Problem 4 - A (terse) introduction to Lebesgue integration

Question

Find the Fourier series for the function \(f(x)=x\) on \([-\pi, \pi]\). Use it with a suitable value of \(x\) to show $$ \frac{\pi}{4}=\sum_{n=0}^{\infty}(-1)^{n} \frac{1}{2 n+1} $$

Step-by-Step Solution

Answer

The Fourier series confirms \( \frac{\pi}{4} = \sum_{n=0}^{\infty}(-1)^{n} \frac{1}{2n+1} \) for \( x = \frac{\pi}{2} \).

Step 1: Understanding the Function and Interval

The function given is \( f(x) = x \) defined on the interval \([-\pi, \pi]\). To determine the Fourier series, we consider the periodic extension of this function, which is odd. So, the Fourier series will only contain sine terms.

Step 2: Identify the Fourier Coefficients

For an odd function, the Fourier series is constructed using the sine terms, and the coefficients \( b_n \) are calculated using the integral:\[b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx\]For the function \( f(x) = x \), the coefficients are calculated as:\[b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \sin(nx) \, dx\]

Step 3: Calculate the Integral for Coefficients

Using integration by parts, we solve:\( u = x \), \( dv = \sin(nx) \, dx \)\( du = dx \), \( v = -\frac{1}{n} \cos(nx) \)Substitute these into the integration by parts formula:\[b_n = \frac{1}{\pi} \left( -\frac{x}{n} \cos(nx) \bigg|_{-\pi}^{\pi} + \frac{1}{n} \int_{-\pi}^{\pi} \cos(nx) \, dx \right)\]The boundary terms vanish, and the integral simplifies, yielding:\[b_n = \frac{2}{n\pi} \begin{cases} (-1)^{n+1}, & \text{if } n \text{ is odd} \ 0, & \text{if } n \text{ is even} \end{cases}\]Thus, \( b_n = \frac{2}{n}(-1)^{n+1} \text{ for odd } n. \)

Step 4: Construct the Fourier Series

Since the function is odd, the Fourier series for \( f(x) = x \) is:\[f(x) = \sum_{n=1, \text{odd}}^{\infty} \frac{2}{n}(-1)^{(n+1)/2} \sin(nx)\]

Step 5: Find Suitable Value of x and Verify Identity

To show that \( \frac{\pi}{4} = \sum_{n=0}^{\infty}(-1)^{n} \frac{1}{2n+1} \), set \( x = \frac{\pi}{2} \) which gives the series:\[\frac{\pi}{2} = \sum_{n=0}^{\infty} (-1)^{n} \frac{2}{2n+1} \]Dividing both sides by 2 results in:\[\frac{\pi}{4} = \sum_{n=0}^{\infty} (-1)^{n} \frac{1}{2n+1}\] which matches the given identity.

Key Concepts

Fourier coefficients

In the world of Fourier series, Fourier coefficients play a crucial role in expressing any function as an infinite series of sine and cosine functions. For the function \(f(x) = x\) defined on the interval \([-\pi, \pi]\), we utilize the properties of odd functions to determine the Fourier coefficients. Here, the function is odd, so the series comprises only sine terms.
The Fourier coefficient, \(b_n\), for sine terms is given by the integral formula:

• \( b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx \)

This formula effectively reduces to finding the integral of \(x \sin(nx)\) over the interval \([-\pi, \pi]\), which gives the weights (coefficients) needed in the sine series.
These coefficients capture information about the function at different frequencies \(n\). Specifically, it captures how much of the \(n\)-th sine wave is present in the function. For odd \(n\), the coefficients conform to a specific pattern, adjusting the series terms to correctly reconstruct the function from its sine components.

integration by parts

Integration by parts is a technique often used to solve integrals that are products of two functions, much like the one encountered in determining the Fourier coefficients for \(b_n\). When we have an integral of the form \(\int u \, dv\), integration by parts provides a way to transform it:

• \( \int u \, dv = uv - \int v \, du \)

Here, for \(f(x) = x\) and the sine function, you select \(u = x\) and \(dv = \sin(nx) \, dx\). Differentiating \(u\) and integrating \(dv\) gives us:

• \(du = dx\)
• \(v = -\frac{1}{n} \cos(nx)\)

Substituting these into the formula allows transformation of the integral into a more manageable form. The boundary terms vanish when computed over the symmetric interval \([-\pi, \pi]\), leaving only the simplified integral of \(\cos(nx)\). This simplification is key in obtaining the concise expression for the Fourier coefficients.

odd function

An odd function is characterized by the property \(f(-x) = -f(x)\), and this symmetry about the origin implies several important aspects when expanding the function into a Fourier series. In this exercise, the function \(f(x) = x\) is an odd function, and thus its Fourier series will contain only sine terms.
The sine function is also odd and resonates well with \(f(x)\)'s properties. This is why the cosine terms, which would appear in an even function or a non-symmetric function's Fourier series, are absent in our final series expression.
This odd function behavior simplifies the series expansion greatly, since:

• The cosine coefficients \(a_n\) are zero due to symmetry.
• The constant term \(a_0\) is also zero because the average value of the function over one period is zero.

Understanding and exploiting the properties of odd (and similarly even) functions allow for more efficient computations and form the basis for correctly choosing terms in a Fourier series.