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Researchers thought this was a bug (Borwein integrals) [video]
- selimthegrim 4y agoI’m waiting for the convolution video he promises in the video and the comments.
- selimthegrim 4y agoIt’s posted: https://youtu.be/KuXjwB4LzSA https://youtu.be/KuXjwB4LzSA
- wallfacer120 4y agoGrant Sanderson redeems the entire Internet.
- cvoss 4y agoFor 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 4y agoOne-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 4y agoDon't you mean convolution in the time domain is multiplication in the "Fourier" domain?
- bonoboTP 4y agoBoth 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 4y agoI’m shocked how succinctly you were able to summarize the ideas here. Bravo. Any chance you have a blog?
- Gordonjcp 4y agoI 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 4y ago> 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
- acchow 4y agoI love all the videos on this channel. The partial differential equations tour is incredible too.
- gugagore 4y agoI'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 https://en.m.wikipedia.org/wiki/Bump_function
- eutectic 4y agoI guess it's a Gaussian, by the central limit theorem.
- gugagore 4y agoNo, 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 4y agoDo 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 https://www.chebfun.org/examples/approx/BSplineConv.html) [1] https://math.stackexchange.com/questions/1254392/the-maximum-value-peak-of-multiple-self-convolution-of-rectangular-function?noredirect=1&lq=1 https://math.stackexchange.com/questions/1254392/the-maximum...
- gugagore 4y agoYes that is what I meant, thanks!
- bheadmaster 4y agoI 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 4y agoI 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.