5. Consider Newton's method for solving the nonlinear equation f(x)=x²-a=0 where a > 0 whose two roots are √a. (a) Should the iterates {n} for satisfy 1 In÷1 (b) Let a = √a. Show that (In+a/xn). In 1-α = (n − a)²/(2x₂). (c) Suppose that To > 0. Show that for n ≥ 1, i. a < Int1 < In ii. In+1-α(Tn - α) iii. limno In = a.
5. Consider Newton's method for solving the nonlinear equation f(x)=x²-a=0 where a > 0 whose two roots are √a. (a) Should the iterates {n} for satisfy 1 In÷1 (b) Let a = √a. Show that (In+a/xn). In 1-α = (n − a)²/(2x₂). (c) Suppose that To > 0. Show that for n ≥ 1, i. a < Int1 < In ii. In+1-α(Tn - α) iii. limno In = a.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![5. Consider Newton’s method for solving the nonlinear equation
\[ f(x) = x^2 - a = 0 \]
where \( a > 0 \) whose two roots are \( \pm \sqrt{a} \).
(a) Should the iterates \(\{x_n\}\) satisfy
\[ x_{n+1} = \frac{1}{2} (x_n + a/x_n). \]
(b) Let \(\alpha = \sqrt{a}\). Show that
\[ x_{n+1} - \alpha = (x_n - \alpha)^2 /(2x_n). \]
(c) Suppose that \( x_0 > 0 \). Show that for \( n \geq 1 \),
i. \(\alpha < x_{n+1} < x_n\)
ii. \( x_{n+1} - \alpha < \frac{1}{2} (x_n - \alpha) \)
iii. \(\lim_{n \to \infty} x_n = \alpha. \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9b8f2041-e50a-4e46-9284-5dc0f6456a77%2F9cd6f527-df65-4f3e-b329-284a497c9306%2F012xk5_processed.png&w=3840&q=75)
Transcribed Image Text:5. Consider Newton’s method for solving the nonlinear equation
\[ f(x) = x^2 - a = 0 \]
where \( a > 0 \) whose two roots are \( \pm \sqrt{a} \).
(a) Should the iterates \(\{x_n\}\) satisfy
\[ x_{n+1} = \frac{1}{2} (x_n + a/x_n). \]
(b) Let \(\alpha = \sqrt{a}\). Show that
\[ x_{n+1} - \alpha = (x_n - \alpha)^2 /(2x_n). \]
(c) Suppose that \( x_0 > 0 \). Show that for \( n \geq 1 \),
i. \(\alpha < x_{n+1} < x_n\)
ii. \( x_{n+1} - \alpha < \frac{1}{2} (x_n - \alpha) \)
iii. \(\lim_{n \to \infty} x_n = \alpha. \)
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