Question

Find the Fourier series expansion of the periodic function defined on its fundamental cell, \((-\pi, \pi)\), as \(f(\theta)=\theta\).

Short answer

The Fourier series expansion of the function \(f(\theta)=\theta\) defined on the interval \((-\pi, \pi)\) is \[f(\theta) = \sum_{n=1}^{\infty} \frac{(-1)^n}{n} sin(n\theta)\].

Step-by-step solution

Identify the Functions Period

Firstly, identify the function's period. The problem states that the function \(f(\theta)=\theta\) is defined on the interval \((-\pi, \pi)\), which means its fundamental period is \(2\pi\). This means that outside this interval, the function's values repeat and the Fourier series will represent the function on this fundamental period.

Compute the Fourier Coefficients

The Fourier series of a periodic function is given by the formula: \[f(\theta) = a_0 + \sum_{n=1}^{\infty} (a_n cos(n\theta) + b_n sin(n\theta))\] where the Fourier coefficients \(a_n\) and \(b_n\) are given by: \[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(\theta) cos(n\theta) d\theta\] and \[b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(\theta) sin(n\theta) d\theta\]. For \(a_0\), \(n\) equals to 0. Integral calculations are needed for every coefficient and \(f(\theta)=\theta\) is substituted into the integrations. This step involves a lot of calculus, so you need to be familiar with integration techniques.

Compute a_0

First, compute the coefficient \(a_0\), which is the average value of function over the period: \[a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} \theta d\theta\]. The result is 0 as the function is odd and is being integrated over a symmetric interval, so \(a_0 = 0\).

Compute a_n

Consider the coefficients \(a_n\), \[a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} \theta cos(n\theta) d\theta\]. By considering that \(\theta\) is an odd function and \(cos(n\theta)\) is an even function, and the product of an odd and an even function is an odd function, the integration over the symmetric interval will give 0, so \(a_n = 0\) for \(n \geq 1\).

Compute b_n

Lastly, consider the coefficients \(b_n\), \[b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} \theta sin(n\theta) d\theta\]. This integration can be solved using integration by parts, where one then finds that \(b_n = (-1)^n/n\) for \(n \geq 1\). This works because the \(\theta\) function can be interpreted as a ramp function, increasing linearly from \(-\pi\) to \(\pi\), and its Fourier transforms are generally well-understood.

Write Out the Fourier Series

Finally, replace Fourier coefficients in series representation. As we have found that \(a_0 = a_n = 0\), and \(b_n = (-1)^n/n\), the Fourier series of \(f(\theta)=\theta\) can be written as: \[f(\theta) = \sum_{n=1}^{\infty} \frac{(-1)^n}{n} sin(n\theta)\]. This is the Fourier series expansion of the function given in the problem.

Key concepts

Fourier Coefficients

Fourier coefficients are essential elements in Fourier Series, representing the weights of the sinusoidal components that make up a periodic function. When you decompose a function like the one in our example, \(f(\theta) = \theta\), into its Fourier series, you need to find these coefficients to express the function as a sum of sines and cosines in terms of its period.
Each coefficient, \(a_n\) and \(b_n\), is calculated through specific integrals. The use of integrals allows us to 'pick out' the components of the function that match these sinusoidal basis functions.
For our exercise, the calculation of the coefficients relies on the integration over one period, here from \(-\pi\) to \(\pi\). As the solution shows:

• \(a_0\), the average value of the function over one period, turns out to be zero, owing to the function's symmetry.
• \(a_n\) coefficients vanish because of the odd and even function properties within the integral.
• For \(b_n\), we apply integration by parts to get the result \((-1)^n/n\), a process that leverages deeper integration techniques.

These coefficients enable the synthesis of the function using the Fourier series.

Integration Techniques

Integration Techniques are crucial when calculating Fourier coefficients. They involve understanding and applying several methods to solve integrals that emerge in Fourier analysis.
For instance, computing these integrals in the given exercise involves recognizing properties of odd and even functions. This insight simplifies integration over symmetric intervals, particularly useful when determining that \(a_n\) is zero.
An additional technique employed is integration by parts, utilized in calculating \(b_n\), which simplifies handling products of functions, such as \(\theta sin(n\theta)\). Here’s how it works in the context of our exercise:

• Identify parts of the function to differentiate and integrate.
• Apply the formula: \(\int u \, dv = uv - \int v \, du\).

Through integration by parts, the transformation of the function simplifies the calculation, allowing solutions that describe complex sinusoidal behavior efficiently.

Periodic Functions

Periodic Functions repeat their values in regular intervals, known as periods. In the exercise at hand, the function \(f(\theta) = \theta\) is periodic on the interval \((-\pi, \pi)\). This indicates that the function replicates its behavior beyond this interval, which is fundamental to Fourier series.
Understanding a function's periodicity aids in compressing complex functions into simpler terms using sinusoidal components, quite like breaking down a sound wave into individual tones.
Key characteristics include:

• Repetitive pattern: The function's values recur, which the Fourier series approximates using sines and cosines.
• Symmetry: As seen with the odd function property in our example, symmetry simplifies computations and reveals the inherent structure.

By leveraging periodicity, Fourier series provides a powerful tool for analyzing functions built on repeated structures, something seen across physics and engineering disciplines.