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Why does Newton's method not converge to a root $\alpha$ of $f(x) = 0$ if $\alpha$ is an inflection point (i.e. $f^{\prime\prime}(\alpha) = 0$)?

Prasiortle
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1 Answers1

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It may converge in this case. For example, with $f(x)=x^3$, starting with $x_0=1$, successive iterations give $x_n=(\frac23)^n$. However, this convergence is slow compared with the quadratic convergence of Newton's method expected in the general case; that is, when the error of each approximation is of the order of the square of the previous error.

As Oscar Lanzi points out, if the gradient is not zero at the root, for example with $f(x)=x^3-x$ at $x=0$, the convergence may actually be accelerated—in this case to cubic order: when $x_n$ is close to $0$, the next approximation (and error) $x_{n+1}$ is about $-2x_n^3$.

John Bentin
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  • That depends on the details of the inflection point. If you solve $x^3-x=0$ with an initial value close enough to $0$, you converge to $0$ faster than would ordinarily be expected with Newton's Method. – Oscar Lanzi May 20 '20 at 16:31
  • @OscarLanzi : Good example. With your permission, I will add it to the answer. – John Bentin May 20 '20 at 16:42
  • Go right ahead! – Oscar Lanzi May 20 '20 at 16:44
  • Although it may converge, I think it should be pointed out that it may not, as with the classical example of $\arctan(x)$. Ultimately it depends on the sign of $f'''(x)/f'(x)$ in a neighborhood of the root. – Simply Beautiful Art Aug 29 '20 at 01:36
  • @SimplyBeautifulArt : I tried Newton's method with $\arctan x$, starting with $x=1$, and it converged after six iterations to $0$ (to within my calculator's limit of precision). – John Bentin Aug 29 '20 at 11:45
  • @JohnBentin Whoops, my bad, think I was thinking of something else. – Simply Beautiful Art Aug 29 '20 at 13:30