How can you rewrite the function as a power series?
f(x) = x/ (1 + 2x^2)

I know I have to take out the x so x ( 1 / (1 + 2x^2)
however I'm stuck afterwards

Question

How can you rewrite the function as a power series?
 f(x) = x/ (1 + 2x^2)

I know I have to take out the x so x ( 1 / (1 + 2x^2)
however I'm stuck afterwards

jhannybean · Answer

@terenzreignz

anonymous · Answer

$$\frac{1}{1-x}=1+x+x^2+x^3+...$$
$$\frac{1}{1+x}=1-x+x^2-x^3+...$$
$$\frac{1}{1+2x^2}=1-2x^2+(2x^2)^2-(2x^2)^3+...$$

anonymous · Answer

So would I write the sum as $$\sum_{n=1}^{\infty} (-1)^{n} (x)^{2n+1}$$

anonymous · Answer

you need a 2 inside the parentheses

anonymous · Answer

Oh yeah my bad. But how can I find the radius of convergence if I can't use the ratio test afterwards? I mean there's no division going on. Only thing I can think of is the root test but its negative, so the root test won't work perfectly.

anonymous · Answer

I thought power series was just sum x^n

anonymous · Answer

well the representation

anonymous · Answer

Well I took the x and added it to the 2x^(2n+1) because the x was "outside" of the sum. So i just added it and  thats what I did. I don't feel like writting the equation. I don't know if you follow what I'm saying.It was my guess really.

anonymous · Answer

converges if $|2x|<1$ i.e if $|x|<\frac{1}{2}$

anonymous · Answer

Ah  So i just take it right away. Makes sense. Thanks a lot^^

anonymous · Answer

also you should start your summation at $n=0$ i think

anonymous · Answer

to be more precise i guess if converges if $|2x^2|<1$ but same answer