Wikipedia:Reference desk/Archives/Mathematics/2010 August 10

= August 10 =

Different groups with the same order signature
For a finite group G, let gn be the number of elements of G with order n; call this sequence its "signature". For example, the signature of S3, with trailing zeroes omitted, is [1, 3, 2], and the signature of Z8 is [1, 1, 0, 2, 0, 0, 0, 4]. Isomorphic groups have the same signature, but the converse is not necessarily so. What are the smallest distinct groups with the same signature? —Mark Dominus (talk) 05:13, 10 August 2010 (UTC)
 * This first occurs at |G|=16. $$\mathbb{Z}_4^2$$ and $$Q_8 \times \mathbb{Z}_2$$ each have signature [1, 3, 0, 12]. Algebraist 17:52, 10 August 2010 (UTC)
 * Thank you very much! —Mark Dominus (talk)

E-books
Hello does anyone know a place where I can find e-books? I'm at a university so I can access things like springer journals and other stuff that require you to login via some institution. I'm looking for the book An introduction to homological algebra by Weibel and my library doesn't have it. I never buy books because of the insane price. For example Hatcher and J P May have online versions of their books. Thanks Money is tight (talk) 06:04, 10 August 2010 (UTC)
 * If you want to know what e-books you can access through your institution and you can't find the information on the institution's website I'd suggest asking one of the institution's librarians. Qwfp (talk) 06:18, 10 August 2010 (UTC)
 * Have you thought about getting Saunders Maclane's Homology? (See this.) While there are many other excellent books on homological algebra, the novel feature of this book is the way in which the author has managed to provide applications of the theory to numerous branches of mathematics. (My copy is not with me at the moment but if I remember correctly, the book discusses homological algebra in the context of algebraic topology (especially; for example, a treatment of spectral sequences in this context is given), group theory (especially in relation to group cohomology which is an important tool in number theory and group theory itself (e.g., the Schur-Zassenhaus theorem)), commutative algebra and algebraic geometry.) The book is also accessible if you "know undergraduate mathematics" and at $48, it is definitely worth the money (at least in my opinion). (If you so desire, you can borrow it from your library and see whether you like it but I have little doubt that you will!) PS  T  08:15, 10 August 2010 (UTC)
 * Both Homology and An introduction to homological algebra are on gigapedia.com -Shahab (talk) 08:20, 10 August 2010 (UTC)
 * (Edit Conflict) Not to say that you should not get Weibel; this is an excellent book as well and does resemble Maclane in some ways. However, Maclane can be found in most libraries (at least it can be found in my library, but I believe that it can be found in other libraries as well), and so could be an option if no other decent book can be found. PS  T  08:23, 10 August 2010 (UTC)
 * Also, most university libraries do Inter-Library Loan. It can't hurt to ask. Septentrionalis PMAnderson 00:10, 16 August 2010 (UTC)

Name of function
Is there a name for this function: f(n) = Number of connected graphs possible on n vertices. Is a formula for this function. Thanks -Shahab (talk) 12:28, 10 August 2010 (UTC)


 * See Graph enumeration and Dmcq (talk) 12:41, 10 August 2010 (UTC)

Transforming a principal stress tensor to maximum shear stress tensor?
Hi,

I'm feeling a little stupid here, because I thought I understood stress tensors, but here we go... All notation is MatLab, but it should be fairly easy to understand.

I have a stress tensor, A, stored as a 3x3 matrix on MatLab. It's a standard stress tensor, 6 independent components.

I do:

[V,D] = eig(A)

to obtain the eigenvalues, i.e. principal stress tensor, D and eigenvectors, i.e. direction cosines, V.

Such that performing:

inv(V)*D*V = V\D*V returns the original stress tensor A

I thought (and the wikipedia page on stress confirms this) that the maximum shear stress plane lies at 45 degrees to the principal stress plane.

So I tried:

V = V + cos( acos(V) + pi/4 ) and then V\A*V again to give the maximum shear tensor, but that didn't seem to work?

or, alternatively:

DIRCOSINES = zeros(3,3)

DIRCOSINES = cos(pi/4)

DIRCOSINES\D*DIRCOSINES

which should also work if I understand it correctly, but it doesn't.

