There is a sequence which satisfies
$$x_{1} = a$$
$$x_{n+1} = x_{n} + \sin x_{n}$$
where a = 1 .
Why does $$\lim_{n \rightarrow \infty} x_{n} = \pi$$ hold ??
(the first version of question was with 2 misprints!!)
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2$\begingroup$ It doesn't. Take $a=0$. $\endgroup$– TZakrevskiyJul 11, 2014 at 15:40
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$\begingroup$ take $a=0$, the limit is thus $0$ $\endgroup$– cirpisJul 11, 2014 at 15:40
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$\begingroup$ Also, what have you tried? If you tell us this then we will be better able to help you. And it helps us feel that we are not just doing your homework for you. $\endgroup$– user1729Jul 11, 2014 at 15:40
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$\begingroup$ If $x_n \to x$, we have $x = x+ \sin x$ by continity, hence $\sin x = 0$, so $x\in \pi\mathbb Z$. $\endgroup$– martiniJul 11, 2014 at 15:41
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1$\begingroup$ @KayneWest If your question has already three(!) answers, you should consider accepting one and asking the corrected question as a new one, leaving this one here in the old state. $\endgroup$– martiniJul 11, 2014 at 15:51
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4 Answers
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If $f(x)=x+\sin(x)$ then $f$ is strictly increasing on $[0,\pi)$ and $f((0,\pi))\subset(0,\pi)$, so if $x_1=a\in(0,\pi)$ then $x_n\in(0,\pi)$. Moreover, it is clear that $f(x)>x$ for $x\in(0,\pi)$. That is $$\forall\,x\in(0,\pi),\qquad 0<x<f(x)<\pi$$ So, if $x_1=a\in(0,\pi)$ then by an easy induction we get $$\forall\,n\geq1,\qquad 0<x_n<x_{n+1}<\pi$$ This proves that $(x_n)_{n\geq1}$ is bounded and increasing. So, it must convege to some limit $\ell\in(0,\pi]$ with $f(\ell)=\ell$. this implies that $\ell=\pi$ and we are done.
Remark. The convergence of this sequence is remarkably fast. Indeed, if $e_n=\pi-x_n$ then $$0\leq e_{n+1}=e_n-\sin e_n\leq \frac{e_n^3}{6}$$ So the rate of convergence is cubic. For instance, when $x_1=1$ we have $e_6=5.6 \times 10^{-17}$, while for $x_1=3$ we have $e_4<10^{-16}$. $$ 0<\pi-\left( 3+\sin 3+\sin (3+\sin 3)+\sin (3+\sin 3+\sin (3+\sin 3))\right)<10^{-16}.$$
answered Jul 11, 2014 at 16:00
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$\begingroup$ Using the same argument with decreasing instead of increasing, we see that $x_n\to\pi $ for all $a\in(0,2\pi)$. $\endgroup$ Jul 11, 2014 at 16:08
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$\begingroup$ @HagenvonEitzen, Generally, for $a\in(2k\pi,2(k+1)\pi)$ the sequence converges to $(2k+1)\pi$ $\endgroup$ Jul 11, 2014 at 16:21
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If we assume that the limit exists, let $\lim_{n\to\infty} x_n = x$. Then:
$$\left(\lim_{n\to\infty}x_{n+1} = \lim_{n\to\infty}x_n+\sin(x_n)\right) \implies x=x+\sin(x)$$
Thus: $$0=\sin(x) \implies x = k\pi, \quad k\in\Bbb{Z}$$
Thus, your stated equality does not hold.
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$\begingroup$ Ugh. I wasn't watching the comments above while typing this... $\endgroup$– apnortonJul 11, 2014 at 15:43
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$\begingroup$ the first version was definitely wrong , for example a = 0 )))) i have updated question $\endgroup$ Jul 11, 2014 at 15:53
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$\begingroup$ Kanye, please read this. No matter your initial a, Anorton's answer holds and explains why it converges. $\endgroup$– JamJul 11, 2014 at 15:54
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2$\begingroup$ @oliveeuler No, it does not, it explaines that if it converges, then to a multiple of $\pi$. $\endgroup$– martiniJul 11, 2014 at 15:56
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$\begingroup$ but it is easy to show that if a = 1 the sequence converges. so thanks for everyone answered $\endgroup$ Jul 11, 2014 at 15:59
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It doesn't necessarily hold for all a. For example
$$x_{1} = {n\pi} \implies x_{n} = x_{1} \forall n$$
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Do you mean $x_{n+1}=x_n+\cos x_n$?
We know $\sin y<y$ when $0<y<\pi/2$, so substitute $x=\pi/2-y$ and get $\cos x< \pi/2-x$ when $0<x<\pi/2$.
So adding $\cos x_n$ will not quite reach $\pi/2$.
On the other hand, $\sin y$ and $y$ get very close when $x$ is small, so adding this term gets very close to $\pi/2$.
