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 $(0,0)$).