For the iteration

$$x_{n+1}=f(x_n)\equiv \sin(x_n) \text{ with initial value } x_0=1,$$

I know it converges since $x_{n+1}\le x_n$ for all $n$ and the limit is zero, so the iteration converges to zero, but how do I know the rate of convergence? Also, what are the effects of rounding errors?

  • $\begingroup$ @par Actually, $x_{n+1} \leq x_n$ and the sequence being bounded is a sufficient condition for convergence. Moreover, any limit must be a fixed point of $\sin$, and there is only one of those. $\endgroup$ – Ian Jun 6 '16 at 22:18
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    $\begingroup$ I would say the rate's not very good, since the derivative of $\sin$ around zero is close to $1$, so we do not have any contraction phenomenon which could make the convergence geometric, for example. $\endgroup$ – Beni Bogosel Jun 6 '16 at 22:18
  • $\begingroup$ @par No, it does imply the limit is zero. For a sequence of iterates of a continuous function, any limit is a fixed point. $\endgroup$ – Ian Jun 6 '16 at 22:22
  • $\begingroup$ Removing myself from this thread as I am on an error-making streak. $\endgroup$ – parsiad Jun 6 '16 at 22:24
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    $\begingroup$ Round-off error effects could lead to $0<x_n=x_{n+1} ,$ depending on your program. For $x>0$ we have $1-x^2/6<(\sin x)/x<1-x^2/6+x^4/120.$ $\endgroup$ – DanielWainfleet Jun 6 '16 at 23:11

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