Solutions of the functional equation $f(x+1)= xf(x)$

Question

I have come across a question of functional equation $$f(x+1)= xf(x).$$ My question is how can we solve these type of functions. I have tried to use substitutions like $x= x-1$ and some manipulations but could't find the function. I also have tried to use differentiation's basic definition. I am still unable to solve the functional equation. Please try to clear my query.

Answer 1

This can be seen as a first order recurrence relation, which can be recast to a first order difference equation by taking the logarithm.

$$f(x+1)=xf(x)$$

can be written

$$g(x+1)=g(x)+\log x.$$

The solution of such recurrences is made of the general solution of the homogeneous equation (obviously a constant), and a particular solution. In this particular case, we see that for integer $x$,

$$g(n)=\sum_{k=0}^{n-1}\log(k)+g(0)=\log((n-1)!)+g(0).$$

More generally, for real $x$,

$$g(x)=g(\lfloor x\rfloor+\{x\})=\sum_{k=0}^{\lfloor x\rfloor-1}\log(k+\{x\})+g(\{x\})$$

and

$$f(x)=\prod_{k=0}^{\lfloor x\rfloor-1}(k+\{x\})\,e^{g(\{x\})}=\frac{\Gamma(x)}{\Gamma(\{x\})}e^{g(\{x\})}=\frac{\Gamma(x)}{\Gamma(\{x\})}f(\{x\}).$$

Without more specification, $f$ is arbitrary on $[0,1)$ or any other unit interval.

Answer 2

The first idea is always to plug in special values for $x$ - but what is "special" may depend on the functional equation. Here, $x=0$ suggests itself as it leads to $f(1)=0$. After that, one readily finds by induction that $f(x)=0$ for all $x\in\Bbb N$. Incidentally, the recursive nature of the functional equation allows us to describe the quite large set of solutions:

Let $g\colon[0,1)\to\Bbb R$ be an arbitrary function and define $f\colon \Bbb R\to \Bbb R$ recursively as $$f(x)=\begin{cases}g(x)&0\le x<1\\ (x-1)(x-2)\cdots(x-\lfloor x\rfloor)\cdot g(x-\lfloor x\rfloor)&x\ge1\\ \frac{g(x+\lceil -x\rceil)}{x(x+1)\cdots (x+\lceil -x\rceil)}&x<0\end{cases} $$ Then $f$ solves the functional equation, and evey solution of the functional equation is of this form.