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April 28[edit]

Iterative method: Cost of moving and cost of evaluation[edit]

I am curious if there is any approach to iterative methods when the cost of evaluation at a point or the cost of moving from one point to another is not fixed. Suppose I want to get to the top of an otherwise-grassy hill but I know there's a thicket at the top. It would be fastest to find the thicket, find the highest point on the perimeter of the thicket, then bushwhack from there to the top, rather than looking only at the slope. Are there well-founded approaches to dealing with this sort of optimization problem? (See my full question on Talk:Iterative method) —Ben FrantzDale 02:17, 28 April 2007 (UTC)[reply]

solvable subgroup[edit]

if S and T are solvable subgroups of G with S normal subgroup in G , Is ST solvable subgroup of G , why? 86.108.104.68 09:30, 28 April 2007 (UTC)[reply]

Do your own homework. Solvable group should help, as should your course notes/textbook. Algebraist 16:53, 28 April 2007 (UTC)[reply]

cyclic group[edit]

Let F be afinite field and let G be the multiplicative group consisting of its nonzero elements. Prove that G is cyclic.86.108.104.68 09:40, 28 April 2007 (UTC)[reply]

Do your own homework. Or if you're feeling lazy, look at any abstract algebra textbook and find this is in there. Algebraist 16:54, 28 April 2007 (UTC)[reply]

Initial value problems[edit]

I'm looking at solving initial value problems using computational methods. However, I'm having trouble doing them analytically to check that my computational efforts are correct.

I have an example where the equation is dx/dt = 0.5x + 0.5exp(t), with the condition x(t=0) = 1. The analytical solution is x(t) = exp(t). I can't see where this is coming from.

Similarly, I have to solve dC/dt = 1 - C^2 where C(t=0)=0, and don't really know how to start.

If someone could give me some pointers, that would be great. →Ollie (talk • contribs) 17:55, 28 April 2007 (UTC)[reply]

I'm guessing you've not previously studied ordinary differential equations. Actually, linear differential equations, like your first one,

Dx-{\tfrac {1}{2}}x={\tfrac {1}{2}}e^{t},

are a nice place to start. Your second equation is not linear (because of the C²), but it is still a relatively simple first order separable equation, which does not involve t at all. If we let h(C) = 1−C², we arrive at

\int {\frac {dC}{1-C^{2}}}=\int dt,

where the right-hand side is trivially t. Performing the left-hand side integral gives us

{\tfrac {1}{2}}(\log(C(t)+1)-\log(C(t)-1))=t+K

(where K is a constant of integration), or

{\frac {C(t)+1}{C(t)-1}}=e^{2(t+K)}.

The rest should be manageable. --KSmrq^T 20:23, 28 April 2007 (UTC)[reply]

I think you mean the right-hand side is trivially t. x42bn6 Talk 20:24, 28 April 2007 (UTC)[reply]

Good catch; thanks. Fixed. --KSmrq^T 20:39, 28 April 2007 (UTC)[reply]

Brilliant, thanks very much for your help. Worryingly, I have studied ODEs before, but I seem to be having a brain fart at the moment.

Could I just ask which integration technique I should be using to solve $\int {\frac {dC}{1-C^{2}}}$ ?

Should I be substituting for $C^{2}$ , or do I need to split $1-C^{2}$ into $(1-C)(1+C)$ ?

The latter: rewrite

{\frac {1}{1-C^{2}}}

in the form

{\frac {A}{1+C}}+{\frac {B}{1-C}}

for suitable expressions A and B (an application of the method of partial fractions). --Lambiam ^Talk 21:27, 28 April 2007 (UTC)[reply]

Many thanks. I'm really having a bad day - I was trying to multiply the two fractions rather than add them! Time for a break I think. Thanks once again. →Ollie (talk • contribs) 21:34, 28 April 2007 (UTC)[reply]

meaning of k in combinations and permutions[edit]

Why is the letter k used in the Combinations and Permutation algorithms? k-set in these articles points to n-set, which redirects to hypergraph, none of which talk about the letter k. I have this question about letter in math/physics often, is there a list of letters and symbols and their origins? —The preceding unsigned comment was added by Steeltoe (talk • contribs) 22:54, 28 April 2007 (UTC).[reply]

Because it's a cool letter. Seriously, you really could use any symbol you wanted, if I'm not mistaken. Splintercellguy 23:23, 28 April 2007 (UTC)[reply]

Often we use letters i through n (the first two letters of the word "integer") as indices. Still, sometimes those are not enough, sometimes i (or j) is used for √(−1), and sometimes we prefer other conventions. Theoretical physics often has elaborate conventions; the following is an example from Supersymmetry and Supergravity, ISBN 978-0-691-08326-1.

Before we begin, however, we first present the supersymmetry algebra:

{\begin{aligned}\{{Q_{\alpha }}^{A},{\bar {Q}}_{{\dot {\beta }}B}\}_{+}&=2{\sigma _{\alpha {\dot {\beta }}}}^{m}P_{m}{\delta ^{A}}_{B}\\{\text{and so on}}.\end{aligned}}

The Greek indices

(\alpha ,\beta ,\ldots ,{\dot {\alpha }},{\dot {\beta }},\ldots )

run from one to two and denote two-component Weyl spinors. The Latin indices (m,n,…) run from one to four and identify Lorentz four-vectors. The capital indices (A,B,…) refer to an internal space; they run from 1 to some number N ≥ 1.

Be happy with the combinatorial k, or run screaming. --KSmrq^T 05:37, 29 April 2007 (UTC)[reply]

I think I'll add that k also has a minor advantage in that it is not likely to be confused with another letter. Depending on someone's handwriting or font or a reader's eyesight, the letters i and j could be confused with each other, and the lowercase letter l could be confused with the numeral 1. The letter k doesn't suffer those risks for misreading the letter. Dugwiki 16:40, 30 April 2007 (UTC)[reply]