All right, I think you're gonna like this. I want to show you a beautiful result that reveals a surprising connection between a simple series of fractions and the geometry of circles. But unlike some other results like this that you may have seen before, this one involves multiplying things instead of adding them up. Now the video you're about to watch is particularly exciting for us at 3Blue1Brown because it came about a little differently for most of the videos that we've done. If you step back and think about it, the value of any kind of math presentation comes from a combination of the underlying math and then all of the choices that go into how to communicate it, and for almost all of the content on this channel, the underlying math is something that's well-known in the field. It's either based on general theory or some particular paper, and my hope is for the novelty to come from the communication half, and with this video the result we're discussing, a very famous infinite product for pi known as the Wallis Product, is indeed well known math. However, what we'll be presenting is, to our knowledge, a more original proof of this result. For context, after watching our video on the Basel problem, Sweether (?), the new 3b1b member who some of you may remember from the video about color and winding numbers, well, he spent some time thinking about the approach taken in that video as well as thinking about the connection between the Basel problem and the Wallis product and he stumbled into a new proof of the relationship between the Wallis product and pi. I mean, I'll leave open the possibility that an argument of this style is hidden somewhere in the literature beyond what our searching pulled up, but I can at least say that it was found independently, and that if it does exist out there, it has done a fantastic job hiding itself from the public view. So without further ado, let's dive into the math. Consider the product 2/1 times 4/3 times 6/5 on and on and on Or what we're doing is including all the even numbers as the numerators and odd numbers as the denominators. Of course, all the factors here are bigger than 1, so as you go through the series, multiplying each new factor in one by one, the result keeps getting bigger and bigger. In fact, it turns out that it eventually gets bigger than any finite limit, so in that sense, it's not super interesting. It just blows up to infinity. And on the other hand, if you shift things over slightly, looking at 2/3 times 4/5 times 6/7 on and on, all of those factors are less than 1, so the result keeps getting smaller and smaller and this time the series turns out to approach 0. But what if we mix the two? If you looked at 2/1 times 2/3 times 4/3 times 4/5 on and on like this where now the partial products along the way keep going up and then down and then up and then down then up a little bit and then down a little bit less until all of these jumps and falls are of almost no change at all so now it must be converging to some kind of positive finite value, but what is that value? Believe it or not, we'll discover that this equals pi divided by 2 and to understand the connection between this product, apparently unrelated to circles and pi, we're going to need to take a slight digression through a few geometric tools. It's a productive digression though since these are some useful ideas to have in your problem solving tool belt for all kinds of other math. The setup here involves a circle with many different points evenly spaced around it and then one additional special point. This is similar to what we had in the video on the Basel problem where we pictured these evenly spaced points as lighthouses and thought of that special point as an observer. Now back then, the quantity we cared about involved looking at the distance between the observer and each lighthouse, then taking the inverse square of each of those distances and adding them all up. This is why we had the whole narrative with lighthouses in the first place since the inverse square law gave a really nice physical interpretation to this quantity: it was the total amount of light received by that observer. But despite that nice physical interpretation, there's nothing magical about adding inverse squared distances that just happened to be what was useful for that particular problem. Now to tackle our new problem of 2/1 times 2/3 times 4/3 times 4/5 and so on, we're going to do something similar, but different in the details. Instead of using the inverse squared distances, just look at the distances themselves directly, and instead of adding them up we'll be multiplying them, giving a quantity that I'll be referring to as the distance product for the observer. That'll be important. And even though this distance product, no longer has a nice physical analogy I still kind of want to illustrate it with lighthouses and an observer because well I don't know. It's pretty and also it's just more fun than abstract geometric points Now for this proof of the Wallis product, we're going to need two key facts about this distance product; two little lemmas. First, if the observer is positioned halfway between two lighthouses on the circle, this distance product, the thing that you get by multiplying together the lengths of all these lines, works out to be exactly two. No matter how many lighthouses there are. And second, if you remove one of those lighthouses and put the observer in its place this distance product from all of the remaining lighthouses happens to equal the number of lighthouses that you started with. Again, no matter how many lighthouses there are. And if those two facts seem crazy, I agree. I mean it's not even obvious that the distance product here should work out to be an integer in either case and also, it seems super tricky to actually compute all of the distances and then multiply them together like this. But it turns out there is a, well, a trick to this tricky calculation that makes it quite simple. The main idea is that the geometric property of these points being evenly spaced around a circle Corresponds to a really nice algebraic property if we imagine this to be the unit circle in the complex plane with each of those lighthouses now sitting on some specific complex number Some of you might recognize these as the roots of unity But let me quickly walk through this idea in case any of you are