The functional equation $\frac{f(x)}{f(1-x)} = \frac{1-x}{x}$

Question

One set of solutions to this is $f(x) = \frac{c}{x}$ for constant $c$. Are these the only solutions?

Answer 1

Let $f$ be a solution to the equation, and let $g(x):=(x+ \frac{1}{2})f(x + \frac{1}{2})$. Then $g$ is an even function (i.e. $ g(x)=g(-x)$). Conversly, let $h$ be any even function. Then by setting $f(x):=\frac{h\left(x-\frac{1}{2}\right)}{x}$, we get $xf(x)=(1-x)f(1-x)$. Thus the set of all solutions to this equation is exactly the set of functions of the form: $$f(x):=\frac{h\left(x-\frac{1}{2}\right)}{x}$$ Where $h$ is an even function.

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Some examples:

\begin{align*} f(x) &= \frac{c}{x} \\ \\ f(x) &= \frac{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}{x} = 1-x \\ \\ f(x) &=\frac{\cos\left(x-\frac{1}{2}\right)}{x} \\ \end{align*}

Answer 2

No, since $f(x)=1-x$ is also a solution, as I wrote in the comments.

Answer 3

Among degree one polynomials, you can find solutions. Suppose $f(x)=a+bx$. Then you require

$$\frac{a+bx}{a+b(1-x)}=\frac{1-x}{x}$$ $$(a+bx)x=(1-x)(a+b-bx)$$ $$ax+bx^2=a+b -bx-(a+b)x+bx^2$$ $$0 = (a+b)(1-2x)$$ so that $b=-a$. You therefore have a family of functions $$f_a(x)=a\cdot(1-x)$$ which are solutions.

You can try the same procedure with other types of functions with arbitrary coefficients and see what you might get.