There is a straight solution for difference equation $f(x+1)=\beta(x)f(x)$ such that $\beta(1)=a$ and $x>1$?

Question

Recently, I stock in find a general or special solution of a difference equation $f(x+1)=\beta(x)f(x)$ such that $\beta(1)=a$ and $x>1$. I do not know much about this group of equations so any name, link, paper,... can helpful to me. I want to know more about this equations.

Answer 1

If you know nothing about $\beta$, then $f(n)$ can be just about any sequence you want, so long as $f(2)=a\cdot f(1)$.

More strictly, the following is true:

Let $f_n$ be a piecewise constant nonzero function on all intervals $[n,n+1)$, that is, let $a_n$ be a sequence of real numbers, and let $f(x)$ equal $a_n$ on $[n,n+1)$. Let $a_2=a\cdot a_1$. Then, there exists a function $\beta$ such that $f(x+1)=\beta(x)\cdot f(x)$ for all $x$.

You can easily see this is true by simply defining $\beta(x)$ to equal $\frac{a_{n+1}}{a_n}$ on $[n,n+1)$