Let $f: \Bbb R \to \Bbb R$ be an odd function.

Is the compositon of $ \underbrace{f \circ f \circ f \circ \cdots \circ f}_{\text{$n$ times}}$ odd or even?

Do I need to prove it with separate cases for $n$ even and $n$ odd?

  • $\begingroup$ You don't need to consider separate cases for $n$ even and odd. Use induction to prove the result. $\endgroup$ – Math Lover Nov 26 '17 at 18:02
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    $\begingroup$ Hint: If $f$ is odd then $f(f(-x)) = f(-f(x))=-f(f(x))$. $\endgroup$ – Rafael Gonzalez Lopez Nov 26 '17 at 18:04

For simplicity of notation, let $f^n$ denote the $n$-fold composition of $f$ with itself, i.e. $$ f^n := \underbrace{f\circ f \circ \dotsb \circ f}_{\text{$n$ times}}. $$ We claim that if $f:\mathbb{R} \to \mathbb{R}$ is odd, then $f^n$ is odd for all $n$. The proof is by induction. As a basis for induction, note that $f^1 = f$, which is odd.

Now suppose that $f^k$ is odd. Then for any $x\in \mathbb{R}$, we have \begin{align} f^{k+1}(-x) &= f(f^k(-x)) \\ &= f(-f^k(x)) &&\text{(induction hypothesis)} \\ &= -f (f^k(x)) &&\text{($f$ is odd)} \\ &= -f^{k+1}(x). \end{align} That is, $f^{k+1}(-x) = -f^{k+1}(x)$ for all $x\in\mathbb{R}$. Therefore $f^{k+1}$ is odd, which completes the induction proof.


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