I know how to extract the maximum shear stress (0.5*(sigma1-sigma3)), but what I really need are the direction cosines to go from A to that tensor where that maximum shear stress exists.

Any help would be really really appreciated.

Thank you! —Preceding unsigned comment added by 89.204.153.131 (talk) 16:27, 10 August 2010

Edit: Well, I've sorted it now, so if anyone else ever has to do this...

Firstly, my MatLab notation above was off, because it was from memory and I haven't been using it for long. It's now been corrected.

Secondly, a\D*a works, were a is a standard rotation matrix about an axis, i.e. for the 1-3 plane (if sigma1 was the highest principal stress and sigma3 was the lowest), it would be:

a = [0.7071 0 -0.7071; 0 1 0; 0.7071 0 0.7071]

(where one of those is a negative due to the -sin(angle) term, though I might have written the wrong one down there by mistake, but it's just the standard rotation matrix..) —Preceding unsigned comment added by 89.204.137.172 (talk) 15:46, 11 August 2010

Irreducible polynomial over the rationals
Hi there,

I want to show that x^3 + x^2 - 2x + 1 is irreducible over the rationals, or at the very least justify why it should be - I tried using Eisenstein's criterion with some sort of trick like the cyclotomic polynomials where you set X = Y-1, but I couldn't find any sort of substitution which gave the right coefficients. Suggestions, anyone?

Thankyou! :) 86.30.204.236 (talk) 17:51, 10 August 2010 (UTC)
 * If it were reducible, it would have to have a linear factor, i.e., a rational root. Since it is a monic integer polynomial, its roots are algebraic integers, and the only rational algebraic integers are true integers. It's easy to see that all roots of the polynomial have to have absolute value at most 4 or so, and then it's a matter of checking that no integer of absolute value at most 4 is a root of the polynomial.—Emil J. 18:00, 10 August 2010 (UTC)
 * You can make that slightly easier by, instead of thinking about absolute values, noting that any root must be a factor of 1 (the constant term), and so must be ±1. This is the idea of the rational root test. Algebraist 18:05, 10 August 2010 (UTC)

Ahh, you are a star, thanks :D 86.30.204.236 (talk) 18:04, 10 August 2010 (UTC)

Divergence Theorem
I've been battling with the following question all day and, after a wide variety of different answers, seem have to achieved the same answers enough times to confirm that I'm wrong.

"Let $$ \mathbf{F} (\mathbf{r}) = (x^3+3y+z^2, y^3, x^2+y^2+3z^2)$$ and let S be the open surface $$ 1-z=x^2+y^2$$, 0≤z≤1. Use the divergence theorem (and cylindrical polar coordinates) to evaluate $$\iint\limits_{S} \mathbf{F}\cdot d\mathbf{S}.$$ Verify your result by calculating the integral directly."

So, using the divergence theorem, I calculate the divergence of F integrated over V and get 4.5π. I then close the surface with the unit disc in the z=0 plane, define the unit normal as (0,0,-1), integrate over the unit disc and get -π/2, giving me an answer for the original surface integral of 5π. But when I calculate the integral directly, using cylindrical polars, I get 2π. These all seem like plausible answers so it's very difficult to tell which is wrong. Can someone help me out? Thanks asyndeton   talk  18:46, 10 August 2010 (UTC)


