Convergence Rate

Question

So I have the polynomial $f(x)=8x^{4}-12x^{3}+6x^{2}-x$

And I was trying to find the convergence rate of the newton iteration when started close to the root r=0

My attempt:

Since $f'(r) \neq 0$ we have quadratic convergence with the convergence rate given $e_{i+1} \approx Me^{2}_{i}$

From Newton iteration I get that $x_{i+1}= \frac{6x^{2}_{i}}{6x_{i}-1}$ However I am not sure this is necessary..

So $f'(0)=-1$ and $f''(0)=0$ thus $M=0$ so the convergence rate is 0? is that possible?

The convergence of newton iteration when started close to the root r=0 — , May 01 '18 at 10:25
How do you get $f''(0)=0$? It is quite obvious that $f''(0)=2\cdot 6=12$. How did you get your iteration equation from $x_{i+1}=x_i-\frac{f(x)}{f'(x)}$? — Lutz Lehmann, May 01 '18 at 13:51

score 1 · Answer 1 · edited May 01 '18 at 14:27

The Newton iteration is, for $x\approx0$, \begin{align} x_+=N_f(x) = x-\frac{f(x)}{f'(x)} &=\frac{(64x^4−36x^3+12x^2-x)-(8x^4−12x^3+6x^2−x)}{64x^3−36x^2+12x-1} \\[1em] &=\frac{56x^4−24x^3+6x^2}{64x^3−36x^2+12x-1} \\[1em] &\approx-6x^2\frac{1-4x}{1-12x}\approx -6x^2. \end{align} Thus you find that $-6x_{i+1}\approx (-6x_i)^2$ or $$x_i\approx -\frac16(6x_0)^{\large 2^i}.$$ To get convergence you need $|x_0|\le\frac16-\varepsilon$ where the small $ε$ accounts for all the terms that were neglected in the derivation.

Convergence Rate

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