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8 votes

Existence of function satisfying $f(f'(x))=x$ almost everywhere

Looking for a solution of the form $f(x)=ax^b$, $x>0$, one finds $$ a = \phi^{-\phi/(\phi+1)}, ~~~ b=\phi $$ where $\phi=\frac{\sqrt{5}+1}{2}$ is the Golden ratio.
Marc Chamberland's user avatar