Question

For \(f(x)=\sum_{k=0}^{n} a_{k} x^{k}\), show that $$\tilde{f}(u)=\sqrt{2 \pi} \sum_{k=0}^{n} i^{k} a_{k} \delta^{(k)}(u), \quad \text { where } \delta^{(k)}(u) \equiv \frac{d^{k}}{d u^{k}} \delta(u) .$$

Short answer

The Fourier transformation of the given polynomial function \(f(x)=\sum_{k=0}^{n} a_{k} x^{k}\) is \( \tilde{f}(u)=\sqrt{2\pi}\sum_{k=0}^{n} i^{k} a_{k} \delta^{(k)}(u) \). The core of the solution involves understanding the relationship between the derivative of the Dirac delta function and the integral.

Step-by-step solution

Write the Fourier Transformation

The most common way to solve such a problem is by starting from the definition of Fourier Transformation, defined as: \( \tilde{f}(u)= \int_{-\infty}^{+\infty} f(x) e^{-iux} dx \). Replace \(f(x)\) with the given polynomial \(f(x)=\sum_{k=0}^{n} a_{k} x^{k}\).

Apply Integral to Each Term

Applying the integral to each term of the polynom: \( \tilde{f}(u)= \sum_{k=0}^{n} a_{k} \int_{-\infty}^{+\infty} x^{k} e^{-iux} dx \).

Solve the Integral

By using the relationship between the derivative of a Dirac delta function and the integral (i.e, \( \int x^{k} e^{-iux} dx= i^{k}\sqrt{2\pi}\delta^{(k)}(u) \)) we can solve the integration part for each term in the series.

Replace Integral Solution into the Equation

Replacing the integral solution into the previous equation gives us: \( \tilde{f}(u)=\sqrt{2\pi}\sum_{k=0}^{n} i^{k} a_{k} \delta^{(k)}(u) \).

Key concepts

Dirac Delta Function

Understanding the Dirac delta function can be a bit challenging, but it's a crucial pillar in the field of signal processing and physics. Think of the Dirac delta function, denoted by \( \( \delta(u) \) \), as a tool that perfectly 'picks out' the value of a function at a particular point. Although not a function in the traditional sense, it behaves like a function that is zero everywhere except at zero, where it's infinitely high in such a way that the total area under this 'spike' is 1.

Now, when talking about the Dirac delta function in the context of Fourier Transformations, it's often linked with the concept of frequency components. Picture a signal that's entirely localized – that’s what \( \delta(u) \) represents in frequency space. It is the Fourier Transform of a constant function, symbolizing that all frequencies are present equally.

For the exercise, we deal with the derivatives of the delta function, expressed as \( \delta^{(k)}(u) \) where \(k\) indicates the number of times the derivative is taken. This becomes critical when we link it back to the Fourier Transform of polynomials, where the action of these derivatives is similar to a 'filter' that can extract different components of a polynomial function when multiplied by powers of \(x\).

Polynomial Functions

Now, let's tackle polynomial functions. Polynomials are among the most frequently encountered types of functions in mathematics. They are built from variables and coefficients, with the variables raised to whole number exponents and the coefficients being constants.

In the exercise, \(f(x)\) is a polynomial expressed as a sum of terms of the form \( a_{k}x^k \). Each term contributes to the shape of the polynomial, with higher powers of \(x\) creating steeper curves in the graph of the function. Polynomials are particularly important because they can be used to approximate many other types of functions and have properties that make them friendly for analytical manipulation.

A Fourier Transform essentially decomposes functions (including polynomials) into their frequency components. Each polynomial term of \(f(x)\) corresponds to a component in the frequency domain, and the Fourier Transform of \(f(x)\), denoted \( \tilde{f}(u) \), reveals how these components are distributed across different frequencies.

Integration

Integration is a fundamental tool in calculus that is used to find areas under curves, among other things. In the context of Fourier Transformations, integration is heavily utilized when we want to transform a function from its original domain (usually time or space) into the frequency domain.

The complexity starts when you face integrals that involve products of polynomial terms and exponential functions, such as in the exercise with \( f(x)e^{-iux} \). This particular integration task is tricky because the integral of this product doesn't have a straightforward analytic solution. But here's where an amazing mathematical trick comes into play – you can relate complicated integrands to the Dirac delta function or its derivatives to simplify the problem.

Through the exercise, you saw that integration is not just about finding an area, but about understanding how a function in one domain (a polynomial in this case) gets translated into another domain (frequency domain). By using the properties of derivatives of the delta function, we can integrate each term of the polynomial, which is a process that uncovers the profound connections between time and frequency, and ultimately, leads to the solution \( \tilde{f}(u) \) of the Fourier Transform.