Problem 22. Tom For a 2 P-Penodic piecewise continuous function, the Parneval Identity is given by to So (f(x) ³ dix = 20² + ½ € (a + b) n=l 2P J-P a) The Fourier Series of the 257-Periodic function f(x) given interval [-11₂ 113 for f(x) = x²³ - 11x (i)n pinnx 3 is 671 n3 Use the Parseval Identity to evaluate Ans: 00 ट n=l SIN 징 쿄M8 하 76 ;) 4774 945 on the
Problem 22. Tom For a 2 P-Penodic piecewise continuous function, the Parneval Identity is given by to So (f(x) ³ dix = 20² + ½ € (a + b) n=l 2P J-P a) The Fourier Series of the 257-Periodic function f(x) given interval [-11₂ 113 for f(x) = x²³ - 11x (i)n pinnx 3 is 671 n3 Use the Parseval Identity to evaluate Ans: 00 ट n=l SIN 징 쿄M8 하 76 ;) 4774 945 on the
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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Question
022
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![**Problem 22:**
For a \(2P\)-periodic piecewise continuous function, the Parseval Identity is given by:
\[
\frac{1}{2P} \int_{-P}^{P} (f(x))^2 \, dx = \frac{a_0^2}{4} + \frac{1}{2} \sum_{n=1}^{\infty} (a_n^2 + b_n^2)
\]
a) The Fourier series of the \(2\pi\)-periodic function \(f(x)\) given on the interval \([- \pi, \pi]\) for \(f(x) = x^3 - \pi^2 x\) is:
\[
6\pi \sum_{n=1}^{\infty} \frac{(-1)^n \sin n x}{n^3}
\]
Use the Parseval Identity to evaluate:
\[
\sum_{n=1}^{\infty} \frac{1}{n^6}
\]
**Answer:**
\[
\sum_{n=1}^{\infty} \frac{1}{n^6} = \frac{4 \pi^6}{945}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9c55fd55-ae67-4b97-a36c-91359ff73a6f%2F762f9113-4134-40f7-ab98-8c9534a9a140%2Fvjmfysm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 22:**
For a \(2P\)-periodic piecewise continuous function, the Parseval Identity is given by:
\[
\frac{1}{2P} \int_{-P}^{P} (f(x))^2 \, dx = \frac{a_0^2}{4} + \frac{1}{2} \sum_{n=1}^{\infty} (a_n^2 + b_n^2)
\]
a) The Fourier series of the \(2\pi\)-periodic function \(f(x)\) given on the interval \([- \pi, \pi]\) for \(f(x) = x^3 - \pi^2 x\) is:
\[
6\pi \sum_{n=1}^{\infty} \frac{(-1)^n \sin n x}{n^3}
\]
Use the Parseval Identity to evaluate:
\[
\sum_{n=1}^{\infty} \frac{1}{n^6}
\]
**Answer:**
\[
\sum_{n=1}^{\infty} \frac{1}{n^6} = \frac{4 \pi^6}{945}
\]
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