Convolving a 'box' with a smaller box

When a rectangular signal (box) is convolved with a smaller box, only a part of the plateau is 'eroded'. For example, when a unit box is convolved with a 'one-third-width' box, it loses 1/3 of its middle plateau, resulting in a trapezoid

Answer by Job Bouwman:

The first seven Borwein Integrals are exactly [math]\frac{\pi}{2}[/math], but unexpectedly, the eighth integral is about one billionth of a percent smaller.

This unexpected c-c-combo breaker seems counterintuitive, but luckily Fourier comes to the rescue. From his perspective we just see the erosion process of a rectangular mountain: initially only the corners are affected, however, inevitably it will lose some height.[math]\large{\color{blue}{\bf \text{Explanation}}} [/math]

Five Familiar Fourier Facts

- A sinc function can be regarded as the Fourier spectrum of a rectangular signal / 'block-pulse' / 'boxcar function'
(seeFourier Transform–Rectangle Function)- The Uncertainty principle says: 'the wider the sinc, the narrower the block'. Thus the decreasing frequencies of each new sinc factor, translates in a decreasing width of the rectangular functions.
- The Convolution theorem tells that a multiplication of sinc-shaped spectra is equivalent to a convolution of rectangular shaped signals.
- The integral of a function in the Fourier domain, is the central value in the time domain. Therefore the integral of the product of sincs is proportional to the central value of the convolution outcome of their corresponding rectangular block pulses.
- The convolution of two boxes having equal width results in a triangular function (an isosceles): the left and right corner of the rectangle are completely eroded:

Animation showing the triangular outcome of two rectangular functions. Source:convolution(Wikipedia)

Convolving a 'box' with a smaller boxWhen a rectangular signal (box) is convolved with a smaller box, only a part of the plateau is 'eroded'. For example, when a unit box is convolved with a 'one-third-width' box, it loses 1/3 of its middle plateau, resulting in a trapezoid:

The trapezoidal convolution outcome has a residual plateau of 2/3. Why? Because the kernel was completely integrated when 'traversing the interval [-1/3, 1/3], and the convolution outcome will therefore be constant in this region. Outside this domain, however, an increasing part of the kernel is not 'matched', resulting in two downward slopes moving out of the center.

The result after some more convolutions:In the second convolution, the resulting trapezoid is convolved with the 1/5 kernel, which takes away another 20% of the original plateau width. Therefore the new residual plateau is reduced to 1 – 1/3 – 1/5 = 7/15 part of the initial plateau:

With each new convolution, the residual plateau gets smaller and smaller:

(which is not that well visible, because the erosion also 'smoothens' the outcome, similar how the probability distribution of a coin flipping experiment gets smoother when more coins are flipped)

Not enough plateau left to completely match the 1/15 boxOnce the residual (1/11)-convolution outcome is convolved with the (1/13) box, the remaining plateau equals:

[math]1 – \frac{1}{3}-\frac{1}{5}-…-\frac{1}{13}= \frac{2021}{45045}[/math]

which is smaller than 1/15.

So for the last convolution, the remaining plateau is too small to completely match the 1/15 kernel in the center, resulting in a slightly lower integration outcome for the main frequency component. And since this main frequency component is proportional to our integral:

[math]\int_0^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}\cdots\frac{\sin(x/15)}{x/15}dx[/math]

we now see why the resulting integral is not 'complete' anymore, and thus below [math]\frac{\pi}{2}[/math].

1. Borwein, David; Borwein, Jonathan M. (2001), "Some remarkable properties of sinc and related integrals",

The Ramanujan Journal5(1): 73–89,PS 1) This is not meant as an official proof, I am aware that the wording is sloppy, and that the heights of all kernels are normalized to unity, and that the width of the kernel was initiated to unity…

PS 2) With some further research I found that a similar visual proof was already given in: http://schmid-werren.ch/hanspete…

What are some of the most counterintuitive mathematical results?