How can I determinine all functions or a function f?

Question

Determine all functions $f$ in $f(x+1)=2f(x)$, for all $x$ in real number.

So I let $x$ be $x+1$. Then I have $f((x+1)+1)=2f(x+1)$. But since there is a new function $f(x+2)$, I couldn't determine the function $f$ using elimination. Any idea would be a great help.

Answer 1

There are very many such functions if those constraints you have given are the only constraints.

For example, you can define an arbitrary function $F:[0,1)\to \mathbb R$. Then you can "extend" it to the whole real axis: on $[1,2)$ make it twice $F$, on $[2,3)$ make it $4$ times $F$, on $[-1,0)$ make it half $F$ etc. The way to write it in one formula:

$$f(x)=2^{\lfloor x\rfloor}F(x-\lfloor x\rfloor)$$

where $\lfloor x\rfloor$ denotes "the biggest integer not bigger than $x$".

One can prove that $f(x)=F(x)$ for $x\in[0,1)$, and any such function $f$ must be obtained this way, by extending its own restriction on $[0,1)$ - thus the above construction produces all such functions $f$.

Answer 2

Define $\,g(x) := f(x)2^{-x}.\,$ Then $\,g(x+1) = g(x)\,$ for all real $x$. Thus, $\,g(x)\,$ is a period $1$ real function. Conversely, any period 1 real function $\,g(x)\,$ determines uniquely the correspoinding $\,f(x)\,$ satisfying $\,f(x+1) = 2f(x).$