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As mentioned in the post What is the solution of $\cos x=x$, one can obtain the solution by this interesting procedure: start with a 0 on a calculator and keep pressing the cosine button. The sequence of displayed numbers will stabilize pretty quickly and that gives the solution. My question is: why in the first place is this sequence convergent? I can tell from the graph of $y=\cos x$ and $y=x$ but I would like to see an analytical proof.

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No_way
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  • Possible duplicate: http://math.stackexchange.com/questions/46934/what-is-the-solution-of-cosx-x?lq=1 – Atvin Jun 08 '15 at 12:30
  • @Atvin: No, that question asks for the value of the fixpoint. This one asks for proof that iterating the function will converge towards the fixpoint. That is two different things. – hmakholm left over Monica Jun 08 '15 at 12:31
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    See http://en.wikipedia.org/wiki/Fixed_point_%28mathematics%29#Attractive_fixed_points and http://en.wikipedia.org/wiki/Banach_fixed-point_theorem – nabla Jun 08 '15 at 12:31

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There's a general theory of function iteration which (among other things) states that if $f$ has a fixpoint $f(x)=x$ with $|f'(x)|<1$, then iterating the function in a neighborhood of the fixpoint will converge towards it.

Since the derivative of the cosine is always less than $1$ except at a few particular points that we can quickly see are not fixpoints, if follows that its fixpoint (which it is also easy to see it must have) must be attractive.

hmakholm left over Monica
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Let this sequence be $\{x_n\}_n^\infty$. It is easy to see that $|x_n|\leq 1$. In the following we need the well-known fact that $|\sin x|\leq |x|$.

Claim: $|x_{n+1}-x_{n+2}|<c|x_n-x_{n+1}|$ for some positive $c<1$.

Proof: \begin{align*} |\cos x_n-\cos(\cos x_n)|&=\left|2\sin \frac{x_n+\cos x_n}{2}\sin\frac{\cos x_n-x_n}{2}\right|\\ &\leq \left|\sin\frac{x_n+\cos x_n}{2}\right|\cdot|x_n-\cos x_n|\\ &\leq \sin 1\cdot|x_n-\cos x_n| \end{align*}

Using the claim, we can show that the sequence is Cauchy and hence convergent.

\begin{align*} |x_n-x_m|&\leq |x_n-x_{n+1}|+\cdots+|x_{m-1}-x_m|\\ &\leq (c^{n-1}+\cdots+c^{m-2})|x_1-x_2|\\ &<\frac{c^{n-1}}{1-c}|x_1-x_2| \end{align*} The latter can be made arbitrarily small if $n, m$ are sufficiently large.

Alex Fok
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