# Recursive sequence convergence with trigonometric terms

If I have a recursive sequence like: $$x_{n+1} = \frac{cos(x_n)}{2}$$ with$$x_0=\frac{\pi}{4}$$ And I would like to show that is converges, I tried to show that is would be monotone and bounded, so looking at the first terms $$x_1 =\frac{\sqrt2}{4}$$ which is smaller than x_0, so I could guess that my function is decreasing and since $$\frac{-1}{2} <= \frac{cos(x_n)}{2} <=\frac{1}{2}$$ it is also bounded from below thus it must converge and consequently be cauchy.

Is this type of reasoning correct to determine the convergence of this recursive sequence?

• Try Taylor expanding $\cos {x_n}$ to two terms $(1 - \frac{x^2}2)$ and see if this converges. You will see that a subsequence converges, however the whole sequence itself does not converge. – астон вілла олоф мэллбэрг Jan 11 '17 at 12:16
• The sequence converges indeed, and pretty quickly too, but it is not monotonic. – Ivan Neretin Jan 11 '17 at 12:26
• The sequence is not monotonic. It converges to the fixed point of $f\colon x \mapsto \frac{1}{2}\cos x$, but since that function is strictly decreasing in a neighbourhood of its fixed point, the $x_n$ are alternatingly smaller and larger than the fixed point. – Daniel Fischer Jan 11 '17 at 12:27
• How can I see it converges, if a subsequence converges that won't tell me the sequence converges – SoHCahToha Jan 11 '17 at 12:34
• @SoHCahToha look at my answer it shows explicitly why this sequence must converge. – Francesco Alem. Jan 11 '17 at 23:04

Perhaps this proof is more intuitive: The equation $$x=cos(x)/2$$ has a solution $x^* \in \left]0,\frac{1}{2}\right[$ without loss of generality we can define the sequence $x_n$ in terms of $\delta x_n$ such that: $$x_n=x^*+\delta x_n$$ so now the equation $$x_{n+1}=\frac{\cos(x_n)}{2}$$ becomes after substitution: $$\delta x_{n+1}+x^*=\frac{\cos(\delta x_n+x^*)}{2}$$ we also have the equality $$x^*=\frac{\cos(x^*)}{2}$$ subtracting the two equations leaves: $$\delta x_{n+1}+x^*-x^*=\frac{\cos(\delta x_n+x^*)}{2}-\frac{\cos(x^*)}{2}$$ simplifying $$\delta x_{n+1}=\frac{\cos(\delta x_n+x^*)-\cos(x^*)}{2}=-\frac{1}{2}\int_{x^*}^{\delta x_n+x^*}\sin(x)dx$$ so we have: $$|\delta x_{n+1}|=\frac{1}{2}\left|\int_{x^*}^{\delta x_n+x^*}\sin(x)dx\right|\le\frac{1}{2}\left|\int_{x^*}^{\delta x_n+x^*}\left|\sin(x)\right|dx\right|\le\frac{1}{2}\left|\int_{x^*}^{\delta x_n+x^*}1dx\right|=\frac{1}{2}|\delta x_n|$$ finally: $$\frac{|\delta x_{n+1}|}{|\delta x_n|}\le \frac{1}{2}\implies \lim_{n\to \infty}\delta x_n=0$$ this proves that $x_n = x^* +\delta x_n\to x^* +0=x^*$.
Hint: let $f(x)= \frac{1}{2}\cos x$. Then , by the Mean Value Theorem:
$|f(x)-f(z)| \le \frac{1}{2}|x-z|$
Now its your turn to show, that $(x_n)$ is a Cauchy-sequence.
This problem is usually done with Banach's fixed point theorem. Consider $f : \Bbb R \to \Bbb R$ given by $f(x) = \dfrac {\cos x} 2$. Notice that $f'(x) = - \dfrac {\sin x} 2$, so $|f'(x)| \le \dfrac 1 2$ for all $x \in \Bbb R$. Using Lagrange's mean value theorem it follows then that $f$ is a contraction (of Lipschitz constant $\dfrac 1 2$) and therefore, according to Banach's theorem, it has a unique fixed point (i.e. a point $x^*$ with $f(x^*) = x^*$).
Letting $x_0 \in \Bbb R$ arbitrary and $x_{n+1} = f(x_n)$, one shows that $x_n \to x^*$ (proofs can be found in the Wikipedia article on Banach's theorem mentioned above).