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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)

March 21

explicit rule

Can anyone give a step-by-step description on writing a rule in equation form of the following pattern?

2,6,24,120

• Well, without doing all of the work for you, check out factorial. Let us know if you're still stuck after that. :) --Kinu ^t/_c 00:40, 21 March 2008 (UTC)

lol I can't believe I didn't get that (the factorial part)! Still, I would appreciate it greatly if you can solve it for me as my brain is working a little more slowly than usual today.

ah (n+1)! is the answer...is there any other way to express this?

No, not really, that's the answer. You can write "(n+1)n(n-1)(n-2) ... 3.2.1", but that's just defining the factorial notation. --Tango (talk) 01:10, 21 March 2008 (UTC)

recursive rule

Speaking of rules and equations, can anyone help me write a recursive rule for the following equation?

2,5,14,41,122

Calculate the differences between the terms (5-2, 14-5, etc) and see if you spot a pattern. --Tango (talk) 01:11, 21 March 2008 (UTC)

yes i see a pattern...a_n=a_n-1+3^n-1

Yes. You can also search integer sequences in OEIS. "2,5,14,41,122" gives a non-recursive formula (and more complicated rules leading to other continuations). PrimeHunter (talk) 05:05, 21 March 2008 (UTC)

Yes, but that's not really very useful. What's useful is knowing how to determine the pattern yourself, having a website do your homework for you is pointless. --Tango (talk) 15:42, 21 March 2008 (UTC)

One issue with this question is that there are quite many patterns that have these first five numbers; how do you decide which is `the' pattern? SmaleDuffin (talk) 17:23, 21 March 2008 (UTC)

It's an issue with all such questions - I think to be sure you need to know the context. It's presumably a homework question and the class will have been studying certain types of sequence, so it's going to one of those types. --Tango (talk) 18:13, 21 March 2008 (UTC)

Occam's razor. If there is not enough information to determine an object uniquely, the one with the simplest representation should be chosen. ${\frac {3^{n}+1}{2}}$ is simpler than $n^{4}-4n^{3}+8n^{2}-2n+2\;\!$. -- Meni Rosenfeld (talk) 00:17, 24 March 2008 (UTC)

Only if it's absolutely essential that you come to a conclusion at all. If your life, livelihood, or peace of mind depends on deciding quickly, that's a good rule of thumb. If not, it's a good idea to slow down a bit, and accept that you just don't have enough information yet. Black Carrot (talk) 00:39, 25 March 2008 (UTC)

Exponential of an exponential

Can an equation of the form y = k(a + bc^x)^x be simplified? NeonMerlin 04:57, 21 March 2008 (UTC)

You could expand the xth power binomial, but I would say that's unsimplifying the above equation. A math-wiki (talk) 08:37, 21 March 2008 (UTC)

Peano Curve

Does anyone know where I could get hold of Peano's original paper on the curve, or a translation if it's in another language? Also, a question about the curve itself: Did Peano himself come up with these pictures of the curve, or did they come later when someone decided they wanted a concrete example? Black Carrot (talk) 05:50, 21 March 2008 (UTC)

Peano curve cites 'G. Peano, Sur une courbe, qui remplit toute une aire plane. Math. Ann 36 (1890), 157-160.', in case you missed that. The original text is available here. The article appears to contain a concrete example, but no pictures (at least in the pages I can get to work). I don't know about translations, but at least the article's in French, not a language of his own invention. Algebraist 11:04, 21 March 2008 (UTC)

Thanks! And apparently I can read it, which surprised the heck out of me. :) Black Carrot (talk) 07:37, 22 March 2008 (UTC)

Mathematical French is dead easy (possibly because we stole most of our terminology from them). Algebraist 13:49, 22 March 2008 (UTC)

Happy Easter!

Another pagan festival approaches and all those wishing to receive a 'happy easter' message get one.

My question is - what simple formulas give a good 'egg' shape. Links please, and it's just for curiousities sake, so don't exert yourselfs too much.87.102.16.238 (talk) 12:08, 21 March 2008 (UTC)

This doesn’t quite answer the question, but both Ellipsoid and Oval have a section on the egg shape. GromXXVII (talk) 12:35, 21 March 2008 (UTC)

Look here for all your oval curve needs. — Kieff | Talk 13:02, 21 March 2008 (UTC)

Super!87.102.16.238 (talk) 13:53, 21 March 2008 (UTC)

simple question

can we finde,or do we have afunction where,limit[f(x)] approches to infinity when ,x,approches to zero,while,f(0)=known value?it sounds no big deal but i think,yes there is such afunction exists.Husseinshimaljasimdini (talk) 12:23, 21 March 2008 (UTC)

Define f by f(0)=0, f(x)=1/x elsewhere. Bo Jacoby (talk) 12:35, 21 March 2008 (UTC).

Well 1/|x| elsewhere I believe, else you’d have to talk about one handed limits. GromXXVII (talk) 12:39, 21 March 2008 (UTC)

I wonder something like the differential of the dirac delta function?87.102.16.238 (talk) 15:17, 21 March 2008 (UTC)

as such take a look at Dirac_delta#Representations_of_the_delta_function the differential of the limits (n to infinity) of these functions.

eg y=d/dx( lim (e^{-x^2n} ))

when x=0 y=(1 error) =0, when x=0+delta x y=big

?87.102.16.238 (talk) 15:21, 21 March 2008 (UTC)

Also maybe consider the differential of the function y=e^{1/x^2} .At x=0 dy/dx = 0?87.102.16.238 (talk) 18:06, 21 March 2008 (UTC)

e^{1/x^2} isn't defined at 0, so can't have a derivative there. Even if you define y(0)=k though, for some k, it won't be continuous (because the limit at that point is positive infinity), so there'll be no derivative. Black Carrot (talk) 06:05, 22 March 2008 (UTC).Well, i was not expecting all these answers.As amatter

of fact i was thinking of the numbers of sectors we can get by dividing acircle.Now if the function ,F(θ)=2pi\θ,describes numbers of sectors we get by dividing acircle where θ is the centeral angle,then how many sectors we will get if,θ=0?is it zero?Husseinshimaljasimdini (talk) 11:46, 22 March 2008 (UTC)

I'd say you get infinitely many. The angle is always positive, and the limit as it tends to 0 from above is positive infinity. --Tango (talk) 15:41, 22 March 2008 (UTC)

Of course if the sector angle is zero - the answer (in the real world) is 'doesn't work' ie impossible - hence the answer 'infinitely many'. It's possible that you are confusing the answer 'zero' with 'not possible' - in this case the two aren't the same. There are of course examples of things were the answer = 0 means - 'not possible' or 'doesn't happen'.87.102.16.238 (talk) 15:58, 22 March 2008 (UTC)

Algebra with probabilistic events

I'm a college freshmen, and I'm no further into my mathematical education than vector calculus, but recently I picked up a book on probability theory. Even if it's mostly out of my league, it sure looks interesting. Anyway, I've encountered a statement that seems to me intuitively correct but algebraically wrong. Surely I'm mistaken; the question is, how?

Since I don't know to what degree the notation that the author uses is standard, I'll briefly recapitulate it. An event $A_{i}$ is a set of one or more of the elementary events within a sample space. $\Omega$ is the set of all elementary events, and $O$ is the empty set. $A_{1}+A_{2}$ is defined as the union of $A_{1}$ and $A_{2}$; $A_{1}A_{2}$ is the intersection of $A_{1}$ and $A_{2}$. As usual, addition and multiplication are associative and commutative, and addition is distributive over multiplication. Lastly, ${\overline {A_{1}}}$, the complement of $A_{1}$, is $\Omega -A_{1}$.

The statement that puzzles me is:

${\overline {A_{1}+A_{2}}}={\overline {A_{1}}}\;{\overline {A_{2}}}$.

That makes sense to me insofar as the event that neither $A_{1}$ nor $A_{2}$ happens is the same as the combined event that $A_{1}$ doesn't happen and $A_{2}$ doesn't happen. But:

${\begin{aligned}\Omega -A_{1}-A_{2}&=(\Omega -A_{1})(\Omega -A_{2})\\\Omega -A_{1}-A_{2}&=\Omega ^{2}-\Omega A_{1}-\Omega A_{2}+A_{1}A_{2}\\\Omega -A_{1}-A_{2}&=\Omega -A_{1}-A_{2}+A_{1}A_{2}\\O&=A_{1}A_{2}\\\end{aligned}}$

which clearly isn't necessarily true. —Saric (Talk) 17:26, 21 March 2008 (UTC)

Your deduction of the second line from the first is flawed, as is your deduction of the last from the third. You have been misled by the author's non-standard and ghastly notation into thinking that the things he has called +, - and * behave as you expect things with those names to behave: in the first case, you've assumed - interacts properly with * (it doesn't), and in the second you've used a cancellation law which doesn't hold in this setting. I recommend finding a better book. Algebraist 17:43, 21 March 2008 (UTC)

Oh, and FYI the statement in question is (one of) De Morgan's laws. Algebraist 17:47, 21 March 2008 (UTC)

(ec)I think the problem you're having is because set difference (ie. complement) and union aren't inverses of each other. $A=(A\cup B)-B$ is only true if $A\cap B=\emptyset$. You need to be careful when using "+" and "-" that you don't assume it works in the way as addition and subtraction of numbers. --Tango (talk) 17:51, 21 March 2008 (UTC)

The usual notation for this set subtraction, known as relative complement, is not A − B but A \ B. Using that notation helps one to avoid the pit you fell in. The intersection of two such complements satisfies the identity

(P \ A) ∩ (Q \ B) = ((P ∩ Q) \ A) \ B.

