Solving for a function

Question

How can I find a general solution to following equation, $$ f\left(\frac{1}{y}\right)=y^2 f(y). $$ I know that $f(y) = \frac{1}{1 + y^2}$ is a solution but are there more? Is there a general technique that I can read up about for problems of this kind?

Answer 1

To add to Olivier's comment:

Defining $g(y)=yf(y)$, the equation is $g(y)=g(1/y)$. One can ensure this by simply taking $g$ to be a constant function, i.e. taking $g(y)=k$ for all $y\neq 0$ for some constant $k$. Then $$f(y)=\frac{g(y)}{y}=\frac{k}{y}$$ for all $y\neq 0$. This satisfies the equation.

Another solution is $f(y)=\frac{k}{1+y+y^2}$ for any $k$.

Answer 2

A family of solutions to $g(y)=g(1/y)$ then would be made by any rational function $P(y)/y^n$, where $P(y)$ is a polynomial that has zeros in $\left\{ {z_1 ,\, \ldots ,\,z_n \,} \right\} \cup \left\{ {1/z_1 ,\, \ldots ,\,1/z_n \,} \right\}\quad \;\left| {\;1 \leqslant z_k } \right.$ .

Answer 3

Let $g(x)$ be any even function. Then $f(x)=\frac{1}{x}g(\ln|x|)$ and satisfies given equation.

Proof: $$f\left(\frac{1}{y}\right)=y^2f(y),\quad y\not=0$$ $$\frac{1}{1/y}g\left(\ln\left|\frac{1}{y}\right|\right)=y^2\frac{1}{y}g(\ln|y|)$$ $$yg(-\ln|y|)=yg(\ln|y|)$$ $$g(-\ln|y|)=g(\ln|y|),\quad \ln|y|=z$$ $$g(-z)=g(z)$$ Last equation is true, because $g(x)$ is even.