Question

Power series question

Answer 1

isn't the geometric series part supposed to be?
\[\sum_{n = 0}^\infty x^n = \frac{1}{1-x}\]

Answer 2

Yep typo sorry.

Answer 3

ok, just making sure I wasn't losing it. :)

Answer 4

Use the geometric series:

\[\sum_{n = 0}^\infty x^n = \frac{1}{1-x}\]

To find the power series about x=0 for:

\[\frac{ 1 }{ (1+x^2)^2 }\]

Answer 5

I'm stuck on the manipulation honestly >.> .

Answer 6

I try to avoid that since i'm terrible at noticing patterns. But okay :P .

Answer 7

ah... I see, that makes these questions a bit tricky I'm sure :)

Answer 8

Hmm... Lets see...

Answer 9

I'm stepping back and trying to see where to go with this myself. :/

Answer 10

OK, let's forget about the expansion and start over since you said you don't prefer that method anyway... sorry :)

Answer 11

It's okay :) .

Answer 12

Alright, instead let's first examine what we are trying to get \[\frac{1}{(1 + x^2)^2}\] and try to find a way to make it look more like a geometric series so something like \[\frac{1}{1 - stuff}\]

Answer 13

SHoudnt it be the other way around?

Answer 14

We take the geometric series and make it the fuction.

Answer 15

You can do that, but it might be a bit tricky to figure out how to get a power of 2 in the geometric series.

Answer 16

But either way should theoretically work. :)

Answer 17

Here's what I did, see if the steps make sense to you: \[\frac{1}{(1 + x^2)^2} = \frac{1}{1+ 2x^2 + x^4} = \frac{1}{1 - (-2x^2 - x^4)}\]

Answer 18

Ohh that's brilliant :) .

Answer 19

Now just substitute all that for x in the sigma right?

Answer 20

Exactly! :D

Answer 21

Thanks!

Answer 22

not a problem, I figured it out by trying to get rid of the power of 2 which was the pesky thing about the problem. :)

Answer 23

For your final answer then, you should have \[\sum_{n=0}^\infty (-2x^2 - x^4)^n \] or a more tricky way of saying this \[\sum_{n =0}^\infty (-1)^n (2x^2 + x^4)^n\]

Answer 24

Yep I saw that :) .

Answer 25

Nice, you did great!