \{w \in M \mid d(w,y) = d(x,z)\}$$
I wonder what conditions one must impose on $(M,d)$ -- aside from the requirement that $M$ is a subspace of the Euclidean plane -- in order to guarantee that there exists an isometry $f:M \
to M$ which is a selection of $F$, i.e., $f(z) \in F(z)$ for all $z \in M$. All the selection theorems which I can think of crucially require convexity of $F(z)$, but maybe there are
experts who know other ways.
5 The correction is here: ams.org/mathscinet-getitem?mr=268782 . Is it obvious that the OP is using the same notion of "homogeneous" as this paper? – Francois Ziegler Jul 12 '13 at 0:55
@Vidit Nanda: Their definition of metric homogeneity is not equivalent to the one in my question. They say "A subset of a metric space is called metrically homogeneous if the set of
distances from a chosen point of the subset to all the other points of the subset is independent of the chosen point." – Wlodek Kuperberg Jul 12 '13 at 1:14
I would not call it a "complete classification" since it is restricted to compact sets. (In the context of the OP, their answer is: Circles or vertex sets regular polygons.) – Misha Jul
12 '13 at 1:30
1 @WlodzimierzKuperberg: True, but one can still extract a partial answer to your question from their result, after a bit of extra work. Once you know that the set is a smooth convex
curve, the problem reduces to the classification of compact Lie subgroups of Isom(R^2). – Misha Jul 12 '13 at 1:32
If you assume compactness and convexity, the problem becomes easy. Assuming connectedness helps a lot, but does not make it trivial. I know that the only closed connected metrically
homogeneous (MH) subsets of the plane are: single points, lines, and circles (oh, and the whole plane and also the empty set, if you insist). But this has not been published yet.
Classifying all MH subsets seems to be much more difficult. – Wlodek Kuperberg Jul 12 '13 at 16:59
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Not the answer you're looking for? Browse other questions tagged mg.metric-geometry plane-geometry or ask your own question.
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The n-Category Café
Kernels in Machine Learning III
Posted by David Corfield
The use of kernel methods in machine learning often goes by the name nonparametric statistics. Sometimes this gets taken as though it’s the opposite of finite parametric statistics, but that’s not
quite right. A typical piece of parametric statistics has a model like the two-parameter family of normal distributions, and the parameters of this model are then estimated from a sample.
What happens in the case of kernel methods in machine learning is that a model from a possibly infinite-dimensional family is specified by a function on the (finite) collection of training points. In
other words you are necessarily restricted to a subfamily of models of dimension the size of the training set. So, in contrast to the usual parametric case, you are dealing with a data-dependent
finite dimensional subfamily of models.
Now we can see why regularisation is important. Fitting a normal distribution is choosing a point in a two-dimensional manifold within the space of all distributions on $\mathbb{R}$. The choice is
made of the member of the family whose first and second moments match those of the empirical distribution. Now in the nonparametric case, before we know the training data, we are dealing with an
infinite-dimensional space dense within the space of all distributions. Where in the Normal distribution case we have two constraints to match the two dimensions of the family of models, now we are
able to satisfy as many constraints as we like, and must end by matching the empirical distribution if we opt simply for the maximum likelihood model.
Instead of this so-called maximum likelihood estimation, what we can do instead is impose a bias so that we end up in our family by favouring ‘smaller’ functions on the training data. An intuitive
way to think of this is by considering the Gaussian radial basis function as kernel on a Euclidean space $X$:
$K(x, y) = (4 \pi t)^{-n/2}exp\left(-\frac{\|x - y\|^2}{4t}\right).$
This is known as a ‘heat kernel’ in Euclidean space. Gaussian process classification with this kernel amounts to selecting negative and positive ‘temperatures’ on the training points, setting the
temperature of the rest of $X$ to zero, and allowing heat to flow for a designated time. After this time, points in $X$ will be classified according to the sign of their temperature, the magnitude of
the temperature determing the certainty of the classification.
Now if we’re allowed to place any temperature values at the training points, we can select very hot ones for those labelled ‘$+1$’ and very cold ones for those labelled ‘$-1$’, so that they’re still
hot or cold after heat flow has taken place. This is a classic case of overfitting, and we should have little confidence that our estimates on test points will be accurate. If on the other hand we
manage to select modest temperatures in such a way that after the heat flow has taken place the classifier is accurate on training data, we should have greater confidence.
Posted at July 6, 2007 10:47 AM UTC
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Fuel and financial savings for operators of small fishing vessels
Annex 4
This annex presents a procedure for estimating the correct propeller diameter and pitch for a given vessel and engine. It is based on an empirical method and formulae developed by George Crouch,
although some of the procedures have been simplified by the integration of formulae derived by Dave Gerr (Gerr, 1989). The charts should be of assistance in a quick check of an existing or proposed
propeller design - they are not intended to be part of a detailed design process. Their application is limited to three-bladed propellers, of ogival section (flat-faced with a symmetrical curve on
the back) and a blade mean width ratio of 0.33.
Only basic information concerning the installation and the vessel is necessary to perform a preliminary propeller check. This is limited to:
• the operating propeller RPM;
• the propeller RPM at MCR;
• the required cruising speed;
• the delivered shaft horsepower at the propeller at MCR.
Annex Figures 17 and 18 present charts for the estimation of pitch based on vessel speed and propeller RPM. Both figures present the same information but cover different RPM ranges. The charts
include a correction for slip, which can be estimated as a function of vessel speed (for more details, see Gerr, 1989). It is very important that the required operating speed reflect the installed
power and the type of vessel (see Figure 4, p. 7, and the section Engines). If the vessel is an existing vessel, according to the graphs in this Annex, the chosen operating speed for use should be
the speed that the vessel currently achieves.
Figure 17: Propeller pitch chart (400-1 500 RPM)
The graphs should be read by entering along the horizontal axis at the RPM corresponding to the propeller's operating RPM at cruising speed. A vertical line should then be drawn until intersecting
the curve corresponding to the required cruising speed. From that point of intersection, a horizontal line is then drawn to the left-hand axis where the pitch can be read.
Suppose we have a 15 m vessel with an engine delivering a maximum of 150 HP (at the propeller), at an engine speed of 1 800 RPM through a 3:1 reduction gearbox. The desired service speed is 8 kt at
an engine speed of 1 650 RPM. Figure 7 should be read by entering at the propeller operating speed, 550 RPM (= 1 650 ¸ 3, due to the reduction gearbox). A line is then drawn vertically at this point
to meet the 8 kt curve. At this intersection the pitch is read off on the vertical axis at 31 inches.
Figure 18: Propeller pitch chart (1 400-2 500 RPM)
Estimation of propeller diameter
The correct propeller diameter is estimated in a similar manner as the pitch. Figures 19 and 20 show the graphs for diameter estimation; however these should be entered using the RPM at the propeller
when the engine delivers maximum power. A vertical line is drawn from this point to meet the curve corresponding to the delivered horsepower at the propeller. The propeller diameter is then read off
the vertical axis at the level of this intersection.
In the case outlined above, the graph is entered at 600 RPM (= 1 800 RPM ¸ 3), and a line is drawn up to the 150 HP curve. At this intersection, the corresponding diameter is 38 inches.
Figure 19: Propeller diameter chart (400-1 500 RPM)
Table 11
│Pitch and diameter adjustments for two- and │
│four-bladed propellers │
│ │Diameter │Pitch │
│Two-bladed propeller │1.05 │1.01 │
│Four-bladed propeller │0.94 │0.98 │
Source: Gerr, 1989.
Figure 20 : Propeller diameter chart (1 400-2 500 RPM)
Adjustments for two- and four-bladed propellers
To find the pitch and diameter for a two- or four-bladed propeller, perform the estimation as outlined above and then multiply the results by the factors given in Table 11. In the case above for a
four-bladed propeller, pitch =
31 x 0.98 = 30.4 inches, diameter = 38 x 0.94 = 35.7 inches.
Faced with the task of changing an existing propeller to try to reduce or increase engine loading, there are a few rules of thumb that can prove as useful guides:
• 1 inch of diameter absorbs the torque of 2 to 3 inches of pitch.
• 2 inches of pitch decreases engine speed by 450 RPM (very rough).
• A square propeller (pitch = diameter) is not special and is not necessarily the best.
• With the propeller RPM reduced by 1/2 and diameter increased by 1/3, the efficiency increases by 1/4.
Sources: Gerr, 1989, and Aegisson and Endal, 1992.
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the first resource for mathematics
A self-adaptive method for solving general mixed variational inequalities.
(English) Zbl 1074.49001
Summary: The general mixed variational inequality containing a nonlinear term is a useful and an important generalization of variational inequalities. The projection method cannot be applied to solve
this problem due to the presence of the nonlinear term. To overcome this disadvantage,
M. A. Noor
[Appl. Math. Comput. 141, No. 2–3, 529–540 (2003;
Zbl 1030.65072
)] used the resolvent equations technique to suggest and analyze an iterative method for solving general mixed variational inequalities. In this paper, we present a new self-adaptive iterative method
which can be viewed as a refinement and improvement of the method of Noor. Global convergence of the new method is proved under the same assumptions as Noor’s method. Some preliminary computational
results are given.
49J40 Variational methods including variational inequalities
47J20 Inequalities involving nonlinear operators
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Huybrechts on Branes in K3, I
Posted by Urs Schreiber
Yesterday we had Daniel Huybrechts from Bonn talking in the ZMP seminar about stability conditions on derived categories of coherent sheaves over K3 surfaces.
Daniel Huybrechts
Derived and Abelian Equivalence of K3 Surfaces
which is based on work by Bridgeland, who defined and studied stability conditions on triangulated categories in
Tom Bridgeland
Stability Conditions on Triangulated Categories
and applied that to the case of derived categories of coherent sheaves on K3 surfaces in
Tom Bridgeland
Stability conditions on K3 surfaces
math.AG/0307164 .
This is pure math, but there is a nice dictionary which maps it 1-1 to the physics of D-branes. For a good overview of how this works see
Paul S. Aspinwall
D-Branes on Calabi-Yau Manifolds
which I once tried to summarize here.
Roughly, this dictionary is as follows:
A “geometric brane” for a B-model topological string on $X$ (i.e. a brane which can be regarded as a submanifold of $X$ with a vector bundle on it) is encoded in a coherent sheaf on $X$.
The general brane for the B-model string is obtained by stacking a collection of geometric branes and anti-branes on top of each other and turning on tachyon condensates between them (which in part
mutually annihilates them, the remaining piece being a general brane). In the formalism, this corresponds to a bounded complex of coherent sheaves on $X$ (with the differential of the complex
encoding the tachyonic string condensates stretching between the branes).
But ultimately we are interested not in the boundary conditions (= branes) of the topological B-model string, but of the physical type II string (i.e. of a CFT instead of a TQFT).
It turns out that all the branes of the topological B-string correspond to branes of the physical string - but not all of them are “stable” for the physical string. Instead, every brane for the
topological string is supposed to decay into a collection of stable branes - the physical BPS branes.
More precisely, the derived category of coherent sheaves is what is called “triangulated”, which means that it contains lots of certain triangle diagrams (i.e. collections of three of its objects
with certain morphism between them, satisfying some conditions). These triangles precisely encode the “brane chemistry”. Roughly, a triangle
(1)$A\to B\to C$
says (as familiar from exaxt sequences), that the brane $B$ is an “extension” of the brane $C$ by the brane $A$. In other words, the existence of this triangle encodes the potential brane reaction
This is very much like in chemistry. (And of course oversimplified, see section 6.2 of Aspinwall’s review for the details.)
Which branes are stable and which are not is encoded by a “stability condition”, which, on the physics side, is called $\Pi$-stability. As far as I am aware, the entire motivation for Bridgeland to
define stability conditions on triangulated categories comes from the desire to axiomatize this piece of physical input. The abstract definition of a stability condition on a triangulated category
may look completely ad hoc, its natural meaning becomes manifest once you think of distinguished triangles as describing reaction processes of fusion and decay of branes.
Hence, whether one is interested in D-branes or not, when reasoning about stability condtions on derived categories of coherent sheaves it helps a lot to keep the above dictionary in mind. It makes
many of the constructions and results better memorizable. (For more on the physics side see Eric Sharpe’s encyclopedia entry $\to$).
All, right, below I reproduce a transcript of the talk.
The talk consisted of two parts:
I) Introduction to the concept of derived categories of coherent sheaves.
II) Equivalences of categories versus isomorphism of target spaces.
The main result presented (theorem 0.1 in Huybrechts’ paper mentioned above) says, in terms of the physics side of the dictionary, that two K3 surfaces are indistinguishable as target spaces for the
topological string if and only if there are complexified Kähler classes on them which make them indistinguishable for the physical string.
The following are the notes that I took in the talk (I have included some links and some personal comments, set in italics). The notes on part II) are given in a seperate entry.
I) Introduction to the concept of derived categories of coherent sheaves.
In the following, $X$ denotes an algebraic variety ($\to$), more precisely, a smooth projective variety sitting in some projective space. Later we restrict attention to $X$ being a K3 surface ($\to$
Denote by $\mathrm{Coh}\left(X\right)$ the category of coherent sheaves ($\to$) on $X$.
Examples for coherent sheaves on $X$ are
i) Holomorphic vector bundles ($\to$) on $X$ are coherent sheaves (or rather, their sheaves of sections are). (This corresponds to stacks of space-filling geometric branes.)
ii) If $C\subset X$ is a holomorphic curve, then its structure sheaf ${O}_{C}$ is a coherent sheaf. (This would be a lower-dimensional geometric brane.)
iii) For every point $x\in X$, the skyscraper sheaf $k\left(x\right)$ is a coherent sheaf. (This is the sheaf whose fiber over $U$ is nonempty precisely if $x\in U$). (These skyscrapersheaves encode
0-branes, localized at points (we do not consider a temporal deriction in this sort of game).))
iv) Let $E$ be a vector bundle on $X$, or rather its sheaf of sections, and let $E\to k\left(x\right)$ be the obvious projection onto the skyscraper sheaf at $x$, then
(1)$F:=\mathrm{ker}\left(E\to k\left(x\right)\right)$
is a torsion free coherent sheaf.
For $E$ trivial this is the “ideal sheaf” (with “ideal” in the sense of ideal of a ring) ${I}_{x}$, namely the sheaf of holomorphic functions vanishing at the point $x$.
(This corresponds to a space filling stack of D-branes having reacted with an anti-0-brane at position $x$.)
Once we understand these examples, we understand all coherent sheaves, in the sense that every coherent sheaf can be constructed by performing direct sums, quotients, kernels, etc. (etc?) of the
sheaves in the above examples.
Next, we pass from $\mathrm{Coh}\left(X\right)$ to ${D}^{b}\left(\mathrm{Coh}\left(X\right)\right)$, its derived category ($\to$, $\to$).
This works as follows.
Denote by ${K}^{b}\left(X\right):={K}^{•}\left(\mathrm{Coh}\left(X\right)\right)$ the category of bounded complexes of coherent sheaves. Its objects are bounded complexes, its morphisms are chain
homotopy classes of chain maps between morphisms.
(BTW, what would happen if we did not divide out by chain homotopies? If we even considered the full 2-category of complexes, chain maps and chain homotopies?)
(Given any complex, we can compute its cohomology in every degree. We can regard the collection of the cohomologies as a complex themselves, with all differentials being trivial. Given a chain map
between two complexes, we get a chain map between the corresponding trivial complexes of cohomologies. Hence cohomology is really a functor from the category of complexes of coherent sheaves to to
the category of complexes of abelian groups. If a morphism in ${K}^{b}\left(X\right)$ is not an isomorphism, but maps to an isomorphism under this functor, then we call it a “quasi isomorphism”.)
The derived category ${D}^{b}\left(X\right):={D}^{b}\left(\mathrm{Coh}\left(X\right)\right)$ is like ${K}^{b}\left(X\right)$, but with all quasi-isomorphisms regarded as true isomorphisms.
This is a special case of “localization of a category” ($\to$).
Now, ${D}^{b}\left(X\right)$ is no longer abelian ($\to$) - but it is triangulated ($\to$)!
In fact, every derived category of any abelian category is triangulated. This is hence not a special property of coherent sheaves. On the other hand ${D}^{b}\left(X\right)$ is actually $ℂ$-linear
triangulated (which hopefully means the obvious thing).
