Consider the cosine function $f = \cos : \Bbb R \to \Bbb R$.

Is it true that the set of iterates $$\left\{f_n := \cos \circ \dotsb \circ \cos, \; n \text{ times } \mid n \geq 1\right\}$$ is linearly independent over $\Bbb R$ ? That is, I am wondering if, for any $r \geq 1$ and any real numbers $a_k$, we have : $$\sum_{k=1}^r a_k f_k = 0 : \Bbb R \to \Bbb R \implies a_k=0 \;\forall k.$$

I know that this true if we consider the powers of $\cos( \cdot )$, but I don't know how to deal with compositions. What I tried is to take derivative, or induction on the minimal length of linear dependence relation.

  • $\begingroup$ You could use Taylor expansion, but I'm not certain that there will be a nice pattern allowing for an easy proof with arbitrary $n$. $\endgroup$ – Arnaud Mortier Mar 21 '18 at 17:53

If you can use the fact that $\cos(x)$ and its iterates are entire functions of a complex variable, you can use the following idea (I use your notations): We proceed by inductioon, the case $n=1$ is obvious.

Let $n\geq 2$, and suppose that $a_1f_1(x)+a_2f_2(x)+\cdots+a_nf_n(x)=0$ for all $x\in \mathbb{R}$. Then this imply that $g(z)=a_1z+a_2f_1(z)+\cdots+a_nf_{n-1}(z)$ is zero for all $z\in [-1,1]$ (because $g(\cos(x))=0$, we have put $z=\cos(x)$). As $g$ is entire, this imply that $g(z)=0$ for all $z\in \mathbb{C}$, and that $a_1z$ is periodic with period $2\pi$. Hence $a_1=0$. Now , putting $b_1=a_2,...$ etc, we have $b_1f_1(x)+\cdots+b_{n-1}f_{n-1}(x)=0$ for all $x$. The induction hypothesis apply, and we are done. .

  • 1
    $\begingroup$ OK. I see it now. It is not very obvious that in the second line you used the change of variable $z=\cos(x)$, you should probably indicate that. $\endgroup$ – Arnaud Mortier Mar 21 '18 at 18:20
  • 4
    $\begingroup$ This appears to be a rare case when all previous downvoters checked on the edited post and either removed or reversed their downvotes. Nice answer. $\endgroup$ – davidlowryduda Mar 21 '18 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.