I have recently studied calculus of finite differences and I was wondering how to solve anti-differences for transcendental functions. Because I want to solve some interesting sums.
In this case, I want to solve the following functional equation below.
$f(x+1) – f(x) = e^{ax}$
Is there any general method in solving problems like this? I will highly appreciate a detailed answer!
Best Answer
Call $T_a$ the set of functions such that $f(x+1)-f(x)=e^{ax}$ for all $x\in\Bbb R$, and consider the map $\Phi_a:T_a\to \Bbb R^{[0,1)}$ defined by $\Phi_a(f)=\left.f\right\rvert_{[0,1)}$. $\Phi_a$ is a bijection, and specifically $$[\Phi_a^{-1}(g)](x)=\begin{cases}g(\{x\})+\frac{e^{a\{ x\}}-e^{ax}}{1-e^a}&\text{if }a\ne 0\\ g(\{x\})+\lfloor x\rfloor&\text{if }a=0\end{cases}$$
Where $\{x\}$ is the real number in $[0,1)$ such that $x-\{x\}\in\Bbb Z$, and $\lfloor x\rfloor=x-\{x\}$ is the largest integer smaller or equal to $x$.
It is clear that $\Phi_a$ is injective and that $\left.\Phi_a^{-1}(g)\right\rvert_{[0,1)}=g$, therefore it is only a matter of verifying that $\Phi_a^{-1}(g)(x+1)-\Phi_a^{-1}(g)(x)=e^{ax}$ for all $x$, which is just algebra.
Added: Another approach is to notice that if you set $g(x)=f(x)+\frac{e^{ax}}{1-e^a}$ if $a\ne 0$ and $g(x)=f(x)-1$ if $a=0$, then you obtain that $g(x+1)=g(x)$, i.e. that the solutions are the functions in the form $\text{(1-periodic function)} +\frac{e^{ax}}{e^a-1}$.