Problem 2 - Advanced Engineering Mathematics

Question

$$\begin{array}{l} a_{0}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) d x=\frac{1}{\pi} \int_{-\pi}^{0}(-1) d x+\frac{1}{\pi} \int_{0}^{\pi} 2 d x=1 \\ a_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x d x=\frac{1}{\pi} \int_{-\pi}^{0}-\cos n x d x+\frac{1}{\pi} \int_{0}^{\pi} 2 \cos n x d x=0 \\ b_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x d x=\frac{1}{\pi} \int_{-\pi}^{0}-\sin n x d x+\frac{1}{\pi} \int_{0}^{\pi} 2 \sin n x d x=\frac{3}{n \pi}\left[1-(-1)^{n}\right] \\ f(x)=\frac{1}{2}+\frac{3}{\pi} \sum_{n=1}^{\infty} \frac{1-(-1)^{n}}{n} \sin n x \end{array}$$

Step-by-Step Solution

Answer

The Fourier series is: \(f(x) = \frac{1}{2} + \frac{3}{\pi} \sum_{n=1}^{\infty} \frac{1 - (-1)^n}{n} \sin(nx)\).

Step 1: Calculate the Constant Term

The constant term, known as the [a_0](https://latex.codecogs.com/svg.latex?%5Ca_0) term in the Fourier series, is calculated as follows: \[a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx = \frac{1}{\pi} \left( \int_{-\pi}^{0} (-1) \, dx + \int_{0}^{\pi} 2 \, dx \right)\]Evaluation of each integral results in:- \( \int_{-\pi}^{0} (-1) \, dx = -\pi \) and- \( \int_{0}^{\pi} 2 \, dx = 2\pi \).Thus, substituting back, the calculation becomes:\[a_0 = \frac{1}{\pi} (-\pi + 2\pi) = 1\]

Step 2: Calculate the Cosine Coefficients

The coefficients for \( \cos(nx) \), denoted as \(a_n\), are calculated by:\[a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx = \frac{1}{\pi} \left( \int_{-\pi}^{0} -\cos(nx) \, dx + \int_{0}^{\pi} 2\cos(nx) \, dx \right)\]Both integrals result in terms that equal zero when evaluated, hence:\[a_n = 0\]This means there are no cosine terms in the Fourier series for \( f(x) \).

Step 3: Calculate the Sine Coefficients

The coefficients for \( \sin(nx) \), denoted as \(b_n\), are calculated by:\[b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx = \frac{1}{\pi} \left( \int_{-\pi}^{0} -\sin(nx) \, dx + \int_{0}^{\pi} 2\sin(nx) \, dx \right)\]After evaluating both integrals, it becomes:\[b_n = \frac{3}{n\pi} \left[ 1 - (-1)^n \right]\]This implies that when \(n\) is odd, \(b_n\) contributes to the series, and when \(n\) is even, \(b_n\) is zero.

Step 4: Formulate the Fourier Series

Substituting \(a_0\), \(a_n\), and \(b_n\) into the Fourier series formula, we have:\[f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right)\]Given that \(a_n = 0\) for all \(n\), the series simplifies to:\[f(x) = \frac{1}{2} + \frac{3}{\pi} \sum_{n=1}^{\infty} \frac{1 - (-1)^n}{n} \sin(nx)\]

Key Concepts

Cosine Coefficients

When dealing with Fourier series, cosine coefficients, denoted as \( a_n \), play a crucial role. They help to describe the even symmetry part of a periodic function. These coefficients are calculated using integrals that involve the function \( f(x) \) and the cosine function \( \cos(nx) \).

The formula for finding the cosine coefficients is given by:

• \( a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx \)

Upon solving these integrals, if we find that \( a_n \) is zero for all \( n \), it indicates that there are no cosine components in the Fourier series representation of \( f(x) \). This result was observed in the solution provided, which means that \( f(x) \) is made up entirely of sine terms and a constant.

Understanding the role of cosine coefficients is vital as they provide insight into the nature of the function being represented. In this case, the absence of cosine terms helps identify specific symmetrical properties of \( f(x) \).

Sine Coefficients

Sine coefficients, denoted as \( b_n \), are key components in a Fourier series that reflect the odd symmetry of a function. They are calculated using sine functions \( \sin(nx) \) in the integration process.

The formula used to determine these coefficients is:

• \( b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx \)

During the solving process in the given exercise, the resulting expression illustrates that \( b_n \) is non-zero only when \( n \) is odd. This is mathematically represented by \( b_n = \frac{3}{n \pi} [1 - (-1)^n] \). As a consequence, sine terms emerge in the Fourier series for odd \( n \), contributing to the overall structure of the function \( f(x) \).

Sine coefficients are significant in capturing the oscillatory nature of \( f(x) \). Their presence indicates the importance of odd harmonics in the function's periodic behavior. Recognizing this helps us better understand how sinusoidal components build complex functions.

Integration

Integration is a fundamental tool used in deriving both cosine and sine coefficients within the Fourier series framework. It involves calculating the integral of the product of the function \( f(x) \) with either \( \cos(nx) \) or \( \sin(nx) \) over the interval \([-\pi, \pi]\). This process effectively extracts the amplitude of each corresponding trigonometric function present in \( f(x) \).

There are some key points to remember when performing these integrations:

• The integration limits always span from \(-\pi\) to \(\pi\), reflecting the periodicity of trigonometric functions.
• When integrating step-wise across sub-intervals where \( f(x) \) has distinct values or behaviors, it’s essential to treat each section carefully and add the results.
• If a trigonometric function multiplies with similar terms (like \( \sin(nx) \) with \( \sin(nx) \)), orthogonality might help simplify the integrals.

In this exercise, accurate integration allowed for the isolation of the different coefficients (\( a_0, a_n, \) and \( b_n \)), which were crucial in constructing the full Fourier series expression of \( f(x) \). Integration in this context is not just computation; it's a means to understand the distribution of frequency components in periodic functions.