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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.