Question

Find the Fourier series of \(f\) on \([-L, L]\) and determine its sum for \(-L \leq x \leq L .\) Where indicated by \(C\), graph \(f\) and $$ F_{m}(x)=a_{0}+\sum_{n=1}^{m}\left(a_{n} \cos \frac{n \pi x}{L}+b_{n} \sin \frac{n \pi x}{L}\right) $$ on the same axes for various values of \(m\). $$ L=1 ; \quad f(x)=\left\\{\begin{array}{cc} 0, & -1

Short answer

# Short Answer # To find the Fourier series of the given function \(f(x)\) on \([-L, L]\), first compute the coefficients \(a_0\), \(a_n\), and \(b_n\). For the given function, we find that \(a_0 = \frac{2}{\pi}\), \(a_n = \frac{\sin((1+n)\pi/2) + \sin((1-n)\pi/2)}{(1+n)\pi}\), and \(b_n = 0\). Therefore, the Fourier series representation of the function on the interval \([-L, L]\) is given by: $$ F_m(x) = \frac{2}{\pi} + \sum_{n=1}^m\left(\frac{\sin((1+n)\pi/2) + \sin((1-n)\pi/2)}{(1+n)\pi}\cos(n\pi x)\right) $$ To graph both the function \(f(x)\) and its Fourier series \(F_m(x)\), plot the original piecewise function \(f(x)\) and the Fourier series \(F_m(x)\) for various values of \(m\). The more terms included in the sum, the closer the graph of \(F_m(x)\) will be to the graph of \(f(x)\).

Step-by-step solution

Compute the coefficient a_0

First, let's find the coefficient \(a_0\). Since \(f(x)\) is piecewise defined, we need to split the integral into the different intervals where the function is defined. The function \(f(x)\) is defined on three intervals, and we have \(L=1\). So the formula for finding \(a_0\) now looks like: $$ a_0 = \int_{-1}^{-\frac{1}{2}} 0 dx + \int_{-\frac{1}{2}}^{\frac{1}{2}} \cos(\pi x) dx + \int_{\frac{1}{2}}^{1} 0 dx $$ The integral of 0 is 0, so the calculation simplifies to: $$ a_0 = \int_{-\frac{1}{2}}^{\frac{1}{2}} \cos(\pi x) dx $$ Now we evaluate the integral: $$ a_0 = \left[\frac{1}{\pi}\sin(\pi x)\right]_{-\frac{1}{2}}^{\frac{1}{2}} $$ $$ a_0 = \frac{1}{\pi}\left(\sin\left(\frac{\pi}{2}\right)-\sin\left(-\frac{\pi}{2}\right)\right) $$ $$ a_0 = \frac{1}{\pi}(1 + 1) = \frac{2}{\pi} $$

Compute the coefficients a_n

Now let's find the coefficients \(a_n\). Similar to the calculation for \(a_0\), we will split the integral for the piecewise defined function. The formula for \(a_n\) is: $$ a_n = \int_{-1}^{-\frac{1}{2}} 0\cos\left(\frac{n\pi x}{1}\right) dx + \int_{-\frac{1}{2}}^{\frac{1}{2}} \cos(\pi x)\cos\left(\frac{n\pi x}{1}\right) dx + \int_{\frac{1}{2}}^{1} 0\cos\left(\frac{n\pi x}{1}\right) dx $$ The integral of 0 times any function is 0, so we get: $$ a_n = \int_{-\frac{1}{2}}^{\frac{1}{2}} \cos(\pi x)\cos(n\pi x) dx $$ Now we evaluate the integral using the product-to-sum formula \(\cos A\cos B=\frac{1}{2}\left(\cos(A-B)+\cos(A+B)\right)\): $$ a_n = \frac{1}{2}\int_{-\frac{1}{2}}^{\frac{1}{2}}\left(\cos((1-n)\pi x)+\cos((1+n)\pi x)\right)dx $$ $$ a_n = \frac{1}{2}\left[\frac{1}{(1-n)\pi}\sin((1-n)\pi x)+\frac{1}{(1+n)\pi}\sin((1+n)\pi x)\right]_{-\frac{1}{2}}^{\frac{1}{2}} $$ We can now plug in the bounds of integration: $$ a_n = \frac{1}{2}\left(\frac{\sin((1+n)\pi/2) - \sin((1-n)\pi/2)}{(1+n)\pi} - \frac{\sin((1+n)(-\pi/2)) - \sin((1-n)(-\pi/2))}{(1+n)\pi}\right) $$ Since \(\sin(-x) = -\sin(x)\), the expression simplified to: $$ a_n = \frac{1}{2}\left(\frac{\sin((1+n)\pi/2) - \sin((1-n)\pi/2) + \sin((1+n)\pi/2) + \sin((1-n)\pi/2)}{(1+n)\pi}\right) $$ And finally, $$ a_n = \frac{\sin((1+n)\pi/2) + \sin((1-n)\pi/2)}{(1+n)\pi} $$

Compute the coefficients b_n

Since \(f(x)\) is an even function, all \(b_n\) coefficients will be 0.

