# Issues with Fourier Transform of $f(t)=(1-t^2)^4$ on $[-1,1]$ – Real Analysis and Fourier Analysis

finite-durationfourier analysislp-spacesreal-analysisupper-lower-bounds

Issues with the Fourier Transform of $$f(t)=(1-t^2)^4$$ on $$[-1,\,1]$$, should be analytical but looks like having a singularity with noise-like rippling

Intro

I was trying to made a compact-supported approximation of a Gaussian Envelope
$$g(t)= e^{-4t^2} \tag{Eq. 1}\label{Eq. 1}$$
So I was trying to use instead:
$$f(t)= (1-t^2)^4\cdot\theta(1-t^2) \cong \left(\frac{1-t^2+|1-t^2|}{2}\right)^4 \tag{Eq. 2}\label{Eq. 2}$$
where $$\theta(t)$$ is the Heaviside step function.

Both which can be seen here:

As is shown in the image, at "my taste" the match "looks good", but accurately speaking, it fulfill that:

1. The function $$f(t) = 0,\,|t|\geq 1$$, so it is of compact support.
2. At the edges of the domain $$\partial t =\{-1;\,1\}$$ the function $$s(t)=(1-t^2)^4$$ is already zero: $$s(-1)=s(1)=0$$, so shouldn't be much issues related with the term $$\theta(1-t^2)$$, since it is going to be at most an avoidable discontinuity: $$\lim\limits_{t \to \partial t^{\pm}} f(t) = \lim\limits_{t \to \partial t} f(t) = 0$$, so the function is continuous on $$\mathbb{R}$$.
3. Using Wolfram-Alpha, the following norms shows to be bounded: $$\|f\|_\infty = 1 < \infty$$, $$\|f\|_1 = \frac{256}{315} \approx 0.813 < \infty$$, $$\|f\|_2^2 = \frac{65536}{109395} \approx 0.599 < \infty$$ so is bounded, absolute integrable, and energy limited.
4. Also its derivative is:
$$\begin{array}{r c l} \frac{d}{dt}\left((1-t^2)^4\cdot\theta(1-t^2)\right) & = & \frac{d}{dt}\left((1-t^2)^4\right)\cdot\theta(1-t^2)+(1-t^2)^4\cdot\frac{d}{dt}\left(\theta(1-t^2)\right)\\ & = & -8\,t\,(1-t^2)^3\cdot\theta(1-t^2)+(1-t^2)^4\cdot\frac{d}{dt}\left(\frac{1+\text{sgn}(1-t^2)}{2}\right)\\ & = & -8\,t\,(1-t^2)^3\cdot\theta(1-t^2)+2\,t\,(1-t^2)^4\cdot\delta(1-t^2)\\ & = & -8\,t\,(1-t^2)^3\cdot\theta(1-t^2)+2\,t\,(1-t^2)^3\cdot\underbrace{(1-t^2)\cdot\delta(1-t^2)}_{=\,0,\,\text{since}\,(1-x)\delta(1-x)=0} \\ \end{array}$$
$$\Rightarrow \frac{df(t)}{dt} = -8\,t\,(1-t^2)^3\cdot\theta(1-t^2) \tag{Eq. 3}\label{Eq. 3}$$
so its derivative is also continuous by the same reasons of point (ii), and $$\|f'\|_\infty = \frac{1798}{343\sqrt{7}} \approx 1.9 < \infty$$, $$\|f'\|_1 = 2 < \infty$$, and $$\|f'\|_2^2 = \frac{131072}{45045} \approx 2.9 < \infty$$.

So with all these point the approximation $$f(t)$$ look quite "well-behaved": note that is not a "smooth function", but same analysis of point $$(4)$$ could be done successfully for $$f''$$ also.

From now on I will continue labeling the observations so you can accurately point where I am having a misconception.

1. Since the function $$f(t)$$ is of compact-support and also is squared-integrable (since $$\|f\|_2^2 < \infty$$ as show in point $$(3)$$), I where expecting that the Paley–Wiener theorem to be fulfilled, so the Fourier Transform of $$f(t)$$ was going to be an analytical function.

