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In each of Problems 19 through 24 : (a) Sketch the graph of the given function for three periods. (b) Find the Fourier series for the given function. (c) Plot \(s_{m}(x)\) versus \(x\) for \(m=5,10\), and 20 . (d) Describe how the Fourier series seems to be converging. $$ f(x)=x^{2} / 2, \quad-2 \leq x \leq 2 ; \quad f(x+4)=f(x) $$

Short Answer

Expert verified
Provide a short answer discussing the convergence of the Fourier series for the function \(f(x) = \frac{x^2}{2}\). The Fourier series for the function \(f(x) = \frac{x^2}{2}\) converges pointwise to the original function as the value of \(m\) increases. This is observed in the graph of the partial sums \(s_{m}(x)\) for different values of \(m\) as they become closer to the original function with the increase in \(m\).

Step by step solution

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01

(1. Sketching the Graph)

To sketch the graph of \(f(x) = \frac{x^2}{2}\) for three periods in the range \(-2 \leq x \leq 2\), we need to plot the function from \(-2\) to \(2\) and observe its periodic behavior. Since \(f(x+4)=f(x)\), we know it will have the same shape for \(-6 \leq x \leq -2\), \(-10 \leq x \leq -6\), \(2 \leq x \leq 6\), \(6 \leq x \leq 10\), and so on. The graph is a parabola with its vertex at \((0, 0)\), opening upwards.
02

(2. Finding the Fourier Series)

To find the Fourier series of \(f(x)\), we first need to compute the coefficients \(a_0\), \(a_n\), and \(b_n\). Since \(f(x)\) is an even function, \(b_n\) will be 0. We can compute \(a_0\) and \(a_n\) using the following formulas: $$ a_0 = \frac{1}{L} \int_{-L}^{L} f(x) dx $$ $$ a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos(\frac{n\pi x}{L}) dx $$ Here, \(L=2\) as the function repeats after an interval of 4. Now, computing \(a_0\) and \(a_n\): $$ a_0 = \frac{1}{2} \int_{-2}^{2} \frac{x^2}{2} dx = \frac{1}{3} $$ $$ a_n = \frac{1}{2} \int_{-2}^{2} \frac{x^2}{2} \cos(\frac{n\pi x}{2}) dx = \frac{4[(-1)^n - 1]}{n^2 \pi^2} $$ So, the Fourier series for \(f(x)\) is given by: $$ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(\frac{n\pi x}{2}) $$ Substituting values, we get: $$ f(x) = \frac{1}{6} + \sum_{n=1}^{\infty} \frac{4[(-1)^n - 1]}{n^2 \pi^2} \cos(\frac{n\pi x}{2}) $$
03

(3. Plotting partial sums \(s_m(x)\))

To plot the partial sums \(s_m(x)\) for \(m=5,10,\) and 20, we need to compute and plot the sum of the Fourier series up to the m-th term. For \(m=5\): $$ s_5(x) = \frac{1}{6} + \sum_{n=1}^{5} \frac{4[(-1)^n - 1]}{n^2 \pi^2} \cos(\frac{n\pi x}{2}) $$ For \(m=10\): $$ s_{10}(x) = \frac{1}{6} + \sum_{n=1}^{10} \frac{4[(-1)^n - 1]}{n^2 \pi^2} \cos(\frac{n\pi x}{2}) $$ And for \(m=20\): $$ s_{20}(x) = \frac{1}{6} + \sum_{n=1}^{20} \frac{4[(-1)^n - 1]}{n^2 \pi^2} \cos(\frac{n\pi x}{2}) $$ Plot \(s_{5}(x)\), \(s_{10}(x)\), and \(s_{20}(x)\) for the interval \(-2 \leq x \leq 2\).
04

(4. Convergence of Fourier series)

As we increase the value of \(m\), the Fourier series converges to the original function \(f(x)\). From the graph of \(s_{m}(x)\) for different values of \(m\), we can see that the partial sums get closer to the original function as \(m\) increases; thus, the Fourier series converges pointwise to the function \(f(x)\).

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Sum Plots
Partial sum plots are a critical visual tool in understanding Fourier series. They show the approximation of a function by a Fourier series up to a certain number of terms, denoted by the variable 'm'. In problems like the exercise provided, plotting partial sums such as sm(x) is an excellent way to observe how the Fourier series progresses towards the original function as more terms are included.

