Analytic Continuation for a Product

Question

I was trying to solve the functional equation $$\phi(x)^2\phi(2x)=x^2+2x+1$$ and by assuming that $\phi(1)=1$, and setting up a recurrence relation, I found the solution $$\phi(x)=\prod_{i=0}^{\log_2(x)-1} (2^i+1)^{x-i}$$ However, this only makes sense for values of $x$ that are perfect powers of two. How can I extend this to non-powers of $2$ but still satisfy the functional equation?

Answer 1

$\phi$ is non-negative.

$f(x)=\log(\phi(x))$ satisfies

$$2f(x)+f(2x)=\log\left((x+1)^2\right)$$

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Since you are talking about analytic continuation

For analytic solutions we can compute the derivatives at $x=0$

$$2f^{(n)}(x)+2^nf^{(n)}(2x)=\frac{d^n}{dx^n}\log\left((x+1)^2\right)$$

Therefore $$f^{(n)}(0)=\frac{1}{2+2^n}\frac{d^n}{dx^n}\log\left((x+1)^2\right)|_{x=0}=\frac{(-1)^nn!}{n\left(1+2^{n-1}\right)}$$

Therefore $$f(x)=\sum_{n=1}^{\infty}\frac{(-1)^n}{n\left(1+2^{n-1}\right)}x^n$$

and

$$\phi(x)=\exp\left(\sum_{n=1}^{\infty}\frac{(-1)^n}{n\left(1+2^{n-1}\right)}x^n\right)$$