5

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.

Xsy
  • 837
  • Hm...lots of exercises from the last exercise in Diff. Calculus $;\aleph;$ in the Tel-Aviv University Mathematics Dept.... – DonAntonio Jan 04 '14 at 00:05
  • About the MVT (Lagrange's Theorem) Hint: how exactly did you prove the sequence is Cauchy?? – DonAntonio Jan 04 '14 at 00:06
  • I have recently asked several questions about what I call the "cosine fixed point", which I have baptized with the name "$\alpha$" - a very interesting transcendental number. @user44197 has a very good insight into this number. any question which asks you to use a particular method signals itself as an exercise rather than an invitation to creative thought. to say this is not to downgrade exercises. but when "doing mathematics" you should always be attempting to "do mathematics". the task of satisfying examiners can safely be left to those who pay their salaries – David Holden Jan 04 '14 at 00:19
  • 1
    ...and for those whose studies/jobs might depend on those examiners, @DavidHolden . Don't you think? – DonAntonio Jan 04 '14 at 00:24
  • 1
    @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 04 '14 at 00:35
  • Maybe you need to prove that the sequence is convergent. – gaoxinge Jan 04 '14 at 00:37
  • @Don I like a good controversialist! human life is a collective endeavour. all play their part. both examiners and those who are critical of the over-influence of the examiner thought-paradigm – David Holden Jan 04 '14 at 00:38
  • 1
    @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 04 '14 at 00:44
  • Thank's @DonAntonio. Never thought i'd waste so much time because of a hint, heh. – Xsy Jan 04 '14 at 00:55
  • @Xsy also see Stephen Montgomery-Smith's comment on my question from a few days ago: http://math.stackexchange.com/questions/621993/convergence-of-the-iterated-cosine (convergence of the iterated cosine) – David Holden Jan 04 '14 at 11:59

1 Answers1

3

$g(x):=\cos x-x\Rightarrow g(-{\pi\over 3})>0 ,g({\pi\over 3})<0$. 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 $\alpha$.

Jack's wasted life
  • 8,849