Question

Find the required Fourier series for the given function and sketch the graph of the function to which the series converges over three periods. $$ f(x)=x, \quad 0 \leq x<1 ; \quad \text { series of period } 1 $$

Short answer

Question: Determine the Fourier series of the given function \(f(x) = x\) over the interval \(0 \leq x < 1\) with a period of 1, and describe the graph of the function to which the Fourier series converges. Answer: The Fourier series for the function \(f(x) = x\) over the given interval and with a period of 1 is: $$ f(x) = \frac{1}{4} + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\pi n} \sin(2\pi nx) $$ The graph of the function to which the Fourier series converges will look like the original function \(f(x) = x\) within the interval \(0\leq x <1\) and will repeat itself periodically. The graph will appear discontinuous at the transition between periods due to the periodicity requirement of Fourier series representation.

Step-by-step solution

Fourier Series Definition

The general formula for the Fourier series of a periodic function of period \(L\) is given as: $$ f(x) \thickspace = \thickspace \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(2n\pi\frac{x}{L}) + b_n \sin(2n\pi\frac{x}{L})) $$ where $$ a_0 = \frac{1}{L} \int_{-L/2}^{L/2} f(x) dx $$ $$ a_n = \frac{2}{L} \int_{-L/2}^{L/2} f(x) \cos(2n\pi\frac{x}{L}) dx $$ $$ b_n = \frac{2}{L} \int_{-L/2}^{L/2} f(x) \sin(2n\pi\frac{x}{L}) dx $$ In our case, \(f(x) = x\), and the period \(L = 1\). Therefore, we need to find \(a_0\), \(a_n\), and \(b_n\) by evaluating the corresponding integrals.

Compute the Coefficients

Since our function f(x) is a non-even function, we must compute all coefficients \(a_0\), \(a_n\), and \(b_n\). To compute \(a_0\), we evaluate: $$ a_0 = \int_{0}^{1} x dx = \frac{1}{2} x^2\Big|{_0^1} = \frac{1}{2} $$ This means, $$ \frac{a_0}{2} = \frac{1}{4} $$ Similarly, we'll compute the coefficients \(a_n\) and \(b_n\) using the formulas mentioned earlier. To compute \(a_n\), we evaluate: $$ a_n = 2\int_{0}^{1}x\cos(2\pi nx)dx $$ To compute \(b_n\), we evaluate: $$ b_n = 2\int_{0}^{1}x\sin(2\pi nx)dx $$ Using integration by parts, we obtain \(a_n = 0\) and $$ b_n = \frac{(-1)^{n+1}}{\pi n} $$

Write the Fourier Series

Now, we write the Fourier series using the coefficients \(a_0\), \(a_n\), and \(b_n\): $$ f(x) = \frac{1}{4} + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\pi n} \sin(2\pi nx) $$

Sketch the Graph

To sketch the graph of the function to which the Fourier series converges, we can use the partial sums of the Fourier series and visualize how the function tends to look like as we increase the number of terms included in the sum. For practical purposes, we'll use the first 5 terms. Plotting the summation of the first 5 terms of the Fourier series (i.e., \(x = \frac{1}{4} + \sum_{n=1}^{5} \frac{(-1)^{n+1}}{\pi n} \sin(2\pi nx)\)) over three periods, we will see that it converges to the given function \(f(x) = x\) within the interval \(0\leq x <1\). Since the Fourier series converges to the given function, the graph will look like the original function for a single period (straight inclined line over the interval \(0\leq x <1\)) and will repeat itself periodically. Remember that the graph will appear discontinuous at the transition between periods due to the periodicity requirement of Fourier series representation.

Key concepts

Understanding Periodic Functions

Periodic functions are essentially the heartbeat of the Fourier series. These functions repeat their values at regular intervals, which we refer to as the period. Imagine the hands of a clock, rotating around again and again, resetting after each revolution—that's periodicity.

For a function to be classified as periodic, it must satisfy the condition: \( f(x+P) = f(x) \), where \( P \) is the period. In the Fourier series context, the period \( L \) plays a crucial role because it determines the argument of the sine and cosine functions that compose the series. Fourier series capitalize on this repetitive nature to break down complex, periodic wave shapes into simpler sine and cosine waves, summing them up to recreate the original waveform.

When given a task to find a Fourier series of a function like \( f(x)=x \) over a certain interval, you are essentially deciphering the language of this function into a series of repeating waves. The exercise provided asks this of us over a specific period of 1, making it critical to grasp the periodic nature of the function in order to understand and compute its Fourier series.

Integration by Parts - The Tool for Fourier Coefficients

The transformation from function to Fourier series often involves the artful calculation technique called 'integration by parts'. This is a fundamental tool when finding the coefficients \( a_n \) and \( b_n \), which are the building blocks of the Fourier series.

The formula for integration by parts is derived from the product rule of differentiation and can be expressed as: \( \int u dv = uv - \int v du \). When applied to calculate Fourier coefficients, we often face integrals of products of functions—that's where this trick shines. By judiciously choosing \( u \) and \( dv \), we simplify the problem into more manageable parts.

In our exercise example, the integration by parts method was key to finding the coefficients \( b_n \), which, due to the sine function and its properties, led to a pattern: \( b_n = \frac{(-1)^{n+1}}{\pi n} \). Without integration by parts, we wouldn't be able to uncover this relationship and progress in synthesizing the Fourier series.

Convergence of Series - The Path to Function Reconstruction

Now, let's dive into the convergence of series, a concept pivotal in ensuring the sum of infinite terms approaches a finite value—the function we're trying to depict. A series converging means that as we sum more and more terms, the total gets closer to a specific value, rather than diverging to infinity.

Understanding the convergence is vital, particularly in the context of Fourier series, as it reassures us that by summing an infinite number of sine and cosine functions, we can accurately reconstruct periodic functions. In your exercise, spotting the convergence of the Fourier series to the function \( f(x) = x \) for \( 0 \<= x <1 \) is both fascinating and necessary. We depend on properties like the Dirichlet conditions to guarantee convergence.

The fascinating aspect of Fourier series is observing how adding more terms in the series gets us closer to the original function, making it a powerful tool in both theoretical and applied mathematics. In the provided exercise, we can graph the partial sums to see this visually, as the series begins to mimic the linear segment of \( f(x) = x \) over its domain, forming a step towards encapsulating the function's essence in its entirety.