Recent content by PiRGood

as n approaches infinity x/n approaches zero, which means e^(x/n) approaches one which leaves me with x on the bottom! Incredible, i never would have seen that. Thanks so much Dick! You are a lifesaver!

So i am are applying L'Hôpital's rule to the fraction in the denominator of our function i.e applying it to (1-e^(x/n))/(1/n) So i get -xe^(x/n)/(n^2) on top and -1/n^2 on the bottom which simplifies to: xe^(x/n)?

But wouldn't taking the derivative of 1/n give us zero, because n is some constant approaching inf? So the derivative of (1-e^(x/n)) would be (xe^(x/n))/n and the derivative of 1/n would be 0, giving me ((xe^(x/n))/n )/0?

If i recall correctly that formula is I tried that but i hit a roadblock. I came up with 1-exn/n/ 1-ex/n. You can cancel the n's in the numerator's exponent of course. But that is where i hit my dead end. I couldn't manipulate it from there in any meaningful way, even using the 1/n factor in...

Homework Statement Prove that: lim n->inf1/n*Ʃn-1k=0ekx/n = (ex-1)/x x>0 Homework Equations That was all the information provided. Any progress i have made is below. I didn't want to add any of that to this section because this is all speculation on my part so far. The Attempt at a...

I do! The lightbulb just went on! thank you so much for bearing with me! :) :)

a_2? I apologize i think you lost me. you want to know what b(a_1)b^-1 is? However i thinking like this may have just turned a lightbulb on for me. What i must show is that for any cycle r of length k, and any b in S. brb^-1 is also a cycle of the same length k. Am i getting somewhere?

a_2?

Well b will permute at to some number (or set, whichever you like) and b^(-1) will permute it back, so won't it just be a_1?

I thought b, (a_1) and b^(-1) were all permutations? It means that b^(-1)b is e, or the identity element

Is it b(a_1)? the problem says b(a_1)b^(-1)=b(a_1)

isn't it simply b^-1(a_1)? Or (b(a_1))^-1?

I apologize, I am not sure if this is a question or a hint. Either way i do not understand. ß is an element of the group S_n. a_1 is an element of r, which is also a set of transpositions contained in S_n. ß(a_1) is a composition of transpositions, i believe Does that help?

Hi all, long time reader first time poster! Just need a hand on this problem I've been stuck on for a few days Homework Statement Let r=(a_1,a_2...a_k) be in S_n. Suppose that ß is in S_n. Show that: ßrß^-1=(ß(a_1), ß(a_2)...ß(a_k)). Homework Equations The Attempt at a...