Math Problem Statement

Find the Fourier series of f(x) = x^2 , x ∈ (0,2L). Hence deduce that

1/1^2 + 1/3^2 + 1/5^2

  • ⋯ ∞ = π^2/8

Solution

Finding the Fourier Series of f(x)=x2f(x) = x^2 on (0,2L)(0, 2L)

The Fourier series of a function f(x)f(x) over the interval (0,2L)(0, 2L) is given by:

f(x)=a02+n=1(ancos(nπxL)+bnsin(nπxL))f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right) \right)

Where the coefficients a0a_0, ana_n, and bnb_n are given by:

a0=1L02Lf(x)dxa_0 = \frac{1}{L} \int_0^{2L} f(x) \, dx

an=1L02Lf(x)cos(nπxL)dxa_n = \frac{1}{L} \int_0^{2L} f(x) \cos\left(\frac{n\pi x}{L}\right) \, dx

bn=1L02Lf(x)sin(nπxL)dxb_n = \frac{1}{L} \int_0^{2L} f(x) \sin\left(\frac{n\pi x}{L}\right) \, dx

For the function f(x)=x2f(x) = x^2:

1. Calculate a0a_0:

a0=1L02Lx2dxa_0 = \frac{1}{L} \int_0^{2L} x^2 \, dx

a0=1L[x33]02L=1L(2L)33=8L23a_0 = \frac{1}{L} \left[ \frac{x^3}{3} \right]_0^{2L} = \frac{1}{L} \cdot \frac{(2L)^3}{3} = \frac{8L^2}{3}

Thus, a02=4L23\frac{a_0}{2} = \frac{4L^2}{3}.

2. Calculate ana_n:

an=1L02Lx2cos(nπxL)dxa_n = \frac{1}{L} \int_0^{2L} x^2 \cos\left(\frac{n\pi x}{L}\right) \, dx

This integral can be solved by parts, or using a known formula:

an=2L2n2π2(2cos(nπ))=4L2n2π2a_n = \frac{2L^2}{n^2\pi^2} \left( 2 - \cos(n\pi) \right) = \frac{4L^2}{n^2\pi^2}

3. Calculate bnb_n:

Since x2x^2 is an even function, bn=0b_n = 0.

Fourier Series of x2x^2 on (0,2L)(0, 2L):

f(x)=4L23+n=14L2n2π2cos(nπxL)f(x) = \frac{4L^2}{3} + \sum_{n=1}^{\infty} \frac{4L^2}{n^2\pi^2} \cos\left(\frac{n\pi x}{L}\right)

Deduction of the Series Sum

Now, consider L=π2L = \frac{\pi}{2}, and f(x)=x2f(x) = x^2 on (0,π)(0, \pi). The Fourier series becomes:

x2=π23+n=14n2cos(nx)x^2 = \frac{\pi^2}{3} + \sum_{n=1}^{\infty} \frac{4}{n^2} \cos(nx)

Evaluate at x=0x = 0:

0=π23+n=14n20 = \frac{\pi^2}{3} + \sum_{n=1}^{\infty} \frac{4}{n^2}

So:

n=11n2=π24\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{4}

Now, considering only the odd terms n=1,3,5,n = 1, 3, 5, \ldots:

112+132+152+=π28\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots = \frac{\pi^2}{8}

This is because the series sum for only odd terms is half the sum of all terms for nn:

k=01(2k+1)2=12n=11n2\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^2}

Thus, the required series sum:

112+132+152+=π28\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots = \frac{\pi^2}{8}


Would you like to explore the integration details, or have any questions? Here are some further topics to consider:

  1. Deriving the general formula for Fourier coefficients for any f(x)f(x).
  2. Exploring the convergence properties of Fourier series.
  3. Understanding the difference between even and odd extensions.
  4. Investigating the Gibbs phenomenon in Fourier series.
  5. Applying Fourier series to solve differential equations.

Tip: The Fourier series provides powerful insights into periodic functions, with applications ranging from signal processing to heat conduction.

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Math Problem Analysis

Mathematical Concepts

Fourier Series
Integration
Infinite Series

Formulas

Fourier series coefficients
Summation of series

Theorems

Fourier series representation

Suitable Grade Level

Advanced Undergraduate