Newton method - speed of convergence

Question

Let $x_0 \neq 0$ lies in the convergence ball of the solution $\alpha=2$ for the equation:

$f(x)=0$ where $f(x):=x^3-5x^2+8x-4$

What will be the convergence speed of Newton's method?

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I did it this way:

Let $\phi(x)=x-\frac{f(x)}{f'(x)}=\frac{2x^3-5x^2+4}{3x^2-10x+8}$

$x_{n+1}=\phi(x_n)$

$e_n=x_n-2$

we have:

$e_{n+1}=x_{n+1}-2=\phi(x_n)-\phi(2)=\phi(e_n+2)-\phi(2)=\phi(2)+\phi'(2)e_n+o(e_n)-\phi(2)=\phi'(2)e_n+o(e_n)$

After calculations:

$\phi'(x)=\frac {2(3x^2-8x+5)}{(4-3x)^2}$

$\phi'(2)=\frac 12$

Thus: $e_{n+1}=\frac12 e_n +o(e_n)=e_n(\frac12+\frac {o(e_n)}{e_n})$

when $n \to \infty$ then $e_{n+1} \approx \frac12 e_n$

So we have linear convergence.

It is correct?

Answer 1

For convenience, we shift the variable with $t:=x+2$, and $f(t):=t^3+t^2$. Newton's iterations are

$$t'=\frac{2t^3+t^2}{3t^2+2t}=t\frac{2t+1}{3t+2}\sim\frac t2.$$

For small $t$, convergence is indeed linear (one more exact bit per iteration). This is because the root is double.