Given the Fourier sine series of $(x) = x as show that the series can be integrated b) Find The Fourier cigine Series for 27²/2. Find the constant of integration. c) By setting e (1) he n=1 n² n2 on (0₁1) term by term. x=0, find the sum
Given the Fourier sine series of $(x) = x as show that the series can be integrated b) Find The Fourier cigine Series for 27²/2. Find the constant of integration. c) By setting e (1) he n=1 n² n2 on (0₁1) term by term. x=0, find the sum
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
010
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![**Problem 10**
Given the Fourier sine series of \( \phi(x) = x \) on \( (0, \ell) \),
a) Show that the series can be integrated term by term.
b) Find the Fourier cosine series for \( x^2/2 \). Find the constant of integration.
c) By setting \( x = 0 \), find the sum
\[
\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}
\]
**Solution:**
\[
x = \sum_{m=1}^{\infty} (-1)^{m+1} \frac{2\ell}{m\pi} \sin\left(\frac{m\pi x}{\ell}\right)
\]
Integration of both sides gives
\[
\int x \, dx = x^2/2 = C + \sum_{m=1}^{\infty} (-1)^m \frac{2\ell^2}{m^2\pi^2} \cos\left(\frac{m\pi x}{\ell}\right)
\]
Constant of integration is the missing coefficient:
\[
C = \frac{A_0}{2} = \frac{1}{\ell} \int_0^\ell \frac{x^2}{2} \, dx = \frac{\ell^2}{6}
\]
**d)** By setting \( x = 0 \), we have
\[
0 = \frac{\ell^2}{6} + \sum_{m=1}^{\infty} (-1)^m \frac{2\ell^2}{m^2\pi^2}
\]
or
\[
\frac{\pi^2}{12} = \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m^2}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9c55fd55-ae67-4b97-a36c-91359ff73a6f%2F7dc99968-df43-49d2-8c7e-439fcf067284%2F9f2r1v_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 10**
Given the Fourier sine series of \( \phi(x) = x \) on \( (0, \ell) \),
a) Show that the series can be integrated term by term.
b) Find the Fourier cosine series for \( x^2/2 \). Find the constant of integration.
c) By setting \( x = 0 \), find the sum
\[
\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}
\]
**Solution:**
\[
x = \sum_{m=1}^{\infty} (-1)^{m+1} \frac{2\ell}{m\pi} \sin\left(\frac{m\pi x}{\ell}\right)
\]
Integration of both sides gives
\[
\int x \, dx = x^2/2 = C + \sum_{m=1}^{\infty} (-1)^m \frac{2\ell^2}{m^2\pi^2} \cos\left(\frac{m\pi x}{\ell}\right)
\]
Constant of integration is the missing coefficient:
\[
C = \frac{A_0}{2} = \frac{1}{\ell} \int_0^\ell \frac{x^2}{2} \, dx = \frac{\ell^2}{6}
\]
**d)** By setting \( x = 0 \), we have
\[
0 = \frac{\ell^2}{6} + \sum_{m=1}^{\infty} (-1)^m \frac{2\ell^2}{m^2\pi^2}
\]
or
\[
\frac{\pi^2}{12} = \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m^2}
\]
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