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

Question

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...

Answer 1

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 $\cos(x)=x$.

Answer 2

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.

Answer 3

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 $f$.

Answer 4

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

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