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For

$x_{n+1} = f(x_n)$ [1]

and

$z_{n+1} = f(f(z_n))$ [2]

Show that all fixed points of [1] are also fixed points of [2]. If the converse is true, describe what some fixed points of [2] are in terms of [1].

I think I understand how to show that all fixed points of [1] are fixed points of [2]:

Substitute $\bar{z} = f(\bar{z})$ into $\bar{z} = f(f(\bar{z}))$ to get $\bar{z} = f(\bar{z})$

But I'm not sure how to do the rest.

Joshua
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1 Answers1

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I do not think the converse is true. Consider $f$ that maps $3$ to $4$ and $4$ to $3$. Then $3$ is a fixed point of $f\circ f$ but not of $f$.

Jan
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