Show that Newton's method used to find $\sqrt[m]{R}$ applied to $x^m-R$ and to $1-\frac{R}{x^m}$ converges at different rates.

Question

Show that Newton's method used to find $\sqrt[m]{R}, (m\ge2)$ applied to $f_1(x)=x^m-R=0$ and to $f_2(x)=1-\frac{R}{x^m}=0$ converges at different rates.

I have a feeling that it converges faster for $f_2(x)$, but I'm not sure why or how to prove it. Thanks!

Answer 1

hint

the sequence $(x_n) $ which converges to the root is such that

$$x_{n+1}=\phi_1 (x_n)$$ $$=x_n-\frac {x_n^m-R}{mx_n^{m-1}} $$ in the first case and

$$x_{n+1}=\phi_2 (x_n) $$ $$=x_n-\frac {1-\frac {R}{x_n^m}}{m\frac {R}{x_n^{m+1}}} $$ $$=x_n-\frac {x_n (x_n^m-R)}{mR} $$

for the second. Compute $\phi_1''(x^*) $ and $\phi_2''(x^*) $ and compare. where $x^*$ is the root.