Question

Let \(f(x)=x\) be a periodic function defined over the interval \((0,2 a)\). Find the Fourier series expansion of \(f\).

Short answer

The Fourier series for the given function \(f(x) = x\) over the interval \((0,2a)\) is \(f(x) = a + \sum_{n=1}^∞ \frac{2a}{nπ}(1-(-1)^n) sin(\frac{nπx}{a})\).

Step-by-step solution

Step 1-Experience of configuration of Fourier series

Configuring Fourier series, the following integral expressions give the coefficients: \(a_0 = \frac{1}{T}\int_{0}^{T} f(t) dt\), \(a_n = \frac{2}{T}\int_{0}^{T} f(t) cos(\frac{2πnx}{T}) dt\), \(b_n = \frac{2}{T}\int_{0}^{T} f(t)sin(\frac{2πnx}{T}) dt\). The period \(T\) in this case is given as \(2a\).

Step 2-Calculate \(a_0\) coefficient

First calculate the \(a_0\) term: \(a_0 = \frac{1}{2a}\int_{0}^{2a} x dx = \frac{1}{2a} [\frac{x^2}{2}]_{0}^{2a} = a\). The \(a_0\) term is the average value of the function over one period.

Step 3-Calculate \(a_n\) coefficient

Now for \(a_n\), we have \(a_n = \frac{2}{2a}\int_{0}^{2a} x cos(\frac{nπx}{a}) dx\). Integrating by parts gives \(a_n = 0\) \(∀n\). Since the function is odd, \(a_n\) terms will have to be zero. Which makes sense because cosine is an even function, the integral over an entire period is zero.

Step 4-Calculate \(b_n\) coefficient

Finally, the \(b_n\) term is calculated as follows: \(b_n = \frac{2}{2a}\int_{0}^{2a} x sin(\frac{nπx}{a}) dx\). Which, upon integrating by parts, gives \(b_n=\frac{2a}{nπ}(1-(-1)^n)\).

Step 5-Combine all the Fourier coefficients

The Fourier series expansion can now be constructed using the coefficients obtained. The expansion is: \(f(x) = a + \sum_{n=1}^∞ \frac{2a}{nπ}[1-(-1)^n] sin(\frac{nπx}{a})\).

Key concepts

Periodic Functions

Periodic functions are the heartbeat of various physical phenomena, ranging from the vibrations of a guitar string to the fluctuations of electronic signals. These functions repeat their values at regular intervals, known as periods. In the mathematical sense, a function f(x) is considered periodic if there exists a positive number p such that f(x + p) = f(x) for all values of x in the function's domain.

The example of f(x) = x being defined over the interval (0, 2a) suggests that the function repeats itself every 2a units, making it periodic with a period of 2a. Understanding this concept is paramount when diving into the world of Fourier series because these series are designed to represent periodic functions through a combination of simple trigonometric functions.

Fourier Coefficients

Fourier coefficients are the DNA of a Fourier series. They are the crucial numbers that, when plugged into sines and cosines, reconstruct our original periodic function. These coefficients are not arbitrary; they are systematically calculated using specific integration formulas.

For instance, to find the constant term a₀, which represents the average value of the function over one complete cycle, we use the formula \[a_0 = \dfrac{1}{T}\int_{0}^{T} f(t) dt\] where T is the period of the function. Similarly, to obtain the coefficients a_n and b_n for the cosine and sine terms respectively, we integrate the product of the function with the corresponding trigonometric function over one complete cycle:

• \[a_n = \dfrac{2}{T}\int_{0}^{T} f(t) \cos(\dfrac{2\pi nx}{T}) dt\]
• \[b_n = \dfrac{2}{T}\int_{0}^{T} f(t)\sin(\dfrac{2\pi nx}{T}) dt\]

It's worth noting that the even or odd nature of the original function greatly influences the Fourier coefficients, largely simplifying the series when the function has symmetry properties.

Integration by Parts

Integration by parts is a technique based on the product rule for differentiation and is integral (pun intended) for finding Fourier coefficients that involve products of functions. According to this method, if you identify two functions u(x) and v(x), then the integral of their product can be calculated as \[\int u(x)dv(x) = u(x)v(x) - \int v(x)du(x)\]

In the context of calculating Fourier coefficients, like a_n and b_n, this method often simplifies the process when the product of the function and a trigonometric function is considered. For the given periodic function, integration by parts was used to calculate the b_n terms, transforming a potentially complex integration task into a series of simpler steps and resulting in an expression for the coefficients that is easier to work with.