nd appi given equ (Kound yuul answe places.) x1 = -3
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
![**Newton's Method for Approximating Roots**
**Objective:**
Use Newton's method with the specified initial approximation \( x_1 \) to find \( x_3 \), the third approximation to the root of the given equation. (Round your answer to four decimal places.)
**Equation:**
\[
\frac{1}{3}x^3 + \frac{1}{2}x^2 + 7 = 0
\]
**Initial Approximation:**
\( x_1 = -3 \)
**Explanation:**
Newton's method is an iterative numerical technique used to find approximate solutions to equations, specifically roots. Given an initial estimate of the root \( x_1 \), subsequent estimates \( x_2, x_3, \ldots \) are generated using the following iterative formula:
\[
x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
\]
Where:
- \( f(x) \) is the original function.
- \( f'(x) \) is the derivative of the function.
**Procedure:**
1. **Calculate the derivative** of the given function:
\[
f'(x) = \frac{d}{dx}\left(\frac{1}{3}x^3 + \frac{1}{2}x^2 + 7\right)
\]
2. **Evaluate the function and its derivative** at the given \( x_1 \).
3. **Apply Newton's formula** to find \( x_2 \), and repeat to find \( x_3 \).
4. **Round the final result** to four decimal places.
This information and the sequence of calculations will help you understand and apply Newton's methodology to solve for approximate roots of polynomial equations effectively.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F59267761-b459-48b3-8663-77485520fda1%2Fc13a3674-42dc-49b9-9598-0c7c707598b8%2Fn3531ro_processed.png&w=3840&q=75)
Transcribed Image Text:**Newton's Method for Approximating Roots**
**Objective:**
Use Newton's method with the specified initial approximation \( x_1 \) to find \( x_3 \), the third approximation to the root of the given equation. (Round your answer to four decimal places.)
**Equation:**
\[
\frac{1}{3}x^3 + \frac{1}{2}x^2 + 7 = 0
\]
**Initial Approximation:**
\( x_1 = -3 \)
**Explanation:**
Newton's method is an iterative numerical technique used to find approximate solutions to equations, specifically roots. Given an initial estimate of the root \( x_1 \), subsequent estimates \( x_2, x_3, \ldots \) are generated using the following iterative formula:
\[
x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
\]
Where:
- \( f(x) \) is the original function.
- \( f'(x) \) is the derivative of the function.
**Procedure:**
1. **Calculate the derivative** of the given function:
\[
f'(x) = \frac{d}{dx}\left(\frac{1}{3}x^3 + \frac{1}{2}x^2 + 7\right)
\]
2. **Evaluate the function and its derivative** at the given \( x_1 \).
3. **Apply Newton's formula** to find \( x_2 \), and repeat to find \( x_3 \).
4. **Round the final result** to four decimal places.
This information and the sequence of calculations will help you understand and apply Newton's methodology to solve for approximate roots of polynomial equations effectively.
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