# Proof that the solution to cosx = x, is the limit of a recursive sequence.

So I've got this question. Exists a sequence $a_n$ such that: $$a_0 = \frac \pi4, a_n=\cos\left(a_{n-1}\right)$$ Prove that $\lim_{n\rightarrow\infty} a_n = \alpha$ Where $\alpha$ is the solution to $\cos x = x$.

There is also a small hint saying I should prove its a Cauchy sequence and then use MVT.

Well, I've proven its Cauchy, and then used the fact that there exists a $L \in \mathbb{R}$ such that:

$$\lim_{n\rightarrow\infty} \cos(a_{n-1}) = L.$$

Because cos is continuous, it can be written as: $\cos(\lim_{n\rightarrow\infty} a_{n-1}) = L$ which is $\cos(L) = L$. I've proven with IVT and the derivative that $\alpha$ from before is unique, therefore $L = \alpha$.

It looks fine to me, but the hint said to use MVT, which I don't really see why, or where to use it. Hopefully you guys will know more.

• Hm...lots of exercises from the last exercise in Diff. Calculus $\;\aleph\;$ in the Tel-Aviv University Mathematics Dept.... – DonAntonio Jan 4 '14 at 0:05
• About the MVT (Lagrange's Theorem) Hint: how exactly did you prove the sequence is Cauchy?? – DonAntonio Jan 4 '14 at 0:06
• ...and for those whose studies/jobs might depend on those examiners, @DavidHolden . Don't you think? – DonAntonio Jan 4 '14 at 0:24
• @DonAntonio I did prove its Cauchy using MVT, but the 'hint' suggests I should use it afterwards. I have quite low self - confidence when it comes to math, so if a 'hint' suggests to do something and I do it differently, I always assume its wrong somewhere. That's why I browse this site a lot, to reassure myself i'm not doing it wrong :) – Xsy Jan 4 '14 at 0:35
• @Xsy , if you already proved the sequence is convergent $\;\iff\;$ it is Cauchy, then the end of the exercise is just as you did. The order of the hints in the exercise can indeed be misleading, but imo the MVT is used only for the first part. – DonAntonio Jan 4 '14 at 0:44

$g(x):=\cos x-x\Rightarrow g(-{\pi\over 3})>0 ,g({\pi\over 3})<0$. As $g$ is continuous over $\mathbb{R}$, $g$ vanishes somewhere between $(-{\pi\over 3},{\pi\over 3})$. Note that as $\pi\gt 3$, $g$ can't vanish outside $(-{\pi\over 3},{\pi\over 3})$ otherwise it would mean $|\cos (x)|\gt 1$ for some $x$. Also notice $|{d\over dx}\cos x|=|\sin x|< {1\over \sqrt 2}\;\forall\; x\in (-{\pi\over 3},{\pi\over 3})$. As $\cos x$ is positive in $({-\pi\over 3},{\pi\over 3})$, the sole solution of $\cos x=x$, let's call it $\alpha$, must lie in the interval $(0,{\pi\over 3})$ (we can rule out $x=0$ by direct checking).

As cosine is differentiable over $(0,{\pi\over 3})$, using $\mathtt{MVT}$ for $x,y \in (0,{\pi\over 3}) \text{ with } x\neq y$ we see that $\exists c: x< c< y$ and $\cos x-\cos y=(x-y)\sin c\Rightarrow |\cos x-\cos y|\le \left({1\over\sqrt{2}}\right)|x-y|$. As $|\cos x|\le1$, for $n>1,x_n\in(0,{\pi\over 3})$. Therefore for $n>1$ we have

$$|\alpha-x_{n}|=|\cos\alpha-\cos x_{n-1}|\le\left({1\over\sqrt{2}}\right)|\alpha-x_{n-1}|\le\left({1\over\sqrt{2}}\right)^2|\alpha-x_{n-2}|\\ \le...\le\left({1\over\sqrt{2}}\right)^{n}|\alpha-x_0|:=\left({1\over\sqrt{2}}\right)^n\delta \cdots(1)$$

So, for $m>n>1$

$$|x_m-x_n|\le|\alpha-x_m|+|\alpha-x_n|\le\left[\left({1\over\sqrt{2}}\right)^m+\left({1\over\sqrt{2}}\right)^n\right]\delta<2\delta\left({1\over\sqrt{2}}\right)^n\cdots(2)$$

As $2\delta$ is a constant and $\left({1\over\sqrt{2}}\right)< 1$, $(2)$ proves that $\{x_n\}$ is Cauchy and $(1)$ proves that it converges to $\alpha$.

• can you explain why $\alpha$ is the sole solution for the equation: $\cos(x) = x$? – Jneven Jan 21 at 13:39