# The functional equation $f(-x+b)=f(x)$

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?)

• 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 $(-\infty,\infty)$. – GEdgar Oct 14 '18 at 18:00
• Define $y = x - b/2$. Then every even function in $y$ satisfies your equation. – M. Wind Oct 14 '18 at 18:31
• What is general form of the even functions? – M.H.Hooshmand Oct 14 '18 at 18:49
• 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 $x$. – M. Wind Oct 14 '18 at 19:03
• Thanks, I know, I said the general form. Also, $y=x-b/2$ doesn't work. – M.H.Hooshmand Oct 15 '18 at 13:55

## 1 Answer

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.