Using Wolfram Alpha I calculate the Fourier Transform (under the "electricians" definition), in 3 different ways showing each of them the same result so I believe is right calculated –way 1, way 2, and way 3:

$$\hat{f}(w) = \int\limits_{-\infty}^{\infty} f(t)\,e^{-iwt}\,dt = \int\limits_{-1}^{1} (1-t^2)^4\,e^{-iwt}\,dt = \frac{768 \Big(5w\,(2 w^2-21)\cos(w)+(w^4-45w^2+105) \sin(w)\Big)}{w^9} \tag{Eq. 4}\label{Eq. 4}$$

So far so good, since \eqref{Eq. 4} looks like the classical Fourier Transform made by a polynomial mixed with some trigonometric functions, but when trying to calculate $$\|\hat{f}\|_1$$ and $$\|iw\hat{f}\|_1$$ the integrals get stuck on Wolfram-Alpha, so I decide to plot it to see what is going on.

I am going to graph it against the Fourier Transform of $$g(t)$$, which under the same definition of the transform is going to be:
$$\hat{g}(w) = \frac{\sqrt{\pi}}{2}e^{-\frac{w^2}{16}}\tag{Eq. 5}\label{Eq. 5}$$

They look similar, but something "weird" is happening near the DC component $$w \approx 0$$ for the solution $$\hat{f}(w)$$ of \eqref{Eq. 4}.

To have a better insight I plot $$\hat{f}(w)$$ also in Desmos and there exist an "horrible noise-like rippling" that also "looks" to be diverging as a singularity!

I tried to find the value of $$\|\hat{f}\|_\infty$$ with Wolfram-Aplha but I believe are numerical errors since the value change when changing the domain limits, but are numbers of the order of $$10^{93}$$!!!.

1. I believe that a Analytical function is smooth (as it should be the Fourier Transform $$\hat{f}$$ following point $$(5)$$), so it should be bounded on every closed interval of its domain: maybe here I am mistaken, but since analytic functions are smooth, they should be continuous on every closed interval of the domain (since are already differentiable), and since is continuous on a closed interval it should achieve a bounded maximum and minimum because of the Extreme value theorem.

But even if I am wrong, I wasn't expecting an analytical function to be so "ill-behaved" as is being $$\hat{f}$$ on $$w \in [-0.2,\,0.2]$$.

Questions

So the main questions are:

Q1: Is $$\hat{f}(w)$$ of \eqref{Eq. 4} properly obtained? Also related quantities as $$\|\hat{f}\|_1$$ and $$\|\hat{f}\|_{\infty}$$

Q2: If right, Why is this "rippling" happening? It is right? or is a miscalculation of the graphic software? or a numerical issue? Actually its looks like "Brownian" or "White Noise".

Q3: Is truly having a singularity at some angular frequency $$w$$? (doing limits in Wolfram Alpha doesn't work, at least it is saying that $$\lim_{w\to 0} \hat{f}(w)=\frac{256}{315}\approx 0.813$$ which is near the main tendency, but surely below the rippling peaks – maybe the singularity is near but not in $$w=0$$).

Q4: Is this rippling a known result like the "Gibbs' Phenomenon? How is called? If it is a known phenomenon I would like to search for any references – maybe the is a way to avoid it.

Q5: Where I am misleading my analysis? (since this singularity shouldn't exists under my assumptions and my actual knowledge – which is quite basic by the way)

Motivation

After seeing this video about the absorption of light, temporal dispersion, and Kramers-Kronig relations, required for the complex-valued modeling of the refraction index to preserve causality (is quite a good video, simple and short), I see why dispersion must happen or other way the absorbed frequency will lead to a "time-wide" signal which will become non-causal in the analysis (it will be entering the absorptive medium before the frequency-component subtraction had happen on the first place). But since a Gaussian Envelope is used, the "causality-issue" where already present since this function only vanishes at infinity, but since spectra decays faster than $$1/|w|$$, the Kramers-Kronig relation still holds "hidding" this issue.

This is why I was trying to find another envelope but with an accurate compact-support, since it is multiplied by a trigonometric function it Fourier Transform is only a displaced version of the Transform of the Envelope function, and here is where I found the "problem" since the approximation I take has this "issue" in the spectrum – don't knowing now if its a "real-life issue" or just a numeric problem of the calculation algorithms.

