2

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?

parsiad
  • 25,154
bowen.jane
  • 361
  • @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. – Ian Jun 06 '16 at 22:18
  • 1
    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. – Beni Bogosel Jun 06 '16 at 22:18
  • @par No, it does imply the limit is zero. For a sequence of iterates of a continuous function, any limit is a fixed point. – Ian Jun 06 '16 at 22:22
  • Removing myself from this thread as I am on an error-making streak. – parsiad Jun 06 '16 at 22:24
  • @Ian $x_{n+1}\le x_n$ must have a lower bound for it to converge. – Simply Beautiful Art Jun 06 '16 at 22:27
  • @SimpleArt Yes, but I mentioned that, and in this case the presence of such a bound is quite obvious ($-1$ would do the job). – Ian Jun 06 '16 at 22:27
  • See your answer here: http://math.stackexchange.com/questions/1537909/fixed-point-iteration-convergence-of-sinx-in-java – Beni Bogosel Jun 06 '16 at 22:37
  • 1
    Round-off error effects could lead to $0<x_n=x_{n+1} ,$ depending on your program. For $x>0$ we have $1-x^2/6<(\sin x)/x<1-x^2/6+x^4/120.$ – DanielWainfleet Jun 06 '16 at 23:11

0 Answers0