unfamiliar think about squaring one of these numbers It has a magnitude of one so that's going to stay the same, but the angle it makes with the horizontal will double That's how square and complex numbers works Similarly cubing this number is going to triple the angle that it makes with the horizontal And in general raising it to the nth power multiplies the angle by n So for example on screen right now there are seven evenly spaced points around the unit circle which I'll call l0 l1 l2 and so on and They're rotated in such a way that l0 is sitting at the number 1 on that right hand side So because the angle that each one of these makes with the horizontal is an integer multiple of 1/7 of a turn Raising any one of these numbers to the seventh power Rotates you around to landing on the number one in Other words. These are all solutions to the polynomial equation X to the seventh minus one equals zero But on the other hand we could construct a polynomial that has these numbers as roots a totally different way by taking X minus l0 Times X minus l1 on and on and on up to X minus L6 I mean you plug in any one of these numbers and that product will have to equal 0 and Because these two degree seven polynomials have the same seven distinct roots and the same leading term It's just X to the seventh in both cases. They are in fact one and the same now take a moment to appreciate just what a marvelous fact that is this right hand side looks like it would be an Absolute nightmare to expand not only are there a lot of terms But writing down what exactly each of those complex numbers is is gonna land us in a whole mess of sines and cosines But because of the symmetry of the set up we know that when all of the algebraic dust settles it's going to simplify down to just being X to the seventh minus one all of the other terms will cancel out and Of course there's nothing special about 7 here if you have n points evenly spaced around a circle like this they are the Brutes of X to the n minus 1 equals 0 and Now you might See why this would give a nice simplifying trick for computing the distance product that we defined a moment ago if you consider the observer to be any other complex number not necessarily on the circle and Then you plug in that number for X that right hand side There is giving you some new complex number whose magnitude is the product of the distances between the observer and each lighthouse but look at that left hand side it is a dramatically simpler way to understand what that product is ultimately going to simplify down to Surprisingly this means that if our observer sits on the same circle as the lighthouses the actual number of lighthouses well It won't be important. It's only the fraction of the way between adjacent lighthouses that describes our observer, which will come into play if This fraction is f then observer to the power n lands F of the way around a full circle so the magnitude of the complex number observer to the N minus 1 is the Distance between the number one and a point F of the way around a full unit circle For example on screen right now we have seven lighthouses and the observer is sitting one-third of the way between the first and the second so when you raise the complex number associated with that observer to the seventh power They end up one third of the way around the full circle so the magnitude of observer to the seven minus one Would be the length of this chord right here? Which for one third of the way around the circle that happens to be about one point seven three and remember this value is quite remarkably the same as the full distance product that we care about We could increase or decrease the number of lighthouses and no matter what so long as that observer is one third of the way between Lighthouses we would always get the length of this same chord as our distance product in General let's define a special function for ourselves chord of F. Which will mean for any fraction F the length of a chord corresponding to that fraction of a unit circle So for example what we just saw was chord of one-third Actually, it's not so hard to see that chord of F. Amounts to the same thing as two times the sine of F halves times two pi Which is 2 times the sine of f pi, but sometimes it's easier to just think of it as chord of f So the result we've just shown is that for an observer f of the way between two lighthouses The total distance product as complicated as that might seem works out to be exactly Chord of F. No matter, how many lighthouses there are so in particular think about chord of 1/2 This is the distance between two points on the opposite ends of a unit circle which is 2? So we see that no matter how many lighthouses there are equally spread around the unit circle putting an observer exactly halfway along the circle between two of them results in a distance product of precisely 2 and That's our first key fact, so just tuck that away for the next key fact imagine putting the observer right on one of the lighthouses Well, then of course the distance product is 0 the distance 0 lighthouse ends up annihilating all other factors But suppose we just got rid of that one troublesome lighthouse and considered only the contributions from all of the other ones What would that distance product work out to be? Well now instead of considering the polynomial observer to the N minus 1, which has a root at all of these n roots of unity We're looking at the polynomial observer to the N minus 1 Divided by observer minus 1 which has a root at all of the roots of unity except for the number 1 itself And a little algebra shows that this fraction is the same thing as 1 plus observer plus observer squared on and on and on up to observer to the N minus 1 and So if you plug in observer equals 1 since that's the number he's sitting on what do you get? all of the terms here become one so it works out to be n Which means the total distance product for this setup equals the number of original lighthouses? Now this does depend on the total number of lighthouses, but only in a very simple way. I mean think about this This is incredible the total distance product that an observer sitting at one of the lighthouses receives from all other lighthouses is precisely n Where n is the total number of lighthouses including the ignored one that is our second key fact And by the way proving geometric facts with complex