Hope you find it has interesting I am, beforehand thanks you very much.

I have noted now another interesting thing:

Integrating $$\hat{f}$$ as is shown here, and integrating $$\hat{g}$$ as is shown here gives the same result:
$$\int\limits_{-\infty}^{\infty}\hat{f}(w)\,dw = \int\limits_{-\infty}^{\infty}\hat{g}(w)\,dw = 2\pi$$

I have tried another slightly different approximation:
$$q(t) = \left(\frac{1-t^2+|1-t^2|}{2}\right)^{\pi} \tag{Eq. 6}\label{Eq. 6}$$

And following Wolfram-Alpha its Fourier Transform is:

$$\hat{q}(w) = \int\limits_{-\infty}^{\infty} q(t)\,e^{-iwt}\,dt = \sqrt{\pi}\,\Gamma(1+\pi)_0\tilde{F}_1\left(;\,\frac{3}{2}+\pi;\,-\frac{w^2}{4}\right) \tag{Eq. 7}\label{Eq. 7}$$
which plot doesn't show the noise-like-rippling-singularity, so with this, as it was noted on the comments and answers, the situation was a numerical issue. Even so, if you directly replace all the numbers $$\pi$$ by the number $$4$$ in \eqref{Eq. 7} the rippling appears again.

Later I noted that the result of \eqref{Eq. 4} is equivalent to:
$$\hat{f}(w) = \sqrt{\pi}\,\Gamma(1+4)_0\tilde{F}_1\left(;\,\frac{3}{2}+4;\,-\frac{w^2}{4}\right) \tag{Eq. 8}\label{Eq. 8}$$

So obviously again the numerical issues arises, but for a "good" approximation I am using on Wolfram-Alpha:
$$\hat{f}^*(w) = \sqrt{\pi}\,\Gamma(1+4)_0\tilde{F}_1\left(;\,\frac{3}{2}+4.0001;\,-\frac{w^2}{4}\right) \tag{Eq. 9}\label{Eq. 9}$$

and suddenly the problem is gone… it is kind of ridiculous that "just" for the exact numbers the problem happen.

Since the regularized confluent hypergeometric function is not quite extended on every software library I find useful the following property shown by Wolfram Alpha to make approximated plots of these kind of functions, by taking an arbitrary parameter $$a$$:
$$\sqrt{\pi}\,\Gamma(1+a)_0\tilde{F}_1\left(;\,\frac{3}{2}+a;\,-\frac{w^2}{4}\right) = \text{sgn}(w)\sqrt{\pi}\,\left(\frac{2}{w}\right)^{a+\frac{1}{2}}\Gamma(1+a)J_{a+\frac{1}{2}}(w) \tag{Eq. 10}\label{Eq. 10}$$
with $$J_n(x)$$ the Bessel function of the first kind.

Unfortunately, I wasn't able to find $$\|\hat{f}\|_1$$ under any of these approximations (both shown on the last image), "as it where diverging" (I don't know if this is the case), even when $$\|\hat{g}\|_1=2\pi << \infty$$, which is counter-intuitive for me giving its similarity (the lobes should be adding even less area than a Gaussian in principle), but comparing their graphs shows that the compact-support function has lobes much more spread than the Gaussian, as is shown here.

But nevertheless, the tails are decreasing faster than $$|\frac{1}{w^2}|$$ so I don´t know why the result is not finite (see here).

As example, if I used the other aporoximarion:
$$\hat{f}^*(w) \approx \frac{962.612\,J_{4.5001}(w)}{w^{4.5001}}\tag{Eq. 11}\label{Eq. 11}$$
Wolfram-Alpha is unable to find $$\|\hat{f}^*(w)\|_1$$, but it is possible to bound it as:
$$\|\hat{f}^*(w)\|_1 < \underbrace{\int\limits_{-10}^{10}\left|\frac{962.612\,J_{4.5001}(w)}{w^{4.5001}}\right|\,dw}_{6.34416}+\underbrace{\int\limits_{10}^{\infty}\left|\frac{2}{|w^{2}|}\right|\,dw}_{0.2} <\infty$$
as can be seen here and here. And since the difference with the approximation and the main function with numerical issues is in the bounded domain $$[-1,\,1]$$, it should also be bounded, so the norm $$\|\hat{f}(w)\|_1<\infty$$ but somehow Wolfram-Alpha can't find it.

