Question

Let $$ f(x)= \begin{cases}0, & -\pi \leq x<0 \\ e^{i x}, & 0 \leq x<\pi\end{cases} $$ Find the complex Fourier series of \(f\).

Short answer

The complex Fourier series of \( f(x) \) is \( \sum_{n=-\infty}^{\infty} \frac{ (-1)^{n-1} - 1}{2\pi i(n-1)} e^{inx} \).

Step-by-step solution

Define the Interval

The function \( f(x) \) is defined on the interval \( [-\pi, \pi) \), but since it is periodic, we will consider its behavior on \( [-\pi, \pi] \) and extend it periodically.

Determine the Fourier Coefficients

For a complex Fourier series, we use the formula for the coefficients \[ c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) e^{-inx} \, dx \]We need to compute \( c_n \) by dividing the integral into two parts, according to the function's definition: \[ c_n = \frac{1}{2\pi} \left( \int_{-\pi}^{0} 0 \cdot e^{-inx} \, dx + \int_{0}^{\pi} e^{ix} \cdot e^{-inx} \, dx \right) \]

Evaluate the Integral from \( -\pi \) to 0

Since \( f(x) = 0 \) in the interval \( [-\pi, 0) \), the first integral is zero:\[ \int_{-\pi}^{0} 0 \cdot e^{-inx} \, dx = 0\]

Evaluate the Integral from 0 to \( \pi \)

Now compute the integral for the second part:\[ \int_{0}^{\pi} e^{i(1-n)x} \, dx = \left. \frac{e^{i(1-n)x}}{i(1-n)} \right|_{0}^{\pi} \]Evaluating this, we get: \[ \left[ \frac{e^{i(\pi-n\pi)} - 1 }{i(1-n)} \right]\]

Simplify the Expression

Evaluate the expression: \( e^{i(n-1)\pi} = (-1)^{n-1} \). Therefore, \[ \frac{(-1)^{n-1} - 1}{i(n-1)} \]So the coefficient \( c_n \) becomes:\[ c_n = \frac{1}{2\pi} \cdot \frac{(-1)^{n-1} - 1}{i(n-1)} \]

Construct the Fourier Series

The series \( f(x) \) can be written as:\[ f(x) = \sum_{n=-\infty}^{\infty} c_n e^{inx} \]The series includes the terms:\[ f(x) = \sum_{n=-\infty}^{\infty} \frac{ (-1)^{n-1} - 1}{2\pi i(n-1)} e^{inx}\]

Key concepts

Fourier Coefficients

Understanding Fourier coefficients is crucial for constructing a Fourier series. In this problem, we calculate these coefficients to express a periodic function using complex exponentials. The coefficient, often denoted as \( c_n \), provides a measure of how much a specific frequency contributes to the overall function. Given the formula for the coefficients:

• \[ c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) e^{-inx} \, dx \]

This integral takes the function \( f(x) \), multiplies it by a complex exponential \( e^{-inx} \), and averages it over a period \( 2\pi \). This operation isolates the frequency component \( n \), allowing us to express the periodic function as a sum of these individual components.
To compute \( c_n \), two main parts arise from the nature of the given function: from \(-\pi\) to 0 and from 0 to \( \pi \). The first part vanishes due to the function's value (zero), while the second part involves a more intricate calculation of integrals.

Periodic Function

A periodic function is one that repeats its values in regular intervals or periods. In this context, our function \( f(x) \) is defined on the interval \( [-\pi, \pi) \), indicating a periodic behavior with period \( 2\pi \).
Periodic functions are pivotal in Fourier analysis as they allow for decomposition into simpler components. By breaking down a function into its periodic characteristics, we can analyze, convey, and reconstruct the signal more effectively using Fourier techniques. For this function, outside the interval \([-\pi, \pi)\), \( f(x) \) repeats itself periodically. Using this repetition property, we can extend our analysis and apply the Fourier series method comprehensively.
Understanding periodicity helps in segmenting the function for integral evaluation, an essential step for finding Fourier coefficients.

Integral Evaluation

Integral evaluation is a critical step in determining the Fourier coefficients. It involves calculating the integrals over specified intervals, which helps in extracting the coefficients needed for the series representation.
For this exercise, the integral evaluation is split into two parts. The first one involves the interval \([-\pi, 0)\), contributing zero to the final sum because the function equals zero there:

• \[ \int_{-\pi}^{0} 0 \cdot e^{-inx} \, dx = 0 \]

The more involved part is the integral from 0 to \( \pi \), where the function \( e^{ix} \) is active. Here, the evaluation is:

• \[ \int_{0}^{\pi} e^{i(1-n)x} \, dx = \left. \frac{e^{i(1-n)x}}{i(1-n)} \right|_{0}^{\pi} \]

Through careful calculation, this gives us an expression which, once simplified, contributes to the determination of \( c_n \). Simplification involves applying algebraic identities and properties of complex numbers to arrive at manageable expressions.

Complex Exponential Form

The complex exponential form is a powerful tool in Fourier series. It represents sinusoidal functions using complex exponentials, allowing for a more streamlined and efficient computation, especially when dealing with oscillatory phenomena. This method exploits Euler's formula:

• \( e^{ix} = \cos(x) + i\sin(x) \)

By using complex exponentials, you convert trigonometric-based series into a form that is easier to manipulate analytically.
In our exercise, the given function is already in a complex exponential form \( e^{ix} \). This instantly aligns with the Fourier series requirement, allowing direct application of integral and coefficient evaluation:

• \( e^{ix} \cdot e^{-inx} = e^{i(1-n)x} \)

This exponential product simplifies the process of integration by reducing it to straightforward exponential handling. Ultimately, understanding the complex exponential form allows for the Fourier series expansion to be handled efficiently, revealing both the amplitude and phase shift of varying sinusoidal components.