# How to solve for the interval of convergence in Newton's Method?

How do I solve for $\delta$ in $[r−\delta,r+\delta]$ where Newton's method will surely converge? For example in:

Explain Newton’s method for $$f(x) = x^3+x−2 = 0.$$ Show that Newton’s method converges if $$x_0 \in [1−1/30 , 1+1/30 ]$$ to a limit $L$. Find an error estimate for the error $$e_n = |x_n−L|.$$ (Hint: $x^3 −3x^2 +2 = (x−1)(x^2 −2x−2)$ and $|x^2 − 2x − 2| ≤ 10$ if $0 ≤ x ≤ 2$.)

How was the $1/30$ obtained?

• See the comment of related question for how to determine $\delta$ in general math.stackexchange.com/questions/1652978/… – Carl Christian Feb 20 '16 at 11:39
• I still don't know how 1/30 was obtained. If you can solve it for me I would be so thankful – Guppy_00 Feb 20 '16 at 12:02
• @ Carl Christian Help me! – Guppy_00 Feb 20 '16 at 12:12

Following the theory explained in https://math.stackexchange.com/a/1653829/115115, determine over $[0,2]$ $$m_1=\min_{x\in[0,2]} |f'(x)|=\min_{x\in[0,2]} 3x^2+1=1$$ and $$M_2=\max_{x\in[0,2]} |f''(x)|=\max_{x\in[0,2]}6x=12$$ and determine the "contraction" constant $$C=\frac{M_2}{2m_1}=6.$$ From $$|x_{n+1}-L|\le C·|x_n-L|^2=(C·|x_n-L|)·|x_n-L|\\ \implies |x_n-L|\le C^{-1}· (C·|x_0-L|)^{2^n}$$ one sees that the method is contractive and quadratically convergent for $$|x_0-L|<\frac16.$$
Starting with the smaller interval $[\frac12,\frac32]$ these estimates give $m_1=\frac74$, $M_2=9$, $C=18/7<3$ leading to the greater radius $$|x_0-L|<\frac13$$ for the initial interval of good starting points.
• Not the maximum interval, since $[1-1/3, 1+1/3]$ is larger, and even that is smaller than what $C=18/7$ leading to a radius of $δ=\min(1/2, 7/18)=7/18=1/3+1/18$ actually gives. But yes, $[1-1/30,1+1/30]$ is contained in these larger intervals. – Dr. Lutz Lehmann Feb 20 '16 at 13:45