Find the Taylor series of f(x) = Choose the Taylor series. O 1-x 1 1- x 1 1-x - 1-x = = = || = ∞ Σ (-1)"+1 n=0 ∞ n=0 Σ(-1)+1. ∞ Σ(-1)". n=0 ∞ 1 1-x (x-7)" 8n+1 (x-8)" 7n+1 (x − 8) n+1 7" Σ(1)", n=0 7n+1 (x-8)" centered at c = 8.
Find the Taylor series of f(x) = Choose the Taylor series. O 1-x 1 1- x 1 1-x - 1-x = = = || = ∞ Σ (-1)"+1 n=0 ∞ n=0 Σ(-1)+1. ∞ Σ(-1)". n=0 ∞ 1 1-x (x-7)" 8n+1 (x-8)" 7n+1 (x − 8) n+1 7" Σ(1)", n=0 7n+1 (x-8)" centered at c = 8.
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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![### Determining the Taylor Series
To find the Taylor series of \( f(x) = \frac{1}{1 - x} \) centered at \( c = 8 \), follow these instructions:
#### Choose the Taylor Series
Select the correct Taylor series representation from the following options:
- \( \bigcirc \quad \frac{1}{1 - x} = \sum_{n=0}^{\infty} (-1)^{n+1} \frac{(x - 7)^n}{8^{n+1}} \)
- \( \bigcirc \quad \frac{1}{1 - x} = \sum_{n=0}^{\infty} (-1)^{n+1} \frac{(x - 8)^n}{7^{n+1}} \)
- \( \bigcirc \quad \frac{1}{1 - x} = \sum_{n=0}^{\infty} (-1)^n \frac{(x - 8)^{n+1}}{7^n} \)
- \( \bigcirc \quad \frac{1}{1 - x} = \sum_{n=0}^{\infty} (-1)^n \frac{7^{n+1}}{(x - 8)^n} \)
#### Validity of the Expansion
Identify the interval on which the expansion is valid.
- Provide your answer as an interval in the form \((\ast, \ast)\).
- Utilize the symbol \(\infty\) for infinity, the union symbol \( \cup \) for combining intervals, and an appropriate type of parenthesis \(( \, )\), \([ \, ]\) depending on whether the interval is open or closed.
- Enter \(\emptyset\) if the interval is empty.
- Express numbers in exact form. Use symbolic notation and fractions where needed.
**The expansion is valid for:**
\[ \underline{\hspace{50mm}} \]
This section will guide users in choosing the correct Taylor series expansion and understanding the interval in which the series is valid. The options involve different expressions for the Taylor series of the function around the specified center \(c = 8\). The reader is required to analyze and select the correct form. Additionally, understanding the convergence interval of the series is crucial for proper application.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fde326383-8b00-4076-b3f6-1f860c5e4029%2F7e297afe-1e55-43b8-a91d-02f3333b25b1%2Fqw04uta_processed.png&w=3840&q=75)
Transcribed Image Text:### Determining the Taylor Series
To find the Taylor series of \( f(x) = \frac{1}{1 - x} \) centered at \( c = 8 \), follow these instructions:
#### Choose the Taylor Series
Select the correct Taylor series representation from the following options:
- \( \bigcirc \quad \frac{1}{1 - x} = \sum_{n=0}^{\infty} (-1)^{n+1} \frac{(x - 7)^n}{8^{n+1}} \)
- \( \bigcirc \quad \frac{1}{1 - x} = \sum_{n=0}^{\infty} (-1)^{n+1} \frac{(x - 8)^n}{7^{n+1}} \)
- \( \bigcirc \quad \frac{1}{1 - x} = \sum_{n=0}^{\infty} (-1)^n \frac{(x - 8)^{n+1}}{7^n} \)
- \( \bigcirc \quad \frac{1}{1 - x} = \sum_{n=0}^{\infty} (-1)^n \frac{7^{n+1}}{(x - 8)^n} \)
#### Validity of the Expansion
Identify the interval on which the expansion is valid.
- Provide your answer as an interval in the form \((\ast, \ast)\).
- Utilize the symbol \(\infty\) for infinity, the union symbol \( \cup \) for combining intervals, and an appropriate type of parenthesis \(( \, )\), \([ \, ]\) depending on whether the interval is open or closed.
- Enter \(\emptyset\) if the interval is empty.
- Express numbers in exact form. Use symbolic notation and fractions where needed.
**The expansion is valid for:**
\[ \underline{\hspace{50mm}} \]
This section will guide users in choosing the correct Taylor series expansion and understanding the interval in which the series is valid. The options involve different expressions for the Taylor series of the function around the specified center \(c = 8\). The reader is required to analyze and select the correct form. Additionally, understanding the convergence interval of the series is crucial for proper application.
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