There is no such thing here as "minus times minus equals plus".  --Lambiam 22:22, 21 March 2008 (UTC)

So these set operations don't work quite like the numerical equivalents, after all. I think I can guess why the author of the book liked this notation— so long as $A_{1}A_{2}=O$, the probability of $A_{1}+A_{2}$ is the probability of $A_{1}$ plus the probability of $A_{2}$, and likewise for multiplication. But if the analogy isn't perfect, I guess the notation is more harmful than helpful. Yeah, I'll probably want to get a different book. Thanks for helping me out here.

(Me, "a college freshmen", indeed! Obviously my graduation from high school hasn't deterred the typo fairy.) —Saric (Talk) 18:55, 22 March 2008 (UTC)

Likewise for multiplication? If by "$O$" in "$A_{1}A_{2}=O$" you mean the empty set (so that A₁ and A₂ are mutually exclusive), it is not true in general that this implies P(A₁A₂) = P(A₁)P(A₂). For example, take A₁ = heads and A₂ = tails in a fair coin flip.  --Lambiam 21:03, 22 March 2008 (UTC)

Yeah, you need mutually exclusive and independent. If you're only dealing with such simple probability, then the notation works fine, but it rapidly falls apart once you do anything real. --Tango (talk) 22:28, 23 March 2008 (UTC)

Solving "almost" superincreasing knapsacks problem

I know that it is really easy to solve the knapsack problem if all elements are super increasing, however, what if all of the elements were "almost super increasing". By almost super increasing, I mean that instead of having each successively larger integer being greater than the sum of all of the smaller integers. Each successively larger integer will be larger than, say, one-half of the sum all of the smaller integers. For example a valid set of "~1/2 almost super increasing integers" would be {8,11,17,26,38,58,86,130,195,292,438,657,985}; Would this instance of knapsack be as difficult to solve as the general knapsack problem? Or maybe somewhere between the difficulty of general knapsack and super increasing knapsack? 24.250.129.216 (talk) 19:16, 21 March 2008 (UTC)mathnoob

We don't know for sure whether the general knapsack problem is difficult to solve, so a proof that this is "somewhere between" easy and hard, but neither of the two, would imply a proof of P ≠ NP.

My gut feeling is that this is still NP-hard, but I don't readily see an applicable reduction like that of Exact cover to the knapsack problem.  --Lambiam 22:39, 21 March 2008 (UTC)

The only difference between a super increasing sequence and this sequence is that in this sequence each term T(n) is larger than the sum of all terms from T(0) to T(n-3); while a super increasing sequence has each term T(n) larger than the sum of all terms T(0) to T(n-1). A super increasing sequence has a polynomial time solution.

March 22

Uncountable Sets of Measure Zero

To a beginner (in measure theory that is) it seems that a set having (Lebesgue) measure zero is equivalent to that set being denumerable (countable or finite) but I just found out that the Cantor ternary set is both uncountable and it has measure zero. My question is, does anyone know of any other uncountable sets that have measure zero? What about an example of a real set which is not measurable at all? Thanks!
A Real Kaiser (talk) 03:37, 22 March 2008 (UTC)

For uncountable sets of measure 0, a bunch of copies of the Cantor set would work, of another set similarly constructed. As for non-measurable sets, see Vitali set. -GTBacchus^(talk) 03:48, 22 March 2008 (UTC)

The Cantor set already gives you lots of uncountable sets of measure zero: since Lebesgue measure is complete, any uncountable subset of the Cantor set will do. There are 2^c of them, which is already far more than, say, the number of Borel sets. Algebraist 12:53, 22 March 2008 (UTC)

What about another example of an uncountable set with measure zero? Something "different" than the Cantor ternary set? In addition, is there another example of a nonmeasurable real set besides a Vitali set, something that can perhaps be described explicitly (you know something like "all irrational numbers between one and two", which I know is measurable)?
A Real Kaiser (talk) 19:27, 23 March 2008 (UTC)

So for the first question, you might be interested in something like this: almost all real numbers are normal, but certainly uncountably many real numbers are not normal. In fact, to be more explicit, consider the real numbers whose decimal expansions follow some slight predictable bias; say, 11% of their digits are 7 (instead of 10%), and only 9% of their digits are 8. It's "obvious" that there are just as many numbers like that as there are normal numbers, but because almost all numbers are normal, the set of all numbers with this bias (none of which are normal) must have measure zero.

For the second question, it depends a little on what you mean by "described explicitly", but the short answer is that for any explicit description you're likely to think of, unless you know quite a bit of set theory, the set of reals so described is measurable. --Trovatore (talk) 22:09, 23 March 2008 (UTC)

Cool, thanks everyone!A Real Kaiser (talk) 01:03, 24 March 2008 (UTC)

a question for econs nerds

cost function c(y) = y^3 - 2y^2 + 6y + 6.

need to derive the supply curve.

figured out that the shutdown point is where p = 5 (AVC = y^2 - 2y + 6, MC = 3y^2 - 4y +6; when y = 1, the two are both equal to 5, and because profit maximization occurs when MC = p, then the smallest price is 5).

so when p < 5, y = 0; when p > 5, p = 3y^2 - 4y + 6. I have to write this in terms of y, however - wtf I do this?

also, the firm makes positive profits when the price is higher than the average cost ofc, right? AlmostCrimes (talk) 18:18, 22 March 2008 (UTC)

Well to get $p=3y^{2}-4y+6$ in terms of y, just subtract p from both sides and then treat p as part of the constant term c, in ax^2+bx+c. So to be a little more clear once you've moved p to the right side, a=3 b=-4 and c=(6-p). A math-wiki (talk) 21:43, 22 March 2008 (UTC)

After you follow AMW's suggestion, you may find two solutions (functions y(p)) using the quadratic formula; of these two solutions, you must choose the increasing one (satisfying the second-order condition of profit maximization). And yes, AlmostCrimes, you are right in that positive profits happen when price is higher than the average (total) cost, for then the average income (price) is greater than the average cost, which implies that total income is greater than the total expenditures. Pallida  Mors 06:27, 25 March 2008 (UTC)

Error in the Total order article?

I'm not all that proficient at mathematics, but I do enjoy to be dazzled by the shiny notation of set theory, so I often find myself reading articles in that subject and saying "Ohh, pretty!". I happened just now to come across the Total order article, and saw something which I'm fairly certain is an error, but I thought I'd check with you guys first. Under the "Example" heading, this is stated:

"The set of real numbers ordered by the usual less than (<) or greater than (>) relations is totally ordered, hence also the subsets of natural numbers, integers, and rational numbers."

This is not true, right? Both "less than" and "greater than" fails the totality criterion since they are not reflexive (statements like "42 < 42" are most certainly not true). Shouldn't that sentence be:

"The set of real numbers ordered by the usual less than or equals operator (≤) or the greater than or equals operator (≥) relations is totally ordered, hence also the subsets of natural numbers, integers, and rational numbers."

Right? 83.250.207.154 (talk) 19:35, 22 March 2008 (UTC)

A (weak) total order and its associated strict total order are interchangeable for most purposes. The orders don't satisfy the same axioms, but each is obtained from the other in a straightforward way, by adding or removing the diagonal. That is, the strict total order is obtained from the weak total order by the rule x < y if and only if x ≤ y and x ≠ y, while the weak total order is obtained from the strict total order by the rule x ≤ y if and only if x < y or x = y. Michael Slone (talk) 20:11, 22 March 2008 (UTC)

True, but strictly speaking, I think the anon is right - it should say ≤. The ordering is the same, but to be consistent with the way its defined at the top of the article, the weak order should be given. --Tango (talk) 20:26, 22 March 2008 (UTC)

weak order? Taemyr (talk) 01:10, 24 March 2008 (UTC)

Sorry, weak *total* order. I think someone needs to come up with better names for orderings... this is far too confusing. --Tango (talk) 02:12, 24 March 2008 (UTC)

Chicken, fox, and grain problem

Here is the traditional problem, both question and answer. Question: A farmer is standing on one bank of a river, with a fox, a chicken, and a bag of grain. He needs to get to the other side of the river, taking the fox, the chicken, and the grain with him. However, the boat used to cross the river is only large enough to carry the farmer and one of the things he needs to take with him, so he will need to make several trips in order to get everything across. In addition, he cannot leave the fox unattended with the chicken, or else the fox will eat the chicken. And he cannot leave the chicken unattended with the grain, or else the chicken will eat the grain. The fox is not particularly partial to grain, and may be left alone with it. How can he get everything across the river without anything being eaten? Answer: The farmer takes the chicken across first, leaving the fox and grain together on the other side. He returns and gets the fox, but when he deposits the fox on the other side, he takes the chicken BACK across, so that the fox and chicken aren't left alone together. He drops the chicken off back on the other side, picks up the grain, and takes it across to deposit with the fox. Finally, he returns to retrieve the chicken and takes it to the other side. At no time were the fox and chicken left alone together, nor were the chicken and grain. At no time was more than one of them in the boat with the man simultaneously.

My question #1 is: Is there any generalized mathematical formula / set-up / algorithm that essentially explains and solves this problem? And, thus, could be extended into more complex similar problems (e.g., the farmer now has 7 items -- or 24 or whatever -- to carry over with various restrictions upon the 7 -- or 24 -- items, etc.) ...?