A category being triangulated means in particular that it carries the following two crucial structures.
i) There is a shift endomorphism
(2)$\begin{array}{ccc}{D}^{b}\left(X\right)& \stackrel{\sim }{\to }& {D}^{b}\left(X\right)\\ {E}^{•}& ↦& E\left[1{\right]}^{•}\end{array}$
(for the case of derived categories $E\left[1\right]$ is the same complex as $E$, but shifted in degree to the left, by one unit, i.e. $E\left[1{\right]}^{n}={E}^{n+1}$)
ii) there are “distinguished triangles”, which are diagrams
(3)${E}^{•}\to {F}^{•}\to {G}^{•}\to E\left[1{\right]}^{•}\phantom{\rule{thinmathspace}{0ex}}.$
As an example: every “mapping cone” ($\to$) gives rise to such a triangle.
(See math.AG/0001045 for an explanation of the mapping cone and why it has the name it has. The triangle obtained from the mapping cone of the map $E\stackrel{f}{\to }F$ between “geometric branes”
$E$ and $F$ corresponds to the D-brane reaction induced by $f$ when this is regarded as a tachyon condensate of strings stretching between brane $E$ and brane $F$.)
We also need to mention something called a t-structure ($\to$) on a triangulated category. This is a collection of subcategories with some properties. For derived categories, such subcategories
include ${D}^{\le 0}\left(X\right)$ and ${D}^{\ge 0}\left(X\right)$, the subcategories of complexes concentrated in non-positive or in non-negative degree, respectively.
The intersection of these two subcategories is called the core of ${D}^{b}\left(X\right)$. Evidently, this is naturally identified with the original category $\mathrm{Coh}\left(X\right)$.
Posted at April 21, 2006 11:01 AM UTC
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Perko pair knots
From Math Images
{{Image Description |ImageName=Perko pair knots |Image=Perko knots.gif |ImageIntro=This is a picture of the Perko pair knots. They were first thought to be separate knots, but in 1974 it was proved
that they were actually the same knot. |ImageDescElem=In 1899, C. N. Little published a table of 43 nonalternating knots of 10 crossings that listed the two knots shown above as being distinct.
Seventy-five years later, Kenneth Perko, a lawyer and part-time mathematician, discovered that these were actually the same knot^[1].
To say that two knots are the same is to say that one can be deformed into the other without breaking the knot or passing it through itself. To prove that two knots are the same, we can create one of
them out of actual rope, and tug at it and move it around until it looks like the other. As the story goes, that's how Perko figured out that these knots are the same - by working with rope on his
We can also prove that two knots are the same by working with their projections. A projection of a knot is a flat representation of it, essentially a 2D drawing of the knot. There are many ways to
use projections to show that certain knots are distinct from each other, but the main way of using projections to demonstrate that two knots are the same is to use the Reidemeister moves, which are
described below. |ImageDesc===Reidemeister moves==
As was stated above, knots are considered to be the same if one can be rearranged into the other without breaking the string or passing it through itself. This kind of transformation is called an
ambient isotopy. But when we're writing a written proof, we have to work with the knots projection, instead of the knot itself. What manipulations can we make on a knot’s projection that correspond
to ambient isotopies in three dimensions?
The first answer is a planar isotopy. A planar isotopy is the sort of transformation you could make if the projection of a knot was printed on very stretchy rubber. The image can be stretched in all
directions, but none of the crossings are affected:
│ │ │ │
│ │ │ │
│ │ │ │
│ │ │ │
│ The original image. │ These two images are planar isotopies of the original image. │ This is not a planar isotopy of the original image. │
│ │ │ │
The second answer is the Reidemeister moves, a set of three changes we can make to a knot’s projection that do affect the knot’s crossings but are still ambient isotopies. Every change to a knot's
projection that corresponds to an ambient isotopy can be described as some combination of these three moves. In the images below, we imagine that the line segments continue and connect in some sort
of unspecified knot, and only the section of the knot we're looking at changes:
Type I Reidemeister Move: Type II Reidemeister move: Type III Reidemeister move:
The first Reidemeister move allows you to create a twist in a The second Reidemeister move allows you to slide one The third Reidemeister move allows you to slide a strand to
strand that goes in either direction. strand on top of or behind another. the other side of a crossing.
Proving that the Perko knots are equivalent
In his paper "On the Classification of Knots", Kenneth Perko provided an abridged proof that the knots now known as the Perko pair are the same^[2]. This proof is shown below:
Perko's proof relies on the ability of the reader to manipulate the knots in their head and verify that each projection can be manipulated to look like the next. To create a full, rigorous proof, we
need to use planar isotopies and the Reidmeister moves, as described above.
Below is a step-by-step Reidemeister moves proof that follows the outline of Perko's shorter proof. The arrows between each step are labeled to show how we get from one image to the other: p.i. means
we use a planar isotopy, I means we use the first Reidemeister move, II means we use the second move, and III means we use the third move. Mousing over a step will highlight the part of the knot
that's about to move in pink, and display a dotted green line showing where it will move to.
Dowker notation
Dowker notation is a way of describing knots with numbers so that anyone else who knows the system can reconstruct the knot. The Dowker notation for the Perko knots will be determined below.
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Homepage of Andrii Dmytryshyn
Main page Curriculum Vitae Publications Activities Links
Created by
Andrii Dmytryshyn
Papers in peer-reviewed journals:
1. A. Dmytryshyn, B. Kågström, V.V. Sergeichuk, Symmetric matrix pencils: codimension counts and the solution of a pair of matrix equations,
Electron. J. Linear Algebra, 27 (2014) 1-18.
2. A. Dmytryshyn, V. Futorny, V.V. Sergeichuk, Miniversal deformations of matrices under *congruence and reducing transformations,
Linear Algebra Appl., 446 (2014) 388-420.
3. A. Dmytryshyn, B. Kågström, V.V. Sergeichuk, Skew-symmetric matrix pencils: codimension counts and the solution of a pair of matrix equations,
Linear Algebra Appl., 438 (2013) 3375-3396.
4. A.R. Dmytryshyn, V. Futorny, V.V. Sergeichuk, Miniversal Deformations of Matrices of Bilinear Forms,
Linear Algebra Appl., 436 (2012) 2670-2700, arXiv:1004.3584v3.
5. A.R. Dmytryshyn, Miniversal deformations and Darboux's theorem,
Bulletin of University of Kyiv, Series: Physics & Mathematics, 4 (2010) 20-22.
6. G. Belitskii, A.R. Dmytryshyn, R. Lipyanski, V.V. Sergeichuk, A. Tsurkov, Problems of classifying associative or Lie algebras over a field of characteristic not 2 and finite metabelian groups are
Electron. J. Linear Algebra, 18 (2009) 516-529.
Technical reports & preprints:
7. A. Dmytryshyn, B. Kågström, Orbit closure hierarchies of skew-symmetric matrix pencils,
Report UMINF 14.02, Department of Computing Science, Umeå University, 2014.
8. A. Dmytryshyn, S. Johansson, B. Kågström, Codimension computations of congruence orbits of matrices, symmetric and skew-symmetric matrix pencils using Matlab,
Report UMINF 13.18, Department of Computing Science, Umeå University, 2013.
9. A. Dmytryshyn, Miniversal deformations of pairs of skew-symmetric forms, arXiv:1104.2492v1.
10. A. Dmytryshyn, Miniversal deformations of pairs of symmetric forms, arXiv:1104.2530v1.
11. A. Dmytryshyn Skew-Symmetric Matrix Pencils: Stratification Theory and Tools,
Licentiate Thesis, Department of Computing Science, Umeå University, Report UMINF 14.05, 2014.
12. A. Dmytryshyn, A Strong Tits Alternative,
Master Thesis, University of Bordeaux 1, 2011.
13. A. Dmytryshyn, Miniversal deformations of pairs of skew-symmetric forms,
Master Thesis, Taras Shevchenko University of Kiev, 2010 (based on #9 from this list).
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Truth’s and Beauty’s Doom and Date
Monday, March 23, 2009
Truth’s and Beauty’s Doom and Date
On “the sequencing of the mathematical genome”
Mathematics is funnier than it gets credit for, and the best laugh I ever had about math involved a friend in college and a course so intimidating he almost quit his mathematics major after hearing
the name. “Advanced Calculus—A rigorous approach,” it was called, a title that betrayed a martinet attitude. Whereas your average multivariable calc class was flabby and slack-ass, here you’d finally
get some goddamn discipline, boy. You will throw up.
Word around the chalkboard was that every homework problem in “Advanced Calculus—A rigorous approach” required six sheets of paper, because you wrote out every nitpicky step and made every assumption
explicit, no matter how obvious—not even arithmetic was taken for granted. For some reason I got endless delight terrorizing Mark by pointing out all the horrid, spindly-legged theorems in other
books he would have to dissect in “Advanced Calculus—A rigorous approach,” predicting the logic would drive him actually mad. Every time I mentioned the class, I used its full draconian title,
“Advanced Calculus—A rigorous approach,” and fear of it drove him to the brink both of hyperventilation and of dropping his major, which probably would have meant him dropping out of college.
Mark was spared by an administrative overhaul of the department, so he never took the class. For my part, I’d almost forgotten the whole incident until I came across a curious bundle of papers in a
recent issue of the Notices of the American Mathematical Society—four treaties on the future of mathematical proofs, and specifically on how computers were going to take over a large burden of the
work in mathematical proofs. However unpromising that topic sounds, it soon had my thoughts dilating like a marijuana smoker’s thoughts into all sorts of wild conjectures, because it turns out (1)
“Advanced Calculus—A rigorous approach” was flabby and slack-ass compared to what’s coming in formal mathematics, and (2) the idea of mathematical beauty might soon be extinct.
First, a few lines of history (cribbed from the papers, natch): The first great revolution in math came from Pythagoras, Euclid, and the rest of the Greeks, who introduced the concept of proofs. You
saw examples of this in your high-school geometry class. Next, in the 1800s, came the next big thing, rigor. Oddly, rigor in math is most easily recognized as a feeling—the scrotum-shrinking
embarrassment that even people really, really good at college math feel upon realizing that some people are way the hell smarter. Namely, people who do original work in rigorous mathematics.
The next and latest revolution in math was the subject of the NAMS papers—formalization. Formalization means tracing math back to fundamental axioms, the kind of migrainous set theory that takes
pages to explicate just why two is the successor of one. It turns out there’s currently a movement in mathematics—and the authors of the quartet of papers claim that not even most mathematicians
realize at, or at least don’t admit it—but there’s a movement to make all mathematical proofs fully formal. To basically take a six-page homework problem from “Advanced Calculus—A rigorous approach”
and apply even more destructive methods to every single line of that problem, expanding the amount of rigor geometrically if not exponentially.
Why? Because when you tease apart every atom of every line, you can actually convert very hard mathematical concepts into a series of very simple steps. The steps might seem stunted and overly
obvious and useless, but they lo and behold add up to something in the end. It’s like someone explaining how a car engine works by starting with the theory of bolts screwing onto threads and going
into incredibly arcane detail about it, and then repeating that nut-and-bolt explanation every time you came across another screw. You’d get pissed off, but you’d also probably understand how a car
engine worked if he continued to break everything down to concepts that simple. That’s formal mathematics. And once you’ve checked all the ticky-tack steps between lines of a proof, you can be pretty
darn sure it’s correct. One paper’s author called this “the sequencing of the mathematical genome.”
Computers enter the scene because there are so unbelievably many lines to check—in one admittedly extreme formalization scheme, it’s estimated it would take a trillion symbols to define the concept
“1”—that only computers can even think about where to start. That means that computers would really, for the first time in math history, be running the show. There are all sorts of consequences for
mathematicians here, including the mundane consequence of possibly turning over peer review to proof-checking software. (This would at least avoid embarrassments like the publication of the paper in
1993 that announced Andrew Wiles had cracked Fermat’s Legendary Last Theorem. Except he hadn’t. He’d left a gap—a gap that would have been obvious to a computer at least in a fully formalized proof.
It took another year of work to nail the sucker down.)
But of all the issues surrounding computerized proofs, I’d like to focus on beauty.
There’s general consensus that really genius-level mathematics is beautiful—purely and uncorruptedly beautiful, the way colored light is, or angels. More particularly, it’s regarded as beautiful in a
way that science is not. With a few exceptions—Einstein’s theories of relativity, string theory, maybe Newton and Darwin—no matter how much science impresses people, it rarely moves them
aesthetically. Science and mathematics stand in roughly the same relation as journalism and fiction—the latter in each set being more admired because it gives us the sense of having moved in a wholly
different realm of being.
Computers, to be blunt, threaten that beauty. One of the four authors in the bundle of papers, Freek Wiedijk, assures mathematicians that existing computers are good for checking proofs only, simply
tidying up the real work and never, never ever conjuring up original theorems. The current generation of mathematicians will continue to live by their wits alone. But beyond that ...? In a
contradiction to Wiedijk, one of the other authors, Thomas C. Hales, admits in an aside that someday, who knows how soon, computers will be doing original work. Doing the work of human
To understand what work computers will horn in on, think again of the relationship between math and science. The former is uncannily, eerily prescient about the latter, and many seemingly esoteric
mathematical ideas, ideas pursued for the pure fun and beauty of them, turned out to have applications in the real world. That’s partly because mathematics sets out to describe all the “rules” of all
the logically consistent universes that could exist. Given a few select starting rules (axioms), and mathematicians show how to manipulate them. Change the rules, and they’ll show you how things are
different. In this sense, mathematicians are working in the multiverse, and so of course sometimes they’ll have done work that applies to our lonely, local universe, even if it’s not obvious at
Computers doing formalized proofs will go even farther. The key thing to understand is that when you’re writing up a formalized proof, you are only allowed a certain repertoire of moves from any
given configuration, no different than calculating end-games for chess. Some steps are valid, some aren’t. And once computers have learned how to take all the possible ticky-tack steps in a formal
proof, they should be able—simply because of brute processing muscle—to map out all of the even illogical and inconsistent universes that exist. A much larger set. They’ll be able to run in every
direction without stopping to think at the beginning—as any human would—that this angle looks idiotic. Computers don’t get embarrassed. Naturally, the computers will run into errors and
contradictions in their attempts at proofs, at which point they’ll stamp them invalid. And some of what they prove will be truly banal. But sometimes during their exploring, the computers will bust
through some unexpectedly subtle opening in some arcane string of symbols, and something incredible will open up, an underground cave. And when they’ve run this new idea all the way to the end—poof,
a new mathematical proof, ex nihilo.
This isn’t going to make mathematicians happy. Not so much because they’ll be suddenly useless (who’s more useless now?, as they’d be the first to admit!) or because the computers will be somehow
“smarter” than them, but because they’ll have to cede control of beauty. You can imagine a comparably teary-eyed-in-frustration scene with a novelist reading a book that was wholly fabricated from a
few hundred lines of code, and having to admit it’s deeper and richer than anything she could have come up with.
The only solace for the world of mathematics is that computerized proofs may open up the game again for the rest of us schlubs, the people who never even attempted classes like “Advanced Calculus—A
rigorous approach.” For, once proofs are fully formalized—once every last step is s-p-e-l-l-e-d o-u-t on a level that insults the intelligence—then we too can follow along. Just like with the car
engine and the didactic mechanic, we could if we wanted to walk through every step and see how it all works and fits together. And if we’ve forgotten how the engine works overall by the end of the
demonstration ... well, at least we got to see a little, to understand for a moment.
Of course, we may not want to. Though fully formalized proofs really only got going in the very late 1900s, the idea of excessively detailed mathematics traces back to Alfred North Whitehead and
Bertrand Russell and their seminal tome Principia Mathematica, published circa 1910. Russell later admitted that his intellect never recovered from the strain that writing out the Principia put on
him. Going through mathematics on that level of rigor stripped out his gears. All he was good for afterward was winning the Nobel Prize in literature.
Posted by Sam Kean at 12:10 AM | Permalink
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NAG Library
NAG Library Routine Document
Note: this routine uses optional parameters to define choices in the problem specification. If you wish to use
settings for all of the optional parameters, then the option setting routine F12ARF need not be called. If, however, you wish to reset some or all of the settings please refer to Section 10 in F12ARF
for a detailed description of the specification of the optional parameters
1 Purpose
F12AQF is a post-processing routine in a suite of routines consisting of
, F12AQF,
, that must be called following a final exit from F12AQF.
