Problem 5 - A (terse) introduction to Lebesgue integration

Question

Find the Fourier series for the function \(f(x)=x^{2}\) on \([-\pi, \pi]\). Use it with a suitable value of \(x\) to evaluate $$ \sum_{n=1}^{\infty} \frac{1}{n^{2}} $$

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \).

Step 1: Define the Function and Interval

The function given is \( f(x) = x^2 \) on the interval \( [-\pi, \pi] \). We are tasked with finding its Fourier series and then using it to evaluate the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \).

Step 2: Calculate the Fourier Coefficients

The Fourier series for a function \( f(x) \) on \([-\pi, \pi]\) is given by: \[ a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right) \] where \( a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx \), \( a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx \), and \( b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx \).

Step 3: Compute \(a_0\)

Compute \( a_0 \) using \( f(x) = x^2 \): \[ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} x^2 \, dx = \frac{1}{2\pi} \left[ \frac{x^3}{3} \right]_{-\pi}^{\pi} = \frac{1}{2\pi} \left( \frac{\pi^3}{3} - \frac{(-\pi)^3}{3} \right) = \frac{\pi^2}{3} \]

Step 4: Compute \(a_n\) Coefficients

The coefficient \( a_n \) is given by: \[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \cos(nx) \, dx \] This integral evaluates to zero because \( x^2 \cos(nx) \) is an odd function over \([-\pi, \pi]\). So, \( a_n = 0 \) for all \( n \).

Step 5: Compute \(b_n\) Coefficients

For \( b_n \): \[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \sin(nx) \, dx \] Using integration by parts twice, we find: \[ b_n = \frac{2(-1)^{n+1}}{n^2} \]

Step 6: Write the Fourier Series

The Fourier series for \( f(x) = x^2 \) is: \[ \frac{\pi^2}{3} + \sum_{n=1}^{\infty} \frac{2(-1)^{n+1}}{n^2} \sin(nx) \]

Step 7: Evaluate the Series at \( x = \pi \)

To use the Fourier series to evaluate \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), set \( x = \pi \). This ensures \( \sin(n\pi) = 0 \). Thus, \[ x^2 = \frac{\pi^2}{3} \ = \pi^2 \sum_{n=1}^{\infty} \frac{2(-1)^{n+1}}{n^2} \] Divide both sides by \( \pi^2 \) to solve for the series: \[ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \]

Step 8: Conclusion

We have used the Fourier series to evaluate: \( \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \).

Key Concepts

Fourier coefficients

In the realm of Fourier series, understanding Fourier coefficients is key. They determine how a function can be expressed as a sum of sine and cosine terms. These coefficients allow for the transformation of complex periodic functions into simpler trigonometric series.

When given a function like \( f(x) = x^2 \) on the interval \([-\pi, \pi]\), the Fourier series uses coefficients \( a_0, a_n, \) and \( b_n \) to build the series.

• \( a_0 \) represents the average value of the function and is calculated as:
  \[ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx \]
• \( a_n \) measures the amplitude of cosine terms:
  \[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx \]
• \( b_n \) captures the sine terms' amplitudes:
  \[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx \]

The goal is to find these coefficients so the trigonometric series portrays the original function accurately. For the function \( x^2 \), we observed that \( a_n = 0 \) for all \( n \), and \( b_n \) simplifies using integration by parts.

trigonometric series

Trigonometric series are an insightful way to represent functions. They consist of infinite sums of sines and cosines that converge to a specific periodic function.

In physics and engineering, such series can describe oscillating phenomena like sound waves or electrical signals. A Fourier series is a type of trigonometric series, where each term is defined by the Fourier coefficients.

• The generic form of a trigonometric series is:
  \[ a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right) \]

For \( f(x) = x^2 \), this becomes:
\( \frac{\pi^2}{3} + \sum_{n=1}^{\infty} \frac{2(-1)^{n+1}}{n^2} \sin(nx) \).
Such a series allows calculations beyond simple function evaluation, like vector projections in a multi-dimensional space.

definite integrals

Definite integrals play a crucial role in finding Fourier coefficients. They help calculate the area under curves, which informs the amplitude of sine and cosine terms in Fourier series.

For \( f(x) = x^2 \), we compute definite integrals over the interval \([-\pi, \pi]\) to derive \( a_0 \), \( a_n \), and \( b_n \).

• \( a_0 \) is calculated as:
  \[ \frac{1}{2\pi} \int_{-\pi}^{\pi} x^2 \, dx = \frac{\pi^2}{3} \]
• \( a_n \) is zero because \( x^2 \cos(nx) \) is odd, meaning its integral over a symmetric interval like \( [-\pi, \pi] \) equals zero.
• For \( b_n \), integration by parts enables calculating:
  \[ b_n = \frac{2(-1)^{n+1}}{n^2} \]
  This shows that repeating definite integrations tightly weaves the function and its Fourier representation together.
  

Using definite integrals assures accurate conversion from a regular function to its Fourier series expression.