How do I simplify the formal power series $1+2x^2F(x)^2$?

Question

I have this formal power series

$$F(x)=1+2x^2F(x)^2$$

that I want to put into non-recursive form. I can expand,

$$1+2x^2F(x)^2=1+2x^2(1+2x^2F(x)^2)^2= 1+2x^2+8x^4F(x)^2+8x^6F(x)^6$$

and I could keep going so that the first $k$ terms do not have recursion, but this process becomes increasingly tedious and hasn't given me any insight. Are there any techniques I can use to get this formal power series into a nicer form?

Answer 1

Previous answers gave you the way of expressing explicitely $F(x)$ as a function of $x$ and you noticed two possible solutions (at least if $x$ is not equal to $0$). Then, you can expand as Taylor series but this has been done for each solution and wil get you very different results. For $$y = \frac{1 - \sqrt{1 - 8x^2}}{4x^2}$$ you would arrive to $$F(x)=1+2 x^2+8 x^4+40 x^6+O\left(x^7\right)$$ while for $$y = \frac{1 + \sqrt{1 - 8x^2}}{4x^2}$$ you would arrive to $$F(x)=\frac{1}{2 x^2}-1-2 x^2-8 x^4-40 x^6+O\left(x^7\right)$$ Suppose that we consider the case where $F(0)$ is suppose to have a finite value; then, what you could also do is to assume that $F(x)$ write $$F(x)=\sum _{i=0}^{\infty } a_i x^i$$ and replace in the expression $$1+2x^2F(x)^2-F(x)=0$$ If you develop, you have, for the lhs, a polynomial and you want to cancel as many terms as possible. For each degree $k$, this gives you a very simple linear equation to solve for $a_k$.

Answer 2

It is a quadratic equation in $F(x)$.

Answer 3

$$F(x)=1+2x^2F(x)^2$$

For simplicity, and readability let, $y=F(x)$

$$y = 1 + 2x^2y^2 \implies 2x^2y^2 - y + 1=0$$

Applying the quadratic formula,

$$y = \begin{cases}\frac{1 \pm \sqrt{1 - 8x^2}}{4x^2}&x\ne0\\1&x=0\end{cases}$$

Also your expanding should yield an infinite series.