## Computation by hand of the Fourier transform

First, recall that (with your convention for the Fourier transform), $$\mathcal F[\theta(1-t^2)](w) = \mathcal F \unicode{x1D7D9}_{[-1,1]}(w)= 2 \DeclareMathOperator{\sinc}{sinc}\sinc w.$$

Second, recall that for any polynomial $$p(t)$$, and any reasonable function (or even distribution) $$g$$, $$\mathcal F ( p(t) g)(w) =p(D) \mathcal Fg(w), \qquad D:= \frac1i\frac{d}{dw}.$$ Since $$(1-t^2)^4 = 1 - 4t^2 + 6t^4 - 4t^6 +t^8,$$ $$(1-D^2)^4 = 1 +4\frac{d^2}{dw^2} +6\frac{d^4}{dw^4} + 4\frac{d^6}{dw^6} + \frac{d^8}{dw^8},$$ and hence for your function $$f(t) = (1-t^2)^4 \unicode{x1D7D9}_{[-1,1]}(t)$$, $$\mathcal Ff(w)= 2\sinc w + 8\sinc^{(2)}w + 12 \sinc^{(4)}w+8\sinc^{(6)}w+2\sinc^{(8)}w.$$ Using Wolfram to do the tedious calculations (screenshot) gives the same formula as you, so the answer is right. As $$\sinc$$ is entire, all its derivatives are, and hence $$\mathcal F f$$ is as well.

## Verifying the behaviour near $$0$$

Note that in the formula $$\mathcal Ff(w)=\frac{768 \Big(5w\,(2 w^2-21)\cos(w)+(w^4-45w^2+105) \sin(w)\Big)}{w^9},$$ if we want to see that the numerator is $$O(w^9)$$, we should (unless we are super lucky) use at least a 8th order Taylor expansion around 0 for the cosine on top, and a 9th order expansion for the sine. This is because of the low order terms in the polynomial coefficients ($$-21$$ and $$+105$$, respectively.)

Denote the $$n$$th partial Taylor expansion around $$x_0$$ of $$f$$ at $$x$$ by $$T^n f(x_0;x) = \sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}(x-x_0)^n.$$ Since all derivatives of sine and cosine are uniformly bounded by 1, Taylor's theorem gives, $$|\cos(w) - T^8 \cos(0;w)| \le \frac{w^9}{9!},$$ and similarly $$|\sin(w) - T^9 \sin(0;w)| \le \frac{w^{10}}{10!} ,$$ If we just replace $$\cos(w)$$ with $$T^8 \cos(0;w)$$ and $$\sin(w)$$ with $$T^9 \sin(0;w)$$ to get $$\tilde f(w):=\frac2{945} (w^4 - 27 w^2 + 384),$$ we incur an error of size at most $$\frac{768 \Big(5|w|\,(2 w^2\color{red}+21)|w|^9/9! +(w^4\color{red}+45w^2+105) w^{10}/10!\Big)}{|w|^9} = \frac{|w| (1155 + 145 w^2 + w^4)}{4725}.$$ For say $$|w|<0.2$$, this error is at most $$\frac{1160.8016|w|}{4725}<\frac {|w|}4<\frac1{20}.$$ Therefore, $$\mathcal Ff$$ stays in the shaded blue region below. The approximation $$\tilde f$$ is plotted in red. (Desmos link) Of course, using more terms will give a more impressive bound, but this is enough to rule out the spiky behavior you saw.

In fact, in the current situation there is extra cancellation, since $$T^8 \cos(0;w)=T^9 \cos(0;w)$$ and $$T^9 \sin(0;w)=T^{10} \sin(0;w)$$, giving us a much better error estimate. Redoing the above calculations shows that the error is no more than $$(w^2 (w^4 + 155 w^2 + 1260))/51975$$ which is $$\le w^2/30$$ for $$w\le 1$$. This explains why the approximation is a much better match than the above error plot suggests. The true function $$\mathcal Fg$$ is plotted in black (for $$|x|>0.2$$) below: (desmos link)

You have to zoom in quite far in order to even notice that they don't exactly match at $$x=0.2$$: