# Probably easy recursion formula solving $x^x=a$

Let $a,x_0\in\mathbb{C}$ and set $$x_{n+1}=\sqrt{x_n\cdot \sqrt[x_n]{a}}$$ For which $a$ and $x_0$ does $x=\lim\limits_{n\rightarrow\infty} x_n$ exist with $$x^x=a$$ Approximating it with a computer shows that it holds in many cases. Furthermore, $$(\frac{d}{dz}\sqrt{z\cdot \sqrt[z]{a}})x<1$$ for $x^x=a$. However, this does not prove the property.

I could not find a proof for this. Can somebody help me?

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–  leonbloy Apr 28 '12 at 23:47
Could you be a little more specific? I do not really know how to find a Lipschitz-Constant –  schoppenhauer Apr 29 '12 at 17:14
Sorry, I'd missed that we are in $\mathbb{C}$. In reals, it feels a little easier - though not very easy: I think that you can find a suitable interval (perhaps $(\log(a),+\infty)$) and prove that inside that interval the $x=g(x)$ has a single solution and $g'(x)<1$, with $g(x)= \sqrt{x a^{1/x}}$. BTW, $g(x)$ can be understood as a geometric mean between $x$ and $g_0(x)=a^{1/x}$ - $g_0(x)$ is also a candidate for the recursion, but it diverges; the geometric mean works as an relaxation –  leonbloy Apr 30 '12 at 13:05