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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?

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  • $\begingroup$ 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. $\endgroup$ – астон вілла олоф мэллбэрг Jan 11 '17 at 12:16
  • $\begingroup$ The sequence converges indeed, and pretty quickly too, but it is not monotonic. $\endgroup$ – Ivan Neretin Jan 11 '17 at 12:26
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    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Jan 11 '17 at 12:27
  • $\begingroup$ How can I see it converges, if a subsequence converges that won't tell me the sequence converges $\endgroup$ – SoHCahToha Jan 11 '17 at 12:34
  • $\begingroup$ @SoHCahToha look at my answer it shows explicitly why this sequence must converge. $\endgroup$ – Francesco Alem. Jan 11 '17 at 23:04
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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 $x_n = x^* +\delta x_n\to x^* +0=x^*$.

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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.

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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).

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