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What are the periodic points of period 5 for $f(x) = x^2+1$ if $f:\mathbb{R}\rightarrow\mathbb{R}$?

Solving the full equation for $f^5(x)$ seems like a nightmare, but I'm not sure of any better way.

Any help appreciated!

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    Note that $f(x) = x + \bigl(x - \frac{1}{2}\bigr)^2 + \frac{3}{4}$. – 2017-02-07

3 Answers 3

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For all $x \in \Bbb R$ $$ |f(x)| = f(x) = x^2 + 1 = \underbrace{(|x|-1)^2 + |x|}_{> 0} + |x| > |x| $$ therefore $f$ has no real periodic points of any order.

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There are no real periodic points. We have

$$((((x^2+1)^2+1)^2+1)^2+1)^2+1 = f_5 (x)$$

For all real $x$, $f_5(x) \geq f_5(0)= 677$. So if we require $f_5(x) = x$, then $x$ must be at least 677. However for $x \geq 677$, always $f_5(x) > x$.

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Alpha finds no real roots. It is a $32$nd degree polynomial, so there should be $32$ roots. The plot is below, The imaginary cube roots of $1$ are two of them. The others do not seem to have a simple form.

enter image description here

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    Try to check whether $f$ has periodic points of period 3. If yes, Sharkovski's Theorem guarantees the existence of periodic points of all periods. If there are no points of period 3 - everything is possible. – 2017-02-07
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    @szw1710: there are no real periodic points of period $3$. Two of the imaginary ones are the complex cube roots of $-1$. The others seem to have no simple form. – 2017-02-07