My question #2 is: (related to question #1 above) Generally speaking, how would one know that the given scenario is indeed "able" or "impossible" to be executed? So, for example, in the traditional problem above ... how would I know at the onset that this can in fact be effectuated (or not) without going through umpteen "trial and error" attempts first? Thanks. (Joseph A. Spadaro (talk) 21:48, 22 March 2008 (UTC))

I can't answer your question, but that won't stop me posting a link to the definitive discussion of the original problem. AndrewWTaylor (talk) 21:55, 22 March 2008 (UTC)

I'm not sure about question 1. For question 2, I would expect the easiest way to confirm it's possible is to find a solution. If it's impossible, there is probably a way to prove that without testing every possible combination. Whether there's a general method, or you just have to do it on a case-by-case basis, I don't know. --Tango (talk) 21:57, 22 March 2008 (UTC)

Actually, on second thoughts, I can partially answer a restricted version of question 1. If the restrictions on the items are all of the same form as in the original question (ie. items x and y cannot be left alone together), then it will be impossible in all but a small number of cases. Unless there is an item which is included in all the restrictions (in the original problem, the chicken), then it's impossible because there is no available first move. If there are more complicated restrictions (x and y can only be together if z is there too, for example), then it gets more complicated. --Tango (talk) 22:02, 22 March 2008 (UTC)

I suppose you can express the constraints in Linear temporal logic. Maybe this article is of interest to you? Phaunt (talk) 10:00, 25 March 2008 (UTC)

I assumed we must have an article on this puzzle, but I can't find it (anyone ?). Anyway, this type of puzzle can be solved algorithmically by constructing a graph of allowed states and transitions between them.

• We start in state {F,f,c,g | } with Farmer, fox, chicken and grain all on one side of the river. We want to get to state { | F,f,c,g}.
• Initially we have sixteen states (each object can be on either side of the river, so number of states is 2^4) - but once we disallow states such as {F,g | f,c} (fox eats chicken) and {c,g | F,f} (chicken eats grain) and {F | f,c,g} (fox eats well-fed chicken) we are left with 10 allowed states. Each of these states is a vertex in the graph.
• Transitions take F and possibly one other object from one side of the river to the other, so we draw an edge from {F,f,c,g | } to {f,g | F,c}, and an edge from {f,g | F,c} to {F,f,g | c} etc. In this case each transition is reversible, so we have an undirected graph. If transitions were not reversible, we could draw a directed graph.
• Once we have drawn the graph, we look for a route from our initial state {F,f,c,g | } to our final state { | F,f,c,g}.

In the F,f,c,g case the routes (there are two of them) are obvious. For larger problems, some sort of search algorithm could be used to find a solution. Gandalf61 (talk) 10:37, 25 March 2008 (UTC)

That approach works, but because the number of nodes is equal to $k^{n}$ (where n is the number of items, and k is the number of states each item may be in), it is only tractable for relatively moderate values of these parameters (especially n). Phaunt (talk) 10:46, 25 March 2008 (UTC)

Finally found our article on this puzzle, in a different guise - it is the fox, goose and bag of beans puzzle. Gandalf61 (talk) 10:37, 27 March 2008 (UTC)

The Riemann Zeta Function

The Wikipedia article on the Riemann Zeta Function lists Specific Values. One of them says that the Riemann Zeta Function of zero is -1/2. Can anyone explain why this is? I know it should probably be obvious, but I just can't seem to figure it out. Digger3000 (talk) 23:07, 22 March 2008 (UTC)

This can be easily obtained from the infinite product expansions listed under Hadamard product. Now why these hold...  --Lambiam 00:35, 23 March 2008 (UTC)

Sorry, but that all looks like Greek to me. (And some of it really is.) If it's not too much trouble, could someone dumb it down a little (a lot) for me? Digger3000 (talk) 01:34, 23 March 2008 (UTC)

I guess we're talking about the formula:

$\zeta (s)={\frac {e^{(\log(2\pi )-1-\gamma /2)s}}{2(s-1)\Gamma (1+s/2)}}\prod _{\rho }\left(1-{\frac {s}{\rho }}\right)e^{s/\rho }\!$,

where the product is over the non-trivial zeros ρ of ζ and the letter γ again denotes the Euler-Mascheroni constant. (from Riemann zeta function#Hadamard product)

This formula can be useful for calculating the zeta function at certain values where the usual infinite series representation fails to converge. The idea is to just plug 0 in for s in that equation...

• Now, in the numerator of the big fraction, there's a long expression in the exponent that's multiplied by s, so that whole exponent becomes 0, and we have e⁰=1, so the numerator equals 1.
• The denominator is going to equal 2 times -1 times Г(1), where that last function is the gamma function. It's a matter of calculating an integral to evaluate Γ(1)=1. That gives us a denominator of -2.
• Now there's the thing over to the right of the fraction, which is a product - the capital pi (Π) notation for products is a lot like the capital sigma (Σ) notation for sums, but we're multiplying instead of adding. This particular product will consist of one factor for each non-trivial zero of the zeta function, ρ. When s=0, it doesn't really matter what any of the values are for ρ, because each term in that product will just equal 1.

Thus, when we plug in 0 for s, we obtain $\zeta (0)={\frac {1}{-2}}\cdot 1=-{\frac {1}{2}}.$ Does that help? -GTBacchus^(talk) 02:43, 23 March 2008 (UTC)

Well, technically, that shows that it either equals -1/2 or 0 - if it's 0, the product is undetermined, so anything can happen. Presumably there is an alternative way of showing it's non-zero. --Tango (talk) 02:48, 23 March 2008 (UTC)

How's that? Aren't all the factors in the product equal to 1? -GTBacchus^(talk) 02:57, 23 March 2008 (UTC)

If 0 is a zero of the function, then one of the factors will have a 0/0 in it. (Although, I note you said the product is over the non-trivial zeros, I'm not quite sure what trivial means in this context.) --Tango (talk) 03:16, 23 March 2008 (UTC)

After looking it up, trivial means negative even integers, so 0 presumably wouldn't count as trivial. --Tango (talk) 03:17, 23 March 2008 (UTC)

Ah, I see. Thank you. -GTBacchus^(talk) 03:20, 23 March 2008 (UTC)

Well, I still have a lot to learn about things like integrals and such, but that does make it a lot clearer. Thank you. Digger3000 (talk) 03:26, 23 March 2008 (UTC)

March 23

Null day?

Isn't there anything mathematically interesting about this date? --hydnjo talk 15:43, 24 March 2008 (UTC)

Well, if you multiply the day by the month, you get 69. Who would want to do math on such a day?! :D  ARTYOM  16:53, 24 March 2008 (UTC)

It's the birthday of Laplace, Jurij Vega, Emmy Noether, and Ludvig Faddeev. —Bkell (talk) 17:01, 24 March 2008 (UTC)

March 24

Checking number of citations

Newbie question: how can I verify how many citations a math paper has? For example:

http://www.math.princeton.edu/~annals/issues/2004/Sept2004/Agrawal.pdf

Thanks in advance. Mdob | Talk 00:13, 24 March 2008 (UTC)

Google Scholar ([1]) says it has 352 cites. I'm not sure how accurate that is. --Tango (talk) 00:18, 24 March 2008 (UTC)

Try MathSciNet -mattbuck (Talk) 00:24, 24 March 2008 (UTC)

MathSciNet finds 27 (+5 reviews) - that's a lot less than 352... --Tango (talk) 00:36, 24 March 2008 (UTC)

Thanks, Tango (and mattbuck). I already knew of MathSciNet, but completely ignored this feature in Google Scholar. Yeah, 300+ was the number I was specting, since the paper was published in the Annals, wich (together with Journal of the AMS, Inventionnes, etc) is a journal with a very high impact index.

Lebesgue measure of a subset of [0, 1) invariant under translation by Q

Let $\Omega =[0,1)=\mathbb {R} /\mathbb {Z}$; that is, addition is taken modulo Z (so that 0.7 + 0.6 = 0.3). We say that $S\subset \Omega$ is Q-invariant if it is invariant under translation by Q, i.e., for every $q\in \mathbb {Q} ,S+q=S$. My question is, if S is Q-invariant, does that imply that the Lebesgue measure of S is either 0 or 1? I've proven a few basic lemmas (for example, that $m(S\cap [a,b))=(b-a)\cdot m(S)$), and I've gained a better understanding of Lebesgue measure for it, but neither I nor the number theory professor I asked this to got anywhere (the question arose from proving the Thue-Siegel-Roth theorem, if I remember). Eric. 86.152.32.69 (talk) 02:29, 24 March 2008 (UTC)

What if you start with a non-measurable set, and take the union of that set with all Q translations of it? Would that set still be non-measurable? -GTBacchus^(talk) 03:12, 24 March 2008 (UTC)

I thought about that too. Starting with a Vitali set would give you the entire real line, so that's no good. Maybe there's another non-measurable set which works, though. —Bkell (talk) 05:19, 24 March 2008 (UTC)

If S is Lebesgue measurable, then yes. Look at the Lebesgue density theorem. If S has positive measure, then for any ε>0, there's a point x and an arbitrarily small open interval containing x such that the measure of S on the interval is at least 1−ε times the measure of the interval. Now shift the interval around by rationals to show that the measure of S is at least, say, 1-2ε. Since ε was arbitrary, the measure of S is one. This general phenomenon is called ergodicity. --Trovatore (talk) 06:16, 24 March 2008 (UTC)

Thanks for your help, everyone. I see now that I had conjectured the Lebesgue density theorem... it's reassuring to know that it's true, although that puts me no closer to finding a proof of it (the Lebesgue density theorem). Hmmm... is the Lebesgue density theorem easy enough that I should try to figure out a proof for myself, or does it take, say, 20 pages of obscure higher math to prove? Eric. 81.157.254.39 (talk) 15:15, 24 March 2008 (UTC)

Honestly I'm not sure I've ever been through the proof of the full result, but I doubt it's hard. I came across a slightly weaker version of it on my own once before I recall hearing about the standard result. Think about this: Suppose there were a set so "evenly smeared out" on the real line that it picked up, say, 75% of the measure of every interval. Its measure on [0,1) is 0.75, its measure on [0.3,0.7) is 0.3, etc. Well, apply the definition of measure, and cover this set with a countable collection of disjoint open intervals, such that the sum of the lengths of the intervals is, say, 0.8. But now for each of the intervals in the cover, the measure of the set in that interval, by assumption, is 0.75 times the length of the interval, so what's the total measure of the set?

