Wikipedia:Reference desk/Archives/Mathematics/2008 March 20

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March 20

What is the summation formula for a negative power?

What's the solution for this?: $\sum _{i=1}^{n}i^{-1}$

That's the Harmonic series (or, at least, a partial sum of it). The article should tell you quite a lot. If you have any specific questions after reading it, ask away! --Tango (talk) 00:15, 20 March 2008 (UTC)

It is called a harmonic number. Bo Jacoby (talk) 01:03, 20 March 2008 (UTC).

Proving Trigonometric Identities

Normally I'm okay at this, but I've been stuck on this one for a while. Prove the identity: (1+Tan^2x)/tan^2x=Csc^2x

I've got the basic trig identities down, but I just can't get this problem... Any help?

• Try simplifying the left side by separating the numerator... things just might cancel out relatively nicely. :) --Kinu ^t/_c 01:56, 20 March 2008 (UTC)

${\frac {1+\tan ^{2}x}{\tan ^{2}x}}$

Use the identity for tan(x)

$={\frac {1+{\frac {\sin ^{2}x}{\cos ^{2}x}}}{\frac {\sin ^{2}x}{\cos ^{2}x}}}$

Simplify.

$=\left(1+{\frac {\sin ^{2}x}{\cos ^{2}x}}\right)\left({\frac {\cos ^{2}x}{\sin ^{2}x}}\right)$

Change the fraction.

$=\left({\frac {\sin ^{2}x+\cos ^{2}x}{\cos ^{2}x}}\right)\left({\frac {\cos ^{2}x}{\sin ^{2}x}}\right)$

Use the identity sin^2(x)+cos^2(x)=1

$=\left({\frac {1}{\cos ^{2}x}}\right)\left({\frac {\cos ^{2}x}{\sin ^{2}x}}\right)$

Cancel.

$={\frac {1}{\sin ^{2}x}}$

Use the identity for csc(x).

$=\csc ^{2}x$

--wj32 ^t/c 05:57, 20 March 2008 (UTC)

Using Kinu's clue, here's another proof:

${\frac {1+\tan ^{2}x}{\tan ^{2}x}}$

Simplify the fraction.

$={\frac {1}{\tan ^{2}x}}+1$

Since cot(x)=1/tan(x):

$=\cot ^{2}x+1$

Because $\csc x={\sqrt {\cot ^{2}x+1}}$ (see List of trigonometric identities):

$=\csc ^{2}x$

--wj32 ^t/c 06:08, 20 March 2008 (UTC)

• Yes, that.... but I was somewhat hesitant to put the complete solution up, since this is the "help" desk, not the "answer" desk... --Kinu ^t/_c 16:34, 20 March 2008 (UTC)

Fundamental group and de Rham cohomology

Is there a simple relationship between the fundamental group of a manifold and its first de Rham cohomology group? The concepts seem very similar, like the de Rham cohomology group is a "continuous version" of the fundamental group, but I don't know how to quantify that. The Hurewicz theorem sounds very close, but I think I'm missing something (probably because I don't yet understand any kind of cohomology other than de Rham, and that only vaguely). —Keenan Pepper 05:44, 20 March 2008 (UTC)

The answer is indeed given by the Hurewicz theorem, as well as any suitable comparison isomorphism. What you get is that the first de Rham cohomology of X is dual to the abelianization of π₁(X) with scalars extended from Z to R (if you take de Rham cohomology with real coefficients). In other words, consider the mapping:

${\begin{aligned}H_{\textrm {dR}}^{1}(X,\mathbb {R} )\times \pi _{1}(X,x_{0})&\to \mathbb {R} \\(\omega ,\gamma )&\mapsto \langle \omega |\gamma \rangle =\int _{\gamma }\omega \end{aligned}}$

taking a 1-form and a loop to the integral of the 1-form along the loop. This is well-defined (it only depends on the cohomology class of $\omega$ and on the homotopy class of $\gamma$), bilinear in the obvious sense, and the only loops such that $\langle \cdot |\gamma \rangle =0$ are the commutators in the fundamental group. It becomes a non-degenerate pairing on $H_{\textrm {dR}}^{1}(X,\mathbb {R} )\times \pi _{1}(X,x_{0})^{\textrm {ab}}$. Bikasuishin (talk) 18:12, 20 March 2008 (UTC)

Wow, it seems so simple now! One more question: What's the simplest example you can think of of a manifold with a nonabelian fundamental group? I want something a little more concrete to think about. —Keenan Pepper 18:32, 20 March 2008 (UTC)