2 Specification
SUBROUTINE F12AQF ( NCONV, D, Z, LDZ, SIGMA, RESID, V, LDV, COMM, ICOMM, IFAIL)
INTEGER NCONV, LDZ, LDV, ICOMM(*), IFAIL
COMPLEX (KIND=nag_wp) D(*), Z(LDZ,*), SIGMA, RESID(*), V(LDV,*), COMM(*)
3 Description
The suite of routines is designed to calculate some of the eigenvalues, $λ$, (and optionally the corresponding eigenvectors, $x$) of a standard eigenvalue problem $Ax = λx$, or of a generalized
eigenvalue problem $Ax = λBx$ of order $n$, where $n$ is large and the coefficient matrices $A$ and $B$ are sparse, complex and nonsymmetric. The suite can also be used to find selected eigenvalues/
eigenvectors of smaller scale dense, complex and nonsymmetric problems.
Following a call to
, F12AQF returns the converged approximations to eigenvalues and (optionally) the corresponding approximate eigenvectors and/or an orthonormal basis for the associated approximate invariant subspace.
The eigenvalues (and eigenvectors) are selected from those of a standard or generalized eigenvalue problem defined by complex nonsymmetric matrices. There is negligible additional cost to obtain
eigenvectors; an orthonormal basis is always computed, but there is an additional storage cost if both are requested.
F12AQF is based on the routine
from the ARPACK package, which uses the Implicitly Restarted Arnoldi iteration method. The method is described in
Lehoucq and Sorensen (1996)
Lehoucq (2001)
while its use within the ARPACK software is described in great detail in
Lehoucq et al. (1998)
. An evaluation of software for computing eigenvalues of sparse nonsymmetric matrices is provided in
Lehoucq and Scott (1996)
. This suite of routines offers the same functionality as the ARPACK software for complex nonsymmetric problems, but the interface design is quite different in order to make the option setting
clearer and to simplify some of the interfaces.
F12AQF is a post-processing routine that must be called following a successful final exit from
. F12AQF uses data returned from
and options set either by default or explicitly by calling
, to return the converged approximations to selected eigenvalues and (optionally):
– the corresponding approximate eigenvectors;
– an orthonormal basis for the associated approximate invariant subspace;
– both.
4 References
Lehoucq R B (2001) Implicitly restarted Arnoldi methods and subspace iteration SIAM Journal on Matrix Analysis and Applications 23 551–562
Lehoucq R B and Scott J A (1996) An evaluation of software for computing eigenvalues of sparse nonsymmetric matrices Preprint MCS-P547-1195 Argonne National Laboratory
Lehoucq R B and Sorensen D C (1996) Deflation techniques for an implicitly restarted Arnoldi iteration SIAM Journal on Matrix Analysis and Applications 17 789–821
Lehoucq R B, Sorensen D C and Yang C (1998) ARPACK Users' Guide: Solution of Large-scale Eigenvalue Problems with Implicitly Restarted Arnoldi Methods SIAM, Philidelphia
5 Parameters
1: NCONV – INTEGEROutput
2: D($*$) – COMPLEX (KIND=nag_wp) arrayOutput
3: Z(LDZ,$*$) – COMPLEX (KIND=nag_wp) arrayOutput
4: LDZ – INTEGERInput
5: SIGMA – COMPLEX (KIND=nag_wp)Input
6: RESID($*$) – COMPLEX (KIND=nag_wp) arrayInput
7: V(LDV,$*$) – COMPLEX (KIND=nag_wp) arrayInput/Output
8: LDV – INTEGERInput
9: COMM($*$) – COMPLEX (KIND=nag_wp) arrayCommunication Array
10: ICOMM($*$) – INTEGER arrayCommunication Array
11: IFAIL – INTEGERInput/Output
6 Error Indicators and Warnings
If on entry
, explanatory error messages are output on the current error message unit (as defined by
Errors or warnings detected by the routine:
On entry, $LDZ < max1,N$ or $LDZ < 1$ when no vectors are required.
On entry, the option $Vectors = Select$ was selected, but this is not yet implemented.
The number of eigenvalues found to sufficient accuracy prior to calling F12AQF, as communicated through the parameter
, is zero.
The number of converged eigenvalues as calculated by
differ from the value passed to it through the parameter
Unexpected error during calculation of a Schur form: there was a failure to compute all the converged eigenvalues. Please contact
Unexpected error: the computed Schur form could not be reordered by an internal call. Please contact
Unexpected error in internal call while calculating eigenvectors. Please contact
Either the solver routine
has not been called prior to the call of this routine or a communication array has become corrupted.
The routine was unable to dynamically allocate sufficient internal workspace. Please contact
An unexpected error has occurred. Please contact
7 Accuracy
The relative accuracy of a Ritz value,
, is considered acceptable if its Ritz estimate
$≤ Tolerance × λ$
. The default
used is the
machine precision
given by
8 Further Comments
9 Example
This example solves $Ax = λBx$ in regular-invert mode, where $A$ and $B$ are derived from the standard central difference discretization of the one-dimensional convection-diffusion operator $d2u dx2
+ ρ du dx$ on $0,1$, with zero Dirichlet boundary conditions.
9.1 Program Text
9.2 Program Data
9.3 Program Results
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Simpsons 3.8 rule, error help matlab
May 10th 2010, 05:24 AM #1
Junior Member
May 2009
Simpsons 3.8 rule, error help matlab
hi this is my code
function [y]=BJ(n,x)
% Remeber that y(1) is the first element, not y(0).
for i=0:N-1
y4(delta)=((x^4)*((cos(delta))^4)-4*n*(x^3)*((cos(delta))^3)+(4+6*(n^2))*(x^2)*((cos(delta))^2)-4*x*n*(1+(n^2))*cos(delta)+(n^4)-3*(x^2)*((sin(delta))^2))*cos(n*delta-x*sin(delta))+6*sin(n*delta- x*sin(delta))*((n^2)-2*x*cos(delta)*n+(x^2)*((cos(delta))^2)+(1/6))*x*sin(delta);
S = S+((h/8)*y(1))+((3*h/8)*y(2))+((3*h/8)*y(3))+((h/8)*y(4))-((3*(h^5)/80)*(y4(delta)));
can anyone fix it for me, to include the error... i got it right without it, but for some reason putting the error in, stuffs it up, and cant find where
it says something about the matrix must be square????
Last edited by CaptainBlack; May 10th 2010 at 06:24 AM.
also does anyone know why, my 3/8 rule is different to the one on the net??? mine is h/8(f0+3f1+3f2+f3) and theirs is 3h/8(f0+3f1+3f2+f3) and my code works and is right according to matlab's
inbuilt besselj function?
hi this is my code
function [y]=BJ(n,x)
% Remeber that y(1) is the first element, not y(0).
for i=0:N-1
y4(delta)=((x^4)*((cos(delta))^4)-4*n*(x^3)*((cos(delta))^3)+(4+6*(n^2))*(x^2)*((cos(delta))^2)-4*x*n*(1+(n^2))*cos(delta)+(n^4)-3*(x^2)*((sin(delta))^2))*cos(n*delta-x*sin(delta))+6*sin(n*delta- x*sin(delta))*((n^2)-2*x*cos(delta)*n+(x^2)*((cos(delta))^2)+(1/6))*x*sin(delta);
S = S+((h/8)*y(1))+((3*h/8)*y(2))+((3*h/8)*y(3))+((h/8)*y(4))-((3*(h^5)/80)*(y4(delta)));
can anyone fix it for me, to include the error... i got it right without it, but for some reason putting the error in, stuffs it up, and cant find where
it says something about the matrix must be square????
Please post the original question.
here's the original qn
Let us consider an approximation to an integral. Let f(x) be some continuous function on
[a, b]. We wish to find an approximation for the integral
I = int from a to b of f(x)dx
in the following manner:
Subdivide the interval into N intervals of length h = (b−a)/N. Let xi = ih for i = 0, . . . ,N.
Ij = int from 0 to h of f(xj+t)dt
Find a cubic polynomial Pj (x) that goes through (xj , f(xj)), (xj + h/3,f(xj + h/3)),(xj+2h/3, f(xj+2h/3) and (xj+1,f(xj+1))
We form an approximation for the integral by letting
I=sum(j=0 to N-1) of w0*f(xj)+w1*f(xj+h/3)+w2*f(xj+2h/3)+w3*f(xj+1)
Find these weights, wi.
In 2 peices of code, plot the first three Bessel functions, J0(x), J1(x) and J2(x), on the
interval [0, 20]. The first peice of code should be a MATLAB function BJ(x, n) outputing
the approximation for the integral representation of Jn, given by
Jn(x) =(1/pi)int from 0 to pi of cos(nt − x sin t)dt
using the above method for 100 subdivisions of [0, pi]. The second peice of code should call
the function an produce the required plots with 2000 subdivisions of [0, 20].
I dont need to find the error for qn, but i was just curious, on how you would do it? does it even matter?
May 10th 2010, 06:01 AM #2
Junior Member
May 2009
May 10th 2010, 06:24 AM #3
Grand Panjandrum
Nov 2005
May 10th 2010, 06:31 AM #4
Junior Member
May 2009
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Faculty Research Presentations
Department of Mathematics
Faculty Research Presentations
Real Zeros of Random Trigonometric Polynomials with Binomial Coefficients
A. Brania and M. Sambandham, Morehouse College
Invited Talk in Gulf Coast Conference on Probability and Statistics-GCCPS-2010, University of South Florida, Tampa, February 27, 2010
A Polynomials Bounded Integer Programming Algorithm
Benedict Nmah, Morehouse College, Atlanta, GA.
The Fourth International Conference On Neural, Parallel & Scientific Computations, Morehouse College, Atlanta, Georgia, August 11-14, 2010
Comparison Theorem for Hybrid Fractional Differential Equations
A. Brania and M. Sambandham, Morehouse College, GA, and N. G. Medhin, North Carolina State University, NC,
The Fourth International Conference On Neural, Parallel & Scientific Computations, Morehouse College, Atlanta, Georgia, August 11-14, 2010
A Program of Producing Pseudo-Anosov Maps Via Dehn Twists and Point-Push
Chaohui Zhang, Morehouse College, Atlanta, GA.
The Fourth International Conference On Neural, Parallel & Scientific Computations, Morehouse College, Atlanta, Georgia, August 11-14, 2010
On the Formation of Large Entropy Sets
Steven Pederson, Morehouse College, GA.
The Fourth International Conference On Neural, Parallel & Scientific Computations, Morehouse College, Atlanta, Georgia, August 11-14, 2010
Fuzzy Number System and Its Algebraic Properties
Chuang Peng, Morehouse College GA.
The Fourth International Conference On Neural, Parallel & Scientific Computations, Morehouse College, Atlanta, Georgia, August 11-14, 2010
On the Real Zeros of Random Pretrigonometric Polynomials
Sajjad Abdullateef, R. E. Bozeman and M. Sambandham, Morehouse College, Atlanta, GA.
The Fourth International Conference On Neural, Parallel & Scientific Computations, Morehouse College, Atlanta, Georgia, August 11-14, 2010
On The Control of Impulsive Hybrids Systems
N. G. Medhin, North Carolina State University, NC. and M. Sambandham, Morehouse College, Atlanta, GA.
The Fourth International Conference On Neural, Parallel & Scientific Computations, Morehouse College, Atlanta, Georgia, August 11-14, 2010
Method of Vector Lyapunov Functions for Hybrid Fractional Differential Equation
M. Sambandham, Morehouse College
Plenary Lecture sessions of the International Conference on Mathematical and Computational Models (ICMCM, 2009), PSG College of Technology, Coimbatore, India, December 21-23, 2009.
Lyapunov Theory for Hybrid Impulsive Fractional Differential Equations
M. Sambandham, Morehouse College
Plenary Lecture sessions of the International Conference on Recent Advances in Mathematics Sciences and Applications, Institute of Advanced Studies, GVP College of Engineering, Visakhapatnam, India,
December 19-21, 2009.
Comparison Theorem & Stability Properties Of Hybrid Fractional Differential Equation
M. Sambandham, Morehouse College
Invited Lecture, Seventh University of South Florida Interdisciplinary Workshop on Cancer and Hybrid Dynamic Systems, October 30, 2009.
Numerical Solution of Hybrid Fractional Differential Equations
Steve Pederson and M. Sambandham, Morehouse College
Colloquium Talk, PSG College of Technology, India, May 25, 2009.
On the Real Roots of Random Polynomials
M. Sambandham, Morehouse College
Colloquium Talk, Annamalai University, Annamalai Nagar, India, May 21, 2009.
Properties of Random Polynomials
M. Sambandham, Morehouse College
Invited Lecture, 6th University of South Florida (USF) Workshop on ”Stochastic Dynamic Systems and Applications”, Tampa, FL, April 10, 2009.
On the Holonomy of the Coulomb Connection over Manifolds with Boundary
William E. Gryc, Morehouse College
Special Session on Infinite Dimensional Analysis, Path Integrals and Related Fields, AMS National Meeting, Washington DC, January 6, 2009.
Numerical Solution of Hybrid Fractional Differential Equations
S. Pederson and M. Sambandham, Morehouse College
Invited Talk in AMS Special Session on Stochastic, Large-Scale, and Hybrid Systems with Applications, AMS National Meeting, Washington DC, January 5-8, 2009.
Complete listing from 2001 through 2009
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Tarzana Trigonometry Tutor
Find a Tarzana Trigonometry Tutor
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18 Subjects: including trigonometry, geometry, algebra 1, GRE
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28 Subjects: including trigonometry, Spanish, chemistry, French
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E = edge(mdl,X,Y)
E = edge(mdl,X,Y,Name,Value)
E = edge(mdl,X,Y) returns the classification edge for mdl with data X and classification Y.
E = edge(mdl,X,Y,Name,Value) computes the edge with additional options specified by one or more Name,Value pair arguments.
Input Arguments
k-nearest neighbor classifier model, returned as a classifier model object.
Note that using the 'CrossVal', 'KFold', 'Holdout', 'Leaveout', or 'CVPartition' options results in a model of class ClassificationPartitionedModel. You cannot use a partitioned tree for prediction,
so this kind of tree does not have a predict method.
Otherwise, mdl is of class ClassificationKNN, and you can use the predict method to make predictions.
Matrix of predictor values. Each column of X represents one variable, and each row represents one observation.
Y — Categorical variablescategorical array | cell array of strings | character array | logical vector | numeric vector
A categorical array, cell array of strings, character array, logical vector, or a numeric vector with the same number of rows as X. Each row of Y represents the classification of the corresponding
row of X.
Name-Value Pair Arguments
Specify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside single quotes (' '). You can specify several
name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.
'weights' Observation weights, a numeric vector of length size(X,1). If you supply weights, edge computes weighted classification edge.
Default: ones(size(X,1))
Output Arguments
E Classification edge, a scalar that is the mean classification margin (see Margin).
The edge is the mean value of the classification margin.
The classification margin is the difference between the classification score for the true class and maximal classification score for the false classes.
Margin is a column vector with the same number of rows as X.
The score of a classification is the posterior probability of the classification. The posterior probability is the number of neighbors that have that classification, divided by the number of
neighbors. For a more detailed definition that includes weights and prior probabilities, see Posterior Probability.
Construct a k-nearest neighbor classifier for the Fisher iris data, where k = 5.
Load the data.
load fisheriris
X = meas;
Y = species;
Construct a classifier for five-nearest neighbors.
mdl = fitcknn(X,Y,'NumNeighbors',5);
Examine the edge of the classifier for minimum, mean, and maximum observations classified 'setosa', 'versicolor', and 'virginica' respectively.
NewX = [min(X);mean(X);max(X)];
Y = {'setosa';'versicolor';'virginica'};
E = edge(mdl,NewX,Y)
E =
The classifier has no doubt that the Y entries are correct classifications (all five nearest neighbors of each NewX point classify as the corresponding Y entry).
See Also
ClassificationKNN | fitcknn | loss | margin
More About
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Math for Slow Learners
Slow learners are normal students who are simply not interested in studying under traditionally acceptable systems of education. Parents or guardians often do not take their slow learning wards
seriously during their earlier days of growth.Courses like mathematics have great bearing on the child’s frame of mind during primary school education. If the concepts are not clear, then the child
loses interest in the subject and as the subject has continuity and is dependent on basic principles, this interesting subject becomes stressful for him or her. Subsequently it becomes difficult for
parents or guardians or teachers to help such children achieve even passing grades where normal students can achieve almost full marks with little effort.
Understanding how mathematics is applied to everyday life is important to a child that sees no relevance in the subject matter, rather just an obscure mystery.
