8. x2n+1 E(-1)" 2nn n=1
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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First state the power series center, then use either the root or ratio test to find R and I.
Please provide step by step solution using simple steps.
![### Infinite Series Representation
The given mathematical expression describes an infinite series. The series is represented as follows:
\[
\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n+1}}{2^n n}
\]
#### Explanation
- **Summation Symbol (∑):** This symbol indicates that we are dealing with a series, which is a sum of a sequence of terms.
- **Limits of Summation (n=1 to ∞):** The variable \( n \) starts at 1 and approaches infinity (∞), meaning we are summing an infinite number of terms starting from \( n = 1 \).
- **Term of the Series (\(\frac{(-1)^n x^{2n+1}}{2^n n} \)):**
- \( (-1)^n \) oscillates between -1 and 1 for different values of \( n \).
- \( x^{2n+1} \) raises \( x \) to the power of \( 2n+1 \), which means the exponent is always an odd number.
- \( 2^n \) raises 2 to the power of \( n \).
- \( n \) is a simple multiplier for each term.
Therefore, each term in the series is a fraction where:
- The numerator is \( (-1)^n \) times \( x^{2n+1} \),
- The denominator is \( 2^n \) times \( n \).
This series may be used in contexts such as Taylor expansions or Fourier series in mathematics and engineering.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4ef7ad6-120a-468a-b4cf-c9be07d8e3b4%2F2ba020c3-bb25-4dd8-84ad-ff1a139908b8%2Fsa7xg99.jpeg&w=3840&q=75)
Transcribed Image Text:### Infinite Series Representation
The given mathematical expression describes an infinite series. The series is represented as follows:
\[
\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n+1}}{2^n n}
\]
#### Explanation
- **Summation Symbol (∑):** This symbol indicates that we are dealing with a series, which is a sum of a sequence of terms.
- **Limits of Summation (n=1 to ∞):** The variable \( n \) starts at 1 and approaches infinity (∞), meaning we are summing an infinite number of terms starting from \( n = 1 \).
- **Term of the Series (\(\frac{(-1)^n x^{2n+1}}{2^n n} \)):**
- \( (-1)^n \) oscillates between -1 and 1 for different values of \( n \).
- \( x^{2n+1} \) raises \( x \) to the power of \( 2n+1 \), which means the exponent is always an odd number.
- \( 2^n \) raises 2 to the power of \( n \).
- \( n \) is a simple multiplier for each term.
Therefore, each term in the series is a fraction where:
- The numerator is \( (-1)^n \) times \( x^{2n+1} \),
- The denominator is \( 2^n \) times \( n \).
This series may be used in contexts such as Taylor expansions or Fourier series in mathematics and engineering.
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