Can one do anything useful with a functional equation like $g(x^2) = \frac{4x^2-1}{2x^2+1}g(x)$?

Question

I got

$$g(x^2) = \frac{4x^2-1}{2x^2+1}g(x)$$

as a functional equation for a generating function. Is there a way to get a closed form or some asymptotic information about the Taylor coefficients from such an equation?

Here g(0) = 1, g'(0) = 0, and g''(0) = 4.

Edit: Thanks, everyone; as has been pointed out , I've just made a mistake in obtaining the recurrence relation. I'll leave this as-is instead of editing because I think some of the comments will be helpful to others and if I change it now they won't make sense.

Answer 1

I get that the only solution is $g(x) = 0$ if we can write $g(x) =\sum_{n=0}^{\infty} a_n x^n $.

Here is my proof:

We have $g(x^2) = \frac{4x^2-1}{2x^2+1}g(x) $ or $(2x^2+1)g(x^2) = (4x^2-1)g(x) $.

From this, as copper.hat pointed out, $g(0) = 0$.

If $g(x) =\sum_{n=1}^{\infty} a_n x^n $, (since $g(0) = 0$) the left side is

$\begin{align*} \sum_{n=1}^{\infty} (2x^2+1)a_nx^{2n} &=\sum_{n=1}^{\infty} 2a_nx^{2n+2}+\sum_{n=1}^{\infty} a_nx^{2n}\\ &=\sum_{n=2}^{\infty} 2a_{n-1}x^{2n}+\sum_{n=1}^{\infty} a_nx^{2n}\\ &=a_1x^2+\sum_{n=2}^{\infty} (2a_{n-1}+a_n)x^{2n}\\ \end{align*} $

and the right side is

$\begin{align*} \sum_{n=1}^{\infty} (4x^2-1)a_nx^{n} &=\sum_{n=1}^{\infty} 4a_nx^{n+2}-\sum_{n=1}^{\infty} a_nx^{n}\\ &=\sum_{n=3}^{\infty} 4a_{n-2}x^{n}-\sum_{n=1}^{\infty} a_nx^{n}\\ &=-a_1x-a_2x^2+\sum_{n=3}^{\infty} (4a_{n-2}-a_n)x^{n}\\ \end{align*} $

Equating coefficients, $-a_1x = 0$, $-a_2x^2 = a_1x^2$, and $2a_{n-2}+a_n =4a_{2n-2}-a_{2n} $ and $0 =4a_{2n-1}-a_{2n+1} $.

Therefore $a_1 = a_2 = 0$ and $a_{2n} =4a_{2n-2}-2a_{n-2}-a_n $ and $a_{2n+1} =4a_{2n-1} $.

From this second recurrence, since $a_1 = 0$, $a_{2n+1} = 0$ for all $n$.

From $a_{2n} =4a_{2n-2}-2a_{n-2}-a_n $, for $ n=2, a_4 =4a_2-2a_0-a_2 =0 $, for $ n=3, a_6 =4a_4-2a_2-a_3 =0 $.

By strong induction, all the $a_{2n} = 0$, so the only solution is $g(x) = 0$.

Answer 2

Suppose you know that $g(x) = f(x) g(x^2)$ where $f(0) = 1$ (this assumption is crucial). Then applying this recurrence infinitely many times gives

$$g(x) = \prod_{n \ge 0} f(x^{2^n}).$$

The hypothesis that $f(0) = 1$ guarantees that this product makes sense as a formal power series. For example, if $g(x) = \frac{1}{1 - x}$, then $f(x) = 1 + x$, and we obtain the infinite product

$$\frac{1}{1 - x} = \prod_{n \ge 0} (1 + x^{2^n}).$$

This infinite product encodes the existence and uniqueness of the binary repesentation of a nonnegative integer.