Fiddle with this sort of idea long enough and I expect you can prove the full theorem, though I personally have never bothered. --Trovatore (talk) 18:21, 24 March 2008 (UTC)

The question of whether there's a non-measurable set closed under rational translation is also interesting. We want a set A, closed under translation by rationals, such that A has neither measure zero nor measure one. Any measure zero set is a subset of a measure zero G-delta set; any measure one set contains a measure one F-sigma set. So enumerate the measure-zero G-deltas and measure-one F-sigmas together in the smallest possible order type. We'll build up two sets A and B by transfinite recursion. If the next set to be considered is a measure-zero G-delta, we want to refute the claim that A is contained in it. So pick a real number x that's not in the measure-zero G-delta, nor in A or B as constructed so far, and throw x and all its rational translates into A. If the next set is a measure-one F-sigma, we want to refute the claim that A contains this set. So pick a real number x that is in the F-sigma set, but not in A or B so far, and throw it and all its rational translates into B (thus forever preventing them from getting into A). How do we know we can always find such an x? Well, at any point, we've done less than $2^{\aleph _{0}}$ iterations, and at each iteration we've thrown only countably many points into A or B, so both A and B have less than full cardinality. However each measure-one set has full cardinality, so there's a point in it that's not in A or B, and each measure-zero set has a complement with full cardinality, so there's a point not in it and also not in A or B. When we're done, we've guaranteed that the final version of A has neither measure zero nor measure one, and by construction it's closed under translation by rationals --Trovatore (talk) 07:04, 24 March 2008 (UTC)

Interesting. So, picking the numbers x at each step requires the axiom of choice, right? (As I understand it, it is not possible to prove the existence of a non-measurable set from within ZF.) I'll admit to not really understanding transfinite recursion or ordinals, but I think I've managed to wrap my head around your argument. Actually, this is the first time I've seen transfinite-anything make any sense. Eric. 81.157.254.39 (talk) 15:15, 24 March 2008 (UTC)

Right, picking the numbers requires AC, and so does picking the wellordered enumeration of the measure-one F-sigmas and measure-zero G-deltas. --Trovatore (talk) 22:26, 24 March 2008 (UTC)

Non-roots of unity having modulus 1

While refreshing some elementary complex number theory, I realise that I have a problem in proving that a complex number is *not* a root of unity. For instance, it is quite obvious that ${\frac {1}{4}}(\pi +i{\sqrt {16-\pi ^{2}}})$ is not a root of unity, even if it has modulus 1: too much irrational stuff around. But how may I prove it formally? Thanks!

Is there any reason it couldn't be an irrational root of unity? I mean, we define the square roots of unity as 1^(1/2). Surely there's no reason we couldn't define the pi-th root of unity as 1^(1/pi)? -mattbuck (Talk) 15:40, 24 March 2008 (UTC)

Yes, sorry, I meant "is not an n-th root of unity for some natural n".

Irrational roots aren't very useful, since they aren't going to be cyclic under multiplication. You can define them, but pretty much whenever you want to use a root of unity, it's necessary that it be an integer root, otherwise you end up with an infinite number of them. --Tango (talk) 16:31, 24 March 2008 (UTC)

Let's say you wanted to prove it's not an nth root of unity for some given n. Using standard trig identities, in particular the angle sum formulas and the pythagorean theorem, you can get an algebraic formula for the sine of a whole angle in terms of the sine of one nth that angle, and likewise for cosine. Since the sine and cosine of 2pi are whole numbers, the sine and cosine of 2pi/n must be algebraic. Their quotient is then algebraic, but that would have to equal the tangent of 2pi/n, which equals by assumption sqrt(16-pi^2)/pi. There's no algebraic formula for pi, so that's impossible. Black Carrot (talk) 17:53, 24 March 2008 (UTC)

Polarity associated with a conic, defined by Steiner construction?

Hello,

Steiner defined a conic in a projective plane as "the locus of the points of intersection of corresponding lines of two homographic pencils with distinct vertices". This is a very nice definition... but in most cases conics come with (orthogonal) polarities (mapping points to the polar line). Is there a nice way to visualize this polarity?, just using the two homographic pencils? I've been thinking about this for a while now, but I haven't found an answer yet. Does anyone know more? ThanksEvilbu (talk) 21:16, 24 March 2008 (UTC)

Parabola vs Hyperbola

what is difference b/w a parabola and hyperbola? I know the geometric differnce. But can someone explain the difference in terms of mathematics?[2]

Have you read parabola and hyperbola? Those articles should help. I'd say the biggest difference is that a hyperbola has asymptotes, a parabola doesn't. --Tango (talk) 16:51, 24 March 2008 (UTC)

Parabolas have an eccentricity of 1; whereas hyperbolas have eccentricities greater than 1. (Ellipses have eccentricity less than 1.) In the equation for a conic section in Cartesian coordinates, parabolas have $B^{2}-4AC=0$; whereas hyperbolas have $B^{2}-4AC>0$. (Ellipses have $B^{2}-4AC<0$.) --Spoon! (talk) 22:43, 24 March 2008 (UTC)

For one thing, a parabola has no asymptotes. But the question is phrased in a vague way, so beyond that maybe the best thing is to suggest you read the standard articles and then see if you have more specific questions. The "geometric difference" is a "difference in terms of mathematics. Michael Hardy (talk) 23:07, 24 March 2008 (UTC)

The parabola closes in the infinity (it touches the ideal line of the plane), the hyperbola continues smoothly on the other side in the infinity (it intersects the ideal line). – b_jonas 19:23, 25 March 2008 (UTC)

rational expressions

Hello, Take any number (except for 1). Square that number and then subtract one. Divide by one less than the original number. Now subtract your original number. Did you reach 1 for a answer? How does the number game work? (hint Redo the number game using a varible instead of an actual number and rewiten the problem as one rational expression). How did the number game use the skill of simplifying ration expression? state whether you number game uses the skill of simplfying ration expression. Please help I don't get what it is saying!

Go through the process using "x" instead of a number and you'll get a rational expression, you can then simplify that (hint: use difference of 2 squares), and you'll find it equals 1. --Tango (talk) 17:52, 24 March 2008 (UTC)

I'm still not getting it. Could you please give me some example so I can understand a little better thank you

(Please reply in this section, rather than creating a new section for each message, thanks!) I'll start you off: You start with x, you then square it and get x². You then subtract one and get x² - 1 and so on. Once you get to the end of the process, simplify the expression and you'll find it simplifies to simply "1". --Tango (talk) 19:23, 24 March 2008 (UTC)

rational expression

How is doing operations (adding, subtracting, multiplying, and dividing) with rational expressions similiar to or different from doing operations with fractions? Can understanding how to work one kind of problem help understand how to work another type. When might you use this skill in real life?

Fractions are just a special case of rational expressions, so the operations are basically the same. Fractions are easier, since there is less to do, but the basic process is identical. I'm not sure you'll use rational expressions directly in real life, but they come up a lot in various areas of maths, and those areas have real world applications. --Tango (talk) 17:55, 24 March 2008 (UTC)

March 25

whaddya call this functional?

If you take a power series and each term c_n×xⁿ is replaced by {c_n÷n!} × xⁿ, then affine functions are fixed and each subspace <xⁿ> has eigenvalue (1/n!). It would send every series convergent on a disk of nonzero radius to a series convergent everywhere. I would think, though knowing zilch about functional analysis, that it would be called a uniformly continuous functional. Does this already have a name and can anyone tell me more or refer me? Thanks, Rich (talk) 05:05, 25 March 2008 (UTC)

Well, I don't know about you latter questions, in that post, but your series is the Maclaurin series for $e^{x}$ at $a=0$, if the bounds for n are 0 and infinity. A math-wiki (talk) 10:34, 25 March 2008 (UTC)

Only if all the c_n's are 1... --Tango (talk) 11:59, 25 March 2008 (UTC)

Sum of diagonals

In

789
612
543

the sum of numbers on diagonals is 25. I made a program to brute-force compute the sum for 1001x1001 spiral, but now the problem is that the place from which this thing is is down so I can't check the answer I got (669,171,001). Also, is there some short way I could do this with just paper and pen? --212.149.216.233 (talk) 13:25, 25 March 2008 (UTC)

Yes and yes. The answer is indeed 669,171,001. To calculate it manually, first try to find the patterns in the numbers on the diagonal (for example, the sequence starting at the center and going to the bottom-left is given by the formula $4k^{2}+1\;\!$. Then you can express the sum as $1+\sum _{k=1}^{500}(16k^{2}+4k+4)$. To calculate this, the following formulae will be useful:

$\sum _{k=1}^{n}k={\frac {n^{2}+n}{2}}$

$\sum _{k-1}^{n}k^{2}={\frac {2n^{3}+3n^{2}+n}{6}}$

-- Meni Rosenfeld (talk) 13:54, 25 March 2008 (UTC)

I see the light now. Thank you! --212.149.216.233 (talk) 14:31, 25 March 2008 (UTC)

This spiral seems to be a very popular problem. I've first met it in this forum thread. Later, the sum of diagonals have been asked in a Project Euler problem, which I've learnt of in this mailing list thread. Somewhere in these, I give an explicit formula for computing any element of the spiral from its position (though I don't prove it). As that formula is a simple conditional with all branches being quadratic polynomials, you could easily write the sum of it as an explicit formula, though I didn't do that. Once you know that, you could just interpolate the polynomial formula for the sum from small values instead of calculating it from the first formula. Or, since the problem is now well-known, you could just look those small values up in the OEIS. – b_jonas 18:39, 25 March 2008 (UTC)

