Question

Find the Fourier series of \(f_{p}(x)=\cos p x\), for \(0 \leq p \leq \pi\).

Short answer

The Fourier series of \( f_p(x) = \cos(px) \) is \( \cos(px) \).

Step-by-step solution

Fourier Series Definition

The Fourier series of a function \( f(x) \) with period \( T \) is given by: \[ f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos \frac{2\pi nx}{T} + b_n \sin \frac{2\pi nx}{T} \right) \] where \( a_0, a_n, \) and \( b_n \) are the Fourier coefficients.

Calculate Fourier Coefficient \( a_0 \)

\( a_0 \) is calculated by the formula \( a_0 = \frac{1}{T} \int_{0}^{T} f(x) \, dx \). For \( f_p(x) = \cos(px) \), if \( T = 2\pi \), then: \[ a_0 = \frac{1}{2\pi} \int_{0}^{2\pi} \cos(px) \, dx = \frac{1}{2\pi} \left[ \frac{\sin(px)}{p} \right]_{0}^{2\pi} = 0 \] since \( \sin(px) \) is zero at both limits.

Calculate Fourier Coefficient \( a_n \)

\( a_n \) is calculated using \( a_n = \frac{2}{T} \int_{0}^{T} f(x) \cos \frac{2\pi nx}{T} \, dx \). For \( T = 2\pi \): \[ a_n = \frac{2}{2\pi} \int_{0}^{2\pi} \cos(px) \cos(nx) \, dx \] Using the trigonometric identity: \( \cos(A)\cos(B) = \frac{1}{2}[\cos(A-B) + \cos(A+B)] \) and simplification, we find after integrating over the interval \(0\) to \(2\pi\), \( a_n = 1 \) if \( n = p \) and \( a_n = 0 \) otherwise.

Calculate Fourier Coefficient \( b_n \)

\( b_n \) is calculated using the formula \( b_n = \frac{2}{T} \int_{0}^{T} f(x) \sin \frac{2\pi nx}{T} \, dx \). For \( f_p(x) = \cos(px) \), the integral is: \[ b_n = \frac{2}{2\pi} \int_{0}^{2\pi} \cos(px) \sin(nx) \, dx = 0 \] because the product of \( \cos \) and \( \sin \) is an odd function over a symmetric interval and the integral of an odd function over this interval is zero.

Construct the Fourier Series

With the computed coefficients, the Fourier series for \( f_p(x) = \cos(px) \) can be expressed as: \[ f_p(x) = \cos(px) = 0 + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + 0 \cdot \sin(nx) \right) = \cos(px) \] where the only non-zero \( a_n \) when \( n = p \) gives the original function, as the contributions from other terms are zero.

Key concepts

Fourier Coefficients

In the realm of Fourier series, Fourier coefficients are pivotal. They allow a complex periodic function to be analyzed and represented as a sum of simple trigonometric functions. Let's break down the roles of these coefficients:

• **The Coefficient \(a_0\)**: This is the average value of the function over one period. It is calculated using the integral \(a_0 = \frac{1}{T} \int_{0}^{T} f(x) \, dx\), where \(T\) is the period.
• **The Coefficients \(a_n\) and \(b_n\)**: These coefficients dictate how much the cosine and sine waves, respectively, contribute at different frequencies. They are crucial for reconstructing the function from its trigonometric components.

For the specific function \(f_p(x) = \cos(px)\), when the Fourier series is computed, \(a_n\) becomes \(1\) if \(n = p\) and \(0\) otherwise, while \(b_n\) is always \(0\). This alignment in values suggests that the original function is essentially dominated by a single cosine term, emphasizing the key role of correctly calculating these coefficients with integral calculus.

Trigonometric Identities

Trigonometric identities are essential tools when calculating Fourier series, as they simplify the integration of products of sine and cosine functions. A primary identity used in this context is: \[\cos(A)\cos(B) = \frac{1}{2}[\cos(A-B) + \cos(A+B)]\]This identity helps in transforming products of trigonometric functions into simpler forms, which are much easier to integrate over specific intervals. For example, while finding Fourier coefficients for \(f_p(x) = \cos(px)\), this identity is applied to simplify the integrals involved in calculating \(a_n\), resulting in substantial simplification. The identity converts the product of two cosines into a sum of cosines, thus enabling easier integration over a period of \(2\pi\). These simplifications are key in handling more complex periodic functions.

Integral Calculus

Integral calculus is vital when dealing with Fourier series, as it provides the method for computing the Fourier coefficients. These integrals are calculated over a full period of the function. Here are the main integral calculations involved:

• **Computing \(a_0\)**: Involves integrating the function over one full cycle. The integral reflects the average value of the function.
• **Computing \(a_n\)** and **\(b_n\)**: Integral calculus is applied again here to find these coefficients as sums of the respective sine and cosine terms over the interval. Using symmetry properties and identities, the integration process can be simplified.

For \(f_p(x) = \cos(px)\), integral calculus reveals that \(a_0\) is zero, since the average value of a full cosine wave over one period is zero. Similarly, the integral producing \(b_n\) results in zero due to the orthogonality of sine to cosine in the trigonometric context. Understanding how integral calculus interacts with these Fourier components is crucial for interpreting and reconstructing periodic phenomena via Fourier series.