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Why does the recursion $x_{n+1} = f(x_n)$ with $f(x) = p^k\cdot\sqrt[k+1]{x}$ converge to $p^{k+1}$?

In addition, how can you write a function that has rational or integer fixed points? For example, you get $k^2$ as a fixed point of $f(x)=k\sqrt{x}$. Is there any known way to work back from an integer/rational number, and basically construct a function with said number as a fixed point?

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    $\begingroup$ "Why does $f(x)=p^k\cdot\sqrt[k+1]{x}$ converges to $p^{k+1}$?" What makes you think it does, and under what limit? $\endgroup$ – orlp Jan 30 '18 at 15:03
  • $\begingroup$ As for your second question, just $f(x) = x$ is a trivial example. You need to be more specific as to what form the function must have. $\endgroup$ – orlp Jan 30 '18 at 15:05
  • $\begingroup$ I meant, if you keep on iterating the function, you will always get $p^{k+1}$, with any value of $x$ as a starting point $\endgroup$ – user520276 Jan 30 '18 at 15:05
  • $\begingroup$ Generally when people say 'converges' they mean under some limit. If you meant as an attractive fixed point under iterated application you should mention that a bit more explicitly. $\endgroup$ – orlp Jan 30 '18 at 15:07
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    $\begingroup$ Consider the power of $p$ in $f(x)$, $f(f(x))$, $f(f(f(x)))$ etc. and show that it becomes a geometrical series: $k + \frac{k}{k+1}+\frac{k}{(k+1)^2}+\frac{k}{(k+1)^3}+\ldots$. The pattern you observe can be proved by induction if you want. $\endgroup$ – Winther Jan 30 '18 at 15:10
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Apparently, the solution is easier solved algebraically. Just let $x=p^{k}\cdot x^{\frac{1}{k+1}}$. Raising both sides by $k+1$ gives: $$\frac{x^{k+1}}{x}=p^{k(k+1)}$$ $$x^k=p^{k(k+1)}$$ $$x=p^{k+1}$$

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