Question

Given \(\mathcal{F}[f](\omega)=\frac{1}{1+\omega^{2}}\), calculate \(\mathcal{F}\left[x^{2} f^{\prime \prime}(x)+2 f^{\prime \prime \prime}(x)\right](\omega)\).

Short answer

Use properties and calculate derivatives: Fourier transform is \( \text{combined expression from derivatives} + 2 (-i \omega^3 \mathcal{F}[f](\omega)) \).

Step-by-step solution

Understanding the Problem

We are given the Fourier transform of a function, \( \mathcal{F}[f](\omega) = \frac{1}{1+\omega^{2}} \), and need to find the Fourier transform of another expression involving derivatives of \( f(x) \). We denote the required transform as \( \mathcal{F}\left[x^{2} f^{\prime \prime}(x)+2 f^{\prime \prime \prime}(x)\right](\omega) \). We will use Fourier transform properties to solve this problem.

Fourier Transform of Derivatives

The Fourier transform of the second derivative of \( f(x) \), \( f^{\prime \prime}(x) \), is \( (i \omega)^{2} \mathcal{F}[f](\omega) = -\omega^{2} \mathcal{F}[f](\omega) \). The Fourier transform of the third derivative \( f^{\prime \prime \prime}(x) \) is \( (i \omega)^{3} \mathcal{F}[f](\omega) = -i \omega^{3} \mathcal{F}[f](\omega) \). These will be useful in finding the final transform.

Applying the Transform to the Expression

We need to find \( \mathcal{F}\left[x^{2} f^{\prime \prime}(x)+2 f^{\prime \prime \prime}(x)\right](\omega) \). The Fourier transform is linear, so:\[ \mathcal{F}\left[x^{2} f^{\prime \prime}(x)+2 f^{\prime \prime \prime}(x)\right](\omega) = \mathcal{F}[x^{2} f^{\prime \prime}(x)](\omega) + 2 \mathcal{F}[f^{\prime \prime \prime}(x)](\omega). \]We already know the transform for \( f^{\prime \prime \prime}(x) \) from the previous step.

Using Convolution Theorem for x²f"(x)

To find the Fourier transform of \( x^{2} f^{\prime \prime}(x) \), we can use the convolution theorem and the property that multiplication by \( x^2 \) in time domain is differentiation in frequency domain. Thus:\[ \mathcal{F}[x^{2} f^{\prime \prime}(x)](\omega) = \frac{d^2}{d\omega^2}\left[-\omega^{2} \mathcal{F}[f](\omega)\right]. \]First, compute \( -\omega^{2} \mathcal{F}[f](\omega) = -\frac{\omega^{2}}{1+\omega^{2}} \), then differentiate twice with respect to \( \omega \).

Complete the Derivations

Calculate the second derivative from the expression obtained:1. Compute \( \frac{d}{d\omega}\left(-\frac{\omega^{2}}{1+\omega^{2}}\right) \).2. Then compute the second derivative \( \frac{d^2}{d\omega^2}\left(-\frac{\omega^{2}}{1+\omega^{2}}\right) \).3. Employ these derivatives to get \( \mathcal{F}[x^{2} f^{\prime \prime}(x)](\omega) \).

Final Expression

By substituting the results from prior steps, the expression is combined to obtain:\[ \mathcal{F}\left[x^{2} f^{\prime \prime}(x)+2 f^{\prime \prime \prime}(x)\right](\omega) = \text{expression computed from derivatives} + 2 (-i \omega^{3} \mathcal{F}[f](\omega)). \] Combine computed results into a single expression.

Key concepts

Derivative Properties

The Fourier transform has an interesting relationship with derivatives. When we take the Fourier transform of a derivative, it simplifies into multiplication with a power of the imaginary unit 'i' and a power of the frequency variable 'ω'.
For example:

• The Fourier transform of the first derivative, \( f'(x) \), is \( i\omega \mathcal{F}[f](\omega) \).
• The Fourier transform of the second derivative, \( f''(x) \), is \( -(\omega^2) \mathcal{F}[f](\omega) \).
• The third derivative, \( f'''(x) \), is even further transformed into \( -i\omega^3 \mathcal{F}[f](\omega) \).

This means the operation of differentiation in the time domain (with respect to 'x') corresponds to a simple multiplication in the frequency domain (with respect to 'ω'). This property makes calculations involving derivatives straightforward when dealing with Fourier transforms.

Convolution Theorem

The convolution theorem is a powerful tool in Fourier analysis. It relates the Fourier transform of a product of two functions in the time domain to the convolution of their Fourier transforms in the frequency domain.
Here's a simple way to understand it:

• In mathematical terms, if two functions, say \( f(x) \) and \( g(x) \), are convolved in the time domain, their Fourier transforms are multiplied in the frequency domain.
• Conversely, if \( f(x) \cdot g(x) \) is a product in the time domain, the result in the frequency domain is the convolution of \( \mathcal{F}[f](\omega) \) and \( \mathcal{F}[g](\omega) \).

This theorem simplifies problems significantly, especially when we're dealing with squared terms or powers, as seen in manipulations involving \( x^2 f''(x) \) in the puzzle we're solving.
Using the convolution theorem, differentiating in the frequency domain becomes as simple as finding derivatives.

Differentiation in Frequency Domain

When you come across differentiation in the frequency domain, things can look a bit tricky, but the math simplifies nicely thanks to Fourier transform properties.
Here’s what to keep in mind:

• In the problem, \( x^2 f''(x) \) is transformed into the second derivative with respect to \( \omega \) in the frequency domain.
• This is a result of applying the differentiation property related to Fourier transforms, where multiplications by \( x^n \) in the time domain turn into differentiations in the frequency domain.

So, if you need to find the Fourier transform of \( x^2 f''(x) \), you actually differentiate the transformed function twice with respect to \( \omega \). It brings out terms like \( \frac{d^2}{d\omega^2}( \mathcal{F}[f''(x)](\omega)) \).
This two-step derivative process simplifies your work: first, calculate the outcome of the multiplication by \( \omega \) and then sequentially perform the differentiation.

Fourier Analysis

Fourier analysis is a cornerstone of signal processing and many areas of applied mathematics. It involves expressing functions as sums of sinusoidal components, which is especially useful in understanding oscillations and waves.
Some key points to grasp in Fourier analysis include:

• The Fourier transform is a tool used to shift a function from its original domain (often time or space) into the frequency domain.
• It permits complicated phenomena to be dissected into simple, periodic functions, making complex signal analysis and manipulation more straightforward.
• The properties of the Fourier transform, such as linearity and derivative transforms, are essential to analyzing and simplifying mathematical models of real-world systems.

In our specific exercise, Fourier analysis allows us to handle the derivatives and multipliers systematically, reflecting changes from the time domain to frequency domain, illustrating the beauty and utility of this transformative method.