Use Newton's method to solve the equation given the accuracy stated. x2 - 2x - 13 = 0 correct to 3 decimal places. An example solution and guide is given in the picture.
Use Newton's method to solve the equation given the accuracy stated. x2 - 2x - 13 = 0 correct to 3 decimal places. An example solution and guide is given in the picture.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Use Newton's method to solve the equation given the accuracy stated.
x2 - 2x - 13 = 0 correct to 3 decimal places.
An example solution and guide is given in the picture.
![Solution
The functional notation method is used to determine the first approximation to the root.
f(x) = 5x2 + 11x-17
f(0) = 5(0)² + 11(0) -17= -17
f(1) = 5(1)² + 11(1) -17--1
f(2)= 5(2)2 + 11(2) -17= 25
This shows that the value of the root is close to x = 1
Let the approximation of r₁ = 1
f(ri)
f'(ri)
f(x) = 5x² + 11x-17
f(₁) = 5(₁)² + 11(₁) -17 = 5(1)² + 11(1)-17--1
f'(x) is the differential coefficient of f(x),
f'(x) = 10x +11
f'(r₁) 10(r₁) +11 = 10(1)+11 = 21
12 = 1- 21 = 1.05 correct to 3 significant figures.
A better approximation to the root r3 is given by:
f(r₂)
f' (1₂)
[5(1.05)2 +11(1.05)-17]
[10(1.05)+11]
T2 = 11-
13=72-
= 1.05
= 1.047 1.05, correct to 3 significant figures.
Checking our solution using the quadratic equation formula,
-1+√[121-4(5)(-17)]
(2)(5)
x =
= 1.047. The positive root is 1.05 correct to 3 significant figures. This is consistent with the answer obtained from the bisection method.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc2f2459f-eecd-40e0-a021-73d4633f4382%2Fdbc7eeed-9e3d-431f-96b7-0980bcabae8c%2F5wrv5v8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Solution
The functional notation method is used to determine the first approximation to the root.
f(x) = 5x2 + 11x-17
f(0) = 5(0)² + 11(0) -17= -17
f(1) = 5(1)² + 11(1) -17--1
f(2)= 5(2)2 + 11(2) -17= 25
This shows that the value of the root is close to x = 1
Let the approximation of r₁ = 1
f(ri)
f'(ri)
f(x) = 5x² + 11x-17
f(₁) = 5(₁)² + 11(₁) -17 = 5(1)² + 11(1)-17--1
f'(x) is the differential coefficient of f(x),
f'(x) = 10x +11
f'(r₁) 10(r₁) +11 = 10(1)+11 = 21
12 = 1- 21 = 1.05 correct to 3 significant figures.
A better approximation to the root r3 is given by:
f(r₂)
f' (1₂)
[5(1.05)2 +11(1.05)-17]
[10(1.05)+11]
T2 = 11-
13=72-
= 1.05
= 1.047 1.05, correct to 3 significant figures.
Checking our solution using the quadratic equation formula,
-1+√[121-4(5)(-17)]
(2)(5)
x =
= 1.047. The positive root is 1.05 correct to 3 significant figures. This is consistent with the answer obtained from the bisection method.
Transcribed Image Text:The
Newton-Raphson method or more popularly known as Newton's method may be stated as follows:
If r₁ is the approximate value of a real root of the equation f(x)=0, then a closer approximation to the root r2 is given by:
f(ri)
f'(ri)
T2 = 11 -
The advantage of Newton's method over the algebraic method of successive approximations is that it can be used for any type of mathematical equation, and it is usually easier to apply
than the algebraic method.
Example
Use Newton's method to determine the positive root of the quadratic equation 5x² + 11x -17 = 0, correct to 3 significant figures. Check the value of the root by using the quadratic
formula.
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