Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Problem 7: Power Series Representation and Evaluation**
Convert the given function to a power series representation with a restricted domain and compute \( f(2) \).
\[ f(x) = \frac{20}{4 - x} \]
**Solution Strategy:**
1. **Identify the General Form:**
Recognize that \(\frac{1}{1 - u}\) can be expanded into a power series \(\sum_{n=0}^{\infty} u^n\) for \(|u| < 1\).
2. **Transform the Function:**
Rewrite \( f(x) = \frac{20}{4 - x} \) in a similar form to apply the geometric series.
Let \( u = \frac{x}{4} \), then:
\[ f(x) = \frac{20}{4 \left(1 - \frac{x}{4}\right)} = 5 \cdot \frac{1}{1 - \frac{x}{4}} \]
3. **Expand as Power Series:**
Using the geometric series formula:
\[ \frac{1}{1 - \frac{x}{4}} = \sum_{n=0}^{\infty} \left(\frac{x}{4}\right)^n \]
Substitute back:
\[ f(x) = 5 \sum_{n=0}^{\infty} \left(\frac{x}{4}\right)^n \]
\[ = \sum_{n=0}^{\infty} \frac{5x^n}{4^n} \]
4. **Determine the Domain:**
The series converges for \(|\frac{x}{4}| < 1\), hence \(|x| < 4\).
5. **Compute \( f(2) \):**
Substitute \( x = 2 \):
\[ f(2) = \sum_{n=0}^{\infty} \frac{5 \cdot 2^n}{4^n} \]
Evaluate the series:
\[ f(2) = \sum_{n=0}^{\infty} \left(\frac{5}{2}\right) \left(\frac{1}{2}\right)^n \]
\[ = 5 \cdot \frac{1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3dc90927-a4d4-417e-bb21-8f8d729f068b%2Fd8c7b208-6f0e-4e3b-ab89-39a6ab0054ef%2Fc6xzdl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 7: Power Series Representation and Evaluation**
Convert the given function to a power series representation with a restricted domain and compute \( f(2) \).
\[ f(x) = \frac{20}{4 - x} \]
**Solution Strategy:**
1. **Identify the General Form:**
Recognize that \(\frac{1}{1 - u}\) can be expanded into a power series \(\sum_{n=0}^{\infty} u^n\) for \(|u| < 1\).
2. **Transform the Function:**
Rewrite \( f(x) = \frac{20}{4 - x} \) in a similar form to apply the geometric series.
Let \( u = \frac{x}{4} \), then:
\[ f(x) = \frac{20}{4 \left(1 - \frac{x}{4}\right)} = 5 \cdot \frac{1}{1 - \frac{x}{4}} \]
3. **Expand as Power Series:**
Using the geometric series formula:
\[ \frac{1}{1 - \frac{x}{4}} = \sum_{n=0}^{\infty} \left(\frac{x}{4}\right)^n \]
Substitute back:
\[ f(x) = 5 \sum_{n=0}^{\infty} \left(\frac{x}{4}\right)^n \]
\[ = \sum_{n=0}^{\infty} \frac{5x^n}{4^n} \]
4. **Determine the Domain:**
The series converges for \(|\frac{x}{4}| < 1\), hence \(|x| < 4\).
5. **Compute \( f(2) \):**
Substitute \( x = 2 \):
\[ f(2) = \sum_{n=0}^{\infty} \frac{5 \cdot 2^n}{4^n} \]
Evaluate the series:
\[ f(2) = \sum_{n=0}^{\infty} \left(\frac{5}{2}\right) \left(\frac{1}{2}\right)^n \]
\[ = 5 \cdot \frac{1
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