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Suppose that the fixed point iteration for the function $\sin x^{100}$ converges to the fixed point 0. Determine the order of convergence with justification.

I may have misunderstandings for this question.

1) Is it right to let $f(x) = \sin x^{100}$ and the fixed point iteration be $p_n = f(p_{n-1})$ for $n\in \mathbb{N}$

2) Let $f(x)=\sin x^{100}$, then any n-th derivative of $f(x)$ evaluated at $0$ is $0$.

Edit: im wrong, there are non-zero terms, but how should i show that the $100$-th derivative has non-zero term and the first to $99$-th derivatives are $0$ at $0$?

Please help, thanks.

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The derivative of $\sin(x^{100})$ is $100 x^{99}\cos (x^{100})$ by the chain rule. If you differentiate that $99$ more times there will be a term $100!\cos(x^{100}) \neq 0$

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  • $\begingroup$ I see, how should i write it in a proof? Also, can u address my concern (1) as well? Thanks! $\endgroup$ – Little Rookie May 1 '17 at 16:11
  • $\begingroup$ Yes, that is the fixed point iteration the problem is talking about. Did I read the question correctly as referring to $\sin (x^{100})$ and not $(\sin x)^{100}?$ Mine would be the usual way to read it. To do a real proof you would have to differentiate it another $99$ times and show all the other terms have a factor $x$. This is the only one that survives. Informally you can argue what happens with the product and chain rules to say that all the other terms will have a factor $x$. That should be acceptable. $\endgroup$ – Ross Millikan May 1 '17 at 16:19
  • $\begingroup$ Hmm, im not sure about the proper way to show the proof, do you mind if you could write one here? please :) $\endgroup$ – Little Rookie May 1 '17 at 17:01
  • $\begingroup$ I would differentiate $100x^{99}\cos(x^{100})$ by hand and point out that all the terms will look like this. Every time you differentiate the trig function you get another factor $x^{99}$, so every term except this one will have at least one factor of $x$ in the $100$th derivative. Every time you differentiate the power of $x$ it reduces by $1$. $99$ reductions gets you there and $\cos 0=1$ $\endgroup$ – Ross Millikan May 1 '17 at 17:51

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