# The condition for the existence of a symmetric form for the reflection formula $f(1-x)= \chi (x) f(x)$

Suppose we have a functional equation in the form $$f(1-x)=\chi (x) f(x)$$ with given function $$\chi (x)$$. What is the condition on the function $$\chi (x)$$ so that we can write this reflection relation in a symmetric form $$\xi(1-x)=\xi(x) ?$$ I have the only simple answer: the $$\chi (x)$$ function should be of the type:$$\chi (x)= \phi(1-x)/\phi(x)$$ then $$\xi(x)=f(x)/\phi(x)$$. Is it correct?

Edit I want to illustrate my question with an example.

Let first $$\chi (x)$$ be $$\chi (x)= \frac {\pi}{\sin\pi x (\Gamma (x))^2}$$ And we have functional equation $$\Gamma (x) \Gamma (1-x)=\frac {\pi}{\sin\pi x}$$ without symmetric form.

Let then $$\chi (x)$$ be $$\chi (x)= \frac { \pi^{-\frac {x}{2}}\Gamma(\frac{x}{2})} { \pi^{-\frac {1-x}{2}} \Gamma(\frac{1-x}{2})}$$ Here $$\phi(x)= ({ \pi^{-\frac {x}{2}}\Gamma(\frac{x}{2})})^{-1}$$ and we have symmetric form with $$\xi(x)=\zeta(x)/\phi(x)$$

• I have the feeling that you can always write $f(1-x)=\chi (x) f(x)$ for an appropriate $\chi$ and set $\xi(x) = f(x)f(1-x)$ or some other symmetric form. Is there something I'm missing? – G.Carugno Apr 10 at 10:41
• @G.Carugno, I mean the case when we don't know the expliсit expression for $f$ and do have an explicit expression for $\chi$ and if it is in the form $\chi (x)= \phi(1-x)/\phi(x)$ we can get symmetrical (but also unknown) function $\xi (x) = f(x)/ \phi(x)$ – Aleksey Druggist Apr 10 at 12:52
• Taking the log derivative we have $F(x)-F(-x) = G(x)$ where $F(x) = f'/f(1/2+x), G(x) = \chi'/\chi(1/2+x)$ so $G$ is odd so $G(x) = H(x)-H(-x)$ and $F(x)+H(x)$ is odd ie. $f(1/2+x) h(1/2+x)$ is even where $h(x) = \exp(\int H(1/2- x)$. With $f(x) = \zeta(x)$ then $h(x) = \pi^{-x/2} \Gamma(x/2)$ – reuns Apr 11 at 11:08
• I still want to understand the answer to my question As far as I can see the function $\chi$ must be the kind of $\chi (x)= \phi(1-x)/\phi(x)$ to satisfy the original functional equation $f(1-x)=\chi (x) f(x)$ hence the additional condition for the existence of a symmetric form is not required – Aleksey Druggist Apr 12 at 9:46

There is not much going on here. Just to make that clear, without loss of essential generality, suppose that we have any function $$\ f(x)\$$ and define $$\ g(x) := f(x) - f(-x).\$$ Now we have $$\ g(-x) = f(-x) - f(-(-x)) = f(-x) - f(x) = -g(x).\$$ Thus, $$\ g(x)\$$ is an odd function as in Even and odd functions. In fact, it is twice the odd part of $$\ f(x).\$$ Further, it follows that by adding any even function to $$\ f(x)\$$ we can recover the same odd function $$\ g(x).\$$
In your case, addition and subtraction is replaced by multiplication and division, and also the involution $$\ x \to -x\$$ is replaced by $$\ x \to 1-x.$$