Problem 4 - Advanced Engineering Mathematics

Question

$$\begin{aligned} a_{0} &=\int_{-1}^{1} f(x) d x=\int_{0}^{1} x d x=\frac{1}{2} \\ a_{n} &=\int_{-1}^{1} f(x) \cos n \pi x d x=\int_{0}^{1} x \cos n \pi x d x=\frac{1}{n^{2} \pi^{2}}\left[(-1)^{n}-1\right] \\ b_{n} &=\int_{-1}^{1} f(x) \sin n \pi x d x=\int_{0}^{1} x \sin n \pi x d x=\frac{(-1)^{n+1}}{n \pi} \end{aligned}$$ $$f(x)=\frac{1}{4}+\sum_{n=1}^{\infty}\left[\frac{(-1)^{n}-1}{n^{2} \pi^{2}} \cos n \pi x+\frac{(-1)^{n+1}}{n \pi} \sin n \pi x\right]$$

Step-by-Step Solution

Answer

The Fourier series is given by \( f(x) = \frac{1}{4} + \sum_{n=1}^{\infty} \left( \frac{(-1)^{n}-1}{n^{2} \pi^{2}} \cos(n\pi x) + \frac{(-1)^{n+1}}{n \pi} \sin(n\pi x) \right) \).

Step 1: Calculating a₀

The value of the coefficient \( a_0 \) is given as \( a_0 = \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} \). This represents the average value of the function over the interval from 0 to 1.

Step 2: Understanding aₙ expression

The expression for \( a_n \) is given by \( a_{n} = \frac{1}{n^{2} \, \pi^{2}} \left[ (-1)^{n} - 1 \right] \), which suggests that for even \( n \), \( a_n = 0 \), and for odd \( n \), \( a_n = \frac{-2}{n^{2} \, \pi^{2}} \). This comes from integrating \( x \) multiplied by \( \cos(n\pi x) \) and using integration by parts.

Step 3: Calculating bₙ

The coefficient \( b_n \) is calculated as \( b_{n} = \int_{0}^{1} x \sin(n\pi x) \, dx = \frac{(-1)^{n+1}}{n \pi} \). This follows integration by parts, reducing to this neat expression dependent on \( n \).

Step 4: Formulating the Fourier Series for f(x)

The function \( f(x) \) is expressed as a Fourier series: \[ f(x) = \frac{1}{4} + \sum_{n=1}^{\infty} \left( \frac{(-1)^{n}-1}{n^{2} \, \pi^{2}} \cos(n\pi x) + \frac{(-1)^{n+1}}{n \pi} \sin(n\pi x) \right) \]. This incorporates the calculated coefficients \( a_0, a_n, \) and \( b_n \).

Step 5: Analyzing Convergence Behavior

Determine that the Fourier series can be rearranged by considering even and odd \( n \). For the cosine series (\( a_n \)), only odd \( n \) contribute non-zero terms, and for the sine series (\( b_n \)), each term alternates in sign as determined by \((-1)^{n+1}\).

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used to evaluate integrals. This method is especially useful when dealing with products of functions, like the expressions found in Fourier series. Consider the integral \( \int x \cos(n\pi x) \, dx \). To solve this, we apply integration by parts which is based on the formula:\[ \int u \, dv = uv - \int v \, du \]We strategically choose one part of the integrand to be \( u \), typically a polynomial because its derivative simplifies things, and the other to be \( dv \), which will be the remaining part. In our case, we choose:

• \( u = x \), so \( du = dx \)
• \( dv = \cos(n\pi x) \, dx \), then \( v = \frac{1}{n\pi} \sin(n\pi x) \)

This substitution is very helpful because it transforms the integral into a form that is easier to evaluate. By performing these steps correctly, you obtain expressions such as \( a_n \) and \( b_n \) in the Fourier series.

Convergence Behavior

Understanding the convergence behavior of a Fourier series is crucial to predict how the series approaches the function it is meant to represent. A Fourier series converges to the function it approximates if certain conditions are met. The series: \[ f(x) = \frac{1}{4} + \sum_{n=1}^{\infty} \left( \frac{(-1)^{n}-1}{n^{2} \pi^{2}} \cos(n\pi x) + \frac{(-1)^{n+1}}{n \pi} \sin(n\pi x) \right) \]has terms that depend on whether \( n \) is odd or even.

• For \( a_n \), all even \( n \) terms are zero, so the convergence is driven by odd \( n \).
• For \( b_n \), each term alternates in sign as given by \((-1)^{n+1}\), which aids in converging to a smoother path.

The series exhibits a behavior influenced by these alternations and vanishings, ensuring that as more terms are added, it effectively mimics the original function more closely.

Fourier Coefficients

Fourier coefficients form the backbone of a Fourier series, determining how each term contributes to the overall function approximation. These coefficients, \( a_0 \), \( a_n \), and \( b_n \), are derived from integrating a function against sine and cosine components. The calculations:\[ a_0 = \int_{0}^{1} x \, dx = \frac{1}{2} \]provides the series' average value over one period. For higher terms,\[ a_n = \frac{1}{n^{2} \pi^{2}} \left[ (-1)^{n} - 1 \right] \] and\[ b_n = \frac{(-1)^{n+1}}{n \pi} \]were derived using integration by parts.

• For \( a_n \), the function and cosine are integrated, leading to zero when \( n \) is even, maintaining symmetry.
• For \( b_n \), the sine terms alternate based again on the simple integration method.

These coefficients ensure that as you synthesize these components, the more terms you include, the more accurately the series approximates the desired function. Their magnitude controls the amplitude of each frequency component within the signal.