Write down the Fourier series

Now we have all the coefficients we need to create the Fourier series \(F_m(x)\): $$ F_m(x) = a_0 + \sum_{n=1}^m\left(a_n\cos\left(\frac{n\pi x}{1}\right)+b_n\sin\left(\frac{n\pi x}{1}\right)\right) $$ By plugging in the coefficients we found in the steps above, we get: $$ F_m(x) = \frac{2}{\pi} + \sum_{n=1}^m\left(\frac{\sin((1+n)\pi/2) + \sin((1-n)\pi/2)}{(1+n)\pi}\cos(n\pi x)\right) $$ This is the Fourier series representation of the function \(f(x)\) on the interval \([-L, L]\). For graphing \(F_m(x)\) and \(f(x)\) together, plot the original piecewise function \(f(x)\) and the Fourier series \(F_m(x)\) for any chosen \(m\). The more terms you include in the sum, the closer the graph of \(F_m(x)\) will be to the graph of \(f(x)\).

Key concepts

Piecewise Functions

When studying functions in mathematics, we often come across piecewise functions, which are simply functions defined by multiple sub-functions, each corresponding to a specific part of the domain. To grasp the concept, imagine a function as a road that changes its paving materials at different segments: concrete for one part, asphalt for another, and so on.

Working with piecewise functions, especially in Fourier series, involves analyzing each segment separately. This is because the function's behavior changes with each piece. In the original exercise, the given function changes from zero to a cosine function and back to zero. Thus, calculating parameters such as the Fourier coefficients requires us to split the calculation according to the intervals of the function's different expressions.

When dealing with piecewise functions in Fourier series, it's crucial to carefully examine these intervals, as this division into segments will dictate how we evaluate our integrals. Incorrectly identifying these intervals may lead to wrong results, which is why understanding piecewise functions is a fundamental step before diving into more complex calculations.

Trigonometric Integrals

The computation of Fourier coefficients often requires evaluating trigonometric integrals, which are integrals involving trigonometric functions. These integrals are ubiquitous in mathematical analysis and physics, and they can sometimes be quite challenging.

In our case, for example, the integral to find the coefficient \( a_0 \) is straightforward as it only involves the cosine function. However, when computing \( a_n \) coefficients, the integrals get trickier since they involve the product of two cosine functions.

Mastering trigonometric integrals is essential when working with Fourier series. This sometimes calls for using specific techniques such as trigonometric identities or product-to-sum formulas, to turn complex trigonometric expressions into forms that are easier to integrate. Remember, when integrating functions over symmetric intervals around the origin, properties of even and odd functions can greatly simplify the process, which, in our case, nullifies the \( b_n \) coefficients.

Product-to-Sum Formulas

To efficiently calculate trigonometric integrals involving products of trigonometric functions, mathematicians rely on product-to-sum formulas. These are trigonometric identities that transform the product of sine and cosine functions into a sum of trigonometric functions.

For instance, one of the product-to-sum formulas is \( \cos A\cos B=\frac{1}{2}\left(\cos(A-B)+\cos(A+B)\right) \), and it's used to simplify the integral of the product of two cosines in calculating the \( a_n \) coefficients. By converting the product into a sum, we can easily integrate because the resulting components are simpler to handle.

Understanding and applying these formulas correctly can streamline complex calculations and is a powerful tool in the mathematician's arsenal. Knowing when and how to use these identities in Fourier series allows for the efficient computation of coefficients, which are vital for constructing the series representation of a function.

Fourier Coefficients

The heart of the Fourier series lies in the Fourier coefficients. These numbers capture the weights of the sines and cosines in the series that, when combined, can describe virtually any periodic function with stunning accuracy. Each coefficient, \( a_n \) or \( b_n \) for cosine and sine terms respectively, provides information about the amplitude or 'strength' of a particular harmonic component of the function.

In our exercise, we've calculated the \( a_0 \) coefficient directly through integration. For \( a_n \) coefficients, the product-to-sum formula was leveraged to transform a tricky integral into a more manageable one. Notably, the symmetry of the function dictated that all \( b_n \) coefficients are zero, revealing that our function is composed purely of cosine terms, characteristic of an even function.

The coefficients encapsulate the essence of the function in the realm of frequency. And while the series theoretically requires an infinite number of terms to reconstruct the function perfectly, in practice, an approximation using a finite number of terms, known as partial sums, can provide a graph very close to the original function. This is a profound application of Fourier analysis, further bridging the gap between continuous functions and discrete analysis.