Question

find a power series representation for the function and determine the interval of convergence \[f \left( x \right)=\left( x \right)\div \left( 2x ^{2}+1 \right)\]

Answer 1

We're going to need to use the geometric series expansion, that is, Sum x^n = 1/(1-x). What this means is we're going to have to play with f(x) until it looks like 1/(1-something). Why not make it f(x) = x/(1- (-2x^2))?
Then
\[\frac{1}{1-(-2x^2)} = \sum_{n=0}^\infty (-2x^2)^n\]
\[= \sum_{n=0}^\infty (-1)^n 2^n x^{2n}\]
But this isn't quite f(x), we want an x on top, so multiply both sides by x:
\[f(x) = \frac{x}{1-(-2x^2)} = \sum_{n=0}^\infty (-1)^n 2^n x^{2n+1}.\]
For the interval of convergence, use the ratio test on this series. Can you take it from here?

Answer 2

let me see

Answer 3

\[\sum_{0}^{\infty}\left( 2^{n +1}x ^{2n +2} \right)\div \left( 2^{n}x ^{2n +1} \right)\] is this right so far?

Answer 4

Yep, except that it should be x^{2(n+1)+1} or x^{2n+3} on top.

Answer 5

ok im always confused about this next step, do we plug in infinity?

Answer 6

Always simplify as much as you can before taking any limits. So we have
\[\lim_{n\rightarrow \infty} \frac{2^{n+1} x^{2n+3}}{2^n x^{2n+1}}\]
that we can cancel some terms from.

Answer 7

\[\lim_{n \rightarrow \infty}2^{n}x ^{2n +2}\] ?

Answer 8

We subtract exponents on like terms, so 2^{n+1}/2^n = 2^{n+1 - n} = 2, and x^{2n+3}/x^{2n+1} = x^2. We should have
\[\lim_{n\rightarrow\infty} \left| 2 x^2 \right|\]
There are no more n's, so taking the limit doesn't do much. Set this less than 1 and solve for |x| for the radius of convergence.

Answer 9

thank you so much. im having so much trouble with this subject its driving me insane

Answer 10

Keep at it, study and practice, and you'll do fine! You've made it this far, after all.