0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $(0,0)$).

$\endgroup$
  • $\begingroup$ 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

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.