# 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.

• $$f(x)=\beta\,\frac{1-\alpha^x}{1-\alpha}$$
– Did
Sep 21, 2016 at 13:21
• You know this has several possible solutions (think of a little bit wiggling in the interval $[0,1)$ of a solution).
– ctst
Sep 21, 2016 at 13:21
• ((Your example is β = 1 and α = 0.5, not β = 0.5 and α = 1.))
– Did
Sep 21, 2016 at 13:22

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.
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 $$f(x) = \alpha \frac{{1 - \alpha ^{\,x} }}{{1 - \alpha }} + \left( {f(1) - f(0) - \alpha } \right)x + f(0)$$
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).