polynomials like this is pretty standard in math And if you went up to your local mathematician and showed him or her these two facts or other facts like these They'd quickly recognize both that these facts are truth, and how to prove them using the methods We just showed and now so can you so next with both these facts in our back pocket? Let's see how to use them to understand the product that we're interested in and how it relates to pi Take this set up with and lighthouses evenly spaced around a unit circle and Imagine two separate observers what I'll called the keeper and the Sailor Put the keeper directly on one of the lighthouses and put the Sailor halfway between that point and the next lighthouse The idea here will be to look at the distance product for the keeper Divided by the distance product for the Sailor, and then we're going to compute this ratio in two separate ways From the first key fact. We know that the total distance product for the Sailor is 2 and The distance product for the keeper well, it's zero since he's standing right on top of one but if we got rid of that lighthouse, then by our second key fact the remaining distance product for that keeper is n And of course by getting rid of that lighthouse We've also gotten rid of its contribution to the sailors distance product so that denominator now has to be divided by the distance between the two observers and Simplifying this just a little bit it means that the ratio between the keepers distance product and the sailors is N times the distance between the two observers all divided by two But we could also compute this ratio in a different way by considering each lighthouse Individually for each lighthouse think about its contribution to the keepers distance product meaning just its distance to the keeper divided by its contribution to these sailors distance product its distance to the sailor and When we multiply all of these factors up over each lighthouse we have to get the same ratio in the end n Times the distance between the observers all divided by two Now that might seem like a super messy calculation But as n gets larger this actually gets simpler for any particular lighthouse For example think about the first light house after the keeper in a sense of counterclockwise from him This is a bit closer to the Sailor than it is to the keeper specifically the angle from this lighthouse to the keeper is exactly twice the angle from this lighthouse to the Sailor and Those angles aren't exactly proportional to these straight-line distances But as n gets larger and larger The correspondence gets better and better and for a very large n the distance from the lighthouse to the keeper is very nearly twice the distance from that lighthouse to the Sailor and In the same way looking at the second lighthouse after the keeper it has an angle to keeper divided by angle to celular ratio of exactly 4/3 which is very nearly the same as the distance to keeper divided by distance to sailor ratio as n gets large and That third lighthouse l-3 is going to contribute a fraction that gets closer and closer to six fifths as n is approaching infinity Now for this proof We're going to want to consider all the lighthouses on the bottom of the circle a little bit differently Which is why I've enumerated them negative 1 negative 2 negative 3 and so on if you look at that first lighthouse before The keeper it has a distance to keeper over distance to sailor ratio that approaches 2/3 as n approaches infinity, and then the second lighthouse before it L negative 2 here contributes a ratio that gets closer and closer to 4/5 and the third lighthouse L. Negative 3 contributes a fraction closer and closer to 6/7 and so on Combining this over all of the lighthouses we get the product 2 over 1 times 2 over 3 times 4 over 3 times 4 over 5 times 6 over 5 times 6 over 7 on and on and on This is the product that we're interested in studying and in this context each one of those terms reflects what the contribution for a particular lighthouse is as n approaches infinity and When I say contribution I mean the contribution to this ratio of the keepers distance product to the sailors distance product Which we know at every step has to equal n times the distance between the observers divided by 2 So what does that value approach as n approaches infinity? Well, the distance between the observers is 1/2 of 1 over n of a full turn around the circle and Since this is a unit circle its total circumference is 2 pi so the distance between the observers approaches PI divided by N. And therefore n times this distance divided by 2 approaches PI divided by 2 So there you have it our product 2 over 1 times 2 over 3 times 4 over 3 times 4 over 5 on and on and on must approach pi divided by 2 This is a truly marvelous result and it's known as the Wallis product named after a 17th century mathematician John Wallis who first discovered this fact in a way more convoluted way and also a little bit of trivia This is the same guy who discovered, or well rather invented the infinity symbol And actually if you look back at this argument We've pulled a little bit of sleight of hand too deep in formality here, which the particularly mathematically sophisticated among you might have caught What we have here is a whole bunch of factors Which we knew multiplied together to get n times the distance between the observers divided by two And then we looked at the limit of each factor individually as n went to infinity and Concluded that the product of all of those limiting terms Had to equal whatever the limit of n times the distance between the observers divided by two is but what that assumes is that the product of limits is equal to the limit of products even when there's infinitely many factors and This kind of commuting of limits in infinite area arithmetic well It's not always true it often holds, but it sometimes fails Here, let me show you a simple example of a case where this kind of commuting of limits doesn't actually work out So we've got a grid here where every row has a single seven and then a whole bunch of ones So if you were to take the infinite product of each row you just get seven for each one of them So since every one of these products is 7 the limit of the products is also 7 But look at what happens if you take the limits first if you look at each column The limit of a given column is going