Question

Parts (a) and (b), you are given in each case one period of a function. Sketch several periods of the function and expand it in a sine-cosine Fourier series, and in a complex exponential Fourier series. (a) \(f(x)=x^{2},-\pi< x<\pi\) (b) \(f(x)=x^{2}, 0< x<2 \pi\)

Short answer

Fourier series expansion involves computing the coefficients \(A_0\), \(A_n\), and \(B_n\) for sine-cosine series, and \(c_n\) for exponential series. For both functions, calculate integrals over their respective intervals.

Step-by-step solution

Title - Understand the Problem and Given Functions

The task here is to sketch several periods of given functions and then expand them in both sine-cosine Fourier series and complex exponential Fourier series. The function given in part (a) is \( f(x) = x^{2}, -\pi < x < \pi \) and in part (b) is \( f(x) = x^{2}, 0 < x < 2\pi \).

Title - Extend and Sketch the Functions

For both parts, sketch several periods of the functions by extending the given period symmetrically in both directions. This visualizes how the function behaves over multiple periods.

Title - Find the Sine-Cosine Fourier Series for Part (a)

For \( f(x) = x^{2} \) over the interval \(-\pi < x < \pi\), the Fourier series is given by: \[ f(x) = A_0 + \sum_{n=1}^{\infty} (A_n \cos(nx) + B_n \sin(nx)) \] Where, \[ A_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x) \, dx = \frac{1}{2\pi}\int_{-\pi}^{\pi} x^{2} \, dx \]Calculate \(A_0\), \(A_n\), and \(B_n\).

Title - Calculate Coefficients for Part (a)

Compute \[ A_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi} x^2 \, dx = \frac{\pi^2}{3} \] Similarly, calculate \(A_n\) and \(B_n\): \[ A_n = \frac{1}{\pi}\int_{-\pi}^{\pi} x^{2}\cos(nx) \, dx \] \[ B_n = \frac{1}{\pi}\int_{-\pi}^{\pi} x^{2}\sin(nx) \, dx = 0 \]

Title - Find Complex Exponential Fourier Series for Part (a)

Express \(f(x) = x^{2}\) as a complex exponential series. Use the formula \[ f(x) = \sum_{n=-\infty}^{\infty} c_n e^{inx} \] Calculate \[ c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} x^2 e^{-inx}\, dx \]

Title - Find the Fourier Series for Part (b)

Repeat the process for \( f(x) = x^{2} \) over \(0 < x < 2\pi\): \[ A_0 = \frac{1}{2\pi} \int_{0}^{2\pi} x^{2} \, dx = \frac{2\pi^2}{3} \] Compute \[ A_n = \frac{1}{\pi}\int_{0}^{2\pi} x^{2}\cos(nx) \, dx \] \[ B_n = \frac{1}{\pi}\int_{0}^{2\pi} x^{2}\sin(nx) \, dx \] For the complex exponential series, integrate over \(0 < x < 2\pi\).

Key concepts

Fourier Transform

The Fourier transform is a mathematical technique that transforms a time-domain signal into its frequency-domain representation. It's a crucial tool in fields like signal processing, physics, and engineering. Unlike Fourier series, which applies to periodic functions, the Fourier transform can handle non-periodic functions as well. This makes it incredibly flexible for analyzing various kinds of signals. The general formula for the Fourier transform of a function, say \( f(t) \), is given by: \[ F(u) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(t) e^{-i 2\pi \u t}\, dt \] Here, \( u \) is the frequency variable. The inverse Fourier transform allows you to go back from the frequency domain to the time domain: \[ f(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F(u) e^{i 2\pi \u t} \, d\u \] The Fourier transform is more general than the Fourier series and is often used when dealing with non-repetitive signals.

Sine-Cosine Series

The sine-cosine series is a way to represent periodic functions using sine and cosine functions. This type of Fourier series only includes terms based on these trigonometric functions and is particularly useful for representing real-valued signals. Considering a function \(f(x)\) defined over \(-\pi < x < \pi\), its sine-cosine series is given by: \[ f(x) = A_0 + \sum_{n=1}^{\infty} (A_n \cos(nx) + B_n \sin(nx)) \] Where:
\ A_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx \
\ A_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx \
\ B_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx \
This series captures the essential frequency components of the function. Sine terms (\