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I can solve the (periodic) functional equation $f(x+b)=f(x)$ completely ($x\in \mathbb{R}$ and $b\neq 0$). Indeed, its general solution is $f=\phi o (\; )_b$, where $(\; )_b$ is the $b$-decimal (fractional) part function defined by $$(\; )_b(x) =(x)_b:=x-b\lfloor \frac{x}{b}\rfloor,$$ (see http://nntdm.net/papers/nntdm-19/NNTDM-19-4-04-15.pdf) and $\phi$ every real function defined on $b[0,1)$. In general, it is solved on arbitrary groups (see http://www.ijmex.com/index.php/ijmex/article/viewFile/194/115).

Now, can somebody solve the functional equation $f(-x+b)=f(x)$? (is there such a general solution for it?)

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    $\begingroup$ Note that the map $-x+b$ maps the interval $(b/2,+\infty)$ bijectively onto the interval $(-\infty,b/2)$. So: define $f$ any way you like on $[b/2,+\infty)$, and use the functional equation to extend the definition to $(-\infty,\infty)$. $\endgroup$ – GEdgar Oct 14 '18 at 18:00
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    $\begingroup$ Define $y = x - b/2$. Then every even function in $y$ satisfies your equation. $\endgroup$ – M. Wind Oct 14 '18 at 18:31
  • $\begingroup$ What is general form of the even functions? $\endgroup$ – M.H.Hooshmand Oct 14 '18 at 18:49
  • $\begingroup$ A function is called "even" if it has the property $f(-x) = f(x)$. For example the cosine function or $x^2$ or the absolute value of $x$. $\endgroup$ – M. Wind Oct 14 '18 at 19:03
  • $\begingroup$ Thanks, I know, I said the general form. Also, $y=x-b/2$ doesn't work. $\endgroup$ – M.H.Hooshmand Oct 15 '18 at 13:55
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In fact this functional equation belongs to the form of http://eqworld.ipmnet.ru/en/solutions/fe/fe1113.pdf.

The general solution is $f(x)=\Theta(x,b-x)$ , where $\Theta(u,v)$ is any symmetric function.

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