Some tips to develop math for slow learners:
• Explaining addition and subtraction using marbles in different bowls
• Explain division and fractions using the concept of sharing as applied by child while playing games of his/her interest like Pokémon cards
• Explain time management using his/her favorite television programs
Fortunately nowadays there is an abundance of resources available in libraries as well as online in regards to developing effective concepts of math for slow learners.
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all 4 comments
[–]Servaphetic1 point2 points3 points ago
sorry, this has been archived and can no longer be voted on
Using integration by parts is a pain in the ass and imo should be avoided where possible.
If you've done any differential equations, this will probably make more sense, but hey ho, try it even if you haven't!
dy/dx = e^-2x(sin(6x))
A particular solution will be of the form:
y = e^-2x(Asin(6x)+B(cos(6x))
y' = -2e^-2x(Asin(6x)+Bcos(6x) + e^-2x(-6Bsin(6x)+6Acos(6x))
Solve simultaneously:
-2A-6B = 1
Yields A=-1/20, B=-3/20
Solution is therefore -1/20 e^-2x(sin(6x)+3(cos(6x)) +C
[–]bubalue0 points1 point2 points ago
sorry, this has been archived and can no longer be voted on
I saw your other post about this problem and answered it. Not sure if you saw it though.
For the first one, you do have to use integration by parts twice (I used u=sin(6x)). After the second time you should get (-sin(6x))/(2e^2x) + 3[(-cos(6x)/(2e^2x) - 3(integral of the starting
equation)]..As you can see in this type of problem it helps if you set the initial integral to a set variable (let's say A). Therefore A=(-sin(6x))/(2e^2x) + 3[(-cos(6x)/(2e^2x) - 3A]. From there you
can solve the equation for "A". Once you have done that, you have your answer. Just add an arbitrary constant "C".
[–]zn849275937[S] 1 point2 points3 points ago
sorry, this has been archived and can no longer be voted on
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Converting math equations to C#
A while ago, I worked on a product where part of the effort involved turning math equations into code. At the time, I wasn't the person who was allocated the role, so my guess is the code was written
by simply taking the equations from word and translating them by hand into C#. All well and good, but it got me thinking: is there a way to automate this process so that human error can be eliminated
from this, admittedly boring, task? Well, turns out it is possible, and that's what this article is about.
Equations, eh?
I'd guess that barring any special math packages (such as Matlab), most of us developers get math requirements in Word format. For example, you might get something as simple as this:
This equation is easy to program. Here, let me do it: y = a*x*x + b*x + c;. However, sometimes, you end up getting really nasty equations, kind of like the following:
Got the above from Wikipedia. Anyways, you should be getting the point by now: the above baby is a bit too painful to program. I mean, I'm sure if you have an infinite budget or access to very cheap
labour, you could do it, but I guarantee you'd get errors, since getting it right every time (if you've got a hundred) is difficult.
So, my thinking was: hey, there ought to be a way of getting the equation data structured somehow, and then you could restructure it for C#. That's where MathML entered the picture.
Okay, so you are probably wondering what this MathML beast is. Basically, it's an XML-like mark-up language for math. If all browsers supported it, you'd be seeing the equations above rendered using
the browser's characters instead of bitmaps. But regardless, there's one tool that supports it: Word. Microsoft Word 2007, to be precise. There's a little-known trick to get Word to turn equations
into MathML. You basically have to locate the equation options...
and choose the MathML option:
Okay, now copying our first equation onto the clipboard will result in something like the following:
<mml:mi mathvariant="italic">bx</mml:mi>
You can probably guess what this all means by looking at the original equation. Hey, we just ripped out the structure of an equation! That's pretty cool, except for one problem: converting it to C#!
(Otherwise, it's meaningless.)
Syntax tree
Keeping data the way we get it is no good. There's lots of extra information (like that italic statement near bx), and there's info missing (like the multiplication sign that ought to be between b
and x). So, our take on the problem is turn this XML structure into a more OOP, XML-like structure. In fact, that's what the program does – it turns XML elements into corresponding C# classes. In
most cases, XML and C# have a 1-to-1 correspondence, so that an <mi/> element turns into an Mi class. So woo-hoo, without too much effort, we turn XML into a syntax tree. Now, the tree is imperfect,
but it's there. Let us instead discuss some of the thorny issues that we have to overcome.
Single/multi-letter variables
Does 'sin' mean s times i times n, or a variable called 'sin', or the Math.Sin function? When I looked at the equations I had, some of them used multiple letters, some were single-letter. There's no
'one size fits all' solution as to how to treat those. Basically, I made this an option.
The times (×) sign
If you write ab, it might mean a times b. If that's the case, you need to find all the locations where the multiplication has been omitted. On a funny note, there are also different Unicode symbols
used by the times sign in different math editing packages (I was testing with MathML as well as Word). The end result is that finding where the multiplication sign is missing is very difficult.
Greek to Roman
Some people object to having Greek constants in C# code. Hey, I code in UTF-8, so I can include anything, including Japanese characters and those other funny Unicode symbols. It does mess up
IntelliSense because your keyboard probably doesn't have Greek keys - unless you live in Greece, that is. Plus, it's a way to very quickly kill maintainability. So, one feature I had to add is
turning Greek letters into Roman descriptions, so that Δ would become Delta and so on. Actually, Delta is a special case because we are so used to attaching it to our variables (e.g., writing ΔV).
Consequently, I added a special rule for Δ to be kept attached even in cases where all other variables are single-letter.
Correctly treating e, π, and exp
Basically, the letter pi (π) can be just a variable, or it can mean Math.PI. Same goes for the letter e – it could be Math.E, and in most cases, it is. Another, more painful substitution is exp to
Math.Exp. Support for all three of these had to be added.
Power inlining
Most people know that x*x is faster than Math.Pow(x, 2.0), especially when dealing with integers. Inlining powers of X and above is an option in the program. I have seen articles (can't find the
link) where people claim that you lose precision if you avoid doing it the Math.Pow way. I'm not sure though.
Operation reduction
I've been alerted to the fact that some expressions output are inefficient as far as their constituent operations go. For example, a*x*x+b*x+c is not as efficient as x*(a*x+b)+c because it has more
multiplications. Thus, one of the future goals of my solution is to attempt to optimize these scenarios. It will make them less readable though!
There were plenty of other problems in converting from XML to C#, but the main idea stayed the same: correctly implement the Visitor pattern over each possible MathML element, removing unnecessary
information and supplying that information which is missing. Let's look at some examples.
Okay, I bet you can't wait to see an actual example. Let's start with what we had before:
Here's the output we get:
I omitted the initialization steps for variables that the program also creates.
Let's look at the more complex equation. Here it is, in case you have forgotten:
Care to guess what the output of our tool is?
p = rho*R*T + (B_0*R*T-A_0-((C_0) / (T*T))+((E_0) / (Math.Pow(T, 4))))*rho*rho +
(b*R*T-a-((d) / (T)))*Math.Pow(rho, 3) +
alpha*(a+((d) / (t)))*Math.Pow(rho, 6) +
((c*Math.Pow(rho, 3)) / (T*T))*(1+gamma*rho*rho)*Math.Exp(-gamma*rho*rho);
I originally had the above output using Greek letters (reminder: C# is okay with them). However, due to coding, I've let my tool change them to Romanized versions, thus demonstrating yet another
Okay, let's do another example just to be sure – this time with a square root. Here is the equation:
I've turned power inlining off for this one - we don't want the expression with the root being evaluated twice. Here is the output:
a = 0.42748 * ((Math.Pow((R*T_c), 2)) / (P_c)) *
Math.Pow((1 + m * (1 - Math.Sqrt(T_r))), 2);
Is this great or what? If you are ever handed a 100-page document full of formulae, well, you can surprise your client by coding them really quickly.
I hope you like the tool. Maybe you'll even find it useful. I have recently redesigned the tool from the ground up using F#, and you can find the latest version here.
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Integral Questions
March 6th 2009, 09:35 AM #1
Mar 2009
Integral Questions
Can you help me.
integral from 4 to 1
x=( x^2-x+1)/(sqrt(x))
integral from pie to pie/4
integral from 4 to x
This one is easy. Just split up the fraction and use the properties of exponents to rewrite each term as a power of $x.$
$\int_{\pi/4}^\pi\sec x\tan x\,dx$
Well, what is the derivative of the secant? That should make the answer to this one obvious.
$\int_4^x(2+\sqrt u)^8\,du$
Substitute $v=2+\sqrt u\Rightarrow dv=\frac{du}{2\sqrt u},$ and then note that $\sqrt u=v-2.$
This has no elementary antiderivative. Did you write it correctly?
Rewrite, using the identity $\tan^2x+1=\sec^2x\colon$
You should be able to take it from here.
March 6th 2009, 12:05 PM #2
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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.
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the first resource for mathematics
Chebyshev series approximations for the Bessel function
of complex argument.
(English) Zbl 0918.65016
Bessel functions of the first kind
and the second kind
of integer order play an important role in mathematical physics and engineering sciences. Numerical methods for efficiently computing these functions are therefore of interest to computational
physicists and engineers. The authors employ the truncated Chebychev series to approximate the Bessel function of the second kind
$|z|\le 8$
. Detailed manipulations and discussions for
are given. Results of numerical experiments are presented to demonstrate the computed accuracy by using the Chevychev series approximation. The computed accuracy is comparable with that computed by
the tau-method approximations, especially when
is small. Advantages and disadvantages of the Chebychev series approximation compared with the tau-method approximation are discussed.
65D20 Computation of special functions, construction of tables
65E05 Numerical methods in complex analysis
33C10 Bessel and Airy functions, cylinder functions, ${}_{0}{F}_{1}$
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On the stochastic nonlinear Schrodinger equation
Seminar Room 1, Newton Institute
We consider a non linear Schrodinger equation on a compact manifold of dimension d subject to some multiplicative random perturbation. Using some stochastic Strichartz inequality, we prove the
existence and uniqueness of a maximal solution in H^1 under some general conditions on the diffusion coefficient. Under stronger conditions on the noise, the nonlinearity and the diffusion
coefficient, we deduce the existence of a global solution when d=2. This is a joint work with Z. Brzezniak.
The video for this talk should appear here if JavaScript is enabled.
If it doesn't, something may have gone wrong with our embedded player.
We'll get it fixed as soon as possible.
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|
Mutually orthogonal Latin squares
latin square
s are said to be
if no pair of corresponding elements occurs more than once. For example,
1, 2, 3
2, 3, 1
3, 1, 2
is orthogonal to
1, 3, 2
2, 1, 3
3, 2, 1
We can see this most easily by writing them together as follows, and observing that no pair appears twice:
11, 23, 32
22, 31, 13
33, 12, 21
A set of n latin squares is mutually orthogonal if every pair of latin squares from the set is orthogonal.
Euler studied orthogonal latin squares because they can be used to construct magic squares, and he found that, while orthogonal latin squares of any odd order are easy to generate, even orders are
not. He conjectured, but did not prove, that there are no orthogonal latin squares of order 4n+2, for any integer n. In 1960, this was proved incorrect; it turns out that there are orthogonal latin
squares of any size except 1, 2, and, oddly enough, 6.
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"adjoint" =?= "inverse of composite endofunctor is uniform bi-composition"
up vote 2 down vote favorite
Understanding adjoints has always been (and continues to be) a bit of a struggle for me.
Today I stumbled upon a property of adjoint functors which seemed extremely intuitive to me. I was wondering why this property isn't mentioned more often in introductory category theory literature,
and whether or not it completely characterizes adjunctions.
If two functors $F:C\to D$ and $U:D\to C$ are adjoint $F\dashv U$, then for every $f:F(Y)\to X$ in $D$ there exists an $\hat f:Y\to U(X)$ in $C$ such that
$$ U(f)\circ \eta_Y = \hat f$$ $$ \epsilon_X\circ F(\hat f)=f$$
If we substitute the top equation into the bottom, we get
$$ \epsilon_X\circ F(U(f)\circ \eta_Y)=f$$
and by functoriality we get
$$ \epsilon_X\circ F(U(f))\circ F(\eta_Y)=f$$ $$ \epsilon_X\circ (F\circ U)(f)\circ F(\eta_Y)=f$$
What the last equation says is that we can recover any morphism $f$ from the action of the "round trip endofunctor" $F\circ U$ on it by pre-composing with $\epsilon_X$ and post-composing with $F(\
eta_Y)$. These two morphisms are determined only by the domain and codomain of $f$ -- we only needed to know $X$ and $F(Y)$ in order to pick the two morphisms. We would have picked the same two
morphisms for some $g\neq f$ as long as $g:F(Y)\to X$.
So, I believe it is correct to say that "if the domain of a morphism is within the range of a functor which has a right adjoint, then it can be recovered from the action of the composite endofunctor
on it by pre-composition with some morphism and post-composition with some other morphism, where the choice of these two morphisms is completely determined by the domain and codomain of the original
morphism". There is, of course, an equivalent statement for morphisms with a codomain in the range of a functor with a left adjoint.
So, my three questions are: (1) is this correct, (2) if so, why isn't it used to explain adjunctions to beginners (I certainly would have caught on quicker!) and (3) does the condition completely
characterize adjoint functors?
ct.category-theory adjoint-functors
2 I don't understand why you say that "if we want to know the action of FU on f, we just pre- and post- compose with $\epsilon$ and $F(\eta)$". Doesn't your formula (which is true) really say that
"f can be recovered from FU(f), by pre- and post-composing ..."? – Charles Rezk Mar 23 '10 at 20:23
Wow, thank you, yes that was a glaring error. I have modified the question and its title to correct this. Thank you! – Adam Mar 23 '10 at 20:41
The condition you wrote + its dual + the assertion that $\eta:I_C\to UF$ and $\varepsilon:FU\to I_D$ are natural totally characterize the adjunction, because an adjunction is totally characterized
by the triangular identities $U\varepsilon\circ\eta U=I_U$, $\varepsilon F\circ F\eta=I_F$, see part (v) of Theorem IV.2 on page 83 of Mac Lane. To get the triangular identities from yours, just
substitute $f=1_{FY}$ (so that $X=FY$), and similarly in the dual. BTW, the derivation of the triangular identities is very similar to yours, see p. 82 in Mac Lane. – user2734 Mar 23 '10 at 22:55
This might sound dumb, but I never understood the $F\eta$ notation. If $F$ is a functor and $\eta$ is a morphism of functors, what does $F\eta$ mean? It can't be composition because they aren't
both morphisms in the same category... and viewing $\eta$ as an object-indexed family of morphisms doesn't seem to work out either because $F$ isn't an object. Does this mean reindexing? – Adam
Mar 24 '10 at 0:21
$F\eta$ is the natural transformation whose component on an object $x$ is the map $F(\eta_x)$. – Reid Barton Mar 24 '10 at 2:12
show 1 more comment
1 Answer
active oldest votes
(1) Yes. (2) Well, it doesn't give me any additional intuition. You didn't say why it helps you understand, so I can't judge what the advantage of it might be. I think this is really
just a complicated way of giving the "bijection of hom-sets" condition.
(3) No, you need something more. For instance, let $r:B\to A$ be a surjection with section $s$, let $C$ have two objects $x$ and $y$ with $C(x,y)=B$, $C(y,x)=\emptyset$, and $C(x,x)=C
(y,y)=1$ (only identities), let $D$ be similar using $A$ instead, and let $F:C\to D$ and $U:D\to C$ be the identity on objects and with action on arrows given by $r$ and $s$
up vote 3 respectively. Pick $\varepsilon$ and $\eta$ to be identities. Then every morphism in $D$ can be recovered, as you describe, but the components of $\eta$ are not natural, and the dual
down vote condition fails.
The "unknown (google)" comment above explained why if you additionally require the dual condition, plus naturality of $\eta$ and $\varepsilon$, then you do get an adjunction. (Although
it's not clear to me from the condition you stated whether you wanted to require the morphism playing the role of $F(\eta)$ to actually be $F$ of something, which is also necessary for
this argument to work.)
