Recent content by mathgirl1

I know that $$x=(k+1)^{\frac{-1}{k}}$$ but I don't know how to use this to compute $$(x+y)^n$$ since x is in terms of k and not in terms of x. Help please! I am sure this should be simple but I am stuck

Simplify (find the sum) of $${30 \choose 0} + \frac{1}{2}{30 \choose 1}+ \frac{1}{3}{30 \choose 2} + ... + \frac{1}{31}{30 \choose 30}$$. Do this is two ways: 1. Write $$\frac{1}{i+1}{30 \choose i}$$ in a different way then add 2. Integrate the binomial thorem (don't forget the constant of...

Prove that if n is even and r is odd then $$\binom{n}{r}$$ is even. Solution: I know I have these two equalities $$\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}=\frac{n(n-1)...(n-r+1)}{r!}$$ Now if n is even and r is odd then (n-r+1) is even. So it seems that we will have at least one more...

Ah ha! Yes! Thank you very much! I knew I was missing something. Much appreciated!

Let F be a field extension of Q (the rationals) with [F:Q] = 24. Prove that the polynomial $$x^5+2x^4-16x^3+6x-10$$ has no roots in F. Proof: Let $$a$$ be a root of $$x^5+2x^4-16x^3+6x-10$$. Since the polynomial has degree 5 by theorem we know that $$[Q(a):Q]=5$$. If $$a \in F$$ and...

Let G be a group of order pm where p is a prime and p > m. Suppose H is a subgroup of order p. Show that H is normal in G. There is a very similar problem Let |G| = p^nm where p is a prime and n \ge 1, p > m. Show that the Sylow p-subgroup of G is normal in G. Proof: Let n_p be the...

Show the direct sum of a family of free abelian groups is a free abelian group. My first thought was to just say that since each group is free abelian we know it has a non empty basis. Then we can take the direct sum of the basis to be the basis of the direct sum of a family of free abelian...

Thanks so much! That helped a lot!

Thank you! That should get me started somewhere. Our book doesn't have the Cesaro Theorem in it, nor did he cover it so not sure I can use that method without proving it and the proof looks quite lengthy. But I guess I at least have a starting point. Confused why he would give us something...

Assume the sequence of positive numbers ${a_n}$ converges to L. Prove that $\lim_{n \to \infty} \sqrt[n]{a_1a_2...a_n} = L$ (The nth root of the product of the first n terms) Since ${a_n}$ converges we know that for every $\epsilon> 0$ there is an $N$ such that for all $n > N$ $ |a_n -...

Well I feel like an idiot now for part a and was making that way more difficult than it needed to be. Thank you both for the help. If you could help me more on part b that would be great but not necessary. I guess I have had enough help at this point but anyway here goes... So if n= 0 the...

I am not understanding. So (2x)/(1+x^2) < 1 => 2x < 1 + x^2 => 0 < x^2 - 2x + 1 => 0 < (x-1)^2 => 0 < x -1 => 1 < x. So I can understand how the series converges for x > 1 but what about for x < 1 like the problem states?

a) Show that sum_(n=0)^infinity (2^n x^n)/((1+x^2)^n) converges for all x in R\{-1,1} b) Even though this is not a power series show that sum above = 1 + sum_(n=1)^infinity (2nx^n) for all -1<x<=1. For part a by the ratio and root test we get |(2x)/(1+x^2)| but this does not have an n in it...