Problem 36 - Calculus

Question

Give an example of: True or false? If \(f\) is an even function, then the Fourier series for \(f\) on \([-\pi, \pi]\) has only cosines. Explain your answer.

Step-by-Step Solution

Answer

True, an even function has a Fourier series with only cosine terms on \([-\pi, \pi]\).

Step 1: Understanding Even Functions

To determine if a function is even, we must check if it satisfies the property: \( f(x) = f(-x) \) for all \( x \) in its domain. This symmetry around the y-axis is a crucial feature of even functions.

Step 2: Fourier Series Components

A Fourier series of a periodic function \( f \) with period \( 2\pi \) is given by: \[ f(x) = a_0 + \sum_{n=1}^{tilde n = infty} (a_n \cos(nx) + b_n sin(nx)) \] where coefficients \( a_n \) are calculated using the function's even property, and \( b_n \) accounts for the sine components.

Step 3: Exploring Coefficient Effect

For an even function \( f(x) = f(-x) \), the calculation for \( b_n \) involves an integral of an odd function over a symmetric interval about the origin. This results in \( b_n = 0 \) for all \( n \), since integrating an odd function over such an interval evaluates to zero.

Step 4: Conclusion with Cosines

Thus, if \( f \) is an even function, the Fourier series on \([-\pi, \pi]\) reduces to \[ f(x) = a_0 + \sum_{n=1}^{tilde n = infty} a_n \cos(nx). \] The series contains only cosine terms, as sine terms, \( b_n \sin(nx) \), are eliminated.

Key Concepts

Exploring Even Functions

In mathematics, an even function is one that displays symmetry with respect to the y-axis. This means that for every point \(x\) in the function's domain, \(f(x) = f(-x)\). A function that meets this criterion will generally graph a mirrored image on both sides of the y-axis.

Common examples of even functions include \(f(x) = x^2\) and \(f(x) = \cos x\). Such functions maintain the same values whether you approach them from the left or the right of the y-axis. The property of evenness is central when determining the specific components that appear in the Fourier series of a function.

• Even Function Formula: \(f(x) = f(-x)\).
• Graphically mirrors about the y-axis.
• Key examples: \(x^2\), \( ext{cos} x\).

Understanding Sine and Cosine Components in Fourier Series

The Fourier series serves as a powerful tool in breaking down periodic functions into a sum of simple trigonometric functions. These consist of both sine and cosine terms, represented as:\[f(x) = a_0 + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx))\] Here, \(a_n\) represents the coefficients of the cosine components and \(b_n\) signifies the coefficients of the sine components.

In context of even functions, these components behave differently owing to their symmetrical nature:

• Cosine Terms (\(a_n \cos(nx)\)): Since cosines are even functions themselves, they naturally fit into the Fourier series of an even function.
• Sine Terms (\(b_n \sin(nx)\)): Sines, being odd functions, conflict with the even nature of the primary function, often leading to their coefficients, \(b_n\), dropping out to zero for even functions.

Integral Over Symmetric Interval

When considering the Fourier series of an even function over a symmetric interval, such as \([-\pi, \pi]\), certain integral properties come into play. If the function \(f\) is even, the integral to compute \(b_n\), which involves \(f(x) \sin(nx)\), results in evaluating an odd function over a symmetric interval.

Mathematically speaking, the integral of an odd function over a symmetric interval always equals zero. Thus:\[b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx = 0\]This integral results in \(b_n = 0\) for all \(n\), affirming that the sine components vanish in the Fourier series of an even function.

• An odd function integrated over a symmetric interval yields zero.
• This confirms the absence of sine terms for even functions in their Fourier series.