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The iteration as given to me is this:

$x_{n+1}=F(x_n)=sin(x_n), x_0=1$

so I think that the start of the iteration would look something like this:

$x_0=1$

$x_1=sin(1)=.841$

$x_2=sin(sin(1))=sin(.841)=.7456$ and so on.

My professor said this iteration converges, but how do we know that? The only test for convergence I could find in my notes is that the iteration will converge if $|g'(a)|<1$. In this problem, is $g(x_n)=sin(x_n)$ since $g(x_n)=x_{n+1}$? Or is $g(x)=1-sin(x)/cos(x)$?

Also, what is the limit of this iteration, and how fast is the convergence?

bowen.jane
  • 361

1 Answers1

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In this special case you can avoid the criterion of the derivative. Observe that any $x_n$ is of course non negative and that (since $\sin(x)\leq x$ for $x\geq 0$) $$ x_{n+1}\leq x_n. $$ Therefore $\{x_n\}$ is a non negative, monotone non-increasing sequence. As such it has a limit $$ \exists \ell\geq 0:\quad x_n\to \ell. $$ Now you can prove that $\ell=0$. In fact, if $\ell>0$, then $\sin(\ell)\not=\ell$, whereas a necesessary condition for $\ell$ being the limit of $x_n$ is $ \sin(\ell)=\ell$.

guestDiego
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