Question

Find the Fourier sine 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 sine series of the function \(f(x) = \begin{cases} 1 & 0 \leq x \leq \frac{L}{2} \\ 0 & \frac{L}{2} < x < L \end{cases}\) on the interval \([0, L]\). Answer: The Fourier sine series representation of the given function \(f(x)\) is: $$ f_s(x) = \sum_{n=1}^\infty \frac{2}{n \pi} \Bigg[ 1 - \cos{(n \pi)} \Bigg] \sin{\frac{n \pi x}{L}} $$

Step-by-step solution

Split the integral for \(b_n\) into two parts

Since \(f(x)\) is a piecewise function, we will split the integral for \(b_n\) into two parts, one for each segment of the function. This means we are calculating the following integral: $$ b_n = \frac{2}{L} \Bigg[ \int_0^{\frac{L}{2}} f(x) \sin{\frac{n \pi x}{L}} dx + \int_{\frac{L}{2}}^L f(x) \sin{\frac{n \pi x}{L}} dx \Bigg] $$

Calculate the integral

Firstly, replace \(f(x)\) with 1 in the first integral and 0 in the second integral. Then, solve the integrals: $$ b_n = \frac{2}{L} \Bigg[ \int_0^{\frac{L}{2}} 1 \sin{\frac{n \pi x}{L}} dx + \int_{\frac{L}{2}}^L 0 \sin{\frac{n \pi x}{L}} dx \Bigg] $$ $$ b_n = \frac{2}{L} \Bigg[ \int_0^{\frac{L}{2}} \sin{\frac{n \pi x}{L}} dx \Bigg] $$ Next, integrate \(\sin{\frac{n \pi x}{L}}\) with respect to \(x\): $$ b_n = \frac{2}{L} \Bigg[ \frac{-L}{n \pi} \cos{\frac{n \pi x}{L}} \Bigg]_0^{\frac{L}{2}} $$

Evaluate the integral at the limits

Now, evaluate this expression at the limits of integration: $$ b_n = \frac{2}{L} \Bigg[\frac{-L}{n \pi} \cos{\frac{n \pi (\frac{L}{2})}{L}} - \frac{-L}{n \pi} \cos{\frac{n \pi (0)}{L}} \Bigg] $$ Simplify the expression and notice that \(\cos{0} = 1\): $$ b_n = \frac{2}{n \pi} \Bigg[ 1 - \cos{(n \pi)} \Bigg] $$

Write the Fourier sine series

Finally, write the Fourier sine series using the sine coefficients (\(b_n\)), which we found in the previous step: $$ f_s(x) = \sum_{n=1}^\infty \frac{2}{n \pi} \Bigg[ 1 - \cos{(n \pi)} \Bigg] \sin{\frac{n \pi x}{L}} $$ This is the Fourier sine series representation of the given function \(f(x)\).

Key concepts

Piecewise Function

A piecewise function is one that is defined by different expressions based on the range of the input value. In this context, the function \(f(x)\) is defined as "1" for \(0 \leq x \leq \frac{L}{2}\) and "0" for \(\frac{L}{2} < x < L\). This means the function behaves differently in each part of its domain.
When dealing with piecewise functions, it is important to consider each segment separately to apply different mathematical operations.

• Ensure the defined segments are clear and non-overlapping.
• Understand how the function changes at different intervals.

By identifying these segments, we can more easily split integrals or other calculations, which is crucial for operations like finding a Fourier sine series.

Integral

An integral is a fundamental concept in calculus representing the area under a curve. In this exercise, integrating is key to finding the Fourier sine coefficients. Since \(f(x)\) is piecewise, the integral has to be split for each segment of the function.
The expression for \(b_n\) involves integrating the product \(f(x) \sin\left(\frac{n \pi x}{L}\right)\).

• Set up different integrals for each piece of the function.
• The integral \(\int_{0}^{\frac{L}{2}} 1 \sin\left(\frac{n \pi x}{L}\right) dx\) reflects the value of the function in its interval.

This step forms the base for deriving the sine coefficients for each term in the series.

Trigonometric Integration

Trigonometric integration involves integrating expressions involving sine or cosine functions. The focus here is to integrate \(\sin\left(\frac{n \pi x}{L}\right)\), which is a standard type of trigonometric integral.
Knowing the antiderivative of sine functions is crucial:
\[\int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + C\]
Apply this to \(\int_0^{\frac{L}{2}} \sin\left(\frac{n \pi x}{L}\right) dx\):

• Substitute \(\sin\) with its antiderivative.
• Evaluate at the bounds, using properties like \(\cos(0) = 1\).

This approach helps in simplifying the expressions to find sine coefficients efficiently.

Sine Coefficients

Sine coefficients, \(b_n\) in this case, are derived from integrating the product of the function \(f(x)\) and sine terms over the defined domain. These coefficients form the building blocks of the Fourier sine series.
The formula used is:
\[b_n = \frac{2}{L}\int_0^L f(x) \sin\left(\frac{n \pi x}{L}\right) dx\]
In the example provided, after evaluating the integral, we get:
\[b_n = \frac{2}{n \pi}\left[1 - \cos(n \pi)\right]\]

• \(\cos(n \pi)\) simplifies to \((-1)^n\), affecting the series' alternate term behavior.
• Sine coefficients are essential in capturing the odd symmetry of the function.

These coefficients are summed up to construct the complete Fourier sine series representation.