to be one since at some point it's nothing but once But then if you're taking the product of those limits You're just taking the product of a bunch of ones so you get a different answer namely one Luckily Mathematicians have spent a lot of time thinking about this phenomenon And they've developed tools for quickly seeing certain conditions under which this exchanging of the limits actually works In this case a particular standard result known as dominated convergence quickly assures us that the argument We just showed will go through in full rigor for those of you Who are interested sweeter has written up a supplemental blog post to this video? Which covers those details along with many more things And I should also say we need to be a little careful about how to interpret a product like this Remember we have contributions from lighthouses Counterclockwise from the keeper as well as lighthouses clockwise from the keeper and what we did was interleave these in order to get our product Now the lighthouses counterclockwise from the keeper contribute two over one for over three six over five on and on and the ones Clockwise from the keeper contribute two over three four over five six over seven and like I said before if you play around with those Individual series you'll find that the first one gets larger and larger and blows up to infinity and the second one gets smaller and smaller Approaching zero so it's actually pretty delicate to make sense out of this overall product in terms of computing the two halves Separately and then combining them and a deed will find that if you intermix these two halves Differently for example taking twice as many factors from one of them for each factor from the other You could get a different result for the overall product It's only when you specifically combine them in this one-for-one manner that you get a product that converges to pi has this is something that falls out of the way the dominated convergence justifies us and commuting limits the way we did and again for more details see the Supplemental post Still those are just technicalities the conceptual gist for what's going on here is exactly what we just showed And in fact after doing all that work You would be ashamed not to take a quick moment to talk about one more neat result that falls out of this argument Arguably, this is the coolest part of the whole proof we see we can generalize this whole discussion think back to when we discovered our First key fact where we saw that you could not only consider placing the Sailor precisely, halfway between lighthouses but any fraction F of the way between adjacent lighthouses in that more general setting the distance product for the Sailor Wasn't necessarily two, but it was chord of F. Where F Is that fraction of the way between lighthouses and? If we go through the same reasoning that we just did with the Sailor at this location instead and changed nothing else What we'll find is that the ratio of the keeper's distance product to the sailors distance product is now n Times the distance between them divided by chord of F. Which approaches F times 2 PI divided by chord of F as n gets larger And, in the same way as before you could alternatively calculate this by considering the contributions from each individual lighthouse if You take the time to work this out the (k)th lighthouse after the keeper will contribute a factor of K Divided by K Minus f2 this ratio and all the lighthouses before the keeper they contribute the same thing But you're just plugging in negative values for K. If you combine all those contributions over all nonzero integers K Where in the same way as before you have to be careful about how you bundle the positive and negative K terms together? What you'll get is that the product of K divided by K minus F over all nonzero integers K is going to equal F times 2 PI Chord of F. Put another way since chord of F is 2 times the sine of f pi This product is the same as f times two pi divided by two times sine of f pi Which is f pi. Over sine of f pi? Now rewriting this just a little bit more what you get is a pretty interesting fact sine of F Times pi is equal to F Pi times this really big product the product of 1 minus F over K over all nonzero integers K So what we found is a way to express sine of X as an infinite product, which is really cool If you think about it, so not only does this proof give us the wallace product Which is being credible in its own right it also generalizes to give us the product formula for the sine And what's neat about that is that it connects to how euler? Originally solved the basel problem the sum that we saw in the previous video He was looking at this very infinite product for sine. I mean connecting these formulas for pi Two circles is one thing but connecting them to each other is another thing entirely and Once again if you want more details on all of this check out the supplementary blog post
When he explained the product representation of the sine, it reminded me of Euler's reflection formula for the gamma function. These two must be related, though I can't yet see how.
That was awesome
These guys are just awesome
Beautifully done, as usual. I appreciated the shoutout to the DCT at the end, too.
It should be noted that this seems reminiscent of Euler's original approach for the sine-product-formula, as seen here. The key idea of both is to apply Euler's Formula to some ratio to get sines and pi's and then apply the factorization of the polynomials 1+x+x2+...+xN-1 to the ratio to get the product side. There is definitely a difference in how these ideas are permuted/setup within these two proofs, but it seems to me that they rely on the same internal machinery.
The key differences I could tell are how they setup the ratio and how they use Euler's Formula. Might be fun to dig into both to see exactly how similar/different they are.
Between this and the latest Mathologer video, it seems YouTube is shaping up to be the next big mathematical journal.
That was fucking tight
I'm glad he did the generalization, because halfway through, I wondered what would happen if the Sailor was some other fraction away from the Keeper.
Really happy to see "Formalities not discussed", because I was kind of squinting my eyes just before that.
I imagine others here briefly did that, but then immediately went on to "OK, but Dominated Convergence".