Use Newton Raphson Example 1. f(x)= x³ - 20 f'(x) = 3x² i 0 1 2 3 4 5 6 X 3 f(x) 33-20 x³-200 start with initial guess = 3 f'(x) 3* (3²) Xi+1 = Xi - x₁ = 3- f(xi) f'(xi) 33 - 20 3* (3²) = 2.741 12.741-31 2.741 error *100 = 9.45%
Use Newton Raphson Example 1. f(x)= x³ - 20 f'(x) = 3x² i 0 1 2 3 4 5 6 X 3 f(x) 33-20 x³-200 start with initial guess = 3 f'(x) 3* (3²) Xi+1 = Xi - x₁ = 3- f(xi) f'(xi) 33 - 20 3* (3²) = 2.741 12.741-31 2.741 error *100 = 9.45%
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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Question
![**Use Newton Raphson Example 1**
The objective is to solve the equation \( x^3 - 20 = 0 \) starting with an initial guess of 3.
### Functions:
- \( f(x) = x^3 - 20 \)
- \( f'(x) = 3x^2 \)
### Iteration Process:
\[
x_{i+1} = x_i - \frac{f(x_i)}{f'(x_i)}
\]
### Table:
| i | X | \( f(x) \) | \( f'(x) \) | \( x_{i+1} \) Calculation | Error |
|---|-----|--------------|--------------|-------------------------------------------------------------|---------------------------------------------------------|
| 0 | 3 | \( 3^3 - 20 \)| \( 3 \times (3^2) \) | \( x_1 = 3 - \frac{3^3 - 20}{3 \times (3^2)} = 2.741 \) | \(\left| \frac{2.741 - 3}{2.741} \right| \times 100 = 9.45\%\) |
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| 4 | | | | | |
| 5 | | | | | |
| 6 | | | | | |
- **Iteration 0:**
- Initial guess: \( x_0 = 3 \)
- Calculation of \( f(x_0) \) and \( f'(x_0) \)
- Next approximation: \( x_1 = 2.741 \)
- Error calculation: 9.45%
This example demonstrates the first iteration of the Newton-Raphson method to find the root of the equation. Further iterations would involve recalculating using the updated value of \( x \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F353f4476-2375-44e0-a0df-7fa04e9ee69f%2F7e3e0997-e0d6-4401-bc1b-18d98f830730%2Fpy9gtya_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Use Newton Raphson Example 1**
The objective is to solve the equation \( x^3 - 20 = 0 \) starting with an initial guess of 3.
### Functions:
- \( f(x) = x^3 - 20 \)
- \( f'(x) = 3x^2 \)
### Iteration Process:
\[
x_{i+1} = x_i - \frac{f(x_i)}{f'(x_i)}
\]
### Table:
| i | X | \( f(x) \) | \( f'(x) \) | \( x_{i+1} \) Calculation | Error |
|---|-----|--------------|--------------|-------------------------------------------------------------|---------------------------------------------------------|
| 0 | 3 | \( 3^3 - 20 \)| \( 3 \times (3^2) \) | \( x_1 = 3 - \frac{3^3 - 20}{3 \times (3^2)} = 2.741 \) | \(\left| \frac{2.741 - 3}{2.741} \right| \times 100 = 9.45\%\) |
| 1 | | | | | |
| 2 | | | | | |
| 3 | | | | | |
| 4 | | | | | |
| 5 | | | | | |
| 6 | | | | | |
- **Iteration 0:**
- Initial guess: \( x_0 = 3 \)
- Calculation of \( f(x_0) \) and \( f'(x_0) \)
- Next approximation: \( x_1 = 2.741 \)
- Error calculation: 9.45%
This example demonstrates the first iteration of the Newton-Raphson method to find the root of the equation. Further iterations would involve recalculating using the updated value of \( x \).
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