Problem 5 - Advanced Engineering Mathematics

Question

Find the complex Fourier series of \(f\) on the given interval. $$ f(x)=x, 0

Step-by-Step Solution

Answer

The complex Fourier series of \(f(x)=x\) is \(f(x) = \pi + \sum_{n \neq 0} \frac{(-1)^n}{in} e^{inx}\).

Step 1: Define the Fourier Series Formula

The complex Fourier series of a function \( f(x) \) over a period \( [0,2\pi] \) is given by this general formula: \[ f(x) = \sum_{n=-\infty}^{\infty} c_n e^{inx} \] where \( c_n \) are the complex Fourier coefficients, and \( i \) is the imaginary unit.

Step 2: Determine the Complex Coefficients

The complex Fourier coefficients \( c_n \) are obtained using the formula: \[ c_n = \frac{1}{2\pi} \int_{0}^{2\pi} f(x)e^{-inx} \, dx \] For \( f(x) = x \), substitute into the formula to get: \[ c_n = \frac{1}{2\pi} \int_{0}^{2\pi} x e^{-inx} \, dx \]

Step 3: Evaluate the Integral for \( c_n \)

To find \( c_n \), evaluate the integral: \[ c_n = \frac{1}{2\pi} \int_{0}^{2\pi} x e^{-inx} \, dx \] Using integration by parts, where \( u = x \) and \( dv = e^{-inx} \, dx \), thus \( du = dx \) and \( v = \frac{e^{-inx}}{-in} \), we get: \[ c_n = \frac{1}{2\pi} \left[ x \cdot \frac{e^{-inx}}{-in} \bigg|_0^{2\pi} - \int_0^{2\pi} \frac{e^{-inx}}{-in} \, dx \right] \]

Step 4: Simplify the Integral Result

Evaluate the boundary terms and simplify: \[ x \cdot \frac{e^{-inx}}{-in} \bigg|_0^{2\pi} = \left(2\pi (-1)^n - 0\right) \cdot \frac{1}{-in} = \frac{2\pi (-1)^n}{-in} \] For the second integral: \[ \int_0^{2\pi} \frac{e^{-inx}}{-in} \, dx = \left[ \frac{e^{-inx}}{n^2} \right]_0^{2\pi} = 0 \] Thus, \[ c_n = \frac{1}{2\pi} \left( \frac{2\pi (-1)^n}{-in} \right) = \frac{(-1)^n}{in} \] This holds for \( n eq 0 \).

Step 5: Special Case for \( n = 0 \)

For \( n = 0 \), \[ c_0 = \frac{1}{2\pi} \int_{0}^{2\pi} x \, dx = \frac{1}{2\pi} \cdot \frac{x^2}{2} \bigg|_0^{2\pi} = \frac{2\pi^2}{2\pi} = \pi \]

Step 6: Write the Complex Fourier Series

Substituting the coefficients into the series formula gives: \[ f(x) = \pi + \sum_{n eq 0} \frac{(-1)^n}{in} e^{inx} \] Thus, this is the complex Fourier series of \( f(x) = x \) on the interval \( (0, 2\pi) \).

Key Concepts

Fourier Coefficients

In the study of complex Fourier series, Fourier coefficients are essential components. They represent the weights of the exponential functions in the series. Fourier coefficients

• are complex numbers,
• help to recreate the original function using a combination of sinusoidal functions, and
• determine the amplitude and phase of each frequency component.

The formula to calculate these coefficients for a function \(f(x)\) over an interval \([0, 2\pi]\) is:\[c_n = \frac{1}{2\pi} \int_{0}^{2\pi} f(x)e^{-inx} \, dx\]where \( n \) is an integer, and \( i \) is the imaginary unit. For each \( n \), the integration calculates how much of the frequency \( n \) is present in the function. This is how we gather information on the periodic patterns within the function. Understanding these coefficients lets us break a complex function into simpler, understandable elements, facilitating insights into the function's behavior.

Integration by Parts

Integration by parts is a valuable technique in calculus, used particularly to solve integrals of products of functions. Its formula is derived from the product rule of differentiation. The formula is:\[ \int u \, dv = uv - \int v \, du \]Where:

• \( u \) is a function of \( x \),
• \( dv \) is another function of \( x \) multiplied by \( dx \),
• \( du \) is the derivative of \( u \), and
• \( v \) is the integral of \( dv \).

In our problem, when finding the Fourier coefficients, we let \( u = x \) and \( dv = e^{-inx} \, dx \). This choice makes it possible to integrate the function by simplifying one part of the product, \( dv \), while differentiating the other, \( u \). The result reduces the original integral into more manageable components, leading to the solution of the Fourier series integral.

Complex Functions

Complex functions are functions that involve complex numbers. These numbers have a real part and an imaginary part, often written as \( a + bi \), where \( i \) is the imaginary unit with \( i^2 = -1 \). In complex Fourier series, these functions take the form of exponential functions involving imaginary numbers, specifically \( e^{inx} \). Complex functions:

• include both sinusoidal and exponential components,
• are used in solutions of differential equations and signal processing, and
• allow the representation of oscillatory behavior via Euler's formula, \( e^{inx} = \cos(nx) + i\sin(nx) \).

By converting real functions into the complex plane, we simplify trigonometric operations and, importantly, work with them algebraically. This property is essential for signal analysis, as it enables the handling of different frequencies present in signals, offering a valuable tool in both mathematical analysis and engineering applications.