Particularly, for our function f(x) = x^2 / 2, the partial sums of the Fourier series start to build up the shape of the quadratic function by including more and more cosine terms. When m is small, the sum may appear choppy or vastly different from the actual function. However, as m increases to 5, 10, and 20, the plot begins to smooth out and closely resemble the original parabola. This progression is crucial for visual learners, as it helps them concretize the abstract concept of a series approximation.
Fourier Coefficients
The Fourier coefficients, specifically a0, an, and bn, are the scalars that multiply each corresponding sine and cosine wave in the Fourier series. They are essentially the 'DNA' of the Fourier series that determine its shape and behavior.

In our exercise, the Fourier coefficients need to be calculated differently since our original function is even. This influences the coefficients, resulting in all bn coefficients being zero, and simplifies the series to only use cosines. Each coefficient captures a specific feature of the original function — for instance, the constant term a0 represents the average value of the function, while the cosine terms adjust the curves to fit the function more accurately with every added term.
Even Function Properties
An even function has specific properties which heavily influence the computation of Fourier series. Symmetry is the principal characteristic — an even function is mirror-symmetric about the Y-axis, meaning that f(x) = f(-x) for all x in the function's domain.

Due to this symmetry, the integration of odd functions (like the sine function) over symmetric intervals around the origin will always yield zero, hence no sine terms appear in the Fourier series of an even function. Our example function f(x) = x^2 / 2 demonstrates this property, resulting in the simplification of the Fourier series to contain only cosine terms. The application of even function properties optimizes the calculation process, as less computing effort is required when certain terms are guaranteed to be absent.
Convergence of Fourier Series
The convergence of Fourier series is a fundamental aspect in the study of these mathematical representations. Convergence refers to how the infinite sum of terms in a Fourier series approaches the given function as the number of terms tends towards infinity.

In our exercise, we observe that as we increase the number of terms, m, the series better approximates the function f(x) = x^2 / 2. This is a pointwise convergence, meaning that for each x-coordinate, the sequence of partial sums converges to the actual function value. It's a crucial concept because it reassures us that the Fourier series is a valid and reliable method to represent functions over intervals. However, it is also essential for students to learn about where and how different types of convergence apply, such as uniform convergence and convergence in the mean, to thoroughly understand the behavior of Fourier series.

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Most popular questions from this chapter

(a) Find the solution \(u(x, y)\) of Laplace's equation in the rectangle \(0

(a) Find the solution \(u(x, y)\) of Laplace's equation in the rectangle \(0

Consider a uniform rod of length \(L\) with an initial temperature given by \(u(x, 0)=\) \(\sin (\pi x / L), 0 \leq x \leq L\) Assume that both ends of the bar are insulated. (a) Find the temperature \(u(x, t) .\) (b) What is the steady-state temperature as \(t \rightarrow \infty\) ? (c) Let \(\alpha^{2}=1\) and \(L=40 .\) Plot \(u\) versus \(x\) for several values of \(t\). Also plot \(u\) versus \(t\) for several values of \(x .\) (d) Describe briefly how the temperature in the rod changes as time progresses.

More Specialized Fourier Scries. Let \(f\) be a function originally defined on \(0 \leq x \leq L\). In this section we have shown that it is possible to represent \(f\) either by a sine series or by a cosine series by constructing odd or even periodic extensions of \(f,\) respectively. Problems 38 through 40 concern some other more specialized Fourier series that converge to the given function \(f\) on \((0, L) .\) $$ \begin{array}{l}{\text { Let } f \text { be extended into }(L, 2 L] \text { in an arbitrary manner. Then extend the resulting function }} \\ {\text { into }(-2 L, 0) \text { as an odd function and elsewhere as a periodic function of period } 4 L \text { (see }} \\ { \text { Figure }10.4 .6) . \text { Show that this function has a Fourier sine series in terms of the functions }} \\\ {\sin (n \pi x / 2 L), n=1,2,3, \ldots . \text { that is, }}\end{array} $$ $$ f(x)=\sum_{n=1}^{\infty} b_{n} \sin (n \pi x / 2 L) $$ where $$ b_{n}=\frac{1}{L} \int_{0}^{2 L} f(x) \sin (n \pi x / 2 L) d x $$ $$ \text { This series converges to the original function on }(0, L) $$ (Figure cant copy)

Prove that if \(f\) is an odd function, then $$ \int_{-L}^{L} f(x) d x=0 $$

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