A new proof of the Wallis product for pi
Added 2018-04-20 23:51:26 +0000 UTCWhat a delightful example of what happens when you get multiple people working together on math exposition. As mentioned in the video, after making the video on the Basel problem, which was based on a paper that Ben brought to our attention, we were all thinking about nice intuitions for the connection between the Wallis product and circles.
There are some nice proofs out there to this effect. For example, the same author of the paper that the Basel problem video was based on also put out a nice piece highlighting a geometric view of the Wallis product, very different from this one: http://www.math.chalmers.se/~wastlund/monthly.pdf
However, in contemplating the lighthouses-around-a-lake Basel problem proof, and Sridhar happened upon this more novel perspective, the one presented in this video. The core proof is actually only a small part of the video, the rest of it being a nice excuse to discuss some useful tactics for using complex polynomials in describing circle geometry.
Hope you enjoy!
-Grant
Comments
Good question! It's one I addressed in the pinned comment of the video. Copying it down here: It looks like some people are asking about why the segment at 12:33 is okay, given that it feels like taking 0/0. Keep in mind, the actual goal at that spot is to find a polynomial whose roots are L_1, L_2, ... L_{N-1}, so the concrete result being stated is that (x - L_1)(x - L_2)...(x - L_{N-1}) will expand out to become 1+x+x^2+....x^{N-1}. No division by zero issues there. Sure, plugging in x=1 to (x^N - 1)/(x - 1) is undefined (at least before explicitly stating the intention to extend the function via a limit), but the reason for doing that polynomial division was just to see how (x - L_1)(x - L_2)...(x - L_{N-1}) would expand. All that division is asking is (x - 1)(...what?...) = (x^N - 1). Here, to give a really simple example, it's like saying x^2 - 1 has roots at 1 and -1, so dividing it by (x - 1) gives a polynomial with just a root at -1, namely (x^2 - 1) / (x - 1) = x + 1. "But wait!", someone could say, "you can't plug x = 1 into that fraction!". For sure for sure dude, but that doesn't change the fact that x + 1 is legitimately a polynomial which just has -1 as a root. Maybe you justify that division by saying something about limits, or about analytic continuation, or just by reframing to say what you care about is the question (x - 1)(...what?...) = x^2 - 1, but that's all kind of beside the point.
3blue1brown
2018-05-16 16:57:27 +0000 UTCAt 12:19, you are dividing by (O-1). This has me think that this expression is undefined at O=1. Still, you are using O=1 at 12:38. Am I missing something here?
2018-05-16 16:54:57 +0000 UTC<a href="https://www.sciencealert.com/einstein-spooky-action-demonstrated-on-massive-scale-for-first-time" rel="nofollow noopener" target="_blank">https://www.sciencealert.com/einstein-spooky-action-demonstrated-on-massive-scale-for-first-time</a>
seerpea
2018-04-27 02:16:03 +0000 UTCNice! That's really the best way to watch if you have the time/patience.
3blue1brown
2018-04-23 16:10:57 +0000 UTC
