Let p = [a, b] be the root of ƒ € C¹([a, b]), and assume f'(p) = f'(po) for some po € [a, b]. Consider an iteration scheme that is similar to, but different from Newton's method: given po, define Pn+1 = Pn f (Pn) f'(po)' n ≥ 0. Assuming that the iterative scheme converges, i.e. that pn → p as n → ∞, show that this method has order of convergence a = = 1.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Newton's method

Let \( p \in [a, b] \) be the root of \( f \in C^1([a, b]) \), and assume \( f'(p) \neq f'(p_0) \) for some \( p_0 \in [a, b] \). Consider an iteration scheme that is similar to, but different from Newton’s method: given \( p_0 \), define

\[
p_{n+1} = p_n - \frac{f(p_n)}{f'(p_0)}, \quad n \geq 0.
\]

Assuming that the iterative scheme converges, i.e. that \( p_n \rightarrow p \) as \( n \rightarrow \infty \), show that this method has order of convergence \( \alpha = 1 \).
Transcribed Image Text:Let \( p \in [a, b] \) be the root of \( f \in C^1([a, b]) \), and assume \( f'(p) \neq f'(p_0) \) for some \( p_0 \in [a, b] \). Consider an iteration scheme that is similar to, but different from Newton’s method: given \( p_0 \), define \[ p_{n+1} = p_n - \frac{f(p_n)}{f'(p_0)}, \quad n \geq 0. \] Assuming that the iterative scheme converges, i.e. that \( p_n \rightarrow p \) as \( n \rightarrow \infty \), show that this method has order of convergence \( \alpha = 1 \).
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