Problem 10 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

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

Step 1: Determine the Constant Term

The given function starts by calculating the constant term, \( a_0 \), which is the average value of the function \( f(x) \) over the interval \(-\pi/2\) to \(\pi/2\). It is computed as follows: \[ a_0 = \frac{2}{\pi} \int_{0}^{\pi/2} \cos x \, dx. \] Evaluating this integral gives \( a_0 = \frac{2}{\pi} \).

Step 2: Compute the Cosine Coefficients

The next step involves the calculation of cosine coefficients \( a_n \). These coefficients are given by \( a_n = \frac{2}{\pi} \int_{0}^{\pi/2} \cos x \cos 2nx \, dx \). The identity \( \cos A \cos B = \frac{1}{2}(\cos(A-B) + \cos(A+B)) \) is used to split the integral. The integral is then solved, yielding: \[ a_n = \frac{2(-1)^{n+1}}{\pi(4n^2 - 1)}. \]

Step 3: Determine the Sine Coefficients

The sine coefficients \( b_n \) are calculated similarly using the expression \( b_n = \frac{2}{\pi} \int_{0}^{\pi/2} \cos x \sin 2nx \, dx \). Using the identity \( \sin A \cos B = \frac{1}{2}(\sin(A+B) + \sin(A-B)) \), the integral is solved to find: \[ b_n = \frac{4n}{\pi(4n^2 - 1)}. \]

Step 4: Compile the Fourier Series

Finally, the Fourier series for \( f(x) \) is constructed using the constant term \( a_0 \), and the sums involving \( a_n \) and \( b_n \). Thus, the complete Fourier series is: \[ f(x) = \frac{1}{\pi} + \sum_{n=1}^{\infty} \left[ \frac{2(-1)^{n+1}}{\pi(4n^2 - 1)} \cos 2nx + \frac{4n}{\pi(4n^2 - 1)} \sin 2nx \right]. \]

Key Concepts

Cosine Coefficients

Cosine coefficients are crucial in expressing a function as a Fourier series by capturing its periodic nature. For a given function like \( f(x) \), these coefficients are represented by \( a_n \). They are calculated using the formula \( a_n = \frac{2}{\pi} \int_0^{\pi/2} \cos x \cos 2nx \, dx \). The use of trigonometric identities simplifies the integration process. In this case, the identity \( \cos A \cos B = \frac{1}{2}(\cos(A-B) + \cos(A+B)) \) helps in handling the product of two cosine terms.

By applying these trigonometric identities, the integral is simplified and solved, resulting in the cosine coefficients expressed as \( a_n = \frac{2(-1)^{n+1}}{\pi(4n^2 - 1)} \). This shows how each harmonic frequency's contribution diminishes as \( n \) increases, reflecting the frequency's complexity in the original function.

Sine Coefficients

The sine coefficients, \( b_n \), in a Fourier series help to capture the odd and oscillatory portions of the function \( f(x) \). Each coefficient is determined by the integral \( b_n = \frac{2}{\pi} \int_0^{\pi/2} \cos x \sin 2nx \, dx \).

The integration leverages the identity \( \sin A \cos B = \frac{1}{2}(\sin(A+B) + \sin(A-B)) \). This identity is particularly useful because it simplifies the product of sine and cosine functions, breaking them into simpler terms that are easier to integrate. As a result, the sine coefficients are computed as \( b_n = \frac{4n}{\pi(4n^2 - 1)} \).

Just like the cosine coefficients, the sine coefficients reveal more about the function's frequency components. The pattern \( 4n/(4n^2 - 1) \) showcases how these coefficients decrease in magnitude as \( n \) increases, signifying less influence from higher-order sine terms.

Integration Techniques

Integration techniques are foundational for calculating Fourier coefficients. To handle integrations involving products of trigonometric functions, identities like \( \cos A \cos B = \frac{1}{2}(\cos(A-B) + \cos(A+B)) \) and \( \sin A \cos B = \frac{1}{2}(\sin(A+B) + \sin(A-B)) \) are pivotal.

These trigonometric identities transform complex integrands into more manageable expressions. For instance, the product of two cosines \( \cos x \cos 2nx \) or a cosine and sine \( \cos x \sin 2nx \) can be rewritten as the sum of two simpler trigonometric functions, each of which is easier to integrate over a specific interval.

By employing these identities, integration becomes straightforward and systematic, allowing the computation of crucial terms in the Fourier series. Mastering these techniques enables one to efficiently decompose any periodic function into its constituent cosine and sine components, providing a clear mathematical representation of its harmonic content.