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Researchers thought this was a bug (Borwein integrals)

3Blue1Brown · Youtube · 114 HN points · 0 HN comments
HN Theater has aggregated all Hacker News stories and comments that mention 3Blue1Brown's video "Researchers thought this was a bug (Borwein integrals)".
Youtube Summary
A curious pattern of integrals that all equal pi...until they don't.
Next video on convolutions: https://youtu.be/KuXjwB4LzSA
Help fund future projects: https://www.patreon.com/3blue1brown​
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Original paper from David and Jonathan Borwein
https://carma.edu.au/resources/db90/pdfs/db90-119.00.pdf

Other fun coverage of the topic:
http://schmid-werren.ch/hanspeter/publications/2014elemath.pdf

https://johncarlosbaez.wordpress.com/2018/09/20/patterns-that-eventually-fail/

Correction: 4:12 The top line should not be there, as that integral diverges

Timestamps
0:00 - The pattern
4:45 - Moving average analogy
10:41 - High-level overview of the connection
16:14 - What's coming up next

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https://github.com/3b1b/manim
https://github.com/ManimCommunity/manim/

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https://github.com/3b1b/videos/

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https://www.vincentrubinetti.com/

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Hacker News Stories and Comments

All the comments and stories posted to Hacker News that reference this video.
Nov 15, 2022 · 110 points, 18 comments · submitted by andersource
gugagore
I'd like to know what function this converges to:

rect(x) • rect(x/2) • rect(x/4) • ...

Where • is the convolution operator.

Unlike the series in the video, 1 + 1/2 + 1/4 + ... converges. So this function has compact support, and the value at 0 does not dip.

I expect it to be a https://en.m.wikipedia.org/wiki/Bump_function

eutectic
I guess it's a Gaussian, by the central limit theorem.
gugagore
No, it can't be. This function has compact support, the Gaussian clearly does not.

You get a Gaussian by repeated convolution of the same function (and normalizing the width).

The equivalent question to what I asked is what's the pdf of X1 + X2/2 + X3/4 + ...

where Xi is iid unit uniform.

krackers
Do you mean `rect(2x) • 2rect(2x) • 4rect(4x)` .., both so that the limiting function is a dirac delta and the convolutions remain area preserving? Otherwise since the function you're convolving with keeps getting wider wouldn't the result also keep getting wider, so the support is no longer finite?

It might be possible to get a closed-form solution via an approach like [1]. (Out of curiosity, for rect convolved with itself, the intermediaries seem to be knwon as b-splines: https://www.chebfun.org/examples/approx/BSplineConv.html)

[1] https://math.stackexchange.com/questions/1254392/the-maximum...

gugagore
Yes that is what I meant, thanks!
selimthegrim
I’m waiting for the convolution video he promises in the video and the comments.
selimthegrim
It’s posted: https://youtu.be/KuXjwB4LzSA
bheadmaster
I love this.

I sometimes regret not studying pure math in college, and going down the software engineer (ahem, code monkey) route. There's so much mathematical beauty out there to be discovered and admired.

But I guess money's better this way.

jojobas
I went with applied maths route that had a lot of pure math, yet I'm a code monkey anyway. There is beauty in maths but while I was able to graduate I'm too dumb for advancing it, and I'm ok with that.
acchow
I love all the videos on this channel. The partial differential equations tour is incredible too.
wallfacer120
Grant Sanderson redeems the entire Internet.
cvoss
For those who'd like to read a quick description of what this video is about:

It shows a sequence of integrals following a very simple pattern. The first seven integrals in the sequence all evaluate to pi. The eighth integral inexplicably evaluates to pi - 0.0000000000462... and from that point on the pattern deviates from pi.

The video goes on to explain how such a seemingly perfect pattern can suddenly break by relating this sequence of integrands to a simpler one where it's easier to see what's happening.

krackers
One-line summary is to look at the fourier transforms of the integrand, fhat(x), then the integral is just fhat(0). The pattern is to multiply the integrand by sinc(x/k) for increasing k, and by fourier transform this is krect(komega). Since multiplication in time-domain is convolution in fourier domain, convolving rect with itself keeps eroding the edges of the rect, until eventually the erosion reaches the center and fhat*(0) dips.
greesil
Don't you mean convolution in the time domain is multiplication in the "Fourier" domain?
bonoboTP
Both are true, the situation is pretty much symmetric, as the Fourier transform is almost the same as the inverse Fourier, except for a sign change in the formula and some constant factor.
acchow
I’m shocked how succinctly you were able to summarize the ideas here. Bravo. Any chance you have a blog?
Gordonjcp
I understood electronic differentiators and integrators long before we did calculus in high school and I've always had more of a practical than mathematical leaning with that sort of stuff.

I guess what you're saying here is that if I pass a square wave through a lowpass filter - reducing the amplitude of the harmonics and rounding off the corners - then the peak amplitude will stay pretty much constant until I pull the cutoff of the filter down sufficiently close to the fundamental that it starts getting attenuated too.

Makes sense I guess.

krackers
> pass a square wave through a lowpass filter

I suppose: in the original problem we want to see what a sinc multiplied by sinc looks like (in time domain) or rect convolved with rect (analyzing in fourier domain). Also the width of the rects we're convolving with is shrinking after each time.

As you mentioned, to see what a rect convovled with a rect looks like, I think you can treat a convolution with rect as a lowpass filter (this should not be confused with convolving with sinc which gives an ideal" lowpass filter), and this gives intuition for why the erosion occurs.

What's not clear a priori to me is that the erosion will indeed actually reach the center. I think this depends on how fast the rect you're convolving with is shrinking, if it shrunk faster than {1, 1/3, 1/5, ...} then I don't think it would. I guess the easiest way to see this visually is to use the sliding method Grant showed, where the width of the y=1 peak after convolution is the overlapping width of the two rects. Thus we get 1, 1-1/3, 1-1/3-1/5,... which eventually < 0

Nov 08, 2022 · 2 points, 0 comments · submitted by herodotus
Nov 04, 2022 · 2 points, 0 comments · submitted by andersource
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