I think this is really just a complicated way of giving the "bijection of hom-sets" condition. -- Except that it never mentions sets. That's the advantage. The whole notion of hom-
sets (rather than hom-objects) still seems weird and unusual to me, and I'm always afraid that some of my deep-seated intuitions about sets are going to limit my thinking. So I try
to block the concept of hom-set out of my mind. The definition above is something that would make sense to a beginner long before they're ready to learn about enrichment. – Adam Oct
11 '10 at 9:16
1 But it still mentions elements of hom-sets, i.e. single morphisms, and therefore does not carry over to the enriched world. – Mike Shulman Oct 11 '10 at 17:17
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Chemistry and Chemical Reactivity 8th Edition Chapter 14.IC Solutions | Chegg.com
The figure caption for Figure 14d says that the scale bar is equal to 500 nm. If this scale bar is the same size on the picture (5 mm) as the scale bar in Figure 14a, the magnification can be
calculated as a proportion.
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Determinant Calculation Program.... HELP PLZ
11-28-2006 #1
Determinant Calculation Program.... HELP PLZ
Hi guys, I have to make a program to print out the matrix from the file and its determinant. Here's the outline Im supposed to fill out:
#define N 10
#define NMAX 4
double determinant2(double a[N][N]) ;
double determinant3(double a[N][N]) ;
double determinant4(double a[N][N]) ;
void read(double a[N][N], int n) ;
void print(double a[N][N], int n) ;
int count_inversions(int b[N], int n) ;
double a[N][N] ;
double determinant ;
int n ;
Get n, the size of the square array
Read the n by n array and then print it.
Find and print the determinant of the matrix
And heres what I got so far:
#define N 10
#define NMAX 4
void read(double a[N][N], int n)
int i,j;
for (i=0; i<n; i++)
for (j=0;j<n;j++)
void print(double a[N][N], int n)
int i,j;
for (i=0; i<n; i++)
for (j=0;j<n;j++)
printf("%.2lf\n", a[i][j]);
main ()
int n;
FILE *inp;
inp = fopen("matrix.txt", "r");
double a[N][N];
fscanf(inp, "%d", &n);
printf("The size of matrix is %d\n", n);
read(a, n);
print(a, n);
and the sample of the input file must be in the following format:
where the first element (4) indicates the size of square matrix(4x4)
As you can see Im trying to read the matrix from the input file and print it out on the screen, but the errors Im getting are:
assi2-.c: In function `main':
assi2-.c:42: parse error before `double'
assi2-.c:47: `a' undeclared (first use in this function)
If anyone can help me out with this, it be greatly appreciated. Thank you for looking.
Last edited by ss7; 11-28-2006 at 08:23 PM.
For starters, read and print take 2 arguments, and you're only sending them one.
Also, I don't believe the way you declare matrix is legal, but don't quote me on that.
Teacher: "You connect with Internet Explorer, but what is your browser? You know, Yahoo, Webcrawler...?" It's great to see the educational system moving in the right direction
thanks for the reply, I just updated the reworked program and errors, still getting those few.
I got it to compile without any errors using this code:
#define N 10
#define NMAX 4
void read(double a[N][N], int n)
int i,j;
for (i=0; i<n; i++)
for (j=0;j<n;j++)
fscanf(inp, "%lf",&a[i][j]);
void print(double a[N][N], int n)
int i,j;
for (i=0; i<n; i++)
for (j=0;j<n;j++)
printf("%.2lf\n", a[i][j]);
main ()
int n;
double a[N][N];
FILE *inp;
inp = fopen("matrix.txt", "r");
fscanf(inp, "%d", &n);
printf("The size of matrix is %d\n", n);
read(a, n);
print(a, n);
but I am getting this error while running it......
Any ideas?
Last edited by ss7; 11-28-2006 at 11:28 PM.
Your array shouldn't have the ampersand there in your scanf function.
Teacher: "You connect with Internet Explorer, but what is your browser? You know, Yahoo, Webcrawler...?" It's great to see the educational system moving in the right direction
hey thanks for pointing that out, I got rid of that, and it still has that error while running
1. you need an ampersand in scanf
2. you said - you wanted to read a numbers from file? you should use fscanf then
3. In this case - fscanf requires file pointer - you should give 3rd parameter to your read function
The first 90% of a project takes 90% of the time,
the last 10% takes the other 90% of the time.
1. ok put back the ampersand
2. replaced scanf with fscanf in read function
3. how would the pointer work exactly? Something like this?
void read(double a[N][N], int n)
int i,j, x;
for (i=0; i<n; i++)
for (j=0;j<n;j++)
fscanf(inp, "%lf", &a[i*x + j]);
Please correct me if im wrong. Thank you.
Last edited by ss7; 11-28-2006 at 11:31 PM.
void read(double a[N][N], int n,FILE* inp)
int i,j;
for (i=0; i<n; i++)
for (j=0;j<n;j++)
fscanf(inp, "%lf", &a[i][j]);
The first 90% of a project takes 90% of the time,
the last 10% takes the other 90% of the time.
hey thanks a lot for your help vart!
I think it kinda worked, it prints out the numbers, but I thought it would print it out in the form of matrix, is anything Im missing?
here is the result screenshot
you add \n after each number - you should print it only at the end of the line
The first 90% of a project takes 90% of the time,
the last 10% takes the other 90% of the time.
thank you, thank you, thank you, I got it..finally........ (and it was supposed to be easy part of an assignment!).
Figuring out determinants is for tomorrow.
my question now, is for example lets say I have input file with 3 matrices of sizes 2x2, 3x3 and 4x4. How can I input all 3 matrices at the same time from 1 file? I tried putting while loop in
read function but no success:
void read(double a[N][N], int n, FILE* inp)
int i,j;
while(fscanf(inp,"%lf", &a[n][n])!=EOF)
for (i=0; i<n; i++)
for (j=0;j<n;j++)
fscanf(inp, "%lf", &a[i][j]);
a sameple of input file:
> while(fscanf(inp,"%lf", &a[n][n])!=EOF)
If you just had
fscanf(inp,"%d", &n );
Then your two loops, then you would read in just one matrix.
Assuming your N constant is big enough to hold any possible matrix you may have.
Returning the value of 'n' to the calling function would be a good idea, so you know how big it is.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
I support http://www.ukip.org/ as the first necessary step to a free Europe.
ok this is what I changed it to:
void read(double a[N][N], int n, FILE* inp)
int i,j;
fscanf(inp,"%d", &n);
for (i=0; i<n; i++)
for (j=0;j<n;j++)
fscanf(inp, "%lf", &a[i][j]);
But it wont read the other matrices below the first one from the input file as above
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Area Under a Curve
September 2nd 2009, 04:03 AM #1
Jul 2009
Area Under a Curve
Also if some1 can help me in this question that would be great.
Find the equation of the tangent to the parabola y=2x^2 at (1,2). Calculate its point of intersection with the x-axis and the volume of the solid formed when the area between the parabola, the
tangent line and the xaxis is reveloved about the xaxis.
Especially the bit about forming a solid of revolution.
The tangent, has equn of
THe pt of intersection is (1/2,0)
But the area formed i found to be 2/15 pi, which is different to the answer. Much help would be appreciated.
I believe 2pi/15 is correct
Draw a diagram and you'll see you need 2 integrals
between x = 0 and 1/2 you have disks
V= (pi)integral(4x^4dx) = pi/40
between x =1/2 and 1 you have washers
V = pi integral (4x^4 - (4x-2)^2)dx) =13pi/120
Adding the 2 you get 16pi/120 = 2pi/15
See attachment for diagram and set up
September 2nd 2009, 04:22 AM #2
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September 2nd 2009, 11:11 PM #4
Jul 2009
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Euler's Formula
February 13th 2013, 07:07 AM #1
Junior Member
Feb 2013
Euler's Formula
Hello, I have a mid term tomorrow and this question was on the practice mid term and our prof has not posted any solutions. I tried to catch him after class but there was a million other students
crowding him. If there is anyone who could give me a hand with this I would appreciate it.
Show that if f is any homogeneous function of degree n, then it
satises Euler's formula
xD1f(x; y) + yD2f(x; y) = nf(x; y):
Hint: Treat each side of the dening equation as a function of 3
variables t; x and y, and use the chain rule to compute the partials
with respect to t. Then set t = 1.
Re: Euler's Formula
Differentiate both sides of $f(tx,ty)=t^nf(x,t)$ with respect to t using the chain rule, then set t = 1.
Wikipedia calls this fact Euler's homogeneous function theorem. Euler's formula usually refers to either $e^{ix} = \cos x + i\sin x$ or $V - E + F = 2$ where V, E and F are respectively vertices,
edges and faces of a convex polyhedron.
Re: Euler's Formula
With appropriate restraints on the functions involved, the chain rule states:
Let u and v be functions of x, y and t, f a function of u and v and g(x,y,t) = f(u(x,y,t),v(x,y,t). Then
$g_t(x,y,t)=f_1(u(x,y,t),v(x,y,t))u_t(x,y,t)+f_2(u( x,y,t),v(x,y,t))v_t(x,y,z)$
So let f be homogenous; i.e. f(tx,ty)=t^nf(x,y). The partials w.r.t t give the equation:
Evaluate at t = 1, QED.
Re: Euler's Formula
So the question is asking to start with the original definition of Euler's theorem? Then just differentiate wrt t? I understand that but I thought I had to differentiate the equation given wrt t
then show that LS=RS.
Re: Euler's Formula
February 13th 2013, 07:47 AM #2
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boat speed
July 31st 2010, 09:53 PM
boat speed
a boat is trying to cross a 8km wide lake. the speed of the boat in the first 2 kilometers is unknown but it started to raise it speed 0.5 kilometers per hour faster than the first 2 kilometers.
it arrived 10 minutes earlier than expected. find the original speed of the boat.
July 31st 2010, 10:40 PM
a boat is trying to cross a 8km wide lake. the speed of the boat in the first 2 kilometers is unknown but it started to raise it speed 0.5 kilometers per hour faster than the first 2 kilometers.
it arrived 10 minutes earlier than expected. find the original speed of the boat.
Does this mean that the speed on the last 6km was 1/2 kph greater than on the first 2km?
Let the initial speed be $x$ kph. Then it would have taken $8/x$ hours to cross the lake at that speed. It in fact took $[2/x] + [6/(x+0.5)]$ hours to cross the lake.
Now can you finish?
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aggregate(...) with multiple functions
A nice package for doing this sort of thing is doBy. Let's manufacture an
since you didn't provide one:
d <- data.frame(g = rep(letters[1:3], each = 10),
x1 = rnorm(30),
x2 = rnorm(30, mean = 5),
x3 = rnorm(30, mean = 10, s = 4))
# --Case 1: no grouping variables
# If there are no grouping variables, you can define a function to apply
# to each variable (column) with the apply() function.
f <- function(x) c(mean(x), median(x))
# Apply to all numeric variables (not column 1):
apply(d[, -1], 2, f)
x1 x2 x3
[1,] -0.0647788 4.813318 10.21010
[2,] -0.0881492 4.916123 10.68559
# The mean of each variable is in the first row, the median in the second.
# --Case 2: one or more grouping variables
# If you have grouping variables, you can create a function with
# names to apply to each variable groupwise. Notice that I named the
# output variables mean and median, normally a no-no, and watch what
# happens when it is used in summaryBy().
# Define the output function to apply to each variable
f2 <- function(x) c(mean = mean(x), median = median(x))
# The leading dot on the left hand side of the formula in summaryBy()
# indicates that the summary function is to be applied to all variables
# not on the RHS of the formula:
summaryBy(. ~ g, data = d, FUN = f2)
g x1.mean x1.median x2.mean x2.median x3.mean x3.median
1 a 0.04571262 -0.06361278 4.253444 4.223015 11.259677 11.06834
2 b -0.15746011 -0.14223959 4.913657 5.116526 10.037674 11.32120
3 c -0.08258890 -0.06227865 5.272853 5.524493 9.332942 10.14600
You can use multiple grouping variables in the formula if desired. The
function is meant to be applied to each LHS variable in each subgroup.
It is required that the input object of summaryBy() be a data frame.
The doBy package comes with a well-written vignette, wherein all of this
is well described.
On Thu, Jul 15, 2010 at 7:45 PM, Murat Tasan <
[hidden email]
> wrote:
> hi all - i'm just wondering what sort of code people write to
> essentially performa an aggregate call, but with different functions
> being applied to the various columns.
> for example, if i have a data frame x and would like to marginalize by
> a factor f for the rows, but apply mean() to col1 and median() to
> col2.
> if i wanted to apply mean() to both columns, i would call:
> aggregate(x, list(f), mean)
> but to get the mean of col1 and the median of col2, i have to write
> separate tapply calls, then wrap back into a data frame:
> data.frame(tapply(x$col1, f, mean), tapply(x$col2, f, mean))
> this is a somewhat inelegant solution for data frames with potentially
> many columns.
> what i would like is for aggregate to take a list of functions for
> columns, something like:
> aggregate(x, list(f), list(mean, median))
> i'm just curious how others get around this limitation in aggregate().
> do most simply make the individual tapply() calls separately, then
> possibly wrap them back up (as done in the example above), or is there
> a more elegant solution using some function of R that i might be
> unaware of?
> ______________________________________________
[hidden email]
mailing list
> PLEASE do read the posting guide
> and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
[hidden email]
mailing list
PLEASE do read the posting guide
and provide commented, minimal, self-contained, reproducible code.
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The Universe of Discourse
Longshot bets by time travelers
If you're a time traveller, one way to make money is by placing bets on events that you already know the outcomes of. The stock market is only the most obvious example of this; who wouldn't have
liked to have invested in IBM in 1916?
But if making money is only your secondary goal, and your real object is to amaze your friends and confound your enemies, there may be more effective bets to place. I forget who it was who
suggested to me how much fun it would be to go back to 1980 and bet that the cross-dressing star of Bosom Buddies would win back-to-back "Best Actor" Oscars within twenty years. But it's a fine
I never finished this article, and I no longer remember where I planned to take it from there. I think I probably abandoned it because I realized that nothing else I could say would be as interesting
as the Bosom Buddies thing. I don't think I could think of any examples that were less likely-seeming.
Until today. I wonder what sort of odds you could have gotten in 1995 betting that within twenty years, the creators of "What Would Brian Boitano Do" ("Dude! Don't say 'pigfucker' in front of Jesus!
") would win the Tony Award for "Best Musical".
[Other articles in category /misc] permanent link
At a book sale I recently picked up Terence Tao's little book on problem solving for 50¢. One of the exercises (pp. 85–86) is the following little charmer: There are six musicians who will play a
series of concerts. At each concert, some of the musicians will be on stage and some will be in the audience. What is the fewest number of concerts that can be played to that each musician gets to
see the each of the others play?
Obviously, no more than six concerts are required. (I have a new contribution to the long-debated meaning of the mathematical jargon term "obviously": if my six-year-old daughter could figure out the
answer, so can you.) And an easy argument shows that four are necessary: let's say that when a musician views another, that is a "viewing event"; we need to arrange at least 5×6 = 30 viewing events.
A concert that has p performers and 6-p in the audience arranges p(6 - p) events, which must be 5, 8, or 9. Three concerts yield no more than 27 events, which is insufficient. So there must be at
least 4 concerts, and we may as well suppose that each concert has three musicians in the audience and three onstage, to maximize the number of events at 9·4 = 36. (It turns out there there is no
solution otherwise, but that is a digression.)
Each musician must attend at least 2 concerts, or else they would see only 3 other musicians onstage. But 6 musicians attending 2 concerts each takes up all 12 audience spots, so every musician is at
exactly 2 concerts. Each musician thus sees exactly six musicians onstage, and since five of them must be different, one is a repeat, and the viewing event is wasted. We knew there would be some
waste, since there are 36 viewing avents available and only 30 can be useful, but now we know that each spectator wastes exactly one event.
A happy side effect of splitting the musicians evenly between the stage and the audience in every concert is that we can exploit the symmetry: if we have a solution to the problem, then we can obtain
a dual solution by exchanging the performers and the audience in each concert. The conclusion of the previous paragraph is that in any solution, each spectator wastes exactly one event; the duality
tells us that each performer is the subject of exactly one wasted event.
Now suppose the same two musicians, say A and B, perform together twice. We know that some spectator must see A twice; this spectator sees B twice also, this wasting two events. But each spectator
wastes only one event. So no two musicians can share the stage twice; each two musicians share the stage exactly once. By duality, each two spectators are in the same audience together exactly once.
So we need to find four 3-sets of the elements { A, B, C, D, E, F }, with each element appearing in precisely two sets, and such that each two sets have exactly one element in common.
Or equivalently, we need to find four triangles in K[4], none of which share an edge.
The solution is not hard to find:
On stage A B C C D E E F A B D F
In audience D E F A B F B C D A C E
And in fact this solution is essentially unique.
