Problem 2 - Advanced Engineering Mathematics

Question

In Problems, find the Fourier series of \(f\) on the given interval. $$ f(x)=\left\\{\begin{array}{lr} -1, & -\pi

Step-by-Step Solution

Answer

The Fourier series is \( f(x) = \frac{1}{2} + \sum_{n=1, \, n\text{ odd}}^{\infty} \frac{6}{n\pi} \sin(nx) \).

Step 1: Define the Fourier Series

The Fourier series of a function \( f(x) \) on the interval \( [-L, L] \) is given by: \[ f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos \frac{n\pi x}{L} + b_n \sin \frac{n\pi x}{L} \right) \] where, \[ a_0 = \frac{1}{2L} \int_{-L}^{L} f(x) \, dx \] \[ a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos \frac{n\pi x}{L} \, dx \] \[ b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin \frac{n\pi x}{L} \, dx \] For the given problem, the interval is \( [-\pi, \pi] \), so \( L = \pi \).

Step 2: Calculate \(a_0\)

Calculate \( a_0 \) using the formula:\[ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx \]Substitute \( f(x) \) into the integral:\[ a_0 = \frac{1}{2\pi} \left( \int_{-\pi}^{0} (-1) \, dx + \int_{0}^{\pi} 2 \, dx \right) \]This simplifies to:\[ a_0 = \frac{1}{2\pi} \left( [-x]_{-\pi}^{0} + [2x]_{0}^{\pi} \right) \]\[ a_0 = \frac{1}{2\pi} ((0 - \pi) + (2\pi - 0)) = \frac{1}{2\pi} \pi = \frac{1}{2} \]

Step 3: Calculate \(a_n\)

Calculate \( a_n \) using the formula: \[ a_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} (-1) \cos \frac{n\pi x}{\pi} \, dx + \int_{0}^{\pi} 2 \cos \frac{n\pi x}{\pi} \, dx \right) \]For the first integral:\[ \int_{-\pi}^{0} (-1) \cos(n x) \, dx = -\left[ \frac{\sin(nx)}{n} \right]_{-\pi}^{0} = 0 \]For the second integral:\[ \int_{0}^{\pi} 2 \cos(nx) \, dx = 2\left[ \frac{\sin(nx)}{n} \right]_{0}^{\pi} = 0 \]Thus, \( a_n = 0 \) for all \( n \geq 1 \).

Step 4: Calculate \(b_n\)

Calculate \( b_n \) using the formula: \[ b_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} (-1) \sin \frac{n\pi x}{\pi} \, dx + \int_{0}^{\pi} 2 \sin \frac{n\pi x}{\pi} \, dx \right) \]For the first integral:\[ \int_{-\pi}^{0} (-1) \sin(nx) \, dx = \left[ \frac{\cos(nx)}{n} \right]_{-\pi}^{0} = \frac{1 - (-1)^n}{n} \]For the second integral:\[ \int_{0}^{\pi} 2 \sin(nx) \, dx = 2\left[ -\frac{\cos(nx)}{n} \right]_{0}^{\pi} = \frac{2((-1)^{n+1} - 1)}{n} \]Thus,\[ b_n = \frac{1}{\pi} \left( \frac{1 - (-1)^n}{n} + \frac{2((-1)^{n+1} - 1)}{n} \right) = \begin{cases} 0, & \text{if } n \text{ is even} \ \frac{6}{n\pi}, & \text{if } n \text{ is odd} \end{cases} \]

Step 5: Write the Fourier Series

Using the calculated values, the Fourier series becomes:\[ f(x) = \frac{1}{2} + \sum_{n=1, \, n\, \text{odd}}^{\infty} \frac{6}{n\pi} \sin(nx) \]This includes only sine terms for odd \( n \) because \( a_n = 0 \) and even \( n \) leads to \( b_n = 0 \).

Key Concepts

Trigonometric Series

The Fourier series is an essential tool in mathematics that expresses a function as the sum of sine and cosine terms, known as a trigonometric series. These series transform complex functions into simpler, periodic forms, allowing for easier analysis and problem-solving.

• The general form of a Fourier series for a function \( f(x) \), defined on an interval \([-L, L]\), is:

\[ f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos \frac{n\pi x}{L} + b_n \sin \frac{n\pi x}{L} \right) \]Here:

• \(a_0\) is the average or mean value of the function over one period, calculated using constant integration.
• \(a_n\) pertains to the cosine terms, emphasizing how much cosine contributes to the function.
• \(b_n\) is associated with the sine terms, illustrating the function's sine component.
• Each term holds specific coefficients, \(a_0, a_n,\) and \(b_n\), calculated through integral formulas.

Providing a series that matches the original function improves the understanding of its properties, especially periodicity and symmetry.

Orthogonality of Sine and Cosine

In the Fourier series, orthogonality refers to how sine and cosine functions work together matter-of-factly. This property significantly simplifies the calculations of the coefficients \(a_n\) and \(b_n\).

• Sine and cosine functions are orthogonal over an interval \([-L, L]\), meaning their integral products over this interval yield zero, whenever they aren't the same frequency.

This orthogonality is represented mathematically as:

• \( \int_{-L}^{L} \cos(nx) \cos(mx) \, dx = 0, \) for \( n eq m \)
• \( \int_{-L}^{L} \sin(nx) \sin(mx) \, dx = 0, \) for \( n eq m \)
• \( \int_{-L}^{L} \sin(nx) \cos(mx) \, dx = 0 \)

This concept helps in isolating the individual contributions from each harmonic frequency, allowing us to determine the series' coefficients easily and precisely. Bundled with the periodicity of trigonometric functions, orthogonality facilitates the breakdown of complex functions into simple, understandable parts.

Piecewise Functions

Piecewise functions are defined by different expressions in distinct intervals, a common format for real-world applications. The function \(f(x)\) given in the original exercise is an example of this. It takes on different values based on the interval of \(x\):

• \(f(x) = -1\) for \(-\pi < x < 0\)
• \(f(x) = 2\) for \(0 \leq x < \pi\)

When dealing with piecewise functions in the context of Fourier series, continuity might not be present at boundary points where one piece transitions to another.

• The Fourier series of a piecewise function potentially converges differently at these discontinuities, often tending towards the average of the limits from either side due to Gibbs phenomenon.

Calculating the Fourier series involves evaluating the integrals for each piece separately, then combining the results. This approach highlights how piecewise functions can be reconstructed as infinite sine and cosine series, promoting deeper insights into their behaviors across different domains.