Function

Posted July 17th, 2009 by alpha

If $f(x)=(a-x^n)^\frac{1}{n} , a>0 ,n\in N$ , then prove that $f(f(x))=x \forall x$

Comments

On July 18th, 2009 Structure says:

Let $g:[0,a]\rightarrow[0,a]\:g(x)=a-x$
We have

$g(g(x))=g(a-x)=a-(a-x)=x$

Let $h:[0,\sqrt[n]{a}]\rightarrow[0,a]$ $h(x)=x^n$
$f(x)=h^{-1}\circ g\circ h(x)$

$f\circ f=h^{-1}\circ g\circ h\circ h^{-1}\circ g\circ h=h^{-1}\circ g\circ g\circ h=h^{-1}\circ h=id_{[0,a]}$