 * I get 3π/2 for the first one and the same answers as you for the other two. -- BenRG (talk) 06:15, 11 August 2010 (UTC)
 * OK time for a stupid question. Is div dependent on the coordinate system? I attempted to calculate the first integral by finding the divergence and then changing to cylindrical polars and then integrating but it appears changing variables first and then taking the divergence gives a different answer. asyndeton   talk  13:27, 11 August 2010 (UTC)
 * It seems unlikely that div is different for different coordinate systems so if I show my working, perhaps someone can tell me where I've gone wrong. In Cartesians, div F is $$3x^2+3y^2+6z$$. Convert this to cylindrical polars to get $$3r^2+6z$$. Multiply by the Jacobian to give $$3r^3+6rz$$. This has no term in theta so the theta integration just gives 2π$$(3r^3+6rz)$$. Integrating between one and zero for r gives $$2\pi(\frac{3}{4}+3z)$$. Integrating again between one and zero for z gives $$\frac{9\pi}{2}$$, which is not what I want. Where's it going wrong? asyndeton   talk  13:47, 11 August 2010 (UTC)
 * I think the problem is with your integration limits. The region you integrate over is a cylinder. You don't want to integrate all the way to $$r=1$$ for each $$z$$ value. Martlet1215 (talk) 17:31, 11 August 2010 (UTC)
 * I don't understand why I don't want 0≤r≤1. In Cartesians $$ 1-z=x^2+y^2$$ so in Cylindrical polars $$ 1-z=r^2$$ so when z=1, r=0 and when z=0, r=1 and since z is continuous, r must take every value in between. Where have I gone wrong and what limits do I actually want then? asyndeton   talk  18:08, 11 August 2010 (UTC)
 * I think Martlet1215 meant z=1, not r=1. You need to integrate z from 0 to 1−r² first and then integrate r from 0 to 1, or else integrate r from 0 to $$\sqrt{1-z}$$ first and then integrate z from 0 to 1.


 * The divergence formula does depend on the coordinate system. In cylindrical coordinates it's $$\nabla \cdot F = \frac 1 r \frac \partial {\partial r} (r F_r) + \frac 1 r \frac \partial {\partial \theta} F_\theta + \frac \partial {\partial z} F_z$$ (copied from Griffiths, Introduction to Electrodynamics). -- BenRG (talk) 19:42, 11 August 2010 (UTC)


 * Ah, I'm with you on the limits now. I'm just so used to the limits, one transformed into cylindrical or spherical polars, being constants that I didn't even think they might depend on the other variables. Thank you for that. And as for the divergence, I know that there's a different formula for working them out in different coordinate systems but what I meant was will your answer be different if you take the divergence and then change variables rather than change variables and then take the divergence? Thanks asyndeton   talk  20:09, 11 August 2010 (UTC)
 * Oh, I see. It will be the same either way (as I think you already know). -- BenRG (talk) 21:55, 11 August 2010 (UTC)
 * Yeah, just as I expected. Thanks for all your help. It's much appreciated. asyndeton   talk  22:49, 11 August 2010 (UTC)
 * Just to clarify my earlier statement. I admit there are other choices but I meant what I said: I was using the latter of BenRG's limits, integrating $$r=0$$ to $$r=\sqrt{1-z}$$ and then $$z$$ from 0 to 1. So $$ r$$ needn't go all the way to 1 for each $$z$$. Martlet1215 (talk) 23:29, 11 August 2010 (UTC)

Hyperbolas
Hello. For some hyperbola, suppose a is the length of the semi-transverse axis, b is the distance from the centre of the hyperbola to a point on the perpendicular bisector of the transverse axis, and c is the distance from the centre to one focus. Why is $$a^2+b^2=c^2$$? I don’t see a right triangle anywhere. Thanks in advance. --Mayfare (talk) 20:09, 10 August 2010 (UTC)
 * I think you're misstating the question. $$a$$ and $$c$$ are fixed for a given hyperbola, while $$b$$ will change depending on the point on the bisector.  So in general, $$a^2 + b^2 \neq c^2$$.  —Preceding unsigned comment added by 203.97.79.114 (talk) 10:23, 11 August 2010 (UTC)

I think if the equation of the hyperbola is
 * $$ {x^2 \over a^2} - {y^2 \over b^2} = 1 $$

and c is the distance from the center to either of the two foci, then
 * $$ a^2 + b^2 = c^2. \, $$

One should be able to give a nice simple geometric explanation. I'll see what I can do. Michael Hardy (talk) 17:16, 11 August 2010 (UTC)
 * In the case of an ellipse there is a nice geometric interpretation for the analogous equation. I don't know what it would be for the hyperbola unless you start using imaginary values for a, b, and c.--RDBury (talk) 02:28, 12 August 2010 (UTC)