Problem 9 - Advanced Engineering Mathematics

Question

$$\begin{aligned} &a_{0}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) d x=\frac{1}{\pi} \int_{0}^{\pi} \sin x d x=\frac{2}{\pi}\\\ &a_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x d x=\frac{1}{\pi} \int_{0}^{\pi} \sin x \cos n x d x=\frac{1}{2 \pi} \int_{0}^{\pi}(\sin (1+n) x+\sin (1-n) x) d x \end{aligned}$$ $$\begin{aligned} &=\frac{1+(-1)^{n}}{\pi\left(1-n^{2}\right)} \quad \text { for } n=2,3,4, \ldots \\ a_{1} &=\frac{1}{2 \pi} \int_{0}^{\pi} \sin 2 x d x=0 \\ b_{n} &=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x d x=\frac{1}{\pi} \int_{0}^{\pi} \sin x \sin n x d x \\ &=\frac{1}{2 \pi} \int_{0}^{\pi}(\cos (1-n) x-\cos (1+n) x) d x=0 \quad \text { for } n=2,3,4, \ldots \\ b_{1} &=\frac{1}{2 \pi} \int_{0}^{\pi}(1-\cos 2 x) d x=\frac{1}{2} \\ f(x) &=\frac{1}{\pi}+\frac{1}{2} \sin x+\sum_{n=2}^{\infty} \frac{1+(-1)^{n}}{\pi\left(1-n^{2}\right)} \cos n x \end{aligned}$$

Step-by-Step Solution

Answer

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

Step 1: Calculate \( a_0 \)

To calculate the constant term \( a_0 \) in the Fourier series, use the integral formula \( a_{0} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx \). Given \( f(x) = \sin x \) for \( x \in [0, \pi] \) and \( f(x) = 0 \) elsewhere, we have: \[ a_{0} = \frac{1}{\pi} \int_{0}^{\pi} \sin x \, dx = \frac{2}{\pi} \] This result is reached by recognizing the symmetry and properties of sine.

Step 2: Calculate \( a_n \) for \( n \geq 2 \)

To find the coefficients \( a_n \) for \( n \geq 2 \), use the formula \( a_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x \, dx \). This integral simplifies to: \[ a_{n} = \frac{1}{2\pi} \int_{0}^{\pi} (\sin(1+n)x + \sin(1-n)x) \, dx \]Evaluating this, we find: \[ a_{n} = \frac{1+(-1)^{n}}{\pi(1-n^{2})} \] This exploits trigonometric identities for the product \( \cos(nx)\sin(x) \).

Step 3: Evaluate \( a_1 \)

For \( a_1 \), which requires separate evaluation similar to higher \( n \) values: \[ a_1 = \frac{1}{2\pi} \int_{0}^{\pi} \sin(2x) \, dx = 0 \] This is a result of directly solving that specific integral, as \( \sin(2x) \) has symmetric properties over \([0,\pi]\).

Step 4: Calculate \( b_n \) for \( n \geq 2 \)

Calculate \( b_n \) using the formula \( b_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x \, dx \). Simplifying gives: \[ b_n = \frac{1}{2\pi} \int_{0}^{\pi} (\cos(1-n)x - \cos(1+n)x) \, dx = 0 \] For \( n \geq 2 \), this term evaluates to zero due to cosine differences that integrate over zero symmetrically.

Step 5: Evaluate \( b_1 \)

To calculate \( b_1 \), which has a specialized calculation: \[ b_1 = \frac{1}{2\pi} \int_{0}^{\pi} (1 - \cos 2x) \, dx = \frac{1}{2} \] This conclusion comes from evaluating the integral over its limits, considering that \( \int \cos 2x \) resolves back in symmetric periods.

Step 6: Compose the Fourier Series

The resulting Fourier series for \( f(x) \) based on the coefficients is: \[ f(x) = \frac{1}{\pi} + \frac{1}{2} \sin x + \sum_{n=2}^{\infty} \frac{1+(-1)^{n}}{\pi(1-n^{2})} \cos n x \] This represents the periodical repetition of \( f(x) \) over its domain.

Key Concepts

Trigonometric Series

A trigonometric series is a way of expressing a function as an infinite sum of sine and cosine terms. This series forms the basis for Fourier series, where a periodic function can be represented in terms of trigonometric functions. The sine and cosine functions are chosen because they are periodic and have well-understood mathematical properties.

In the case of a Fourier series, each term in the series is defined by coefficients that are determined by integrating the product of the function and a sine or cosine function over a specific period. By adding these terms together, we can approximate periodic functions to any desired level of accuracy.

Learning about trigonometric series provides insight into how complex waveforms can be broken down into simpler sine and cosine components, which is pivotal in fields like signal processing and acoustics.

Orthogonal Functions

When we talk about orthogonal functions in the context of Fourier series, we're referring to functions that, when multiplied together and integrated over a specific interval, give a result of zero. This concept is similar to perpendicular vectors in geometry.

Using orthogonal functions in Fourier series is beneficial because it allows us to uniquely determine the coefficients of each sine and cosine term without interference from the others. The sine and cosine functions are orthogonal over a period of \([-\pi, \pi]\), which is why they are ideal for representing periodic functions in a Fourier series.

Understanding orthogonal functions helps in recognizing why Fourier series work so effectively in separating and analyzing different frequency components of a signal.

Coefficients Calculation

Finding the coefficients in a Fourier series is a critical step in expressing a function as a trigonometric series. These coefficients determine the amplitude of each sine and cosine term that makes up the series.

The calculation process includes integrating the original function multiplied by sine or cosine functions over a specified interval. This ensures that each coefficient accounts for the contribution of the corresponding frequency component to the overall function.

The formulas for these coefficients are given by:

• For the constant term: \(a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx\)
• For cosine terms: \(a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx\)
• For sine terms: \(b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx\)

By accurately determining these coefficients, we can build an effective Fourier series representation of the original function. Each calculated coefficient allows us to analyze and reconstruct the function with precision.