Find a derivable function $f$ for which $f(x) - f(x-1) =\alpha ( f(x-1) - f(x-2) )$ Find a derivable function $f$ for which $f(x) - f(x-1) =\alpha ( f(x-1) - f(x-2) )$.

My initial conditions would be:
$$\begin{align*}
f(0) &amp;= 0\\
f(1) &amp;= \beta
\end{align*}$$
and $\alpha &lt; 1$
and my domain $[0,+\infty[$

Basically, if looping on integers, every increment will be $\alpha$ times the previous increment, but I want a derivable function.

for example, for $\beta = 0.5$ and $\alpha = 1$:

$$\begin{align*}
f(2) &amp;= 1.5\\
f(3) &amp;= 1.75\\
f(4) &amp;= 1.875
\end{align*}$$
I want to be able to evaluate $f(5.7)$ for instance.
If $\alpha=0$ then any periodic function of period $1$ is a solution. If $\alpha\ne0$ write $\alpha=e^\lambda$ for some $\lambda\in{\mathbb C}$. Then any function $f$ of the form
$$f(x):=g(x)+e^{\lambda x} h(x)$$
with $g$ and $h$ periodic of period $1$ is a solution.
If you take a possible solution $f$ to your problem (e.g. Dids) and multiply it by a differentiable function $g$ with $g(x)=g(x+1)$ and $g(0)=1$, e.g. $g(x)=cos(x\cdot 2 \pi)$ you get a new solution $fg$ which also solves your problem. Hence there is no reasonable way to evaluate $f(5.7)$ with only this information (but there is a reasonable way to evaluate $f(n)$ where $n$ is an integer).