Proving $\displaystyle\lim_{n\to\infty}a_n=\alpha$ $a_1=\frac\pi4, a_n = \cos(a_{n-1})$

Let $a_1=\frac\pi4, a_n = \cos(a_{n-1})$

Prove $\displaystyle\lim_{n\to\infty}a_n=\alpha$.

Where $\alpha$ is the solution for $\cos x=x$.

Hint: check that $(a_n)$ is a cauchy sequence and use Lagrange's theorem.

Well I tried to show that it's a cauchy sequence: $|a_m-a_n|<\epsilon , \ \forall m,n$ but I just don't see how it's done with a trig recursion sequence...

• This is a recent duplicate which I'm failing to find. Edit: It's not a duplicate after all, but it is the same problem. See here. – Git Gud Jan 6 '14 at 9:15
• you should have made the question a little better.. may be you write $\displaystyle\lim_{n\to\infty}a_n=\alpha$ and in the next line you write "where $\alpha$ is solution of $\cos x =x$" – user87543 Jan 6 '14 at 9:16

Note that $$\cos(x):\left[\frac1{\sqrt2},\frac\pi4\right]\mapsto\left[\frac1{\sqrt2},\frac\pi4\right]$$ and on $\left[\frac1{\sqrt2},\frac\pi4\right]$, $$\left|\frac{\mathrm{d}}{\mathrm{d}x}\cos(x)\right|=|\sin(x)|\le\frac1{\sqrt2}\lt1$$ Thus, the Mean Value Theorem ensures that $\cos(x)$ is a contraction mapping on $\left[\frac1{\sqrt2},\cos\left(\frac1{\sqrt2}\right)\right]$. The Contraction Mapping Theorem says that $\cos(x)$ has a unique fixed point in $\left[\frac1{\sqrt2},\cos\left(\frac1{\sqrt2}\right)\right]$. Furthermore, for $n\ge3$, $$\left|a_n-a_{n-1}\right|\le\left(\frac1{\sqrt2}\right)^{n-3}\left|a_3-a_2\right|$$ Thus, $a_n$ is a Cauchy sequence. The point to which it converges must be the unique fixed point where $\cos(x)=x$.

• Without using the contraction mapping theorem, how can you conclude that: $$\left|a_n-a_{n-1}\right|\le\left(\frac1{\sqrt2}\right)^{n-3}\left|a_3-a_2\right|$$ ? – GinKin Jan 6 '14 at 17:06
• @GinKin: That is gotten using the Mean Value Theorem: since $|\sin(x)|\le\frac1{\sqrt2}$ on $\left[\frac1{\sqrt2},\frac\pi4\right]$, we have that, for some $\alpha_{n-1}$ between $a_{n-1}$ and $a_{n-2}$, \begin{align} \left|a_n-a_{n-1}\right| &=\left|\cos(a_{n-1})-\cos(a_{n-2})\right|\\ &=\left|\sin(\alpha_{n-1})(a_{n-1}-a_{n-2})\right|\\ &\le\frac1{\sqrt2}\left|a_{n-1}-a_{n-2}\right| \end{align} – robjohn Jan 6 '14 at 19:52

Hints: exists $\;c\;$ between $\;a_n\;,\;a_m\;$ s.t.

$$|a_n-a_m|=|\cos(a_{n-1})-\cos(a_{m-1})|\stackrel{\text{MVT}}= |(\cos c)'||a_{n-1}-a_{m-1}|=q|a_{n-1}-a_{m-1}|$$

since $\;|(\cos(c))'|=|\sin(c)|\;$ cannot be $\;1\;$ (why?) , and now

$$|a_n-a_{n-1}|\le q|a_{n-1}-a_{n-2}|\le\ldots\le q^{n-1}|a_1-a_0|$$

• It cannot be 1 because $c\in[\frac\pi 4 , \frac1{\sqrt2}]$ but why it becomes a constant... Why is it a telescopic sequence ? – GinKin Jan 6 '14 at 14:51
• What becomes constant? You know $\;\sin c=q<1\;$ for any possible $\;c\in\left[\frac\pi4\,,\,\frac1{\sqrt2}\right]\;$ , andthat's all that matters. – DonAntonio Jan 6 '14 at 14:59
• @DonAntonio, why is $c \in \left[ {\pi \over 4},{1 \over \sqrt 2} \right]$? – AnnieOK May 31 '14 at 8:56
• @DonAntonio, Why is the MVT make it $\le$? shouldn't it be an equality? – AnnieOK May 31 '14 at 9:33
• @AnnieOK, this is prehistory! Anyway, I can't remember what I was thinking of when I answered this, but you're right: under that MVT thing there must be an equality sign though this doesn't matter that much. Now, $$\frac1{\sqrt2}\le\cos\frac\pi4=\frac1{\sqrt2}<\frac\pi4$$ , so iterating cosine on these values leave us all the time between $\;\frac1{\sqrt2}\;$ and $\;\frac\pi4\;$ (remember: on $\;[0,\pi/2]\;,\;\;\cos x$ is motone descending) – DonAntonio May 31 '14 at 10:01

Look over the Banach Fixed Point Theorem. Its practically the same deal.

Hint: Show with MVT that: $$| \cos(a_{n}) - \cos(a_{n-1})| <= q^n|a_1 - a_0|$$ Where $q$ is a constant so $0 < q < 1$. Use that in the Cauchy proof.

• It's exactly the same proof as the BFPT's though I doubt it whether it'd be allowed to use that theorem in Calculus I. Yet the MVT allows to prove the sequence is Cauchy and that's almost all is needed. – DonAntonio Jan 6 '14 at 11:29

Over the interval $[0,1]$, the function $f(x) = \cos(x)$ is a strictly decreasing continuous function mapping $[0,1]$ into itself. So $f\circ f$ is strictly increasing there. If one pick $a_1 = \frac{\pi}{4} \in [0,1]$ and generate a sequence by iteration $a_n = f(a_{n-1})$, it is not hard to check

$$a_2 < a_3 < a_1$$

Since $f$ is strictly decreasing, $a_2 < a_3 < a_1 \implies a_3 > a_4 > a_2$ and hence

$$a_2 < a_4 < a_3 < a_1$$

Since $f \circ f$ is strictly increasing, this implies the even sub-sequence $a_{2n}$ is strictly increasing and the odd sub-sequence $a_{2n-1}$ is strictly decreasing. Since all $a_k$ belongs to $[0,1]$, the even and odd sub-sequences are both bounded monotonic sequences. As a result, following two limits exists:

$$a_{even} = \lim_{n\to\infty} a_{2n}\quad\text{ and }\quad a_{odd} = \lim_{n\to\infty} a_{2n-1}$$ Since $f\circ f$ is continuous, both $a_{even}$ and $a_{odd}$ are fixed points for $f \circ f$. If one make a plot of $f\circ f$ over the interval $[0,1]$, one will notice $f \circ f$ has a unique fixed point in $[0,1]$. That point is also the unique fixed point of $f$. This forces $a_{even} = a_{odd}$ and the sequence $a_n$ converges to the unique fixed point of $f$.