Exponential generating function

I've forgotten some of my power series skills. How does one efficiently figure out which function is given by, say, $f(x)=\sum _{n=0}^{\infty }{\frac {n^{2}x^{n}}{n!}}$? (In other words, how do we find the exponential generating function for the sequence {n²}?) I know which function it is, because I've worked it out a couple of different ways (including finding it in a list), but I feel that I'm missing the easy way to calculate it. Thanks in advance. -GTBacchus^(talk) 16:03, 25 March 2008 (UTC)

Either note that the operator x(d/dx) has the effect of multiplying each term in an exponential generating function by n, so your function is

$x{\frac {d}{dx}}\left(x{\frac {d}{dx}}\left(e^{x}\right)\right)$

or use the following identity:

${\frac {n^{2}}{n!}}={\frac {n}{(n-1)!}}=\left(1+{\frac {1}{n-1}}\right){\frac {1}{(n-2)!}}={\frac {1}{(n-2)!}}+{\frac {1}{(n-1)!}}$

Gandalf61 (talk) 16:18, 25 March 2008 (UTC)

You should also check out this link. It is a link to the book "Generatingfunctionology", which is free online. It helped me quite a bit in my extremal combinatorics class, where we discussed generating functions. –King Bee ^{(τ • γ)} 18:51, 25 March 2008 (UTC)

I'll add that rule 2' on page 41 will give you an idea of how to find such an f that you seek. (The rule is basically what is described above by Gandalf.) –King Bee ^{(τ • γ)} 18:54, 25 March 2008 (UTC)

Cool; thank you both very much! -GTBacchus^(talk) 20:34, 25 March 2008 (UTC)

Sum of powers

It is possible to simplify $1^{3}+2^{3}+3^{3}+...+n^{3}$ to $(1+2+3+...+n)^{2}$. (Proved by Mathematical Induction) It is possible to simplify $1^{2}+2^{2}+3^{2}+...+n^{2}$ similarly? How about $1^{m}+2^{m}+3^{m}+...+n^{m}$?

Yes. For any positive integer m, you can (rather easily) find a polynomial P of degree m such that

$\sum _{k=1}^{n}k^{m}=nP(n)$

In the case of $m=2$, you have

$\sum _{k=1}^{n}k^{2}={\frac {2n^{3}+3n^{2}+n}{6}}$

I don't know of a closed form formula for a general m. -- Meni Rosenfeld (talk) 18:39, 25 March 2008 (UTC)

A general approach might be to follow http://mathforum.org/library/drmath/view/56383.html ? 86.130.122.126 (talk) 19:37, 25 March 2008 (UTC)

There's also Faulhaber's formula and Bernoulli number, I guess. x42bn6 Talk Mess 19:39, 25 March 2008 (UTC)

Solving equation with inverse tangents

How would I begin to solve an equation of the following form for x?

$\tan ^{-1}(Ax+B)+C=\tan ^{-1}(Fx+G)+H\,$

--Bavi H (talk) 16:46, 25 March 2008 (UTC)

You could begin by applying tan to both sides and using a trigonometric identity. Algebraist 17:08, 25 March 2008 (UTC)

Or use the identity arctan(a)-arctan(b) = arctan((a-b)/(1+ab)). Collect the arctans on one side, apply the identity, and I believe that you'll get a quadratic in x involving tan(H-C). I assume that A, B, C, F, G and H are constants.—81.132.237.54 (talk) 19:43, 25 March 2008 (UTC)

Thanks to both of you for your help. I used Algebraist's advice and took the tan of both sides, then used the identity $\tan(a+b)={\frac {\tan a+\tan b}{1-\tan a\tan b}}$. I was eventually able to see it would turn out to be a quadratic equation in terms of x, although I didn't complete all of the algebra myself.

In my question above, I used a general form with x representing the variable I'm solving for and A, B, C, F, G, H representing expressions not involving x. In fact, I was actually trying to solve the following equation for h

$\tan ^{-1}\left({\frac {L+h-E}{D}}\right)-\tan ^{-1}\left({\frac {{\tfrac {1}{2}}C}{D}}\right)=\tan ^{-1}\left({\frac {E-h}{D}}\right)-\tan ^{-1}\left({\frac {{\tfrac {1}{2}}E}{D}}\right)$

I created this equation while trying to solve a made-up problem about how high to hang a mirror on a wall so the angles above and below your reflection are equal. To begin solving this equation, I used another identity, -arctan(a) = arctan(-a), to change the differences into sums, then I was able to take the tan of both sides and use the tan(a+b) identity. Or I can see that the arctan a - arctan b identity that 81.132.237.54 mentioned does the same thing in one step.

I had originally tried to plug my equation into Quickmath to get some ideas about the solution. But the first time I entered my equation Quickmath couldn't find a solution. When I entered the general form I used in my question above, Quickmath found a complicated solution involving complex numbers. After the advice I got here, I was able to change my equation from a trig equation to a quadratic equation, but still got bogged down by all the algebra. Although I didn't completely solve it, I was more certain there was a simpler solution, so I tried Quickmath again. I re-entered my equation and Quickmath was able to solve it this time. (The first time I used .5C/D and .5Y/D -- Quickmath reserves E for the natural log base -- and Quickmath couldn't find the solution. This time I used C/2D and Y/2D and Quickmath was able to solve it.)

Thanks again to both of you for your help. Using your advice, I was more certain of the feasibility and method of the solution. I could have solved it myself if I had more determination to wade through the algebraic manipulation of all the variables of my problem, but gave up and went with an automated solution. --Bavi H (talk) 05:04, 26 March 2008 (UTC)

Parabolas and Flashlights - real world situation - moving the focus to change the light output

Ok... so most flashlights have a reflective curve in the shape of a parabola, to focus the light source. the light bulb itself is at the focus of the parabola.

my question: when you twist the top of a common flashlight and the beam narrows or widens, what is happening? clearly the metal reflective mirror is not physically changing, so i assume the light bulb (focus) is moving in and out... how does this in and out movement result in the beam changing?

Thanks...

The focus is a property of the reflector, so moves with it. If the reflector moves, the bulb changes position wrt the focus - if the reflector is unscrewed so that it moves out from the body of the torch, the bulb is then behind the focus and the beam will diverge. If it is possible to screw the reflector in past its normal position, i.e. with the bulb at the focus, the beam will converge to something approximating a point, and then obviously diverge thereafter.—81.132.237.54 (talk) 19:36, 25 March 2008 (UTC)

sphere question

can a 4" sphere fit through a 4" opening?

Depends—is its radius 4 inches? its diameter? its circumference? Strad (talk) 20:51, 25 March 2008 (UTC)

I assume that the opening is a circular hole cut out from a plane in Euclidean space. Let us define "fit through" as: there is a continuous path (curve) for the (centre of) the sphere such that at one point of the path the sphere is wholly on one side of the plane, at another point the sphere is wholly at the other side, and everywhere in between the sphere and the plane-with-a-hole are disjoint (they have no points in common).

Without loss of generality, let the plane be given as the set of points (x,y,z) such that , and let the radius of the circular hole be R. If we define the plane-with-a-hole as the closed set of points (x,y,z) in the plane such that , so that the circular edge of the hole has not been removed, then only (and precisely) spheres with a radius less than R will fit through. If also the edge of the disk cut out is removed, which means that we are left with the set of points in the plane for which , then also the sphere with radius R will fit through, but still none with a larger radius.

The above equally applies if we give the plane a certain thickness, by taking the points such that for some δ > 0.

In physical reality there are no closed sets, and any sphere will fit through any hole if you push hard enough – although one or both may not be quite the same as they were before the push. If there is no measurable difference between the radii within the limits of present technology, however, then a slight push should suffice, and there should be no appreciable change in quality of the sphere and the hole.

In a nice physics experiment, 's Gravesande's ring and ball,[3] a metal ball that would normally just fit through a ring is heated and placed on the ring, which is fixed at some height. Because the ball was heated, it has expanded in size and doesn't fit through. But as you wait, the ball will cool down and shrink, and suddenly it drops through.  --Lambiam 09:41, 26 March 2008 (UTC)

Julian and Gregorian Calendars and Zeller's Congrunce

Hey everyone, recently I discovered Zeller's Congruence as well as Conway's Doomsday Rule. Zeller's Congruence is obviously powerful and useful (because I wanted to write a little program which would give the day of the week given a date and this was perfect). And FYI, my program takes care of identifying which calendar to use (Julian or Gregorian) based on the input date. Now my question is this, if I wanted to find out the day of the week on May 28, 585 B.C., would I plug in -585 for the year? The reason I ask is because as far as I know, there was no "year" numbered zero. We had 1 A.D. and then 1 B.C. but if we correspond years with integers, then what happens to the zero integer. Does Zeller's Congruence take this into account automatically or should I always add a year and plug in -584 instead (so 1 corresponds to 1 A.D., 0 corresponds to 1 B.C., -1 corresponds to 2 B.C., etc...)? Another question which was posed to me was that "A man was nearly 48 years old on celebrating his first birthday. Where, when, and what day of the week was it?" Any help from someone who knows more about Western calendars than me, would be appreciated! Thanks!
A Real Kaiser (talk) 21:41, 25 March 2008 (UTC)

The Julian calendar came into force in 45 BC, just in time to fix the date of Julius Caesar's murder as March 15, 44 BC. To find the day of the week for that date, you would have to treat 44 BC as −43 (making sure you don't compute remainders of negative numbers for modulo arithmetic). So .[4] I don't know whether in general earlier dates anachronistically but standardly use the Julian calendar, but on the NASA website maintained by Fred Espenak they do: [5]. For the Battle of Halys, see the blue swipe labelled "−0584 May 28" in this image.  --Lambiam 10:56, 26 March 2008 (UTC)

"Perfect" partition of the reals

This has been bothering me for years. I thought I would have figured it out by now, but I never did. There must be something completely obvious I am missing. So the question is: Can we find two sets A and B such that:

1. $A\cup B=\mathbb {R}$,
2. $A\cap B=\phi$,
3. $\mathrm {int} A=\mathrm {int} B\ (=\phi )\;\!$ and
4. $|A|=|B|\ (={\mathfrak {c}})$?

