I have a problem with finding solution. I suppose it will be something like $f(x) =G(x)\Re(e^{\frac{x\pi}{a}})$, where $\Re$ is real part of a complex number, $G(x)$ periodic function whith period $\frac{a}{n}$ and $n$ is a natural number. Can you help me? Thanks a lot for any help. :)
Solution of functional equation $f(x)=-f(x-a)$
3 Answers
We may assume $a=\pi$. Simple examples that come to mind are the sine and the cosine function. Unfortunately these function have zeros, but a combination of the two allows the following construction:
If $f(x)\equiv-f(x-\pi)\qquad (*)$ then the function $g(x):=e^{ix}f(x)$ is a $\pi$-periodic complex valued function (check it!). Conversely: If $x\mapsto g(x)\in{\mathbb C}$ is an arbitrary $\pi$-periodic function then $f(x):=e^{- ix} g(x)$ satisfies the functional equation $(*)$.
Now I assume you are interested in solutions of $(*)$ that are real-valued for $x\in{\mathbb R}$. In this respect note that the real part ${\rm Re}f(x)$ of a solution of $(*)$ automatically is a solution of $(*)$ either. Doing the computations one can say the following: Any real solution of $(*)$ can be written in the form $f(x)=a(x)\cos x+b(x)\sin x\ ,$ where the functions $a(\cdot)$ and $b(\cdot)$ are real-valued and $\pi$-periodic; but this representation is not unique.
Have you heard antiperiodic function?
http://mathworld.wolfram.com/AntiperiodicFunction.html
In fact when $a$ is any non-zero real number, the general solution of $f(x)=-f(x-a)$ is $f(x)=\Theta(x)$, where $\Theta(x)$ is an arbitrary antiperiodic function with period $|a|$.
The general solution if $f(x+na)=(-1)^ng(x)$ for every real number $x$ in $[0,a)$, every integer $n$, and every function $g$ defined on $[0,a)$.