Think of the twice-punctured plane. The fundamental group is the (non-abelian) free group on two generators (loops around the two punctures), whereas the cohomology is just R². An example of a loop that is non-trivial in the fundamental group but homologically trivial is the "figure 8" loop around the two punctures. Bikasuishin (talk) 23:38, 20 March 2008 (UTC)

Isn't the figure eight (once round each hole) non-trivial in homology? For a homologically trivial curve, you want to go round each puncture twice in opposite directions, corresponding to the word aba^-1b^-1. Algebraist 11:08, 21 March 2008 (UTC)

Yes, sorry, you're correct of course. Bikasuishin (talk) 12:09, 21 March 2008 (UTC)

Discrete cosine transform

I'm trying to understand the "why" of DCT. (It's part of a larger goal to understand what's going on inside JPEG/MPEG.) In the Discrete cosine transform article these formulas are given for DCT-II and DCT-III:

$X_{k}=\sum _{n=0}^{N-1}x_{n}\cos \left[{\frac {\pi }{N}}\left(n+{\frac {1}{2}}\right)k\right]\quad \quad k=0,\dots ,N-1.$

$X_{k}={\frac {1}{2}}x_{0}+\sum _{n=1}^{N-1}x_{n}\cos \left[{\frac {\pi }{N}}n\left(k+{\frac {1}{2}}\right)\right]\quad \quad k=0,\dots ,N-1.$

It's then stated, but not proven, that they are inverses (with a constant multiplier). Every other source I've found does the same thing. It seems like they expect the inverse relationship to be so obvious it doesn't need proving, but to me it's quite surprising. Intuitively I expect a function full of cosines to be inverted by a function full of arccosines.

Of course I can numerically verify the results (which shows that I didn't misread the formulas) and I even proved N=1, N=2, and N=3 by expanding the sums and substituting the result of one formula into the other. But that gets really tedious by the time you get to N=4, and I'm not seeing any way to generalize it.

Can someone provide the missing proof? --tcsetattr (talk / contribs) 08:03, 20 March 2008 (UTC)

It should be relatively easy to find proofs of the Discrete Fourier transform. The DCT is just the real part of it: apply the operator Re to both sides of the identity, and use Re e^ix = cos x.  --Lambiam 11:29, 20 March 2008 (UTC)

That's the kind of hand-waving that has left the question unresolved after a long time of trying to find the answer. I need details. And doing it without a detour through complex numbers would be a bonus. Here's what happens when I try to confirm what you said:

The Discrete Fourier transform article says

$X_{k}=\sum _{n=0}^{N-1}x_{n}e^{-{\frac {2\pi i}{N}}kn}\quad \quad k=0,\dots ,N-1$

The real part is:

$X_{k}=\sum _{n=0}^{N-1}x_{n}\cos -{\frac {2\pi }{N}}kn\quad \quad k=0,\dots ,N-1$

Because cos(-x) = cos(x) the minus sign can be dropped:

$X_{k}=\sum _{n=0}^{N-1}x_{n}\cos {\frac {2\pi }{N}}kn\quad \quad k=0,\dots ,N-1$

How is that equivalent to the DCT-II formula above? It has 2pi instead of pi, and n instead of n+1/2. They aren't the same thing at all! I've barely got started on this and I'm lost already. Is there no one who will actually write everything out so it can be understood? --tcsetattr (talk / contribs) 23:33, 20 March 2008 (UTC)

You shouldn't expect arccosines in this case because you're not taking the cosine of the input, but rather multiplying it by the cosine of something else. So there might be secants in the inverse, but not arccosines.

Let me add the right normalization factor and rename some variables so that I can substitute one equation in the other:

$X_{n}=\sum _{j=0}^{N-1}x_{j}\cos \left[{\frac {\pi }{N}}\left(j+{\frac {1}{2}}\right)n\right]$

$x_{k}={\frac {2}{N}}\left[{\frac {1}{2}}X_{0}+\sum _{n=1}^{N-1}X_{n}\cos \left[{\frac {\pi }{N}}\left(k+{\frac {1}{2}}\right)n\right]\right]$

Now substituting the first in the second I get:

$x_{k}={\frac {2}{N}}\left[{\frac {1}{2}}\sum _{j=0}^{N-1}x_{j}+\sum _{n=1}^{N-1}\sum _{j=0}^{N-1}x_{j}\cos \left[{\frac {\pi }{N}}\left(j+{\frac {1}{2}}\right)n\right]\cos \left[{\frac {\pi }{N}}\left(k+{\frac {1}{2}}\right)n\right]\right]$

