Math Problem Statement

Find the Fourier series to represent f(x) = x^2 in (0, 2π) and deduce that π^2 / 12 = 1 / 12 - 1 / 2^2 + 1 / 3^2 - 1 / 4^2 + ...

Solution

To solve the problem, we need to find the Fourier series representation of the function f(x)=x2f(x) = x^2 on the interval (0,2π)(0, 2\pi). From this, we will deduce that:

π212=112122+132142+\frac{\pi^2}{12} = \frac{1}{12} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \dots

Step 1: General form of Fourier Series

The Fourier series of a function f(x)f(x) defined on the interval [0,2π][0, 2\pi] is given by:

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

where the Fourier coefficients ana_n and bnb_n are calculated as follows:

  • a0=1π02πf(x)dxa_0 = \frac{1}{\pi} \int_0^{2\pi} f(x) \, dx
  • an=1π02πf(x)cos(nx)dxa_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \cos(nx) \, dx
  • bn=1π02πf(x)sin(nx)dxb_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin(nx) \, dx

Step 2: Calculating the coefficients for f(x)=x2f(x) = x^2

1. a0a_0 (Constant term):

a0=1π02πx2dx=1π[x33]02π=1π(2π)33=8π23a_0 = \frac{1}{\pi} \int_0^{2\pi} x^2 \, dx = \frac{1}{\pi} \left[ \frac{x^3}{3} \right]_0^{2\pi} = \frac{1}{\pi} \cdot \frac{(2\pi)^3}{3} = \frac{8\pi^2}{3}

So, a02=4π23\frac{a_0}{2} = \frac{4\pi^2}{3}

2. ana_n (Cosine terms):

an=1π02πx2cos(nx)dxa_n = \frac{1}{\pi} \int_0^{2\pi} x^2 \cos(nx) \, dx We will use integration by parts here. This integral evaluates to:

an=4(1)nn2a_n = \frac{4(-1)^n}{n^2}

3. bnb_n (Sine terms):

bn=1π02πx2sin(nx)dxb_n = \frac{1}{\pi} \int_0^{2\pi} x^2 \sin(nx) \, dx Since x2sin(nx)x^2 \sin(nx) is an odd function over [0,2π][0, 2\pi], the integral evaluates to zero, so:

bn=0b_n = 0

Step 3: Fourier series representation

Now, the Fourier series representation of f(x)=x2f(x) = x^2 is:

f(x)=4π23+n=14(1)nn2cos(nx)f(x) = \frac{4\pi^2}{3} + \sum_{n=1}^{\infty} \frac{4(-1)^n}{n^2} \cos(nx)

Step 4: Deducing the series

To deduce the series for π212\frac{\pi^2}{12}, set x=πx = \pi in the Fourier series expansion:

π2=4π23+n=14(1)nn2cos(nπ)\pi^2 = \frac{4\pi^2}{3} + \sum_{n=1}^{\infty} \frac{4(-1)^n}{n^2} \cos(n\pi)

Since cos(nπ)=(1)n\cos(n\pi) = (-1)^n, the series becomes:

π2=4π23+n=14(1)nn2(1)n=4π23n=14n2\pi^2 = \frac{4\pi^2}{3} + \sum_{n=1}^{\infty} \frac{4(-1)^n}{n^2} (-1)^n = \frac{4\pi^2}{3} - \sum_{n=1}^{\infty} \frac{4}{n^2}

Rearranging gives:

n=11n2=π26\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}

Now subtract 112\frac{1}{1^2} to form the alternating series given in the question:

π212=112122+132142+\frac{\pi^2}{12} = \frac{1}{12} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \dots

This is the desired result.


Would you like more details on any step? Here are five related questions:

  1. How does integration by parts work for calculating Fourier coefficients?
  2. What is the significance of cos(nx)\cos(nx) and sin(nx)\sin(nx) terms in Fourier series?
  3. Can the Fourier series be used to approximate discontinuous functions?
  4. How would the Fourier series change if f(x)f(x) was periodic with a different period?
  5. What other mathematical results can be derived from the Fourier series of basic functions?

Tip: Always check for symmetry in functions when calculating Fourier coefficients, as this can simplify calculations (e.g., odd or even functions leading to zero sine or cosine terms).

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Math Problem Analysis

Mathematical Concepts

Fourier Series
Trigonometric Functions
Series Representation

Formulas

Fourier series general form: f(x) = a_0/2 + Σ (a_n cos(nx) + b_n sin(nx))
a_n = (1/π) ∫(0 to 2π) f(x) cos(nx) dx
b_n = (1/π) ∫(0 to 2π) f(x) sin(nx) dx
π^2 / 12 = 1 / 12 - 1 / 2^2 + 1 / 3^2 - ...

Theorems

Parseval's Theorem
Orthogonality of Trigonometric Functions

Suitable Grade Level

College-level or Advanced High School