5

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?

Raj
  • 964
  • 11
    If you set $g(x):=xf(x)$, then you are looking for all $g$ such that $g(x)=g(1/x)$: they are plenty of such $g$. – Olivier Oloa Jul 14 '16 at 18:04
  • 3
    $f$ can be arbitrary on $[-1,1]$, then you can define $f$ on $(-\infty, -1) \cup (1,\infty)$ by $f(x) = x^{-2} f(1/x)$ to get a solution. – Robert Israel Jul 14 '16 at 18:14

3 Answers3

4

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 $k$.

smcc
  • 5,694
2

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.

Somnium
  • 1,567
  • 1
  • 13
  • 27
  • This is very nice, but there are some typos here. From the second line on you should have $g$ on each side rather than $f$. Also you need to restrict the domain of $f$ to positive real numbers (i.e. it should be $y>0$ instead of just $y\neq 0$). (Another small thing: in writing your proof perhaps you should probably not start by assuming what you want to prove.) – smcc Jul 15 '16 at 22:44
  • @smcc Corrected typo and changed to be valid for negative $y$ also. – Somnium Jul 16 '16 at 17:32
0

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 $\left\{ {z_1 ,\, \ldots ,\,z_n \,} \right\} \cup \left\{ {1/z_1 ,\, \ldots ,\,1/z_n \,} \right\}\quad \;\left| {\;1 \leqslant z_k } \right.$ .

G Cab
  • 35,272