Fourier transform IV - Derivatives

The Fourier transform and derivatives

We have been slowly analyzing how the Fourier transform interacts with the other operations we can perform on functions, and now it is time to finally ask:

The Fourier transform and derivatives

How does the Fourier transform interact with the derivative operator? In other words, how does $F (f^{'})$ relate to $F f$ ?

Let's begin by making the following assumptions:

The function $f$ is differentiable and $f^{'}$ is continuous.
The function $f$ has a Fourier transform, i.e., the integral $\int_{- \infty}^{\infty} f (t) e^{- 2 π i t s} d t$ converges.
The function $f (t)$ "rapidly decays to $0$ ", i.e., $lim_{t \to \pm \infty} f (t) = 0$ .

The final assumption is not an unreasonable one if we want the Fourier transform of $f$ to have any chance to exist. (We'll see where this assumption helps us shortly.)

Now, by definition we have

(F (f^{'})) (s) = \int_{- \infty}^{\infty} f^{'} (t) e^{- 2 π i s t} d t

Since $f^{'}$ is continuous, by integration-by-parts we have

\begin{aligned} (F (f^{'})) (s) & = \int_{- \infty}^{\infty} f^{'} (t) e^{- 2 π i s t} d t \\ = {[e^{- 2 π i s t} f (t)]}_{- \infty}^{\infty} - \int_{- \infty}^{\infty} f (t) \cdot (- 2 π i s) e^{- 2 π i s t} d t \\ = 2 π i s \int_{- \infty}^{\infty} f (t) e^{- 2 π i s t} d t \\ = 2 π i s \cdot (F f) (s) . \end{aligned}

Note that in the third equality we used the assumption that $lim_{t \to \pm \infty} f (t) = 0$ .

So, under these mild assumptions, we have the very nice property:

F (f^{'}) = 2 π i s \cdot F f

Assuming $f^{'}$ itself has similarly nice properties, we can repeat this argument to deduce that

F (f^{″}) = (2 π i s)^{2} \cdot F f,

and so on.

The Fourier transform and differentiation

Under some mild assumptions (detailed above), we have the following formula:

F (f^{(k)}) = (2 π i s)^{k} \cdot F f .

In other words, the Fourier transform exchanges differentiation for multiplication by $2 π i s$ . Very nice!

An enlightening example

Suppose $f (t)$ is a function that satisfies the mild assumptions above and is a solution to the second-order, linear nonhomogeneous differential equation

f^{″} (t) - f (t) = - Λ (t) .

Applying the Fourier transform then produces the following algebraic equation:

(2 π i s)^{2} \cdot F (s) - F (s) = - {sinc}^{2} (s) .

We can easily solve this equation for $F (s)$ , finding that

\begin{aligned} F (s) & = \frac{- {sinc}^{2} (s)}{(2 π i s)^{2} - 1} \\ = \frac{{sinc}^{2} (s)}{1 + 4 π^{2} s^{2}} . \end{aligned}

In theory, we can now "simply" use the inverse Fourier transform to recover $f$ :

f (t) = (F^{- 1} F) (t) = \int_{- \infty}^{\infty} \frac{{sinc}^{2} (s)}{1 + 4 π^{2} s^{2}} e^{2 π i t s} d s .

But this looks like an incredibly difficult integral to compute! Is there anyway to determine $f$ without computing this monstrous integral? Maybe ...

You might recognize the function $F (s)$ (which is the transform of $f$ ) as the product of the transforms of two recognizable functions. Indeed, first write

F (s) = {sinc}^{2} (s) \cdot \frac{1}{1 + 4 π^{2} s^{2}} .

We know $(F Λ) (s) = {sinc}^{2} (s)$ and $F (e^{- | t |}) = \frac{2}{1 + 4 π^{2} s^{2}}$ , so if we let $g (t) = \frac{1}{2} e^{- | t |}$ , then $(F g) (s) = \frac{1}{1 + 4 π^{2} s^{2}}$ . With these observations in mind, we can view the equality above as the equality

F f = F Λ \cdot F g .

If we can figure out how to rewrite the right-hand side as the Fourier transform of a single function, then our equation above tells us $f$ must equal that function. So let's try to do that.

We begin by inserting the integral definitions of the transforms on the right-hand side:

F Λ \cdot F g = (\int_{- \infty}^{\infty} Λ (t) e^{- 2 π i t s} d t) (\int_{- \infty}^{\infty} g (x) e^{- 2 π i x s} d x)

Here we have used different variables of integration for the two integrals, so that we can use the multivariable calculus trick of rewriting the product of those two integrals as a single double integral:

F Λ \cdot F g = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} Λ (t) g (x) e^{- 2 π i (t + x) s} d t d x .

Now we'll make the substitution $u = t + x$ on the inner integral, obtaining:

F Λ \cdot F g = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} Λ (u - x) g (x) e^{- 2 π i u s} d u d x .

Finally, we'll change the order of integration and introduce some visually suggestive parentheses:

F Λ \cdot F g = \int_{- \infty}^{\infty} (\int_{- \infty}^{\infty} Λ (u - x) g (x) d x) e^{- 2 π i u s} d u .

If we define a function $h (u)$ by

h (u) = \int_{- \infty}^{\infty} Λ (u - x) g (x) d x,

then our previous equation suddenly becomes

F Λ \cdot F g = \int_{- \infty}^{\infty} h (u) e^{- 2 π i u s} d u = F h .

It follows that $f = h$ . In other words, we've proven that our solution is

f (t) = \int_{- \infty}^{\infty} Λ (t - x) g (x) d x .

We've accidentally discovered the convolution product!

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Fourier transform V - Convolution