5
$\begingroup$

Is there a name for an equation that takes the following form? $$F(f(x),f^{-1}(x),x)=0$$ A nice example being $$f(x)-f^{-1}(x)=0$$ because the solutions of this equation are their own inverses. WolframAlpha solved this problem, though I had to type the equation as $f(f(x))=x$. Is there any nice method for solving such equations? How do you think WolframAlpha does it? For instance, what if I wanted to solve $$f(x)={f^{-1}(x)}^2+x$$ Which could also be written as $$f(\sqrt{f(x)-x})-x=0$$

$\endgroup$
  • $\begingroup$ By putting $x=f(y)$ you will get $F(f(f(y)),y,f(y))=0$ and this will possibly be an ordinary functional equation. $\endgroup$ – Mohsen Shahriari Jun 10 '15 at 19:36
  • $\begingroup$ Related. $\endgroup$ – Cosmas Zachos Jul 15 '18 at 21:44
2
$\begingroup$

Yes, basic functional equations theory. It is far more hit-or-miss than other more systematic fields.

Resource recommendation: You might start with something practical, like C Efthimiou's Introduction to Functional Equations, AMS, ISBN: 978-0-8218-5314-6 online; and only then proceed to, e.g. J Aczel's Lectures on Functional Equations and Their Applications, ISBN: 9780486445236 , and on and on...

First, resolve the constraint, as you are doing, and neutralize inverses, as you already did. One cannot reverse-engineer W-Alpha, but it simply looks things up from a lookup list. See it gag, abjectly, on , e.g. , $f(f(x))=-x$.

It looked up $f(f(x))=x$, since this is the celebrated Babbage equation, solved in 1815. Note from the WP article that any workable functional conjugate of a solution, so the basic fractional solution given, produces further solutions... an infinity of them. W-Alpha throws a handful of illustrative examples to the impressionable and uninformed audience, and all but invites them to find "new" solutions.... supplant the 3 by a 5 ... etc. There may be sweet conjugacy orbits of $\Psi(x)$s which you may find practical for your purposes.

Your second example equation is messy, and, since you are asking orientation questions, I might supplant it with a much easier one, instead, $$ f(f(x)-x)=x . $$ Note the fixed point $f(0)=0$. Consider the Taylor expansion around it, $$ f(x)=ax+bx^2+cx^3+... $$ Solve for a,b,c,... recursively, order by order in x.

A particularly central functional equation utilizing functional conjugacy is Schröder's equation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.