Wikipedia:Reference desk/Archives/Mathematics/2008 March 26

= 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)
 * Depends on the question, I guess. Your way won't necessarily find nice factors with integer coefficients, whereas mine can. You can simply factor out the a before you start and not worry about it, but you lose the convenience of working with integers. Your way is somewhere inbetween. --Tango (talk) 23:37, 27 March 2008 (UTC)
 * Actually, I just messed up the formula for the general case (I'm not sure it can even be described simply). If you apply the method to any given case, you will find factors with integer coefficients. -- Meni Rosenfeld (talk) 23:56, 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
 * $$\mathrm dS=r^2\sin\theta\,\mathrm d\theta\,\mathrm d\varphi$$ ?

Thank you! —Preceding unsigned comment added by 71.245.169.69 (talk • contribs) 01:17, 26 March 2008
 * 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 $$dV = r^2\sin\theta dr d\theta d\phi$$ (see Spherical coordinate system), 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 Sn. However, G might be a subgroup of a smaller Sm, 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 Sn.-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 St 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)

Thanks very much. -- Dominus (talk) 05:08, 28 March 2008 (UTC)


 * Finding such an m is in general a difficult computation. What sort of groups are you considering? I have written some basic tools for finding small m, and there are some papers that for instance show that m can be quite a bit smaller than n, or it might be arbitrarily close to n (and the groups where they are equal are classified).  If you are just interested in abelian groups, then the structure theorem gives an answer.  If you are just interested in simple groups, I believe the answer is basically known, though it might take quite a few papers to track down all the cases.  JackSchmidt (talk) 19:54, 31 March 2008 (UTC)