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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![**Finding the Maclaurin Series for the Function**
We are tasked with finding the Maclaurin series for the function \( f(x) = x^8 \sin(x) \).
### Function Representation:
The function \( f(x) \) is given by:
\[ f(x) = x^8 \sin(x) \]
### Maclaurin Series Expression:
The Maclaurin series representation of a function is expressed using the summation notation:
\[ f(x) = \sum_{n = 0}^{\infty} \text{[expression]} \]
The summation index \( n \) ranges from 0 to infinity (\(\infty\)), indicating that the series continues indefinitely. A blank space is provided suggesting where the series expression is inserted after finding the series expansion.
### Explanation:
The Maclaurin series is essentially a Taylor series expansion of a function about the point \( x = 0 \), focusing on polynomial representations of the given function. It involves calculating derivatives of \( f(x) \) at 0 and using them in the series expansion formula:
\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \]
In our scenario, to solve for the series, we need to expand \( \sin(x) \) using its own Maclaurin series:
\[ \sin(x) = \sum_{k = 0}^{\infty} \frac{(-1)^k}{(2k+1)!} x^{2k+1} \]
By substituting this into \( f(x) = x^8 \sin(x) \), we obtain the Maclaurin series for the entire function. The resultant series can be represented in the summation notation provided in the diagram.
This approach is vital for approximations and is a foundational concept in calculus and mathematical analysis.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15d21cd-3875-4270-92b4-6749e1bc7456%2F3b1fbf99-ed3b-4280-b90d-f09ba0c012b1%2F4ucxtrr_processed.png&w=3840&q=75)
Transcribed Image Text:**Finding the Maclaurin Series for the Function**
We are tasked with finding the Maclaurin series for the function \( f(x) = x^8 \sin(x) \).
### Function Representation:
The function \( f(x) \) is given by:
\[ f(x) = x^8 \sin(x) \]
### Maclaurin Series Expression:
The Maclaurin series representation of a function is expressed using the summation notation:
\[ f(x) = \sum_{n = 0}^{\infty} \text{[expression]} \]
The summation index \( n \) ranges from 0 to infinity (\(\infty\)), indicating that the series continues indefinitely. A blank space is provided suggesting where the series expression is inserted after finding the series expansion.
### Explanation:
The Maclaurin series is essentially a Taylor series expansion of a function about the point \( x = 0 \), focusing on polynomial representations of the given function. It involves calculating derivatives of \( f(x) \) at 0 and using them in the series expansion formula:
\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \]
In our scenario, to solve for the series, we need to expand \( \sin(x) \) using its own Maclaurin series:
\[ \sin(x) = \sum_{k = 0}^{\infty} \frac{(-1)^k}{(2k+1)!} x^{2k+1} \]
By substituting this into \( f(x) = x^8 \sin(x) \), we obtain the Maclaurin series for the entire function. The resultant series can be represented in the summation notation provided in the diagram.
This approach is vital for approximations and is a foundational concept in calculus and mathematical analysis.
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