$x_{k}={\frac {1}{N}}\sum _{j=0}^{N-1}x_{j}\left(1+\sum _{n=1}^{N-1}2\cos \left[{\frac {\pi }{N}}\left(j+{\frac {1}{2}}\right)n\right]\cos \left[{\frac {\pi }{N}}\left(k+{\frac {1}{2}}\right)n\right]\right)$

$x_{k}={\frac {1}{N}}\sum _{j=0}^{N-1}x_{j}\left(1+\sum _{n=1}^{N-1}\cos \left[{\frac {\pi }{N}}(j+k+1)n\right]+\sum _{n=1}^{N-1}\cos \left[{\frac {\pi }{N}}(j-k)n\right]\right)$

So it all hinges on whether the parenthesized part, $1+\sum _{n=1}^{N-1}\cos \left[{\frac {\pi }{N}}(j+k+1)n\right]+\sum _{n=1}^{N-1}\cos \left[{\frac {\pi }{N}}(j-k)n\right]$, equals N when j=k and 0 otherwise. And it does, but I admit to not understanding in a deep way why it does. It follows from the weird identity

$\sum _{q=0}^{N-1}\cos \left({\frac {\pi pq}{N}}\right)={\begin{cases}N&p=0\\1&p{\mbox{ odd}}\\0&{\mbox{otherwise}}\end{cases}}$

which I can prove, but again without all that much insight. Let me know if you want the details. This does seem much less elegant than the complex case, which is straightforward to prove and easy to understand geometrically. I don't see any obvious way to adapt a proof of the complex case to the real case. -- BenRG (talk) 02:13, 21 March 2008 (UTC)

I have to disagree with complex numbers being easy to understand geometrically. They double the number of spatial dimensions you have to visualize. That's only making things harder. Thanks for the other ideas though: splitting the product of cosines into a sum was the big step I wasn't coming up with on my own. The last identity is new to me too. --tcsetattr (talk / contribs) 00:10, 22 March 2008 (UTC)

As to that last identity, we can prove it for odd integers p straightforward-ly while remaining in the real numbers... fix any integer q, and let q' = N - q, x = $(\pi pq)/N$, and x' = $-(\pi pq')/N$. Then

$x-x'={\frac {\pi pq}{N}}+{\frac {\pi pq'}{N}}={\frac {\pi p(q+q')}{N}}={\frac {\pi pN}{N}}=\pi p.$

Since x and x' differ by an odd multiple of π, therefore $cos(x)=-cos(x')$. So we have

$\cos \left({\frac {\pi pq}{N}}\right)+\cos \left({\frac {\pi pq'}{N}}\right)=\cos(x)+\cos(-x')=\cos(x)+\cos(x')=0.$

This shows that, for odd p, we can pair up the qs from q = 1 through q = N - 1, by pairing q with N - q, so that the sum of the terms corresponding to q and N - q cancel out. So,

$\sum _{q=0}^{N-1}\cos \left({\frac {\pi pq}{N}}\right)=1+\sum _{q=1}^{N-1}\cos \left({\frac {\pi pq}{N}}\right)=1.$

For even p, I didn't see any easy way in real numbers, but the identity is clear if we use complex numbers. Let p = 2m, and let $\zeta =\exp \left({\frac {2\pi i}{N}}\right)$ be a primitive Nth root of unity; we find that

$\sum _{q=0}^{N-1}\exp \left({\frac {2\pi imq}{N}}\right)=\sum _{q=0}^{N-1}\zeta ^{mq}={\begin{cases}N&{\mbox{if }}N\mid m\\0&{\mbox{otherwise}}\end{cases}}$

In retrospect, these don't provide as much insight as I had hoped. Eric. 86.152.32.69 (talk) 01:36, 24 March 2008 (UTC)

That was confusing. When you're pairing up terms, what about the one in the middle? When N is even, there are an odd number of terms for q=1 through q=N-1 so you've got one left over. Wait, I've got the answer for that: the term in the middle is q=N/2, for which ${\frac {\pi pq}{N}}={\frac {\pi p}{2}}$ which has a cosine of 0 because p is odd. The middle term of the sum is a 0 so it doesn't affect the result.

As for the part with the complex numbers, I have no idea what it means or how it relates to anything else. --tcsetattr (talk / contribs) 22:48, 24 March 2008 (UTC)

Another breakthrough: I went to the Root of unity article to see if that could tell me anything, and ended up reading this (Geometric progression#Complex numbers): "The summation formula for geometric series remains valid even when the common ratio is a complex number." That was my blind spot all along! I never recognized any of the complex sums as geometric series so I had no idea how people were magically extracting values from them. Using this new tip, that $\sum _{k=0}^{n}r^{k}={\frac {1-r^{n+1}}{1-r}}$ even if r is complex, I can finally do something with $\sum _{q=0}^{N-1}\zeta ^{mq}$ other than stare at it and wait for inspiration to strike.

