# On the convergence rate of Newton's method

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?

• 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. – 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.
• 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? – user135671 Mar 11 at 20:57
• 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. – LutzL Mar 11 at 21:00