1 and 2 just means this is a partition, and 3 means the parts are dense in each other (between any two points of one part, there is a point of the other). So far this is satisfied by, say, the rationals and irrationals. But the asymmetry in their cardinalities is an eyesore, and while it seems intuitive that one could satisfy 4 as well, such a construction has so far eluded me. Thanks for any assistance resolving the issue. -- Meni Rosenfeld (talk) 22:45, 25 March 2008 (UTC)

How about the A as repeated copies of cantor set union the rational numbers, and B as it’s compliment? I don’t know if A would have any interior points or not though. GromXXVII (talk) 23:23, 25 March 2008 (UTC)

Hi Meni. I agree that this is tricky. You want a real analogy to the integer concepts of even and odd numbers. Bo Jacoby (talk) 23:32, 25 March 2008 (UTC).

Yes, that's a good way to look at it. -- Meni Rosenfeld (talk) 00:25, 26 March 2008 (UTC)

I think GromXXVII's construction works. All you have to do is add an uncountable number of points to the rationals, and even 1 copy of the cantor set would do it. The cantor set is totally disconnected, so it has no interior points, and throwing a few rationals in isn't enough to give it an interior, because they're measure 0. It's not a particularly symmetric construction, although it does satisfy the four conditions (I think). If you also require that both sets have the same measure, it gets a little harder, but you can still finesse it: of all the points in [0,1), put the rationals and the cantor set values in set A, and the rest in set B. On the interval [1,2), do the opposite. Continue alternating, and you've got two totally disconnected sets of unbounded measure that partition the reals. -GTBacchus^(talk) 00:08, 26 March 2008 (UTC)

So I am an idiot after all. I guess I never thought about this because subconsciously I was actually looking for something else, where there is even more symmetry. The last construction has a "global" symmetry, but I'd like to see something exhibiting more local symmetry - perhaps, the added requirement that even when restricted to any interval, the sets have the same measure. Any ideas? -- Meni Rosenfeld (talk) 00:25, 26 March 2008 (UTC)

Hmmm... another nice symmetry would be that any real translation of one of the two sets is either onto itself or onto its complement. The evens and odds have that property, if you look at integer translations. -GTBacchus^(talk) 00:43, 26 March 2008 (UTC)

What Meni wants can't be done with measurable sets. Suppose you had such a partition. Then the measure of A intersect (0,1) would have to be 0.5. So then cover A∩(0,1) with a countable collection of disjoint open intervals such that the sum of their lengths is less than 0.6. But then the measure of the intersection of A with each of those intervals is half the length of the interval, so the measure of A∩(0,1) as a whole is less than 0.3; this is a contradiction.

What you can get is a partition such that the intersection of A with any interval has inner measure zero and outer measure equal to the length of the interval (and similarly for the complement of A). I think the following works: Let U be an ultrafilter on the natural numbers, and then let x be in A just in case the set of positions in the binary expansion of x (past the binary point) where there's a 1, is an element of U. --Trovatore (talk) 00:58, 26 March 2008 (UTC)

You write "So then cover A∩(0,1) with a countable collection of disjoint open intervals such that the sum of their lengths is less than 0.6." I believe you can cover Q∩(0,1) with disjoint regions having this property, but doesn't countable additivity of Lebesgue measure imply that you can't cover the whole interval? (I know from experience that here be dragons, so I'm prepared to be wrong.) Tesseran (talk) 01:19, 26 March 2008 (UTC)

"The whole interval" is (0,1)? You're right; you can't cover that interval with a countable collection of disjoint open intervals whose lengths sum to less than 1. That's what keeps the whole concept of measure from trivializing. But by hypothesis A has measure 0.5, so by the definition of measure (or one of the definitions, anyway), for any ε>0, you can cover A with a collection whose lengths sum to less than 0.5+ε. --Trovatore (talk) 01:28, 26 March 2008 (UTC)

Note by the way that this is closely related to a question above, #Lebesgue_measure_of_a_subset_of_.5B0.2C_1.29_invariant_under_translation_by_Q, which once this discussion is archived should be reachable from the following link: Wikipedia:Reference desk/Archives/Mathematics/2008_March_24#Lebesgue_measure_of_a_subset_of_.5B0.2C_1.29_invariant_under_translation_by_Q. --Trovatore (talk) 03:05, 26 March 2008 (UTC)

How about letting A be the union of all positive rationals and all negative irrationals, and B the union of all positive irrationals and all negative rationals? And throw 0 into A. SmaleDuffin (talk) 13:46, 26 March 2008 (UTC)

That looks good to me. It's no more symmetric than the cantor set method, but it's a hell of a lot simpler. --Tango (talk) 14:26, 26 March 2008 (UTC)

Ok, thanks for all the replies! -- Meni Rosenfeld (talk) 20:19, 26 March 2008 (UTC)

March 26

Algebra Question

I am having trouble figuring out this problem. I would like if you could explain it to me, but if you can't that is okay. Thankyou. 22z-35=3z to the second power

Do you mean $22z-35=3z^{2}\;\!$? Start by moving everything to the same side - $3z^{2}-22z+35=0\;\!$. Since 22=15+7 you can write this as $3z^{2}-15z-7z+35=0\;\!$. Can you continue from here? -- Meni Rosenfeld (talk) 00:28, 26 March 2008 (UTC)

That seems a rather odd way to solve a quadratic... I think I can see what you're doing, but why not just complete the square? --Tango (talk) 14:28, 26 March 2008 (UTC)

Factorising is easier than completing the square. Having said that stating that 22=15+7 without stating why you chose these numbers (as opposed to 22= 20+2 for example) is likely to confuse the OP I would think.Theresa Knott | The otter sank 14:37, 26 March 2008 (UTC)

Sure it is, but that's not how I factorise (It's equivalent, but it just seems very strange to write that out as an intermediate step). --Tango (talk) 17:14, 26 March 2008 (UTC)

What I suggested was the first method of solving quadratics taught to us at school (with the parts found by guessing), so I hoped it would ring a bell for the OP. While it is definitely not systematic, it is often faster and easier to understand than other methods. -- Meni Rosenfeld (talk) 21:07, 26 March 2008 (UTC)

Yeah, I'm familiar with factorisation, I've just never written it down like that. We just wrote down "(_x+_)(_x+_)" and tried putting different numbers in the gaps until it worked. We never wrote down the intermediate step of a polynomial with two linear terms, which is why it looked really odd to me. (It took me a while to work out that it was simply an intermediate step in factorising it.) --Tango (talk) 01:28, 27 March 2008 (UTC)

Seems harder to have to guess 4 numbers than just 2 (the general method is: To factorize $ax^{2}+bx+c$, find numbers p and q such that $pq=ac$ and $p+q=b$. Then $ax^{2}+bx+c=ax^{2}+px+qx+c=ax(x+{\tfrac {p}{a}})+q(x+{\tfrac {c}{q}})=(ax+q)(x+{\tfrac {p}{a}})$). -- Meni Rosenfeld (talk) 09:05, 27 March 2008 (UTC)

area element

Hi,
Is there a way to systematically find the area element in an arbitrary coordinate system? For example, how would one get (using some sort of algorithm) the surface element in spherical coordinates

$\mathrm {d} S=r^{2}\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \varphi$ ?

Thank you! — Preceding unsigned comment added by 71.245.169.69 (talk • contribs) 01:17, 26 March 2008 (UTC)

Yes. If memory serves, given a parametrisation, such as

$\Phi (\phi ,\theta )=(r\,\sin \theta \,\cos \varphi ,r\,\sin \theta \,\sin \varphi ,r\,\cos \theta )$

the area element is given by

$\left|{\frac {\partial \Phi }{\partial \phi }}\times {\frac {\partial \Phi }{\partial \theta }}\right|$

Algebraist 01:54, 26 March 2008 (UTC)

The volume element in spherical coordinates is $dV=r^{2}\sin \theta drd\theta d\phi$ (see Spherical coordinate system#Applications), which is simply the Jacobian determinant of the substitution from Cartesian to spherical coordinates. --Spoon! (talk) 07:37, 26 March 2008 (UTC)

• Algebraist, why doesn't your parametrisation include the r component? And what would happen if we had a different coordinate system (say, the cylindrical coordinate system)?
• Spoon, the Jacobian is a systematic way of getting the volume element, but not the area element... 71.245.169.69 (talk) 15:44, 26 March 2008 (UTC)

No r because we're parametrising a surface, so r is fixed. For cylindrical polars, with ρ fixed, we have

$\Phi (\phi ,z)=(\rho \cos \varphi ,\rho \sin \varphi ,z)$

and we do the same thing: take partial derivatives (I see I used the wrong symbol above. oops!) wrt φ and z, and take the norm of the cross product. Algebraist 15:56, 26 March 2008 (UTC)

Intuitive understanding of the nCr formula

I'm trying to understand the formula for the number of (unique sized) combinations from a set. Say the set has n elements and I wish to know the number of ways to select r of them.

• I know that n! represents the number of ways to arrange the n elements, and
  
• I would assume that the product of r! and (n-r)! represents the number of ways to arrange the n elements after one such partition. Is this assessment correct?
  

My question is: why does the quotient of these two give the number of ways to perform such a partition i.e. the number of ways to select r elements from a set containing n?
Zain Ebrahim (talk) 15:19, 26 March 2008 (UTC)

Yes, you're right. There are n! ways of arranging the elements, however some of those ways are equivalent, so we don't want to count them more than once. When partitioning elements, you divide them into 2 groups (the ones you're choosing and the ones you're not). Any rearrangement within those groups is irrelevant, so you divide by the number of such rearrangements (r! is the rearrangements of the ones you're choosing, (n-r)! is the rearrangements of the rest). The reason you divide is because when you have 2 different types of rearrangement (for a specific definition of "type", anyway), the total number of arrangements is the product of the number of each type, so if you have to total number and want to find out the number of one type, you have to divide by the number of the other type. Does that make any sense? --Tango (talk) 15:39, 26 March 2008 (UTC)

Okay I get that we need to remove the irrelevant arrangements but why then do we not subtract them? I don't understand what you meant about each type of rearrangement. Sorry!

