Question

Find the Fourier cosine series. $$ f(x)=\left\\{\begin{array}{ll} 1, & 0 \leq x \leq \frac{L}{2} \\ 0, & \frac{L}{2}

Short answer

Question: Find the Fourier cosine series representation of the following function defined on the interval [0, L]: \[f(x) = \left\{\begin{array}{ll} 1, & 0 \leq x < \frac{L}{2} \\ 0, & \frac{L}{2} \leq x \leq L \end{array}\right.\] Answer: The Fourier cosine series representation of the given function f(x) is: $$ f(x) = \frac{1}{2} + \sum_{n=1}^{\infty}\left\\\{\begin{array}{ll} 0 \cdot \cos(\frac{n\pi x}{L}), & \text{if n is even} \\\ \frac{2}{n\pi}\cos(\frac{n\pi x}{L}), & \text{if n is odd} \end{array}\right. $$

Step-by-step solution

Calculate \(a_0\)

Consider the integral for \(a_0\): $$ a_0 = \frac{2}{L}\int_{0}^{L} f(x) dx $$ Since f(x) is piecewise defined, we can divide the integral into two parts based on its definition: $$ a_0 = \frac{2}{L}\left[\int_{0}^{\frac{L}{2}} 1 dx + \int_{\frac{L}{2}}^{L} 0 dx\right] $$ Now, evaluate the integrals and solve for \(a_0\): $$ a_0 = \frac{2}{L}\left[\frac{L}{2}\right] = 1 $$

Calculate \(a_n\)

Now, we move on to calculate \(a_n\): $$ a_n = \frac{2}{L}\int_{0}^{L} f(x)\cos(\frac{n\pi x}{L}) dx $$ Again, divide the integral based on the definition of f(x): $$ a_n = \frac{2}{L}\left[\int_{0}^{\frac{L}{2}} \cos(\frac{n\pi x}{L}) dx + \int_{\frac{L}{2}}^{L} 0\cos(\frac{n\pi x}{L}) dx\right] $$ Now, evaluate the integrals and simplify: $$ a_n = \frac{2}{L}\left[\int_{0}^{\frac{L}{2}} \cos(\frac{n\pi x}{L}) dx\right] $$ To solve the cosine integral, apply the substitution: $$ u=\frac{n\pi x}{L} \quad \rightarrow \quad du=\frac{n\pi}{L} dx \quad \rightarrow \quad dx=\frac{L}{n\pi} du $$ Now replace \(dx\) and x with its equivalent in terms of \(u\): $$ a_n = \frac{2}{L} \int_{0}^{\frac{n\pi}{2}} \cos(u) \frac{L}{n\pi} du $$ Solve the integral: $$ a_n = \frac{2}{n\pi} \left[\sin(u)\right]_0^{\frac{n\pi}{2}} $$ Since the sine function is periodic, \(a_n\) will be different for even and odd n. When n is even, the sine value will be 0, for odd n, the sine value will be 1. So, we can write \(a_n\) as: $$ a_n = \left\\\{\begin{array}{ll} 0, & \text{if n is even} \\\ \frac{2}{n\pi}, & \text{if n is odd} \end{array}\right. $$

Substitute Coefficients into Fourier Cosine Series

Now, that we have the coefficients \(a_0\) and \(a_n\), we can substitute them back into the Fourier cosine series: $$ f(x) = \frac{1}{2} + \sum_{n=1}^{\infty}a_n\cos(\frac{n\pi x}{L}) $$ Substitute the derived \(a_n\) value and simplify: $$ f(x) = \frac{1}{2} + \sum_{n=1}^{\infty}\left\\\{\begin{array}{ll} 0 \cdot \cos(\frac{n\pi x}{L}), & \text{if n is even} \\\ \frac{2}{n\pi}\cos(\frac{n\pi x}{L}), & \text{if n is odd} \end{array}\right. $$ Thus, we have found the Fourier cosine series representation of the given function f(x).

Key concepts

Piecewise Functions

Piecewise functions are unique in that they can have different expressions based on varying intervals of the independent variable. In the exercise provided, the function \( f(x) \) is defined by two different expressions depending on the value of \( x \). For the interval \( 0 \leq x \leq \frac{L}{2} \), \( f(x) = 1 \). For the interval \( \frac{L}{2} < x < L \), \( f(x) = 0 \). This kind of setup is especially useful in applications where a function's behavior changes at certain points.

When solving problems involving piecewise functions, it is crucial to consider each section separately when performing operations like integration. In a Fourier series context, handling piecewise functions entails dividing the integral across the different intervals and solving each one separately. By doing this, we ensure that the distinct behavior of the function on each interval is accurately represented. This is key in the derivation of Fourier series coefficients, which rely on accurately computed integrals spanning the entire function's domain.

Integration Techniques

Integration techniques are essential in finding the coefficients needed for forming a Fourier series. In the exercise, we split the original integral into parts that correspond to the intervals of the piecewise function:

• From \( 0 \) to \( \frac{L}{2} \), we have \( \int_{0}^{\frac{L}{2}} 1 \cdot \cos\left(\frac{n\pi x}{L}\right) dx \)
• From \( \frac{L}{2} \) to \( L \), where the function is \( 0 \), the integral becomes 0.

Splitting integrals like this simplifies the calculation process, focusing only on where the function doesn't vanish.

A common substitution technique simplifies the cosine integral. By substituting \( u = \frac{n\pi x}{L} \), we change variables and simplify the expression. The differentials adjust to be consistent with this variable change, making evaluating the integral straightforward. This results in an easier form to handle, primarily turning the integrate into a more manageable form of sine or cosine functions.

Fourier Series Coefficients

The Fourier series coefficients are the backbone of representing a periodic function as a sum of sine and cosine terms. For the cosine series portion, these coefficients are determined through integrals that consider both the function value and the cosine basis functions. In this problem, two essential coefficients derive from integrals:

• The average term \( a_0 \) calculated as \( \frac{2}{L} \int_{0}^{L} f(x) dx \)
• The specific terms \( a_n \) where the basis function \( \cos(\frac{n\pi x}{L}) \) is included in the integral.

With this function, \( a_0 = 1 \), indicating that the constant offset of the Fourier cosine series is \( \frac{1}{2} \) once it is adjusted for standard form presentation of series.

The sequence of coefficients \( a_n \) gets evaluated over the interval to account for each non-zero part of the piecewise function. As a result, certain values result in coefficients being zero (when sine outcomes cancel the terms), and when \( n \) is odd, \( a_n = \frac{2}{n\pi} \). This exercise exemplifies how Fourier coefficients function to fit periodic functions into cosine series representations, smoothly transitioning behavior across intervals.