The coefficient of (x - 1)5 in the Taylor series for a In x centered at x=1 is Hint: you can't do this one by multiplying the In x series by x, as it would not longer be in the Taylor series format. This one must be worked out the "longer way." 1 5! O 1 5! 1 4! 1 20 1 4!
The coefficient of (x - 1)5 in the Taylor series for a In x centered at x=1 is Hint: you can't do this one by multiplying the In x series by x, as it would not longer be in the Taylor series format. This one must be worked out the "longer way." 1 5! O 1 5! 1 4! 1 20 1 4!
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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![### Coefficient of \((x-1)^5\) in the Taylor Series for \(x \ln x\)
The problem at hand is to determine the coefficient of \((x-1)^5\) in the Taylor series for \(x \ln x\) centered at \(x=1\).
**Hint:** You can't do this one by multiplying the \(\ln x\) series by \(x\), as it would no longer be in the Taylor series format. This one must be worked out the "longer way."
#### Options
- \[
\begin{cases}
\circ \ - \dfrac{1}{5!} & \\
\bullet \ - \dfrac{1}{20} & \leq \text{(Correct Answer)}
\circ \ - \dfrac{1}{4!} & \\
\end{cases}
\]
- \[
\begin{cases}
\circ \ - \dfrac{1}{4!} & \\
\textcolor{red}{\circ \ \dfrac{1}{5!}} & \leq \text{(Incorrect Answer)}
\circ \ \dfrac{1}{4!} & \\
\end{cases}
\]
#### Explanation
Choosing the correct coefficient involves careful calculation. Here are the given options and how they are marked:
1. \( \circ \ - \dfrac{1}{5!} \)
2. \( \bullet \ - \dfrac{1}{20} \) (correct)
3. \( \circ \ - \dfrac{1}{4!} \)
4. \( \textcolor{red}{\circ \ \dfrac{1}{5!}} \) (incorrect)
5. \( \circ \ \dfrac{1}{4!} \)
The correct result is \(-\dfrac{1}{20}\), indicated by the yellow highlighted circle and checkmark. Another option, \(\dfrac{1}{5!}\), indicated by a red circle with an “X” mark, is incorrect.
#### Conclusion
The correct coefficient of \((x-1)^5\) in the Taylor series expansion of \(x \ln x\) centered at \(x=1\) is \(- \dfrac{1}{20}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2bb9483f-e549-4e9c-b819-8ce0593a1865%2Fb39e489f-7c63-4136-a96b-ed73a3752bb9%2F4ntxomp_processed.png&w=3840&q=75)
Transcribed Image Text:### Coefficient of \((x-1)^5\) in the Taylor Series for \(x \ln x\)
The problem at hand is to determine the coefficient of \((x-1)^5\) in the Taylor series for \(x \ln x\) centered at \(x=1\).
**Hint:** You can't do this one by multiplying the \(\ln x\) series by \(x\), as it would no longer be in the Taylor series format. This one must be worked out the "longer way."
#### Options
- \[
\begin{cases}
\circ \ - \dfrac{1}{5!} & \\
\bullet \ - \dfrac{1}{20} & \leq \text{(Correct Answer)}
\circ \ - \dfrac{1}{4!} & \\
\end{cases}
\]
- \[
\begin{cases}
\circ \ - \dfrac{1}{4!} & \\
\textcolor{red}{\circ \ \dfrac{1}{5!}} & \leq \text{(Incorrect Answer)}
\circ \ \dfrac{1}{4!} & \\
\end{cases}
\]
#### Explanation
Choosing the correct coefficient involves careful calculation. Here are the given options and how they are marked:
1. \( \circ \ - \dfrac{1}{5!} \)
2. \( \bullet \ - \dfrac{1}{20} \) (correct)
3. \( \circ \ - \dfrac{1}{4!} \)
4. \( \textcolor{red}{\circ \ \dfrac{1}{5!}} \) (incorrect)
5. \( \circ \ \dfrac{1}{4!} \)
The correct result is \(-\dfrac{1}{20}\), indicated by the yellow highlighted circle and checkmark. Another option, \(\dfrac{1}{5!}\), indicated by a red circle with an “X” mark, is incorrect.
#### Conclusion
The correct coefficient of \((x-1)^5\) in the Taylor series expansion of \(x \ln x\) centered at \(x=1\) is \(- \dfrac{1}{20}\).
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