Zain Ebrahim (talk) 15:53, 26 March 2008 (UTC)

When you have things that are equivalent, it partitions a set into equivalence classes. The r!(n-r)! is the size of each class - it's the number of ways of rearranging the elements after choosing r of them, and any such arrangement given them same initial choice of r elements is equivalent. What you're interested in is the number of classes, and that's the total number of arrangements divided by the size of each class (since all the classes are the same size - that's not the case with all equivalences, but it is the case with this one). --Tango (talk) 16:18, 26 March 2008 (UTC)

(After ec)I just read equivalence class and equivalence relation which involved the type of math I haven't looked at in years (my favourite kind :P).

Please tell me if I'm right:

• In this case X would be the set of all possible arrangements of the original n objects. The size of X is n!
• Given that we want to partion it into r and (n-r), we have created an equivalence relation and thus equivalence classes. So two elements of X would be equivalent if they both have the same r and (n-r) objects but just selected in a different order, right?
• Each class is equal in size because all of them have r in one hand and (n-r) in the other. So the size is r! × (n-r)!.
• We are interested in the number of classes because each class has a different r objects.
• The number of classes is the size of X divided by the size of each class.

I think I have it! Do I?

Zain Ebrahim (talk) 16:47, 26 March 2008 (UTC)

Yes, exactly right. --Tango (talk) 17:12, 26 March 2008 (UTC)

Of course, in a sense you are subtracting the unnecessary arrangements. We know there are n! ways to arrange n elements, so imagine just taking the first r elements of each arrangement as the ones you're choosing. Of course, we've counted every possible choice many times here -- any rearrangement of the first r terms (of which here are r!) and any rearrangement of the last (n-r) terms (there are (n-r)! of them) would have given us the same choice of r elements. So We've counted every choice r!(n-r)! times. We only want 1 of each of these, so we subtract all but one of them. Then we get nCr = n! - nCr(r!(n-r)! - 1). (We've taken the n! arrangements, and removed all but one of the r!(n-r!) copies of each choice that gave us.) You can fairly easily rearrange this equation into nCr = n!/(r!(n-r!)). This is not in any way a sensible way to look at the calculation, but you can see that though we get a division, we are in some sense subtracting the repeated choices. --PaulTaylor (talk) 16:39, 26 March 2008 (UTC)

That was a lot simpler. Thank you, Paul and Tango. Zain Ebrahim (talk) 16:52, 26 March 2008 (UTC)

Yeah, that was a lot simpler - I wanted to explain it like that but I couldn't work out the details. Good job, Paul! --Tango (talk) 17:12, 26 March 2008 (UTC)

Improvements on Cayley's theorem

Cayley's theorem states that every group G of order n is isomorphic to a subgroup of S_n. However, G might be a subgroup of a smaller S_m, where m < n. For example, Z₆ is easily seen to be a subgroup of S₅ because it is generated by the permutations (1 2 3) and (4 5).

For every group G there is a least such m. I would like references to theorems or other information that might be helpful in calculating such m, or in finding better bounds on the least m than the one given by Cayley's theorem.

Thanks for any suggestions. -- Dominus (talk) 21:13, 26 March 2008 (UTC)

I don't have an answer, but here's a rephrasing of your question that might help you with searches (assuming you don't know the words already, of course): you're looking for the least m such that G acts faithfully on a set of size m. To get a special case out of the way, I believe the structure theorem handles the abelian case. Algebraist 21:47, 26 March 2008 (UTC)

I don't know the answer either, and this isn't strictly relevant, but the texts I learned it from never mentioned, as far as I could see, that the isomorphic subgroup that Cayley guarantees is a TRANSITIVE subgroup of S_n.-Rich Peterson130.86.14.88 (talk) 00:19, 27 March 2008 (UTC)

There is also this result, that if G has a subgroup containing no nontrivial normal subgroups of G, then G is isomorphic to a subgroup of S_t where t=[G:H].

Although this doesn’t apply in your example because every subgroup is normal. GromXXVII (talk) 13:09, 27 March 2008 (UTC)

March 27

Algebra problem

I do not understand these kinds of problems. 4c to the 2nd power+4c+1 Thankyou —Preceding unsigned comment added by 64.119.61.7 (talk • contribs) 11:10, 27 March 2008

What's the problem? Finding roots of that quadratic? There are various techniques you can use - I would use factorisation for that one. --Tango (talk) 01:19, 27 March 2008 (UTC)

Algebra problem 2

The problem is p to the 2nd power+14p+49=0. I understand that you use 7,7. Now do I just right 7,7 for my answer or do I do this-(p+7)(p+7)? Thanks —Preceding unsigned comment added by 64.119.61.7 (talk • contribs) 11:13, 27 March 2008

The solution to p^2 + 14p + 49 = 0 is p = -7 (you missed the minus sign, but this time I'll give it to you). You generally don't have to list double roots twice, unless asked to (or if the fact that it's a double root is important). If you're asked to factorise the expression p^2 + 14p + 49, then you write (p + 7)^2 or (p + 7)(p + 7). If you're asked to solve it, then you would probably factorise it first anyway, and then use that to find the solutions - (p + a)(p + b) = 0 means that either p + a = 0 or p + b = 0, and then solve like any linear equation. The form the answer takes depends on the question. Confusing Manifestation(Say hi!) 00:19, 27 March 2008 (UTC)

antiderivative help on ?

A.) It says to first find f ' and then f f ''(x)=x^3+7x^2+8x-4, f '(1)=5, f(1)=-9 I have already figured out that f '(x)=(x^4/4)+7(x^3/3)+8(x^2/2)-4x+2.4166... but i cant figure out what f (x)=________________

B.) also i am having trouble with a similar one. it says: Given f ''(x)=cos(x), f '(pi/2)=8, f(pi/2)=8, find: f '(x):_________(which i have already calculated the correct anwser to be: sinx +7 f (x):__________________(this is the one that i again dont understand how to get.

Can anyone please help...people have tried helping but everytime i do it i still get it wrong. pls help!!

00:54, 27 March 2008 (UTC)Bubbles16x (talk) 00:54, 27 March 2008 (UTC)

(I've added nowiki tags so your double primes don't get turned into italics tags!) To get from f' to f you do exactly the same thing you did to get from f'' to f'. --Tango (talk) 01:14, 27 March 2008 (UTC)

And for doing that, it helps if you realize that

(x⁴/4) + 7(x³/3) + 8(x²/2) − 4x + 29/12 = (1/4)x⁴ + (7/3)x³ + (8/2)x² − 4x + 29/12.

For problem B, here is a method you can use for this and similar problems.

Step 1: Make a reasonable guess for the form of the answer. Since f'(x) has both a trig function part (namely cos(x)) and a polynomial part (namely 7), we make room for both. It takes some practice and trial and error to get good guesses. In this case you can use

f(x) = a sin(x) + b cos(x) + cx + d,

in which form the variables a, b, c and d are still unknown. You have to determine their values.

Step 2. From that formula, compute f'(x) = ... This will again be something with sines, cosines and a polynomial part.

Step 3. Now equate what you computed in Step 2 with what you already have:

... = sin(x) + 7.

Step 4. What does that tell you about a, b, c and d? How can you choose their values so that the two sides of the equation become equal? You should be able to find at least some values.

Step 5. Substitute the known values you have found for the variables in the form we guessed for f(x).

Finally. If you got that far, one unknown variable will still remain. Use the fact that to find its value and finish this off.  --Lambiam 07:05, 27 March 2008 (UTC)

The number line

On the number line, the segment from 0 to 1 are divided into x and y equal parts, respectively. For example, if x=2 and y=3, the points 1/3, 1/2 and 2/3 are marked. Given x=X and y=Y, besides manually calculating all such points, is there a formula that can calculate the least distance between any two marks? Imagine Reason (talk) 01:53, 27 March 2008 (UTC)

In other words, given x, y natural numbers >1, you want to know the minimum value of |ax-by| for a, b natural numbers with a less than y, b less than x. Isn't this just the HCF of x and y? Algebraist 02:46, 27 March 2008 (UTC)

I think the problem is more to find the minimum of |a/x − b/y|. If x and y have a common factor, some "x marks" will coincide with some "y marks". For example, if x = 6 and y = 10, we have 3/6 = 5/10. In general, if x = pz and y = qz, p/x = q/y. That gives a distance of 0 with this formula. Otherwise, the least distance is 1/(xy).

If coinciding "x marks" and "y marks" are not considered to be separate – in other words, we just take the set of all marks – we get a different result for when x and y have a common factor. The case x = y = 2 must be excluded, for in that case there is only 1 mark in total. Otherwise, the least distance is 1/m, where m is the least common multiple of x and y.  --Lambiam 07:39, 27 March 2008 (UTC)

I think you are both saying the same thing, since HCF(x,y) / xy = 1 / LCM(x,y). Algebraist is working in units of 1/xy. Gandalf61 (talk) 10:06, 27 March 2008 (UTC)

Yeah, I just prefered to think in terms of integers only. Algebraist 15:23, 27 March 2008 (UTC)

Three calculus problems

Hi, I'm studying for a test tomorrow and there are three practice problems I haven't been able to do. I was wondering if anyone could show me how so that I will be able to solve analogous problems tomorrow.

1. ∫e^-x/(1 + e^-x)dx. I can't figure out what to make my u. Should I be looking for a du/u kind of pattern so that it will evaluate to ln |u| + c?

