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Is that possible for some function whose convergence rate is linear by using Newton's method?

I am solving the function $$f(x) = \sin^2(x) - x \sin(x) + \frac 14 x^2$$ by Newton's Method. I got the error with $0.11$, $0.05$, $0.024$, $0.012$ in the first few iterations, which looks like it has linear convergence rate with $1/2$. Is that possible?

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  • $\begingroup$ Yes, this is possible and expected when you are approaching a "double" root with Newton iterations. Note that each of the terms in $f(x)$ has a root of multiplicity two at $x=0$, so if that is the root your iterates converge to, linear convergence is quite possible. $\endgroup$ – hardmath Mar 10 at 0:44
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As your function is a square, $f(x)=(\sin x-\frac12x)^2$, all roots will have even multiplicity. And indeed, Newton's method has linear convergence towards multiple roots, with factor $1-\frac1m$ for multiplicity $m$.


Note that as $\frac\pi2>0=\sin(\pi)$ and $\frac\pi4<1=\sin(\frac\pi2)$, there are additional (double) roots inside $[\frac\pi2,\pi]$ and its mirrored interval.

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  • $\begingroup$ Thanks for the help, Can I ask one more question? What is even multiplicity? Does that mean for each $f(x) = 0$, there are two x with the same values? $\endgroup$ – user135671 Mar 11 at 20:57
  • $\begingroup$ It means that first there is an integer multiplicity at any root $a$, meaning that $f(x)=(x-a)^mg_a(x)$ with $g_a$ (at least) continuous and $g_a(a)\ne 0$, and that $m$ is an even number. $\endgroup$ – LutzL Mar 11 at 21:00

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