Question

Suppose \(f(x)= \begin{cases}\frac{\pi}{4}, & -\pi

Short answer

The coefficients are \(a_n = 0\) and \(b_n = \frac{1}{n} (-1)^n\). \(S(x) = -\sum_{n=1}^{\infty} \frac{(-1)^n}{n} \cos(nx)\).

Step-by-step solution

Fourier Series Basics

The Fourier series for a function defined piecewise over (-\pi, \pi) is given by the function's decomposition into sinusoids. The coefficients \(a_n\) and \(b_n\) in a Fourier series describe the function's projection onto these basic wave shapes.

Find \(a_0\)

The coefficient \(a_0\) is calculated from the average value of the function over its entire period, which in this case is (-\pi, \pi). \[a_{0} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx = \frac{1}{\pi} \left(\int_{-\pi}^{0} \frac{\pi}{4} \, dx + \int_{0}^{\pi} \left(\frac{\pi}{4} - x\right) \, dx\right)\]\Evaluating the integrals:\[ = \frac{1}{\pi} \left( \frac{\pi}{4} \cdot (0 + \pi) + \frac{\pi x}{4} \bigg|_{0}^{\pi} - \frac{x^2}{2} \bigg|_{0}^{\pi} \right)\]\[= \frac{1}{\pi} \left(\frac{\pi^2}{4} + \frac{\pi^2}{4} - \frac{\pi^2}{2}\right) = 0 \]

Calculate \(a_n\)

The coefficients \(a_n\) are calculated by multiplying \(f(x)\) with \(\cos(nx)\) and integrating over a full period:\[a_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} \frac{\pi}{4} \cos(nx) \, dx + \int_{0}^{\pi} \left(\frac{\pi}{4} - x \right) \cos(nx) \, dx \right)\] Solving the first integral: For even \(n\), \(a_n\) simplifies to 0 due to symmetry properties of the cosine function. For odd \(n\), the term evaluates but results in a cancelation upon integration. Checking specifics yields:\[a_n=0, \text{ for all } n.\]

Calculate \(b_n\)

The coefficients \(b_n\) are derived similarly by replacing cosine with sine:\[b_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} \frac{\pi}{4} \sin(nx) \, dx + \int_{0}^{\pi} \left(\frac{\pi}{4} - x\right) \sin(nx) \, dx \right)\]The first integral results in zero, as sine is an odd function. The second integral evaluates similarly and simplifies, considering specific terms:\[b_n = \frac{1}{\pi} \left(\frac{-\pi}{4n} (\cos(n\pi) - 1) + \frac{1}{n^2} (\sin(n\pi) - n\pi\cos(n\pi))\right) = \frac{1}{n} (-1)^n\]

Compute S(x)

The function \(S(x)\) uses the coefficients in a different form:\[S(x) = \sum_{n=1}^{\infty} \frac{1}{n} \left(a_n \sin(nx) - b_n \cos(nx)\right)\]Given \(a_n = 0\) and \(b_n = \frac{1}{n} (-1)^n\), the resultant:\[S(x) = -\sum_{n=1}^{\infty} \frac{(-1)^n}{n} \cos(nx)\] Substituting \(b_n\) and under the symmetry property and evaluation, \(S(x)\) evaluates further based on values derived for convergence of Cosine Fourier terms.

Key concepts

Fourier coefficients

In the study of Fourier series, Fourier coefficients are the crucial elements that help us decompose a complicated function into a series of sine and cosine functions. These coefficients, denoted as \(a_n\) and \(b_n\), indicate the weights of the respective cosine and sine terms in the series.

The function \(f(x)= \begin{cases}\frac{\pi}{4}, & -\pi

• \(a_0\) is the average value of the function over one period.
• \(a_n\) is obtained by integrating the product of \(f(x)\) and \(\cos(nx)\) over the function's full range.
• \(b_n\) is calculated in a similar manner, but with \(\sin(nx)\) replacing \(\cos(nx)\).

In this problem, the integration steps show that \(a_0\) is zero, meaning the constant component of our Fourier series vanishes. This also results in \(a_n = 0\) for all \(n\), due to the integral properties with cosine terms when applied to an even function.
The coefficients \(b_n\), however, yield a pattern of \(\frac{1}{n} (-1)^n\), reflecting the function’s symmetry and alternating oscillation captured by the sine terms.

Trigonometric series

A Fourier series is a type of trigonometric series, which essentially breaks down a potentially complex, periodic function into an infinite sum of sines and cosines.

This decomposition allows the approximation of the function in terms of simpler, harmonic components. In the function given here, we have each trigonometric component with its coefficient derived from the function's behavior.

The terms are given as:

• \(\frac{a_0}{2}\)
• \(a_n\cos(nx)\)
• \(b_n\sin(nx)\)

Each component of this series corresponds to a distinct frequency of oscillation. The role of the trigonometric series in this context is to represent the periodic nature of \(f(x)\) using these oscillating terms. The use of cosine and sine allows the series to account for even and odd symmetries in the function, respectively.
This becomes significant as we calculate the Fourier coefficients, which in turn correctly capture the oscillations and discontinuities of piecewise functions like the one presented.

Piecewise functions

Piecewise functions are defined by different expressions for different intervals of the independent variable. This leads to a need for methods that can handle abrupt changes in the function, creating discontinuities.

Our specific function, \(f(x)= \begin{cases}\frac{\pi}{4}, & -\pi
When representing such functions through Fourier series, understanding how to handle these jumps is crucial.

• The Fourier series, by leveraging sine and cosine terms, can smoothly bridge such discontinuities through infinite summation.
• Fourier coefficients adjust to accommodate abrupt changes within the function, capturing essential function details across intervals despite the discontinuity.

This ability to express and approximate piecewise functions makes Fourier series a powerful tool in analyzing real-world signals and functional behaviors that exhibit such discontinuities.