2. ∫₁³((e^3/x)/x²). My confusion is the same as in problem 1.

3. For this one I actually got an answer. I'm supposed to use logarithmic differentiation to evaluate dy/dx. The problem is y = x√(x² - 1). Here is my work:
ln y = ln(x√(x² - 1))
ln y = ln x + ln √(x² - 1)
ln y = ln x + (1/2)ln (x² - 1)
(1/y)(dy/dx) = 1/x + (1/2)(1/(x² + 1))·2x
(1/y)(dy/dx) = 1/x + (1/(x² + 1))·x
(1/y)(dy/dx) = 1/x + x/(x² + 1)
dy/dx = y(1/x + x/(x² + 1))
substituting from the original problem dy/dx = x√(x² - 1)(1/x + x/(x² + 1)) The back of the book says that dy/dx = (2x² - 1)/√(x² - 1). Is this form equivalent to what I got?

Any help is greatly appreciated. Thanks again, anon. —Preceding unsigned comment added by 141.155.126.183 (talk) 04:05, 27 March 2008 (UTC)

• Your intuition on #1 is correct. The key to a u-sub (at least at an introductory level) is usually to make the more complicated part of the integrand your u. #2 doesn't look like it involves du/u at all, actually... think about the exponent on the e and also the fact that the x^2 happens to be in the denominator, meaning it's actually x^-2. Play around with that a bit and see what happens. :) --Kinu ^t/_c 05:49, 27 March 2008 (UTC)
• Oh, and on #3... the process is correct, but when you took the derivative, the x^2-1 on the right became an x^2+1 for the remainder of the problem. Switching the sign to a negative as it should be yields an equivalent answer to the book's. Good work! --Kinu ^t/_c 05:56, 27 March 2008 (UTC)
  • For #1, it can involve substitution by letting u equal the denominator. Then one gets $\displaystyle \int \displaystyle {\frac {u-1}{u}}\displaystyle {\frac {1}{1-u}}\,du$ which can obviously be simplified ($a-b=-(b-a)$). x42bn6 Talk Mess 18:03, 27 March 2008 (UTC)

Factorials and Primes

I've noticed this:

• Prime factors of 3!-2!-1!-0!: 2
• Prime factors of 4!-3!-2!-1!-0!: 2, 7
• Prime factors of 5!-4!-3!-2!-1!-0!: 2, 43
• Prime factors of 6!-5!-4!-3!-2!-1!-0!: 2, 283
• Prime factors of 7!-6!-5!-4!-3!-2!-1!-0!: 2, 2083
• Prime factors of 8!-7!-6!-5!-4!-3!-2!-1!-0!: 2, 17203 (mmm, these all end in 3)
• Prime factors of 9!-...: 2, 11, 37, 389 (this and the three below don't work)
• Prime factors of 10!-...: 2, 1609843
• Prime factors of 11!-...: 2, 853, 21031
• Prime factors of 12!-...: 2, 23, 9457531
• Prime factors of 13!-...: 2, 101, 1523, 18541
• Prime factors of 14!-...: 2, 40214157043 (bingo, it's working again)
• Prime factors of 15!-...: 2, 606873049843
• Prime factors of 16!-...: 2, 9760593625843
• Prime factors of 17!-...: 2, 829, 201063350767 (stopped again)
• Prime factors of 18!-...: 2, 67, 397, 168143, 673499
• Prime factors of 19!-...: 2, 57432357441241843 (works again)
• Prime factors of 20!-...: 2, 1152238261120729843
• so on...

Do you guys have any explanations for this, or why the ones with only two prime factors have 2 and then some number ending in 3?

--wj32 ^t/c 07:05, 27 March 2008 (UTC)

Also, it doesn't seem to work from 21!-... onwards... --wj32 ^t/c 07:10, 27 March 2008 (UTC)

Put f(n) = n! − (n−1)! − (n-2)! ... − 0!. For n ≥ 5, n! ≡ 0 mod 20. That means that , so . So after factoring out 2, you have a number ending in 3. This also works for f(21) = 2·24264807338798809843 (where the larger factor is a multiple of 479) and onwards.  --Lambiam 08:01, 27 March 2008 (UTC)

f(n)/2 never has a prime factor ≤ n. It is a proved prime for n = 4, 5, 6, 7, 8, 10, 14, 15, 16, 19, 20, 32, 90, 353, 544 (1254 digits), and a gigantic probable prime for n = 3479 (10812 digits), 3602 (11249 digits), 3800 (11955 digits), and no other n below 4000. Guess the reason for my username! PrimeHunter (talk) 15:56, 27 March 2008 (UTC)

Finding Non-real solutions of systems of equations

Do any of you know of any methods for finding solutions of the form (a+bi,c+di) for systems of equations? Thanks in advance. A math-wiki (talk) 09:49, 27 March 2008 (UTC)

How about exactly like finding real solutions, only without assuming the solutions are real?

More specifically, methods for solving systems of linear equations like Gaussian elimination work over any field, complexes included. So you really just do everything normally, but if you have an underspecified system, you let the free variables range over the complexes rather than the reals.

If the equations are nonlinear, again you can use any of the known methods, such as extracting one unknown from an equation and substituting it in another. You only have to remember that when inverting a function, you need to keep in mind nonreal preimages. So, for $z^{2}=-1$ you don't conclude there is no solution, and for $e^{z}=2$ the solution is not just $\ln 2$ but $\ln 2+2\pi ki$.

In some cases you can simplify the problem by representing the complex unknown with real numbers, such as $z=a+bi$ or $z=r\mathrm {cis} \theta$, eliminating all instances of i, and proceeding with the mundane real equations.

If none of this helps with what you had in mind, you will have to specify it. -- Meni Rosenfeld (talk) 12:58, 27 March 2008 (UTC)

Weight of paper

Can anyone help me with an estimate of the weight, in kilograms, of a box (2500 sheets, or 5 reams) of paper of 80g/m2 (grammage)? Thanks if you can help. —Preceding unsigned comment added by 213.84.41.211 (talk) 14:19, 27 March 2008 (UTC)

We need to know the size of the paper, but given that it's easy. Assuming A4, which is 210mm × 297mm or 0.21m x 0.297m, so area per sheet = 0.06237 m²; multiply by 2500 sheets = 155.925 m²; at 80 gsm, this weighs 12474g, or 12.474kg, or about 27.5 pounds. (Plus the weight of the box of course.) AndrewWTaylor (talk) 14:38, 27 March 2008 (UTC)

Core 2 Exam Revision

Hello,

I am trying to do a question on my Core 2 exam revision paper (I'm doing A Level maths, first year), and I not sure how to go about it.

The question is

f(x) = x³-2x²+ax+b a and b are constants.

When f(x) is divided by (x-2), the remainder is 1. When f(x) is divided by (x+1), the remainder is 28.

Find the value of a and the value of b.

So I was wondering if anyone would offer a step by step guide on how to complete the question, I think it has something to do with the remainder theorem, but we haven't been taught how to deal wih two variables.

89.242.2.102 (talk) 14:28, 27 March 2008 (UTC)

You are told that "when f(x) is divided by (x-2), the remainder is 1". So you know that

$f(x)=(x-2)p(x)+1$

where p(x) is some polynomial in x. Setting x equal to 2 gives

$f(2)=(2-2)p(x)+1=1$

But we also know that

$f(2)=2^{3}-2(2^{2})+2a+b=2a+b$

so

$2a+b=1$

Carry out the same process with (x+1), setting x equal to -1 this time, and you will get a second equation in a and b. Solve the pair of simultaneous equations to find the values of a and b. Gandalf61 (talk) 14:50, 27 March 2008 (UTC)

Thank you for a super-fast response. Finally I can continue with my maths work, YAY!

89.242.2.102 (talk) 14:55, 27 March 2008 (UTC)

Actually just one more thought, when I put (x+1) in does the equation look like this:

f(-1) = (-1-2)p(x)+28=28  ?

89.242.2.102 (talk) 15:03, 27 March 2008 (UTC)

Not quite. Starting with "when f(x) is divided by (x+1), the remainder is 28", this means that

$f(x)=(x+1)q(x)+28$

where q(x) is some polynomial in x. Now set x equal to -1 ... Gandalf61 (talk)

So I get

f(-1) = (-1+1)q(x)+28 = 28

so what do I do next? I don't understand how you went from step 2:

$f(2)=(2-2)p(x)+1=1$

But we also know that

to step 3:

$f(2)=2^{3}-2(2^{2})+2a+b=2a+b$

so

$2a+b=1$

89.242.2.102 (talk) 16:03, 27 March 2008 (UTC)

We know that f(x) is defined as $f(x)=x^{3}-2x^{2}+ax+b$.

We don't know what a and b are yet, but we can just treat them as unknown constants. Substituting x=2 into this definition gives f(2)=2a+b.

But we also know, from using the remainder theorem, that f(2)=1. So this must mean that 2a+b=1.

If you want to follow this up further, I suggest you post a message on my talk page at User talk:Gandalf61, as other Ref Desk readers have probably lost interest in this topic ! Gandalf61 (talk) 16:15, 27 March 2008 (UTC)

Indeed kudos to you though for helping me out and all.

89.242.2.102 (talk) 16:26, 27 March 2008 (UTC)

Finding the solution

This is absolutely homework, and I can need a wee bit of guidance. If anyone can help by pointing me in the right direction, I would be very delighted.

I need to take the following: 1-1/4+1/27-1/256+1/3125-1/46656+1/823543 - and find a solution for it.

The solution is as follows:

7

(sumsign here) (-1)^(k+1) * (1/k^k)

k=1

How do I go from mere numbers to an algebraic expression? Thank you for your help. 81.93.102.185 (talk) 16:47, 27 March 2008 (UTC)