<s>[INST] 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:
$$\begin{align*}
f(0) &= 0\\
f(1) &= \beta
\end{align*}$$
and $\alpha < 1$
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 $\alpha = 1$:
$$\begin{align*}
f(2) &= 1.5\\
f(3) &= 1.75\\
f(4) &= 1.875
\end{align*}$$
I want to be able to evaluate $f(5.7)$ for instance.
 [/INST] Starting from your equation
$$
f(x) - f(x - 1) = \alpha \left( {f(x - 1) - f(x - 2)} \right)
$$
put
$$
g(x) = f(x) - f(x - 1)
$$
from which you get the general solution for $g(x)$
$$
g(x) = \alpha g(x - 1)\quad  \Rightarrow \quad g(x) = \alpha ^{\,x}  + c
$$
with $c$ being a constant (actually, any periodic function of period $1$).
Then you get $f(x)$ as
$$
\begin{array}{l}
 f(x + 1) - f(x) = g(x + 1)\quad  \Rightarrow  \\ 
  \Rightarrow \quad f(x) = \left( {\sum\limits_{k = 0}^{x - 1} {g(k + 1)} } \right) + d = \left( {\sum\limits_{k = 0}^{x - 1} {\alpha ^{\,k + 1}  + c} } \right) + d =  \\ 
  = \alpha \frac{{1 - \alpha ^{\,x} }}{{1 - \alpha }} + cx + d \\ 
 \end{array}
$$
with $d$ being again a constant or any periodic function of period $1$,
and adjusting for the initial conditions
$$
f(x) = \alpha \frac{{1 - \alpha ^{\,x} }}{{1 - \alpha }} + \left( {f(1) - f(0) - \alpha } \right)x + f(0)
$$
</s>