Proposition
Let $f(x) = k^{x}x$, where the values of both $f(x)$ and $k$ are known.
Let $x_{0} = f(x)$, and:
$$x_{n + 1} = \frac{1}{2}\log_{k}{\left(\frac{k^{x_{n}}x_{0}}{x_{n}}\right)}$$
$$\lim_{n \to \infty}{x_{n}} = x$$
I was wondering how one could prove this. I derived it in rather an odd manner.
Thank you.
I note the following:
$$x_{n+1}=\frac12\log_k\left(\frac{k^{x_n}x_0}{x_n}\right)=\frac12(x_n+\log_k\left(\frac{x_0}{x_n}\right))$$
$$x_0=x_0$$ $$x_1=\frac12x_0$$ $$x_2=\frac12(\frac12x_0+\log_k(2))$$
And so I find it slightly difficult, finding a proof that $\lim_{n\to\infty}x_n=x$, but I will most certainly enjoy seeing it.