If you generalize these arguments to 2m musicians, you find that there is a lower bound of $$\left\lceil{4m^2 - 2m \over m^2 }\right\rceil$$ concerts, which is 4. And indeed, even with as few as 4
musicians, you still need four concerts. So it's tempting to wonder if 4 concerts is really sufficient for all even numbers of musicians. Consider 8 musicians, for example. You need 56 viewing
events, but a concert with half the musicians onstage and half in the audience provides 16 events, so you might only need as few as 4 concerts to provide the necessary events.
The geometric formulation is that you want to find four disjoint K[4]s in a K[4]; or alternatively, you want to find four 4-element subsets of { 1,2,3,4,5,6,7,8 }, such that each element appears in
exactly two sets and no two elements are in the same. There seemed to be no immediately obvious reason that this wouldn't work, and I spent a while tinkering around looking for a way to do it and
didn't find one. Eventually I did an exhaustive search and discovered that it was impossible.
But the tinkering and the exhaustive search were a waste of time, because there is an obvious reason why it's impossible. As before, each musician must be in exactly two audiences, and can share
audiences with each other musician at most once. But there are only 6 ways to be in two audiences, and 8 musicians, so some pair of musicians must be in precisely the same pair of audiences, this
wastes too many viewing events, and so there's no solution. Whoops!
It's easy to find solutions for 8 musicians with 5 concerts, though. There is plenty of room to maneuver and you can just write one down off the top of your head. For example:
On stage E F G H B C D H A C D F G A B D E G A B C E F H
In audience A B C D A E F G B E H C F H D G
Actually I didn't write this one down off the top of my head; I have a method that I'll describe in a future article. But this article has already taken me several weeks to get done, so I'll stop
here for now.
[ Addendum: For n = 1…10 musicians, the least number of concerts required is 0, 2, 3, 4, 4, 4, 5, 5, 5, 5; beyond this, I only have bounds. ]
[Other articles in category /math] permanent link
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[SOLVED] Linear Equation
January 17th 2009, 02:13 AM #1
[SOLVED] Linear Equation
Solve the system of linear equations in complex numbers:
$(2+i) x + (-3i) y = 3+i$
$3i x + (3-2i) y = 3-3i$
How would I do this?
you can easily solve linear equations like this 1 with Gaussian elimination
Mr Fantastic supposed that x and y are real, which is usually the case for unknowns called x and y.
If x and y are complex then you can use substitution to calculate them. But you need of course both equations.
I am still unsure as to how to solve this.
If I put the real and imaginary numbers together I would get
But I do not see how I can solve this with these solutions.
I understood them as two seperate problems and I did the first. Implicit in that is that x and y are both real (which is the usual convention).
If they are meant to be solved simultaneously then x and y are not real. You need to provide clarification: are they two seperate questions (x and y are real) or a single question (x and y are
not real).
Both equations are one It is a "system of linear equations" I think you use matrices to solve but im not sure how to when it has complex numbers.
Im not sure how to do it as if it was real as I dont know what to do when there is i values
Get the inverse of the 2x2 coefficient matrix in the usual way. If you understand the arithmetic of complex numbers there is nothing difficult here. Have you actually tried working through the
process? Post your working of this and state where you get stuck.
$\left[\begin{array} {cc} 2+i&-3i \\ 3i&3-2i \end{array}\right]<br /> \left[\begin{array} {cc} x \\ y \end{array}\right]=<br /> \left[\begin{array} {cc} 3+i\\3-3i \end{array}\right]$
Is this correct?
Yes. Now calculate the inverse matrix. Then left multiply both sides by the inverse matrix. Then read off the answer for x and y.
$\left[\begin{array} {cc} 2+i&-3i \\ 3i&3-2i \end{array}\right]<br /> \left[\begin{array} {cc} x \\ y \end{array}\right]=<br /> \left[\begin{array} {cc} 3+i\\3-3i \end{array}\right]$
$\left[\begin{array} {cc} 2+i&-3i \\ 3i&3-2i \end{array}\right]^{-1}$$\left[\begin{array} {cc} 3+i\\3-3i \end{array}\right]$= $\left[\begin{array} {cc} x \\ y \end{array}\right]$
$\frac{1}{17-i}\left[\begin{array} {cc} 3-2i&3i \\ -3i&2+i \end{array}\right]=\left[\begin{array} {cc} x \\ y \end{array}\right]$
After this it gets really messy and I am unable to get a solution.
January 17th 2009, 02:53 AM #2
January 17th 2009, 08:09 AM #3
Junior Member
Jan 2009
January 17th 2009, 08:18 AM #4
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Nov 2008
January 17th 2009, 12:07 PM #7
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January 17th 2009, 03:03 PM #13
January 17th 2009, 03:06 PM #14
January 17th 2009, 04:32 PM #15
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, a measure of length, divided into 12 inches, and each inch supposed to contain 3 barley-corns in length. Geometricians divide the Foot into 10 digits, and the digit into 10 lines, &c. The French
divide their Foot, as we do, into 12 inches; but their inch they divide into 12 lines.
It seems this measure has been taken from the length of the human Foot; but it is of different lengths in different countries. The Paris royal Foot is to the English Foot, as 4263 to 4000, and
exceeds the English by 9 1/2 lines; the ancient Roman Foot of the Capitol consisted of 4 palms; equal to 11 inches and 7/10 English; the Rhinland, or Leyden Foot, used by the northern nations, is to
the Roman Foot, as 19 to 20. For the proportions of the Foot of several nations, compared with the English, see the article Measure.
Square Foot, is a square whose side is 1 foot, or 12 inches, and consequently its area is 144 square inches.
Cubic Foot, is a cube whose side is one Foot, or 12 inches, and consequently it contains 12^3 or 1728 cubic inches.
Foot-bank, or Foot-step, in Fortification. See BANQUETTE.
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Topic: Mathematica: How can I plot a graph of values returned by FindRoot[]?
Replies: 2 Last Post: Aug 9, 2011 8:23 AM
Messages: [ Previous | Next ]
Mathematica: How can I plot a graph of values returned by FindRoot[]?
Posted: Aug 9, 2011 6:30 AM
I have defined a function Xin, which takes 3 arguments.
Xin[lambda_,lenght_, radius_]: = ....whatever (it does not matter)
I want to find the root of that when the length 'l' is about 0.49.
FindRoot[Xin[1, l, 0.00001], {l, 0.49}]
That's working fine. But I want to plot the roots for various values
of radius - not just 0.00001 m as above.
This does *not* work
since the output of FindRoot[] is not a simple numeric number like
0.488315, but rather of the form "{l->0.488315}".
How can I convert something like "{l->0.488315}" to give just the
numeric part, so I can plot a graph using Plot[]?
Although I know there will be an infinite number of roots of my
equation, there will only be one close to 0.48. The next will be
around 1.02. FindRoot[] seems to be able to find the single root OK,
its just I don't know how to plot the point.
Date Subject Author
8/9/11 Mathematica: How can I plot a graph of values returned by FindRoot[]? David Kirkby
8/9/11 Re: Mathematica: How can I plot a graph of values returned by FindRoot[]? Peter Pein
8/9/11 Re: Mathematica: How can I plot a graph of values returned by FindRoot[]? Nasser Abbasi
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Help with proof using Wilson's theorem
February 22nd 2009, 07:36 PM #1
Feb 2009
Help with proof using Wilson's theorem
Let $p$ be an odd prime and let $a_1, . . . ,a_{p-1}\$ be a permutation of $\{1, 2, . . ., p-1\}$. Prove that there exist $i ot = j$ such that $ia_i\equiv ja_j(\bmod p)$.
How can I use Wilson's theorem to prove this. I appreciate any help.
Say that there was no $iot =j$ so that $ia_i \equiv ja_j(\bmod p)$. Therefore, $1\cdot a_1,2\cdot a_2, 3\cdot a_3, ... , (p-1)\cdot a_{p-1}$ are all incongruent to eachother. There are $p-1$ of
them which means that they are a permutation of $a_1,...,a_{p-1}$ by pigeonhole principle. Therefore, $(a_1)(2a_2)(3a_3)...((p-1)a_{p-1}) \equiv (a_1)...(a_{p-1})(\bmod p)$. Canceling we get, $
(p-1)!\equiv 1(\bmod p)$, but this is a contradiction because $(p-1)!\equiv -1(\bmod p)$.
February 23rd 2009, 03:05 PM #2
Global Moderator
Nov 2005
New York City
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Calculator PRO/EDU
Fraction Calculator PRO/EDU
Fraction calculator with step-by-step operations. Shows results as you type. Graphical display of expressions with history. Use space to enter mixed fractions.
* Workspaces to work on multiple tasks
* Store often used expressions in a library
* No advertisements
* Step-by-step operations (option to turn off)
* Arithmetic operations (+,-,*,/,÷), hold / to enter ÷
* Powers of fractions
* Fractions simplification
* Fractions with complex numbers
* Decimal to fraction conversion and back
* Symbolical fractions and operations
* Integer roots simplification
* Pinch to zoom
* Linear equations x+1=2 -> x=1
* Quadratic equations x^2-1=0 -> x=-1,1
* Approximate roots of higher polynomials
* Systems of linear equations, write one equation per line, x1+x2=1, x1-x2=2
* Polynomial long division
* Polynomial expansion, factoring
* Solving inequalities with one variable, double tap ( and ) to enter less and greater signs
* Linear and polynomial inequalities, x^3-4>4
* Inequalities with absolute values, abs(2x+3)<=5
* Compound inequalities, 1* Rational inequalities, (x+3)/(x-1)<=0
It does everything I want it to and more
Really like but wish one thing Great app and when dividing fractions if you use the ÷ sign between fractions it will show you division by inverting the second fraction and multiplying. Great tool for
Want REFUND Kept failing to enter correct figureS,
Used a lot, helpful, but very limited. Wants to make fractions where they are unwanted.
Good to go An excellent little tool that is simple to use
What's New
Solving inequalities with one variable, double tap ( and ) to enter less and greater signs
Linear and polynomial inequalities, x^3-4>4
Inequalities with absolute values, abs(2x+3)<=5
Compound inequalities, 1<x+1<2
Rational inequalities, (x+3)/(x-1)<=0
Percent mode, enable in settings
This calculator performs the following operations with fractions : multiplication, division , addition, subtraction , and reduces the fractions and converts the answer as a decimal . They are very
easy to use: click on an empty entry field and use the keypad to type in the desired value at the bottom . If you want to enter a negative fraction, click on the small square in front of the input
field and shot it appears the sign "-" (if you click this button again the "- " disappears) . Then select the desired operation by pressing the appropriate button on the left of the keyboard ( if you
do not select the operation calculator will simply reduce fractions with output greatest common divisor - GCD ) . By clicking on the most right button, your answer becomes a value in the first input
field , and you can continue a series of calculations.
This free math calculator is able to simplify fractions. Just enter the fraction, click the button and the app will simplify.
Best mathematical tool for school and college! If you are a student, it will helps you to learn arithmetics and calculations with fraction numbers.
Note: A common fraction can be reduced to lowest terms by dividing both the numerator and denominator by their greatest common divisor.
Basic calculator adds, subtracts, multiplies and divides two fractions.
Android 4.0 and above:
-simplify fraction
-convert to decimal
-convert to vulgar fraction
"OMS Fractions Calculator" is handy tool for addition, subtraction , multiplication and division of fractions and mixed numbers (fractions with integer part).
Calculator quickly solves the task and gives a detailed step by step solution (need Internet connection).
Application gives you all calculations in detail and shows how to:
- convert improper fractions to mixed numbers;
- convert fractions to a common denominator;
- simplify fractions;
- convert mixed numbers to improper fractions.
Calculator with detailed solutions helps you better understand how to solve task with fractions. And always will help parents check homework of their children.
You can always use this and many other mathematical online calculators by visiting my website OnlineMSchool.com.
A fast and very useful list of more then 2000 formulas of:
-Solved esercises
-Interactive GRAPH CALCULATOR AND SYSTEM SOLVER !
Constantly updated! No Ads!
Best on: Android > 4.0
Add, subtract, multiply and divide fractions with Mohiosoft Fraction Calculator. Simplify fractions and converting decimals has never been easier. In-App help system makes using the calculator a
This free math app is a fractions calculator with various functions:
- add, subtract, multiply and divide fractions.
- convert fractions in decimals,
- simplify fractions
- compare fractions
- find the lowest common denominator
- fraction trainer, to learn fraction calculations
Best mathematical tool for school and college! If you are a student, it will helps you to learn arithmetics and calculations with fraction numbers.
Note: A common fraction consists of an integer numerator, displayed above a line (or before a slash in this app), and a non-zero integer denominator, displayed below (or after the slash) that line.
Numerators and denominators are also used in fractions that are not common, including compound fractions, complex fractions, and mixed numerals.
This free mathematical calculator is able to convert decimal numbers into fractions.
Best math tool for school and college! If you are a student, it will help you to learn arithmetic and division!
Simple and easy-to-use all purpose calculator. You can work with whole number, fractions and mixed number with same ease. With this app you can get your calculations in snap. This app can help you in
wide verities of scenario like complex calculation office, trying to find out best bargain while shopping, smarting out others in your class and many more scenarios. Now you can calculate surely
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Large buttons for fast calculation
Simplify results automatically
Works with mix of numbers, fractions and mixed numbers
Separate boards for Real part, numerator and denominator
Easy to use interface
This free mathematical calculator is able to convert:
- Fractions to Decimal numbers
- Decimal numbers to Fractions
- Fractions to Percent
- Percent to Fractions
Best math tool for school and college! If you are a student, it will help you to learn arithmetic and division!
Note: A common fraction consists of an integer numerator, displayed above a line (or before a slash in this app), and a non-zero integer denominator, displayed below (or after the slash) that line.
Hi Guys,
Enter any partial fraction functions and this generates the solution with a step by step outline solution. It basically does just what it says on the tin. So if you have homework that you need to
check out, then look no further.
This tool is taken from the selection of tools that appears on my live worksheet app on Differentiation, that appears on my developer page.
Hope it proves helpful.
This free app is able to calculate the square root of a number. Integer numbers, decimal numbers and fractions are supported.
Very useful math calculator for school and college! If you are a student, it will help you to learn arithmetic and algebra!
The Squeebles are back again, this time to help with your children's fractions.
☆ The Squeebles games have been featured on BBC Radio, as well as in The Daily Telegraph, The Observer, The Guardian and more.
Featuring 4 mini games, a great reward scheme involving cakes (and we all love cakes), several difficulty levels and unlimited players this Squeebles game will make the task of learning fractions a
LOT more fun.
The game is aimed at 7-11 year olds but can be used by anyone who wants to brush up on their fractions in a fun way.
☆ 4 mini games aimed at helping children learn fractions:
- A Piece of Cake: drag different fractions of cakes to each Squeeble as they ask for them. Easy mode covers halves, quarters and thirds with hard mode covering fractions up to tenths.
- Tricky Pairs: match fractions of the same value (i.e.: 1/2 matches with 3/6) as quickly as possible. Easy mode features 4 pairs and hard mode features a whopping 9 pairs!!!
- Think Big: drag the cake to the Squeeble who calls out the largest fraction. Easy mode covers halves, thirds and quarters. Hard mode covers fractions up to twelfths.
- Super Sums: Answer 10 of the Maths Monster's fractions questions as quickly as possible. Easy, medium and hard modes.
☆ UK or US English
☆ Universal app runs on both iPad and iPhone / iPod Touch
☆ Great reward system and storyline. The Squeebles want to enter the Cake Making Contest but The Maths Monster has stolen all their ingredients and their entry tokens. You have to help the Squeebles
win back their ingredients and tokens by working through the four fractions mini-games. You can then make cakes from over 100 ingredients and enter them into the contest!
☆ Over 100 cake ingredients from delicious sweet classics like toffee sauce and chocolate flakes to bizarre Squeeble favourites like curry sponge and sugared carrots!
☆ Covers fractions from halves to twelfths.
☆ Win trophies for doing well in games.
☆ Features 10 of the Squeebles characters, including Whizz, Lex and the nasty Maths Monster.
☆ Set up unlimited players, making this app perfect for classroom use as well as home use.
☆ Makes learning fractions fun!!!
☆ No in-app purchases, advertising or web links mean all our apps are 100% safe for your children to use.
Other Squeebles apps include Times Tables, Spelling, Addition & Subtraction and Division.
