Solve the functions power series on $c=0$ and where it converges:
$f(x)= {x^3 \over {1-2x^2}}$
I'm kind of confused about this whole concept, but was able to solve all this far, I think I have to put it like this first: $x^3 \cdot {1 \over {1-2x^2}}$ but how do I continue from here?
One may use the standard geometric result $$ 1+u+u^2+...+u^N=\frac{1-u^{N+1}}{1-u}, \quad |u|<1, \tag1 $$ then setting $$ u=2x^2, \quad |u|<1, $$ and letting $N \to \infty$, one gets $$ \frac{x^3}{1-2x^2}=\sum_{n=0}^\infty 2^n x^{2n+3}, \quad |x|<\frac1{\sqrt{2}}. \tag2 $$
It is well known that $$\frac{1}{1-u^2}=1+u^2+u^4+u^6+\dots\qquad |u|<1.$$ Thus, if $|\sqrt{2}x|<1$, that is, $|x|<\frac{1}{\sqrt{2}}$, then $$\begin{align} \frac{x^3}{1-2x^2}&=x^3\cdot\frac{1}{1-(\sqrt{2}x)^2}\\ &=x^3\cdot\bigg[1+(\sqrt{2}x)^2+(\sqrt{2}x)^4+(\sqrt{2}x)^6+\dots\bigg]\\ &=x^3+(\sqrt{2})^2x^5+(\sqrt{2})^4x^7+(\sqrt{2})^6x^9+\dots\\ &=x^3+2x^5+2^2x^7+2^3x^9+\dots\\ &=\sum_{n=0}^{\infty}2^nx^{2n+3} \end{align}$$