I'm stuck with this whole question on convergence -
f(x) = xe^(-x) + 3.
For what values of xo ("x naught") does Newton's method converge to the root, and for what values of xo will Newton's method fail to find the root?
Thanks
I'm stuck with this whole question on convergence -
f(x) = xe^(-x) + 3.
For what values of xo ("x naught") does Newton's method converge to the root, and for what values of xo will Newton's method fail to find the root?
Thanks
Hi
$\displaystyle f(x) = xe^{-x} + 3$
$\displaystyle f'(x) = (1-x)e^{-x}$
Newton's method starting from $\displaystyle x_0$ will give
$\displaystyle x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}$
If $\displaystyle f'(x_0) = 0$ you will not be able to calculate $\displaystyle x_1$