# Solution to the functional equation $(x(t))^p(1-t^2)=x\big(\frac{2t}{1-t^2}\big)$ for $x(t).$

Let $p\in\mathbb{R}$ be a fixed real number. I am trying to find any and all functions $x(t)$ such that $$(1-t^2)(x(t))^p=x\bigg(\frac{2t}{1-t^2}\bigg)$$ Is there a solution to this functional equation? If so, what is it, and for which values of $p?$ I am hoping to find a closed form solution as a function of $t$ and $p.$

I have solutions for $p=0$ and $p=2.$ When $p=0,$ I was able to find two functions that solve this equation, and they are $$x(t)=-\frac{2}{t^2}\pm\frac{1}{t}\sqrt{1+\frac{1}{t^2}}.$$ When $p=2,$ I obtain the piecewise function below as a solution: $$x(t)=\left\{ \begin{array}{ll} \frac{1}{\sqrt{1+t^2}} & t\in[-1,1] \\ -\frac{1}{\sqrt{1+t^2}} & |t|>1 \\\end{array} \right..$$ I don't know if these are the only solutions. Please help me. Anything helps. Thank you very much.

• Only a note: $\enspace\displaystyle x(t):=\prod\limits_{k=1}^\infty (1-(\tan\frac{\arctan t}{2^k})^2)^{p^{k-1}}\enspace$ also works (if it's convergent, which I haven't checked) :-) Aug 7, 2017 at 16:19
• Also, if we had $u(t)(x(t))^{p}=x\big(\frac{2t}{1-t^2}\big)$ then $x(t):=\prod_{n=1}^{\infty} \big(u(\tan\big(\frac{\arctan t}{2^n}\big)\big)\big)^{p^{n-1}}.$ You're the best. Just a question, though. Does this require that $|t|<1$? The arctangent identity $$\arctan\Big(\frac{x+y}{1-xy}\Big)=\arctan(x)+\arctan(y)$$ is only applicable when $|xy|<1.$
– user429040
Aug 8, 2017 at 1:06
• Of course, the value ranges must be considered. But even if the $\arctan$ converges, this doesn't mean that the product is convergent. This must be examined separately. – Additional note: $\arctan$ is only a symbol, means: For calculations you can use e.g. the integral ($t\in\mathbb{R}$) or the functional equation (as you have written above) instead of the series ($|t|<1$) and then the value range is optimal. :-) Aug 8, 2017 at 7:09

I solved my original problems! The general solution to $u(t)\big(x(t)\big)^p=(x\circ f)(t)$ is $$x(t):=\prod_{n=1}^{\infty}\big(g_n(t)\big)^{p^{n-1}}$$ where $$g_n:=u\circ f^{-n}.$$ This is, of course, assuming $f$ is invertible and this product of functions converges (point-wise?). I am currently trying to find conditions for this by looking at the the corresponding log-sum. Thank you for your insight and techniques which ultimately led to my solution. I know user90369 is anonymous, but, if you'd like, I am willing to cite you in a paper I plan on publishing which uses this result. I can't thank you enough. Also, thanks to Ronald Blaak whose solution made me realize that if this equation has one solution and $p=1$, then any constant multiple must also be a solution.

• We can relax the condition on invertibility and get the general solution $$x(t):=\prod_{n=1}^{\infty} \big(h_{n}(t)\big)^{-1/p^{n}}$$ where $h_{n}:=u\circ f^{n-1}.$
– user429040
Aug 10, 2017 at 5:16

Solving functional equations is not necessarily impossible, but in general quite bit of work. It would be therefore good idea to give people a more compelling reason to want to invest some time in helping you solve such a problem. Specially since you actually ask for a generic solution with arbitrary $p$. So why do you actually want to solve this particular problem? If it is of the type, let's see whether I/we can you underestimate the difficulties of such a question.

Anyway, for the case $p=1$ there is the general solution $f(x) = c \frac{ \arctan x}{x}$ for any constant $c$.

• Is there anyway you can email me? This problem is related to an original theorem in my research, and I actually love you for helping me out.
– user429040
Aug 7, 2017 at 12:47
• omg it works I LOVE you
– user429040
Aug 7, 2017 at 12:52
• Okay. I read copyright laws, and, technically speaking, writing this post actually counts as publishing my work. So here it goes. I'm just afraid that one of the many math geniuses on here takes my idea and runs with it.
– user429040
Aug 7, 2017 at 12:54
• I'm just going to do a bit more research on copyright procedures before writing my theorem out.
– user429040
Aug 7, 2017 at 13:22
• Just google my name and you will get my homepage in Vienna Aug 7, 2017 at 15:24