$\sum _{q=0}^{N-1}\zeta ^{mq}=\sum _{q=0}^{N-1}(\zeta ^{m})^{q}={\frac {1-(\zeta ^{m})^{N}}{1-\zeta ^{m}}}$ in which the numerator is 0 because $\zeta ^{N}=1$. When m is a multiple of N the denominator is also 0 so this method fails but in that case every term of the sum is 1 so the total is N. And that is what I mean by "details".

I see we're now editing an "archive" so if there's any more trouble I'll start a new section. --tcsetattr (talk / contribs) 00:19, 25 March 2008 (UTC)

Incidence matrix

Hi. I was playing this video on the ocw.mit.edu site related to graphs, networks and incidence matrices and the teacher said that $A^{T}y=0$ is one of the most fundamental relationship in applied maths (A is the incidence matrix of a network). He then went on to show if y is the current vector then $A^{T}y=0$ is Kirchoff's circuit law. What I want to know is in what other sense is this relationship important and why should we consider it fundamental. Thanks.--Shahab (talk) 09:50, 20 March 2008 (UTC)

I don't know about applied math, but it closely resembles homology, which is fundamental in abstract math. I think, though, that he meant to imply a connection to the topics he mentioned at the beginning, like fluid flow in a hydraulic system and balancing of forces in a weight-bearing structure. Black Carrot (talk) 09:42, 22 March 2008 (UTC)

Distribution Theory

If Xi ~ N(0,1), derive the probability density function of Xi^2. Write down the probability density function of SigmaXi^2 ..Jacques

The probability that Xi^2 is within an interval dx of some number x is equal to the probability that Xi is within an interval ${\frac {d{\sqrt {x}}}{dx}}dx$ of either ${\sqrt {x}}$, or $-{\sqrt {x}}$ (because those are the only solutions of Xi^2 = x). Use that together with some calculus, and come back if you're stuck and having trouble with something specific. —Keenan Pepper 18:41, 20 March 2008 (UTC)

By "SigmaXi^2" (is that Σ X_i² or Σ Ξ²?), do you mean a sum of independent and identically-distributed random variables? How many? For a fixed number, you get a p.d.f. that is continuous but piecewise polynomial (just like a piecewise linear function, but replace linear function by polynomial function). As the number of rv's in the sum increases, so does the number of intervals into which the domain needs to be decomposed.  --Lambiam 18:09, 22 March 2008 (UTC)

Chemistry question!

If I have 12 coordinate sphere packing ie rhombic dodecahedral of spheres A

then there are 14 vertices : 6 (4line) vertices and 8 (3line vertices)

the (4line) vertices are surrounded by 6 A spheres (??)

the (3line) vertices are surrounded by 4 A spheres.

So if I place (smaller) spheres B at all the 14 vertices the formula is A1 B (6vertices/6coordinate)+(8verticles/4coodinate) = A1 B3

Is this correct?83.100.183.180 (talk) 14:57, 20 March 2008 (UTC)

duals (in a platonic sense) extended..

the vertices in rhombic dodecahedral packing appear to be the centres of tetrahedron or octahedron (taking the centres of surrounding) rhombic dodecahedra as vertices - is there a name for this type of (octahedron/tetrahedron) / (rhombic dodecahedron) dual relationship

article

Is there a space filling solid article - under another name, if not should there be one?

more

FCC is equivalent to connected rhombic dodecahedra, HCP packing forms a different shape "if the spheres of hexagonal close packing are expanded, they form a second irregular dodecahedron consisting of six rhombi and six trapezoids " from http://mathworld.wolfram.com/HexagonalClosePacking.html does this shape have no generic name?83.100.183.180 (talk) 15:13, 20 March 2008 (UTC) Sorry about the big list of little questions.83.100.183.180 (talk) 15:15, 20 March 2008 (UTC)

Our article on space filling tesellations is called Honeycomb (geometry). We also have an article on the rhombic dodecahedral honeycomb. The equivalent cell for hexagonal close packing is the trapezo-rhombic dodecahedron. Gandalf61 (talk) 15:35, 20 March 2008 (UTC)

Good thanks. Had no idea (or had forgotten) about 'honeycombs'83.100.183.180 (talk) 16:54, 20 March 2008 (UTC)