Question

Let $$ f(x)= \begin{cases}x-a, & a0\), and let \(F(\omega)\) be the Fourier transform of \(f\). Let \(g \in G(\mathbb{R})\) be a continuous function whose Fourier transform \(G(\omega)\) satisfies the equality \(G(\omega)=2 \pi F(\omega) F(-\omega)\). Calculate \(g(2 a)\).

Short answer

Evaluate the inverse Fourier of \(G(\omega)\) and substitute \(x = 2a\).

Step-by-step solution

Define the Fourier Transform

The Fourier transform of a function \(f(x)\) is defined as \[ F(\omega) = \int_{-\infty}^{\infty} f(x) e^{-i \omega x} \, dx. \] We will use this definition to calculate the Fourier transform of \(f(x)\).

Find Fourier Transform of f(x)

Given the function \[ f(x) = \begin{cases} x-a, & a<x<4a, \ 0, & \text{otherwise}, \end{cases} \] we calculate the Fourier transform:\[ F(\omega) = \int_{a}^{4a} (x-a) e^{-i \omega x} \, dx. \] This can be evaluated by integration by parts or seamlessly, resulting in terms of \(a\).

Use Property of G(ω)

The function \(g\) has a Fourier transform given by \[ G(\omega) = 2 \pi F(\omega) F(-\omega). \]Compute \(F(-\omega)\) and then use this property to determine \(G(\omega)\).

Inverse Fourier Transform to Find g(x)

The inverse Fourier transform of \(G(\omega)\) is given by:\[ g(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} G(\omega) e^{i \omega x} \, d\omega. \]After finding \(G(\omega)\), use inverse transform properties to calculate \(g(2a)\).

Evaluate g(2a)

Substitute \(x = 2a\) in the expression for \(g(x)\) obtained from the inverse Fourier transform. Evaluate this integral or expression to find \(g(2a)\).

Key concepts

Integration by parts

Integration by parts is a technique often used to solve integrals that are products of functions, like in Step 2 of the original solution. When dealing with the Fourier transform of piecewise functions like \( f(x) \), this method can simplify the problem significantly. Integration by parts formula is:\[ \int u \, dv = uv - \int v \, du \]Here, you choose functions \( u \) and \( dv \) such that differentiating \( u \) and integrating \( dv \) make the process easier. Typically, \( u \) is chosen to become simpler when differentiated, and \( dv \) is easy to integrate.

• Identify the parts: For this exercise, set \( u = x-a \) (since differentiating a polynomial simplifies it) and \( dv = e^{-i \omega x} \, dx \) (an exponential is straightforward to integrate).
• Compute \( du \) and \( v \): Differentiate \( u \) to get \( du = dx \), and integrate \( dv \) to find \( v = \int e^{-i \omega x} dx = \frac{-e^{-i \omega x}}{i \omega}. \)
• Substitute back into the integration by parts formula: Replace each part in the formula to simplify the integral.
• Solve the resulting equation: This involves substituting the limits and simplifying the resulting terms.

By strategically applying these steps, you manage to evaluate the Fourier transform of \( f(x) \), which involves a product of a linear function and an exponential function.

Inverse Fourier transform

The inverse Fourier transform allows us to recover a time-domain function from its frequency-domain representation. In this exercise, it's used to find \( g(x) \) from its Fourier transform \( G(\omega) \). The inverse Fourier transform formula is:\[ g(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} G(\omega) e^{i \omega x} \, d\omega \]When you have \( G(\omega) = 2 \pi F(\omega) F(-\omega) \), the task involves combining the transformed functions back into \( g(x) \).

• Plug \( G(\omega) \) into the inverse formula: Substitute the expression involving the Fourier transforms into the integral.
• Evaluate the integral: You carry out contour integration or other techniques as suitable for the function \( G(\omega) \).
• Substitute real values if required: In the workout, you specifically evaluate \( g(2a) \), which means substituting \( x = 2a \) into the resulting expression from the inverse transform.

The inverse Fourier transform is crucial for converting back from the frequency domain, where analysis often occurs, to the more familiar time domain.

Continuous functions

Continuous functions are a fundamental aspect of calculus and analysis, especially in the context of Fourier transforms. A continuous function does not have abrupt changes or gaps, making them smooth and ideal for analysis using Fourier series or transforms.In the task at hand, \( g(x) \) is mentioned to be a continuous function. This property implies:

• Smoothness: \( g(x) \) has no sharp corners or discontinuities over its domain. This is essential for the valid application of Fourier theorems.
• Feasibility of integration: Continuous functions are easier to integrate, ensuring that integrals like those involved in Fourier transforms exist and are well-behaved.
• Applicability of Fourier techniques: Fourier transforms assume the integration over the function returns a finite result, possible only if the function does not exhibit any pathological discontinuities.

Understanding that \( g(x) \) is continuous helps assure that methods like the inverse Fourier transform provide reliable and coherent results for values such as \( g(2a) \). Continuous functions enable seamless computation and logical predictions about the behavior of solutions to differential equations in both time and frequency domains.