Question

Find the power series for the following and determine the radius of convergence.
f(x) = x^2/(2+x)
AWARD GIVEN AT END!

Answer 1

\[\frac{1}{2+x}=\frac{1}{2\left(1+\dfrac{x}{2}\right)}=\frac{1}{2\left(1-\dfrac{-x}{2}\right)}\]

Answer 2

\[\sum_{n=0}^{\infty}r^n=\frac{1}{1-r}\text{ for }|r|<1\]

Answer 3

So f'x is 1/(2+x)? @Zarkon

Answer 4

by the above we have ...

\[\frac{1}{2\left(1-\dfrac{-x}{2}\right)}=\frac{1}{2}\sum_{n=0}^{\infty}(-x/2)^n\] for |-x/2|<1

Answer 5

then multiply by \(x^2\) to get the full series

Answer 6

\[\frac{1}{2}\sum_{n=0}^{\infty}(-x/2)^n \times x^2\] ??

Answer 7

yes....then ...

\[\Large\frac{1}{2}\sum_{n=0}^{\infty}(-x/2)^nx^2=\frac{1}{2}\sum_{n=0}^{\infty}(-x)^nx^2/2^n=\sum_{n=0}^{\infty}(-x)^{n+2}/2^{n+1}\]

Answer 8

\[\Large=\sum_{n=0}^{\infty}(-1)^{n+2}x^{n+2}/2^{n+1}\]

\[\Large=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+2}}{2^{n+1}}\]

Answer 9

Awesome! :) Thanks! So the radius of convergence is 1?

Answer 10

no

Answer 11

we need |-x/2|<1

so |x|<2

so the radius of convergence is 2

Answer 12

Ohhh okay. Thanks a bunch! :)