I want to make sure I understand when the secant method will not converge as compared to the Newton's method.

When I look at $\arctan(x)$ and try to determine the initial guesses for which it will converge, and those for which it won't, I've come up with the following:

For Newton's method, when $|x_0| < 2$, the method will converge. and diverges otherwise.

For Secant method, when both $|x_0| < 2$, and $|x_1| < 2$ (since it requires 2 initial guesses) the the method converges, and diverges otherwise.

Can someone help me determine if this is correct? thanks very much.


No, this is not correct.

For $f(x)=\arctan x$, Newton's method is $$ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-(1+x_n^2)\arctan(x_n) $$ so will give $\lvert x_{n+1}\rvert<\lvert x_n\rvert$ if $\lvert x_n\rvert<\xi$, where $\xi\approx 1.39$ is the unique positive solution of $\arctan(\xi)=2\xi/(1+\xi^2)$. For starting value bigger in magnitude that $\xi$, the iterates diverges, changing sign with every iteration.

The secant method can converge for large starting values of $\lvert x_0\rvert, \lvert x_1\rvert$ when, e.g., $x_0=-x_1$ and also when they are close enough in magnitude such as $(x_0,x_1)=(-30,29.9)$ (so eventually some $(x_{n-1},x_n)$ are close to $(0,0)$).

  • $\begingroup$ Thanks. How did you come up with the expression $\arctan(\xi)=2\xi/(1+\xi^2)$? That's brilliant, I actually tried it out. And for the secant method, I was just playing with this and discoveredthat myself but I can't find exactly what distance consecutive starting values must be in order for this to converge. HOw do I determine the upper bound on $|x_0 - x_|1$ such that it still converges? $\endgroup$ – nundo Jun 3 at 4:17
  • $\begingroup$ At first I thought it was concavity that you were computing, but when I compute the concavity of $arctan(x)$ I get $\frac{2x}{(1+x^2)^2}$ whereas you have $\frac{2x}{1+x^2}$ (i.e. with no square in the denominator). Is that a typ0? $\endgroup$ – nundo Jun 3 at 4:48
  • $\begingroup$ How did you get that expression? $\endgroup$ – nundo Jun 3 at 4:50
  • $\begingroup$ Oops, I really do mean $/(1+\xi^2)$. The "critical" case is if the iterates are $\xi,-\xi,\xi,-\xi,\dots$ (because we know the iterates are moving towards $0$ but can overshoot). So feeding it into the defining equation $x_{n+1}=x_n-f(x_n)/f'(x_n)$ yields $2\xi=\arctan(\xi)(1+\xi^2)$, or equivalently $\arctan\xi=2\xi/(1+\xi^2)$. $\endgroup$ – user10354138 Jun 3 at 4:53

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