# infinite sum

• May 16th 2007, 07:49 PM
pakman
infinite sum
So I'm doing some diffy eq, and I'm stuck on this simple algebraic problem.

What does (1+x^2)*x^n equal?

I was thinking x^n + x^(2n)... x^(2n+1)... which then I don't see how I could index shift that to make it only x^n. I'll write out part of the problem since it's the only part I'm confused on... if you can help just from that info stop here otherwise continue.

We'll see E[n=0] is summation of n=0 to infinity... and A>(n+2) is A sub n+2

(1+x^2)*E[n=0] (n+2)(n+1)A>(n+2)x^n

It was originally y'' = E[n=2] n(n-1)A>(n)x^(n-2) which I index shifted so it would be x^n.

Okay so now I realize that what I typed out is very confusing since I don't know LaTex, sorry guys. Basically I just want to index shift it so it's x^n, which would match the rest of the function I did not write out.
• May 16th 2007, 07:51 PM
topsquark
Quote:

Originally Posted by pakman
So I'm doing some diffy eq crap, and I'm stuck on this simple algebraic problem.

What does (1+x^2)*x^n equal?

(a + b)*c = a*c + b*c

So
(1 + x^2)*x^n = 1*x^n + x^2 * x^n

= x^n + x^{n+2}

You're thinking too hard about it. :)

-Dan

Edit: Say you have Sum[x^{n + 2}, n = 0, infinity] and you want to get the exponent back to "n."

Let i = n + 2. Then n = i - 2. So:
Sum[x^{n + 2}, n = 0, infinity] = Sum[x^{i - 2 + 2}, i - 2 = 0, infinity]

= Sum[x^i, i = 2, infinity]
• May 16th 2007, 08:21 PM
pakman
Okay thanks for clearing that up for me. The problem with substituting i for n+2 is that I would have to make all the other x^n's into x^i's, right? There's 3 other summations in this function... and I need to factor out an x^n.

(1+x^2)*E[n=0] (n+2)(n+1)A>(n+2)x^n - 4x*E[n=1] n*A>(n)*x^(n-1) + 6*E[n=0] A>(n)*x^n
• May 17th 2007, 04:57 AM
topsquark
Quote:

Originally Posted by pakman
Okay thanks for clearing that up for me. The problem with substituting i for n+2 is that I would have to make all the other x^n's into x^i's, right? There's 3 other summations in this function... and I need to factor out an x^n.

(1+x^2)*E[n=0] (n+2)(n+1)A>(n+2)x^n - 4x*E[n=1] n*A>(n)*x^(n-1) + 6*E[n=0] A>(n)*x^n

Well, once you have the form of the summation you want the "i" is just a dummy variable so you can change it back to an "n."

-Dan