Analytic Continuation for a Product 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?

$\phi$ is non-negative.
$f(x)=\log(\phi(x))$ satisfies 
$$2f(x)+f(2x)=\log\left((x+1)^2\right)$$

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