Solution of the f_1(x)-f_1(x-pi)=f_2(x) functional equation

[Domdamo]

Homework Statement

We have the equation
f1(x)-f1(x-pi)=f2(x)
where f1, f2 are the unknown and the known functions, respectively.
f1, f2 are periodic functions with period 2pi.

Is it possible to determine the explicit formula of f1?
If yes how should we calculate it?

Relevant Equations

[1]
f1(x)-f1(x-pi)=f2(x)
[2]
f1(x)=f1(x+2pi)
[3]
f2(x)=f2(x+2pi)

Laplace transform of eq. [1]
[4]
F1(p)-exp{-pi*p}*F1(p) = F2(p)
Rearranging eq. [4]
[5]
F1(p) = frac{1}{1-exp{-pi*p}}*F2(p)
Inverse LT of eq. [5]

 

[BvU]

Hi,

You might want to learn a little $\LaTeX$ so that your f1(x)-f1(x-pi)=f2(x) becomes

$$f_1(x)-f_1(x-\pi)=f_2(x)$$ and looks like $$f_1(x)-f_1(x-\pi)=f_2(x)$$
Re:

Is it possible to determine the explicit formula of f1?

No. Lots of $f_1$ can be constructed since e.g. $\sin(\alpha) - \sin(\pi-\alpha) = 0$ and $\sin(n\beta)$ is periodical with period $2\pi$ .
So any multiple of any function (periodical with period $2\pi$ ) that has $f_1(x)-f_1(x-\pi)=0$ can be added to a candidate $f_1$.

[edit]Oops, sign errors...

 

[BvU]

Recant: any function with period $\pi/n$ has $f(x) - f(x-\pi) = 0$.

 

[Domdamo]

Thank you very much.