If I have a semi-recursive equation for a function $f$, namely,
$f(2^n)/f(2^{n+1}) = 11/18 + n/18$ for all real values of n,
How do I convert this (without a programmable computer) to a non-recursive formula for $f(x)$ in terms of $x$?
Assume a value of $f(1) = H$ for some constant $H$.
Edit: It would be best to get a closed formula for $f(x)$ that does not include factorials, because I will eventually differentiate a function that includes this one. Also, factorials do not allow me to include negative numbers or nonintegers in the domain.
Let $g(x)=f(2^x)$. So $$\frac{g(n)}{g(n+1)}=\frac{11+n}{18}$$ In other words, $$g(n+1)=\frac{18}{11+n}g(n)$$ Let $g(x)=18^xk(x)$. So $$k(n+1)=\frac{1}{11+n}k(n)$$ Let $k(x)=\frac{1}{(10+x)!}m(x)$, so $$m(n+1)=m(n)$$
Let $m(n)$ be constant: $m(n)=m$.
So $k(n)=\frac{m}{(10+n)!}$.
So $g(n)=\frac{18^nm}{(10+n)!}$.
So $f(n)=\frac{18^{\log_2(n)}m}{(10+\log_2(n))!}$. aThe factorial applied to general real numbers should be interpreted as the Gamma function, shifted by $1$: $$f(n)=\frac{18^{\log_2(n)}m}{\Gamma(11+\log_2(n))}$$ For $f(1)=H$, you need $m=H\cdot\Gamma(11)$.