**REAL TEACHER TAUGHT LESSONS**
This algebra course teaches basic number operations, variables and their applications. Gain a fundamental sense of equations, inequalities and their solutions. This course offers 11 full chapters
with 6-8 lessons each chapter that present short easy to follow algebra videos. These 5 to 10 minutes videos take students through the lesson slowly and concisely. Algebra is taken by students who
have gained skills like operation with number, rational numbers, basic equations and the basic coordinate plane.
Chapter 1 Algebra Tools
1.1 Variables and Expressions
1.2 Exponents anf Order of Operations
1.3 Exploring Real Numbers
1.4 Adding Real Numbers
1.5 Subtracting Integers
1.6 Multiply and Dividing Real Numbers
1.7 The Distributive Property
1.8 Properties of Numbers
1.9 Number Systems
1.10 Functions and Graphs
Chapter 2 Solving Equations
2.1 Solving Two Step Equations
2.2 Solving Muti Step Equations
2.3 Solving Equations With Variables on Both sides
2.4 Ratios and Proportions
2.5 Equations and Problem Solving
2.6 Mixture Problems
2.7 Percent of Change
2.8 Solving For a Special value
2.9 Weighted Averages
Chapter 3 Solving Inequalities
3.1 Inequalities and their Graphs
3.2 Solving Inequailty by Add Subtract
3.3 Solve an Inequality Mutiplying and Dividing
3.4 Solve Muti Step Inequalities
3.5 Solving Compund Inequal
3.6 Absolute Value And Inequal
3.7 Graphing Systems of Inequalities
3.8 Graphing Inequalities in Two variables
Chapter 4 Graphs and Functions
4.1 Graphing data on the Coordinate Plane
4.2 Greatest Common Divisor
4.3 Equivalent Fractions
4.4 Equivalent Forms of Rational Numbers
4.5 Comparing and Ordering Rational Numbers
4.6 Direct Variation
4.7 Deductive and Inductive
Chapter 5 Linear Equations and Their Graphs
5.1 Rate of Change and Slope
5.2 Slope Intercept Form
5.3 Standard Form
5.4 Point Slope Form
5.5 Parallel and Perpendicular
Chapter 6 System of Equations and Inequalities
6.1 Solve Systems by Graphing
6.2 Solve Systems using Substition
6.3 Solve Systems Using Elimination
6.4 Application of Systems of Equations
6.5 Linear Inequalities
6.6 Systems of Inequalities
Chapter 7 Exponents
7.1 Zero and Negative Exponents
7.2 Scientific Notation
7.3 Multiplication Properties of Exponents
7.4 More on Multiplications of Exponents
7.5 Division Properties of Exponents
Chapter 8 Polynomials and Factoring
8.1 Adding and Subtracting Polynomials
8.2 Multiplying and Factoring Polynomials
8.3 Multiply Binomials (FOIL)
8.4 Multiply Special cases
8.5 Factor Trinomials (a=1)
8.6 Factor Trinomials (a>1)
8.7 Special cases of factoring polynomials
8.8 Factoring polynomials using grouping
8.9 Multiplying Monomials
8.10 Dividing Mononials
8.11 Special Products of Binomials
8.12 Factor Difference of Squares
8.13 Perfect Squares
Chapter 9 Quadratic Equations and Functions
9.1 Exploring Graphing Quadratics
9.2 Quadratic Equation
9.3 Finding and Estimating Square Roots
9.4 Solving Quadratic Equations
9.5 Factor Quadratic ro Solve
9.6 Complete the Square to Solve Quadratics
9.7 Solve Quadratic Equations using the Quadratic Formula
9.8 Using Discriminant
9.9 Graphing Quadratics
9.10 Exponent Functions
9.11 Growth and Decay
Chapter 10 Radical Expressions and Equations
10.1 Simplify Radicals
10.2 The Pythagorean Theorem
10.3 Operations with radical Expressions
10.4 Solve Radical Equations
10.5 Graphing Square Root Functions
10.6 The Distance Formula
11.1 Simplify Rational Expressions
11.2 Multiply and Divdiding Rational Expressions
11.3 Divide Polynomials
11.4 Adding and Subtracting rational Expressions
11.5 Rational Equations
11.6 Inverse Variation
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ihre Mathe-Defizite zu beseitigen. Die App wurde von einer Diplom-Mathematikern entwickelt, welche seit 20 Jahren an einem Gymnasium als Mathe-Lehrerin arbeitet.
Übersicht über die Themen in dieser App:
ALGEBRA 1
• Basics
• Umrechnungen
• Bruchrechnen
• Lineare (Un-) Gleichungen
• Geraden
• LGS
ALGEBRA 2
• Quadratische Gleichung
• Parabel
• Wurzel
• Potenz
• Logarithmus
• Wachstum
• Ableitung
• Kurvendiskussion
• e & ln-Funktion
• Wachstum & DGL
• Trigonometrische Funktionen
• Spez. Werte bei sin/cos/tan
• Integralrechnung
• Spezielle Stammfunktion
Diese App verwendet jQuery mobile (http://www.jquerymobile.com) und PhoneGap (http://www.phonegap.com). Das App-Icon für diese Android-App wurde mit dem "Launcher Icon Generator", einem Tool des
"Android Asset Studio" (http://android-ui-utils.googlecode.com/hg/asset-studio/dist/icons-launcher.html), bearbeitet. Mehr Informationen zu dieser App findest Du auch auf http://www.mathe-vokabeln.de
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Graphing calculator with algebra. Essential tool for school and college. Replaces bulky and expensive handheld graphing calculators.
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* arithmetic expressions +,-,*,/,÷
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Graphing calculator with algebra. Essential tool for school and college. Replaces bulky and expensive handheld graphing calculators.
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derivatives, complex numbers and more. Shows results as you type. Use menu to switch between modes.
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Try free version first.
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Elementary math operations, addition, subtraction, multiplication and division. Displays results in tabular form. Shows carries, borrows and crossouts.
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inconsistent equations
1. Inconsistent equations is defined as two or more equations that are impossible to solve based on using one set of values for the variables.
An example of a set of inconsistent equations is x+2=4 and x+2=6.
inconsistent equations
two or more equations impossible to satisfy by any one set of values for the variables (Ex.: x + y = 1 and x + y = 2)
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Here's the question you clicked on:
Joey starts reading at the top of page 103 and stops at the bottom of page 204. How mant pages has he read? I keep getting this type of question wrong. is there any specific pattern or equation to
• one year ago
• one year ago
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SAT question ? Right :D
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@hba yup!
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102 is the answer :D
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@Zofia1 Its a timed test!! i cant sit there adding -_-
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@hba oh gee thanks that solved my problem :'(
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well,geez alright then ehh!
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@Zofia1 lol i had the addition part figured out xD
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@Areesha.1D I don't know why do you keep putting easy questions lol.
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what grade you in?
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@hba and i dont know why i get stuck on these -_-
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@zofia im in my second year of A levels sooo 12th
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I always think i am smarter than the average students like @Areesha.1D :D
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@hba lmao if you're so smart then give me a shortcut to this!
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@Areesha.1D I gave you a direct answer lol.
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@hba explanation!!
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@Areesha.1D I do not want to tell you the easy method,it is better you figure it out with your rusty mind :D
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Anyways her you go @Areesha.1D To find inclusive numbers take the differences and add 1 204 - 103 = 101; 101 + 1 = 102
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that was the explantion on the site lol thanks
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@Areesha.1D You couldn't find it,So i found it :D
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I am an intelligent Google search user :D
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@hba hahaha well said xD
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204-103 only give u the page in between. Therefore, when you are doing minus, the left out the 1st page. Thus, u have to add 1 more page. If the question is starting at the bottom page of 103,
then u do not need to add 1 =D
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There you GO!
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sorry, i am not an A level student @@
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Your question is ready. Sign up for free to start getting answers.
is replying to Can someone tell me what button the professor is hitting...
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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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Please use this identifier to cite or link to this item: http://hdl.handle.net/2005/1107
Title: Transport In Quasi-One-Dimensional Quantum Systems
Authors: Agarwal, Amit Kumar
Advisors: Sen, Diptiman
High Energy Physics
Quantum Theory
Transport Theory (Physics)
Quantum Wires
Luttinger Liquid Wires
Keywords: Mesoscopic Physics
Quantum Wires - Conductance
Mesoscopic Quantum Wires
Floquet Scattering Theory
Quantum Charge Pumping
Quantum Systems
Luttinger Wires
Submitted Mar-2009
Series/ G23051
Report no.:
This thesis reports our work on transport related problems in mesoscopic physics using analytical as well as numerical techniques. Some of the problems we studied are: effect of
interactions and static impurities on the conductance of a ballistic quantum wire[1], aspects of quantum charge pumping [2, 3, 4], DC and AC conductivity of a (dissipative) quantum Hall
(edge) line junctions[5, 6], and junctions of three or more Luttinger liquid (LL)quantum wires[7]. This thesis begins with an introductory chapter which gives a brief glimpse of the
underlying physical systems and the ideas and techniques used in our studies. In most of the problems we will look at the physical effects caused by e-e interactions and static
scattering processes. In the second chapter we study the effects of a static impurity and interactions on the conductance of a 1D-quantum wire numerically. We use the non-equilibrium
Green’s function (NEGF) formalism along with a self-consistent Hartree-Fock approximation to numerically study the effects of a single impurity and interactions between the electrons
(with and without spin) on the conductance of a quantum wire [1]. We study the variation of the conductance with the wire length, temperature and the strength of the impurity and
electron-electron interactions. We find our numerical results to be in agreement with the results obtained from the weak interaction RG analysis. We also discover that bound states
produce large density deviations at short distances and have an appreciable effect on the conductance which is not captured by the renormalization group analysis. In the third chapter we
use the equations of motion (EOM) for the density matrix and Floquet scattering theory to study different aspects of charge pumping of non-interacting electrons in a one-dimensional
system. We study the effects of the pumping frequency, amplitude, band filling and finite bias on the charge pumped per cycle, and the spectra of the charge and energy currents in the
leads[2]. The EOM method works for all values of parameters, and gives the complete time-dependences of the current and charge at any site of the system. In particular we study a system
with oscillating impurities at several sites and our results agree with Floquet and adiabatic theory where these are applicable, and provides support for a mechanism proposed elsewhere
for charge pumping by a traveling potential wave in such systems. For non-adiabatic and strong pumping, the charge and energy currents are found to have a marked asymmetry between the
two leads, and pumping can work even against a substantial bias. We also study one-parameter charge pumping in a system where an oscillating potential is applied at one site while a
static potential is applied in a different region [3]. Using Floquet scattering theory, we calculate the current up to second order in the oscillation amplitude and exactly in the
Abstract: oscillation frequency. For low frequency, the charge pumped per cycle is proportional to the frequency and therefore vanishes in the adiabatic limit. If the static potential has a bound
state, we find that such a state has a significant effect on the pumped charge if the oscillating potential can excite the bound state into the continuum states or vice versa. In the
fourth chapter we study the current produced in a Tomonaga-Luttinger liquid (TLL) by an applied bias and by weak, point-like impurity potentials which are oscillating in time[4]. We use
bosonization to perturbatively calculate the current up to second order in the impurity potentials. In the regime of small bias and low pumping frequency, both the DC and AC components
of the current have power law dependences on the bias and pumping frequencies with an exponent 2K−1 for spinless electrons, where Kis the interaction parameter. For K<1/2, the current
grows large for special values of the bias. For non-interacting electrons with K= 1, our results agree with those obtained using Floquet scattering theory for Dirac fermions. We also
discuss the cases of extended impurities and of spin-1/2 electrons. In chapter five, we present a microscopic model for a line junction formed by counter or co-propagating single mode
quantum Halledges corresponding to different filling factors and calculate the DC [5] and AC[6] conductivity of the system in the diffusive transport regime. The ends of the line
junction can be described by two possible current splitting matrices which are dictated by the conditions of both lack of dissipation and the existence of chiral commutation relations
between the outgoing bosonic fields. Tunneling between the two edges of the line junction then leads to a microscopic understanding of a phenomenological description of line junctions
introduced by Wen. The effect of density-density interactions between the two edges is considered exactly, and renormalization group (RG) ideas are used to study how the tunneling
parameter changes with the length scale. The RG analysis leads to a power law variation of the conductance of the line junction with the temperature (or other energy scales) and the line
junction may exhibit metallic or insulating phase depending on the strength of the interactions. Our results can be tested in bent quantum Hall systems fabricated recently. In chapter
six, we study a junction of several Luttinger Liquid (LL) wires. We use bosonization with delayed evaluation of boundary conditions for our study. We first study the fixed points of the
system and discuss RG flow of various fixed points under switching of different ‘tunneling’ operators at the junction. Then We study the DC conductivity, AC conductivity and noise due to
tunneling operators at the junction (perturbative).We also study the tunneling density of states of a junction of three Tomonaga-Luttinger liquid quantum wires[7]. and find an anomalous
enhancement in the TDOS for certain fixed points even with repulsive e-e interactions.
URI: http://hdl.handle.net/2005/1107
Appears in Centre for High Energy Physics (cts)
Items in etd@IISc are protected by copyright, with all rights reserved, unless otherwise indicated.
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Working With Lists
Excel is ideal for working with lists of data. This page describes a number of worksheet formulas for extracting basic information about a list -- sums, minimums, maximums, and so on. It also
show you how to restrict these functions to only certain values in the list, such as values greater than zero, or values between two other values.
Also, there are a few formulas for transposing a list (changing a row into a column) and reversing the order of a list, or both. Finally there are a few miscellaneous formulas that didn't see
to fit on other pages.
All of the formulas on this page are array formulas, so you must press Ctrl+Shift+Enter rather than just Enter when you enter the formula, and whenever you edit it later.
SUM, MIN, MAX, and AVERAGE
In this section, we will refer to a list of data named List. For more information about naming a range, click here.
You are already familiar with the basic SUM, MIN, MAX, and AVERAGE formulas, so they won't be explained here. Refer to the on-line help manual for basic information about these functions.
Instead, for each function, use the function to use only the following values from List:
• Values Not Equal To Zero
• Values Greater Than Zero
• Values Between An Upper And Lower Limit
For example, you can find the average of those values in the List that are greater than zero, or between 10 and 20.
In all of these formulas, we will be using a range called List, which refers to the data show in the figure to the left.
Of course, your actual data will be different, and your List can be of any length. The values in this example will clearly show the different results when we're calculating the results using
only non-zero values, or values between an upper and a lower limit.
Non-Zero Values
You can restrict the values used by the functions to only those values (positive and negative) that are not equal to zero. Remember, all these formulas are array formulas.
returns the sum of non-zero values, or 11. Of course, summing non-zero values is never really necessary (since the 0 values don't contribute to the sum in any case), but the formula is
illustrative nonetheless.
returns the average of non-zero values, or 1.1. If we used the AVERAGE function on the entire list, the result would be 0.73, because the 5 zero values would be included. In this function, they
are not included.
returns the minimum of non-zero values, or -4.
returns the maximum of non-zero values, or 6.
Positive Values
You can restrict the values used by the functions to only positive values -- those that are greater than zero. Remember, all these formulas are array formulas.
returns the sum of positive values, or 21. The negative numbers are not included in the sum.
returns the average of positive values, or 3.5.
returns the minimum of positive values, or 1.
returns the maximum of positive values, or 6.
Values In An Interval
You can restrict the values used by the functions to only values between two other values. In these formulas, we will use two more named cells -- LLim which is the lower limit, and ULim, which
is the upper limit. When we say "between" two numbers, we mean inclusively between. In other words, the numbers 2, 3, and 4 are all between 2 and 4. If you want to have the formulas work with
exclusively between intervals (only the number 3 is between 2 and 4), change the <= and >= operators to < and >, respectively. Remember, all these formulas are array formulas. In the examples,
assume that LLim contains 2 and ULim contains 5.
returns the sum of values between 2 and 5, or 14.
returns the average of values between 2 and 5, or 3.5.
returns the minimum of values between 2 and 5, or 2.
returns the maximum of values between 2 and 5, or 5.
Reversing The Order Of A List
You can use array formulas to reverse the order of a list.
Reversing the Order Of A Column
Still using the List from the previous section, we can reverse its order, which will give us the list shown below. This formula uses a named range called RevList, which refers to the range of
the new, reversed list.
Enter this array formula in the first cell of the RevList range, then select the entire RevList range, and use Fill Down from the Edit menu to copy this formula down into all the cells in the
RevList range. Remember this is an array formula.
