Solving for a function 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?
 A: 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 $k$.
A: 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.
A: 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 $\left\{ {z_1 ,\, \ldots ,\,z_n \,} \right\} \cup \left\{ {1/z_1 ,\, \ldots ,\,1/z_n \,} \right\}\quad \;\left| {\;1 \leqslant z_k } \right.$
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