Question

(a) Find the required Fourier series for the given function. (b) Sketch the graph of the function to which the series converges for three periods. (c) Plot one or more partial sums of the series. $$ f(x)=-x, \quad-\pi

Short answer

In this exercise, we analyzed the function \(f(x) = -x\) over the interval \(-\pi < x < 0\) and found its Fourier sine series representation. We calculated the coefficients \(b_n\) and obtained the Fourier sine series as follows: $$ f(x) = \sum_{n=1}^{\infty} \left[\frac{(-1)^{n+1} \pi}{n \pi} + \frac{(-1)^n}{n^2 \pi}\right] \sin(nx) $$ Then, we discussed how to sketch the function over three periods and plot several partial sums to analyze the convergence. As we include more terms, the partial sums will approach the original function within the specified interval.

Step-by-step solution

Finding the Fourier sine series coefficients

To find the Fourier sine series of \(f(x)\), we will need to compute the coefficients \(b_n\), as follows: $$ b_n = \frac{2}{T} \int_{0}^{T} f(x) \sin \left(\frac{2\pi nx}{T} \right) dx $$ Since our function is an odd function and the period is specified as \(T=2\pi\), we can write: $$ b_n = \frac{1}{\pi} \int_{-\pi}^{0} (-x) \sin(nx) dx $$ Now we need to compute the integral and find an expression for \(b_n\).

Computing the integral

We will use integration by parts to compute the integral. Let: $$ u = -x \quad \Rightarrow \quad du = -dx $$ $$ dv = \sin(nx) dx \quad \Rightarrow \quad v = -\frac{1}{n} \cos(nx) $$ The integral becomes: $$ b_n = \frac{1}{\pi}\left[-x \cdot -\frac{1}{n} \cos(nx) |_{-\pi}^{0} - \int_{-\pi}^{0} -\frac{1}{n} \cos(nx) (-dx)\right] $$ Computing this integral, we get: $$ b_n = \frac{1}{\pi}\left[\frac{1}{n}\left(x\cos(nx)\right) |_{-\pi}^{0} + \frac{1}{n} \int_{-\pi}^{0} \cos(nx) dx \right] $$ $$ b_n = \frac{1}{\pi}\left[\frac{1}{n}\left(x\cos(nx)\right) |_{-\pi}^{0} + \frac{1}{n^2}\left(\sin(nx)\right) |_{-\pi}^{0} \right] $$ $$ b_n = \frac{1}{n\pi}\left[(-1)^{n+1}\pi - 0 + \frac{(0-(-1)^n)}{n}\right] $$ Simplify the expression for \(b_n\): $$ b_n = \frac{(-1)^{n+1} \pi}{n \pi} + \frac{(-1)^n}{n^2 \pi} $$

Writing down the Fourier sine series

Now that we have the coefficients \(b_n\), we can write down the Fourier sine series for the function \(f(x)\): $$ f(x) = \sum_{n=1}^{\infty} \left[\frac{(-1)^{n+1} \pi}{n \pi} + \frac{(-1)^n}{n^2 \pi}\right] \sin(nx) $$

Sketching the function and partial sums

The original function \(f(x) = -x\) is a straight line with a negative slope over the interval \((-\pi, 0)\). The graph should be plotted for three periods, which will show the behavior of \(f(x)\) over three adjacent intervals of length \(2\pi\). To visualize the convergence of partial sums, take a few terms of the series and plot them. You will see that for larger values of \(n\), the partial sums approach the original function within the specified interval. To sum up, we have found the Fourier sine series for the given function and discussed how to sketch the function over three periods and plot some partial sums.

Key concepts

Fourier Series Convergence

Understanding the concept of Fourier series convergence is crucial when studying periodic functions and their frequency components. A Fourier series represents a periodic function as an infinite sum of sine and cosine terms. Convergence refers to the idea that as we sum more and more terms of the Fourier series, the approximation of the function improves and ideally, we 'converge' to the actual function.

For a given periodic function, the convergence is dependent on the function's properties such as continuity, differentiability, and the presence of discontinuities. The convergence can be pointwise, where the series converges to the function at each point, or uniform, where the series converges to the function uniformly over an interval.

In the case of the exercise provided, the Fourier sine series, which is a specific type of Fourier series consisting solely of sine terms because of the function's odd symmetry, is used to represent the function over the interval \( -\pi, 0 \). The odd extension of this function outside this interval is implied for the convergence of the sine series.

Integration by Parts

The method of integration by parts is a powerful tool used to solve integrals involving products of functions. This technique is based on the product rule for differentiation and is formally written as \[ \int u dv = uv - \int v du \]. The idea is to strategically choose \(u\) and \(dv\) such that the resulting integral \(\int v du\) is simpler than the original integral.

This technique is extensively used in computing Fourier coefficients where integrals involving products of the target function and trigonometric functions arise. In the step-by-step solution, integration by parts was perfectly demonstrated. The selection of \( u = -x \) and \( dv = \sin(nx) dx \) allowed us to transform the integral involving the product of \( x \) and sine function into simpler integrals that eventually led to an expression for the Fourier sine series coefficients \( b_n \).

Partial Sums Plotting

Plotting partial sums is a visual method to illustrate how well the Fourier series approximates the original function as more terms are added. A partial sum of a Fourier series is simply the sum of the first \( N \) terms of the series. As \( N \) increases, the partial sum becomes a better approximation of the function, especially within one period.

In practice, particularly within computational software or graphing calculators, we can generate plots for the function being analyzed and its Fourier series partial sums with different values of \( N \). Observing the plots, it is generally expected that the more terms we include, the closer the series will be to the original function. This concept reinforces our understanding of convergence. For the given exercise, the requirement to plot partial sums serves two main purposes: to verify the accuracy of computed Fourier coefficients and to visualize the convergence of the series to the function \( f(x) = -x \)over the interval \( -\pi, 0 \).