This formula will work only for reversing the order of a column. To reverse the order of a row, see the next section.
Transposing A Row Into A Column
These functions will transpose a row list, RowList, into a column. To keep the data in the original order, create a named range called TXList referring to the cells in a column that is to
contain the transposed data. Then use the following array formula:
Fill this formula into the entire TXList range.
To reverse the order of the value in RowList, create a named range called TList, and use the following array formula:
Fill this formula into the entire TList range.
Examples of data in RowList, TXList, and TList are shown in the figure below.
Other List Formulas
Most Or Least Common Value In A List
The following formula will return the most frequent value in a range:
where Rng is the range of the list.
The following formula will return the least frequent value in a range:
where Rng is the range of the list.
In both of the these formulas, if there are two different values, each of which occurs the minimum or maximum number of times, the formula will return the one which appears earlier in the list.
Additional information about working with data in lists can be found on the following pages:
Duplicate And Unique Items In Lists
Eliminating Blank Cells From Lists
More advanced techniques of transposing ranges are described on the following page:
Array To Columns
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Second-order total differential
July 18th 2010, 06:17 PM #1
Jun 2010
Second-order total differential
Consider the production function:
y(x1, x2) = x2/1 + x2/2
* i put the "/" signs just to show the first number above the 2nd, its not actually divided*
can someone try to explain how I go about finding the second-order total differential to find if it is concave or convex?
I'd appreciate it if you can explain as simply as possible(in words not math)
thanks a lot in advance!
Do you mean $y(x_1, y_1)= x_1^2+ x_2^2$? If so, a better notation would be y(x_1, x_2)= x_1^2+ x_2^2.
The total differential is $dy= 2x_1 dx_1+ 2x_2 dx_2$ and the "second order" total differential is 0 because there is no term "mixing" the two variables.
If the problem had been $y(x_1, x_2)= x_1^2+ 3x_1x_2+ x_2^2$, then the total differential would have been $dy= (2x_1+ 3x_2)dx_1+ (3x_1+ 2x_2)dx_2$ and the second order differential would have
been $3 dx_1dx_2$
July 19th 2010, 03:53 AM #2
MHF Contributor
Apr 2005
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Random subset of elements from multimap grouped by keys
up vote 0 down vote favorite
I have a list consisting of car brands ID's and associated car models, e.g.:
1 Corolla
1 Yaris
1 Matrix
2 Cherokee
2 Liberty
3 CR-V
3 CR-Z
3 Element
3 Civic
3 Pilot
where 1=Toyota, 2=Jeep, and 3=Honda. Note that the cardinality of car models per car brand differs.
I would like to retrieve random car models per car-brand. The number of cars to be retrieved per car brand depends upon the total number of associated models and an input float parameter:
'nPercentage'. (The 'nPercentage' parameter is the same for all different car-brands). For example, if nPercentage=0.5, a possible random output would be:
1 Corolla
1 Matrix
2 Liberty
3 CR-Z
3 Civic
3 Pilot
I'm currently working with a multimap class, since the keys can be duplicated. So far, I'm able to find non-duplicated keys and count the number of associated elements. Could anyone shed some light
on how to retrieve the random car models per car-brand? Below, the code that I have so far.
//The variable 'm_mapDatasetMapping' is of type: multimap<int, string>
multimap< int, string >::size_type countPerKey;
const int *pLastKey = NULL;
multimap<int,string>::const_iterator it=m_mapDatasetMapping.begin();
// looking for non-duplicated keys.
for( ; it!=m_mapDatasetMapping.end(); it++){
if( (pLastKey!=NULL) && (*pLastKey==it->first) ){
pLastKey = &(it->first);
// count the number of values associated to the given key.
countPerKey = m_mapDatasetMapping.count(*pLastKey);
/* Select 'x' random elements associated with the key '*pLastKey'.
The number of random elements to be extracted
is a percentage of the total number of values per key, i.e.:
x = nPercentage * countPerKey
c++ stl multimap random-sample
add comment
1 Answer
active oldest votes
Perhaps the simplest approach to follow would be to copy all values for the given key out into a new container, say, a vector, random_shuffle it, and resize() it to reduce its size
to x:
int x = nPercentage * countPerKey;
auto range = m_mapDatasetMapping.equal_range(*pLastKey);
up vote 0 down std::vector<std::string> values;
vote accepted for(auto i = range.first; i != range.second; ++i)
std::random_shuffle(values.begin(), values.end());
thanks! It works like a charm! I just needed to change the type of the variable range, i.e. from auto to pair<multimap<int,string>::iterator,multimap<int,string>::iterator> and
also the type for variable i, from auto to multimap<int,string>::iterator. – Javier Mar 23 '11 at 14:14
@Javier for NON-selected, did you mean how to make a new vector initialized with values from values[x] to values.end()? vector<string> other_values(values.begin()+x, values.end
());, before values.resize(x) of course. And yes, auto types save a lot of typing on modern compilers! – Cubbi Mar 23 '11 at 14:14
yes! that's the way I did it. – Javier Mar 23 '11 at 14:16
add comment
Not the answer you're looking for? Browse other questions tagged c++ stl multimap random-sample or ask your own question.
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Velocity and acceleration vectors for a particle at different times (picture prob)
October 4th 2011, 06:35 PM #1
Oct 2011
Velocity and acceleration vectors for a particle at different times (picture prob)
I have several pictures similar to the one provided in that attached word doc. The question is:
The picture shows the velocity and acceleration vectors for a particle at different times. (The particle is moving in the plane of the paper.) Tell whether aT (a subscript T) is negative, zero,
or positive. Also, tell whether aN (a subscript N) is zero or positive. If aN is positive tell whether the velocity vector is rotating clockwise or counterclockwise.
I am not sure how to picture this to determine whats going on in this picture. Or If I should arrange these vectors to see them differently. Any help would be greatly appreciated. Thanks
Re: Velocity and acceleration vectors for a particle at different times (picture prob
I have several pictures similar to the one provided in that attached word doc. The question is:
The picture shows the velocity and acceleration vectors for a particle at different times. (The particle is moving in the plane of the paper.) Tell whether aT (a subscript T) is negative, zero,
or positive. Also, tell whether aN (a subscript N) is zero or positive. If aN is positive tell whether the velocity vector is rotating clockwise or counterclockwise.
I am not sure how to picture this to determine whats going on in this picture. Or If I should arrange these vectors to see them differently. Any help would be greatly appreciated. Thanks
all that I see in your vector sketch is a single velocity vector pointing "east" , and a single acceleration vector pointing roughly "northeast".
what are the meanings of the subscripts T and N ... Tangential and Normal?
Seems to me that there is more to this question that has been left out ... maybe?
Re: Velocity and acceleration vectors for a particle at different times (picture prob
Skeeter, it is Tangential and Normal. Other than that this is all the info I was provided. I apologize for not stating this initially.
Re: Velocity and acceleration vectors for a particle at different times (picture prob
Re: Velocity and acceleration vectors for a particle at different times (picture prob
I am going to assume just a general curve.
Re: Velocity and acceleration vectors for a particle at different times (picture prob
well, you can't tell much if the two given vectors are for different times. had the two vectors been for the same time, then you could sketch the tangential and normal components of acceleration.
October 5th 2011, 04:44 AM #2
October 5th 2011, 08:12 AM #3
Oct 2011
October 5th 2011, 10:24 AM #4
October 5th 2011, 10:40 AM #5
Oct 2011
October 5th 2011, 01:54 PM #6
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Math Forum: Class2Class Project Suggestions
Project Suggestions
C2C Home || Search projects || List projects || Submit projects || Mailing List || About C2C
To find ideas for math collaborations, Mel Sprouse suggests browsing these Internet projects:
Students from all over the globe calculate the circumference of the Earth by measuring the shadow of the sun, then using that measurement in a simple equation. Students and classes are
encouraged to share their results with others by completing an experiment report form. You may view the experimental results of participating students and classes via the WWW. The March 1997,
September 1997, and March 1998, experimental results are also available. See also Eratosthenes' Experiment.
In 1995 there were 7,164 participants reaching a total of 19,546.22 feet! Your students might want to put the information into a database, creating fields to compare the lengths that came in
from each site.
Have you ever noticed that many of the terms in geometry sound like they are the names of some strange animals? Terms like Scalene, Dodecagon, and Heptagon? In this project, students use
their imaginations to create imaginary animals and describe them, while they learn about geometry at the same time!
TEAMS looked for the farthest flying airplanes throughout the nation - the best from each class; classes sent in written directions (diagrams were acceptable) for folding this airplane so it
could be "built" for the "fly-off." An origami exercise.
Students from across the United States took a package of M&M's, predicted the color they thought there would be the most of, graphed them by color, sent their data to TEAMS Distance
Learning... Then they ate their M&M's.
Schools participate by measuring the angle of the sun at Local Noontime during a specific week of the year and submit their results through the Noon Project submission form. If you are
interested in participating in Noon '99, contact Kenneth Cole at kcole@waymark.net.
Learning about mathematical patterns in numbers and in poetry. This activity involves students in writing two forms of poetry - diamante and cinquain - that involve mathematical word
Students work in groups and record the numbers that come up on each of the 4 spinners. As one student spins, the other records the results on the tally sheet. As students record their
results, they begin to see a relationship between the design of the spinner and the outcomes that result from the spins. The purpose of this project is to collect as large a sample as
possible so a comparison can be made between the classroom data and the national data.
A multi-grade, integrated thematic project that begins with 6 activities varying from math to research skills to persuasive writing and focuses on the stock market, with participants in
grades 5 through 12 and from California to New York to Florida.
For more projects, see the Houghton Mifflin Mathematics Project Center
Other suggestions? Send them to Suzanne Alejandre.
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language to decimal to hexadecimal conversion
Disregaring the origion of the date, is this not at least a rarity???
On this site: http://www.greatdreams.com/2012.htm
They say that the Mayan calendar ends on December 21, 2012. Furthermore, they say it happens at 11:11am.
I took Dec. 21/2012 and made it 12/21/2012. If you put 11:11 in front of that, you get 11:11/12/21/2012. I Changed that to 111112212012. I then converted 111111212012 to HEX, and the result is:
Is it just me or does it seem awfully coincidental that the Hexadecimal outcome is almost the exact day of the origional decimal number?
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1. Which of the following is considered a hybrid organizational - JustAnswer
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Write and Solve Proportions by Using Equivalent Rates
4.3: Write and Solve Proportions by Using Equivalent Rates
Have you ever faced a reading challenge? Take a look at this dilemma.
Jamie looked at the clock. She still had 20 minutes to read in the reading challenge.
Jamie looked up at the clock when the bell rang. He had read 15 pages in 20 minutes. If she had read 15 pages in 20 minutes, how many pages would she read in 40 minutes?
Do you know how to figure this out? You can use equivalent rates to help you solve this problem. Pay attention to this Concept and you will understand the solution by the end of it.
A ratio is a comparison between two quantities or numbers. Ratios can be written in fraction form, with a colon or by using the word “to”.
Sometimes, you will compare ratios. Sometimes one ratio will be greater than another, and other times they can be equal or equivalent. When you have two equal ratios, you have a proportion.
A proportion is created when two ratios are equal, or we can say that two equal ratios form a proportion.
We can write a proportion when we know that two ratios are equivalent.
$1 : 2 = 2 : 4$
These two ratios are equivalent. We can say that the two ratios form a proportion.
Do these two ratios form a proportion?
To figure this out, we have to figure out if the two ratios are equivalent. If they are, then we know that they form a proportion. If not, then they don’t. To figure this out, we can simplify the
$& \qquad \ \frac{3}{4} \ is \ in \ simplest \ form.\\& 4 : 24 \ can \ be \ written \ as \ \frac{4}{24} = \frac{1}{6}\\& \qquad \qquad \quad \ \ \frac{3}{4} e \frac{1}{6}$
These ratios do not form a proportion.
These proportions were given to you. You can also write your own proportions.
To write a proportion, set two equivalent fractions equal to each other, using the information in the problem.
If you know the ratio of girls to boys in a class is 2 : 3, and you know there are 24 boys in the class, you can write a proportion in order to find the number of girls in the class.
The most important thing to remember when writing a proportion is to keep the units the same in both ratios.
$\frac{girls}{boys}: \frac{2}{3} = \frac{x}{24}$
You know the fractions are equivalent because each shows the ratio of girls to boys in the class. The first fraction shows the known ratio of girls to boys. The second ratio shows the known number of
boys in the class, 24, and uses a variable to stand for the unknown number of girls.
Now let's use equivalent rates to solve a proportion.
The ratio of teachers to students in a certain school is 2 : 25. If there are 400 students in the eighth-grade class, how many teachers are there?
First set up a proportion. The problem gives a ratio of teachers to students, so set up two equivalent ratios comparing teachers to students.
$\frac{teachers}{students} = \frac{8^{th} \ grade \ teachers}{8^{th} \ grade \ students}$
You can see that we are comparing teachers to students in both ratios. The first one shows the ration in the whole school and the second ratio represents the eighth grade ratios. Next, we fill in the
given information.
$\frac{2}{25} = \frac{x}{400}$
Now use what you know about equivalent ratios to solve the proportion.
Look at the denominators. You know that the first fraction, when the numerator and denominator are multiplied by some number, will equal the second fraction. What number, when multiplied by 25, will
equal 400? Since $25 \times 16 = 400$$x$
$2 \times 16 = 32$$x = 32$
There are 32 teachers in the eighth-grade class.
Note: You can check that your answer is correct by making sure that the two ratios are equivalent.
$\frac{32}{400} = \frac{8}{100} = \frac{2}{25}$
Since the second ratio simplifies to the first, the ratios are equivalent.
Yes it is. Just remember that what you do to the numerator you have to do the denominator. If you can remember to always apply this rule, then you will create equal ratios.
Solve each proportion by using equal ratios.
Example A
$\frac{3}{4} = \frac{6}{x}$
Solution: $x = 8$
Example B
$\frac{9}{50} = \frac{x}{100}$
Solution: $x = 18$
Example C
$\frac{3.5}{7} = \frac{x}{35}$
Solution: $x = 17.5$
Now let's go back to the dilemma from the beginning of the Concept.
Jamie read 15 pages in 20 minutes. She wonders how many pages she will read in 40 minutes.
We can set up a proportion and look for an equivalent rate.
$\frac{pages}{minutes} = \frac{15}{20} = \frac{x}{40}$
Here is our proportion.
Next, we can look at the relationship between the denominators.
$20 \times 2 = 40$
What we do to the bottom, we can do to the top. This will give us the equivalent rate.
$15 \times 2 = 30$
At this rate, Jamie will read 30 pages in 40 minutes.
a comparison between two quantities. Ratios can be written in fraction form, with a colon or by using the word “to”.
means equal.
formed when two ratios are equivalent. We compare two ratios, they are equal and so they form a proportion.
Guided Practice
Here is one for you to try on your own.
Write a proportion to describe this situation.
The proportion of red paper to white paper in a stack is 2 to 7. If there are 32 red pieces of paper, what proportion could be used to find the number of pieces of white paper?
Write the known ratio of red paper to white paper as the first fraction: $\frac{2}{7}$
Now write the second ratio, using $x$
$\frac{\text{red paper}}{\text{white paper}} : \frac{2}{7} = \frac{32}{x}$
The proportion $\frac{2}{7} = \frac{32}{x}$
Video Review
Directions: Solve each proportion using equal ratios.
1. $\frac{3}{4} = \frac{x}{12}$
2. $\frac{5}{6} = \frac{x}{12}$
3. $\frac{4}{7} = \frac{8}{y}$
4. $\frac{2}{3} = \frac{12}{y}$
5. $\frac{4}{5} = \frac{44}{y}$
6. $\frac{12}{13} = \frac{x}{26}$
7. $\frac{9}{10} = \frac{81}{y}$
8. $\frac{6}{7} = \frac{18}{y}$
9. $\frac{7}{8} = \frac{x}{56}$
10. $\frac{12}{14} = \frac{36}{x}$
11. $\frac{6}{4} = \frac{x}{12}$
12. $\frac{12}{14} = \frac{24}{x}$
13. $\frac{13}{14} = \frac{x}{42}$
14. $\frac{1.5}{4} = \frac{x}{8}$
15. $\frac{3.5}{4.5} = \frac{x}{9}$
16. $\frac{9}{14} = \frac{108}{x}$
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