### Video Transcript

All right, I think you’re gonna like this. I wanna 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 our set 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, it’s for the novelty to come from the communication half. And with this video, the result we’re discussing, a very famous infinite product for
𝜋 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, Sridhar, 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
𝜋. 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 two over one times four over three times six over five, on and
on and on. Well, 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 one. 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 now on the other hand, if you shift things over slightly, looking at two divided
by three times four divided by five times six divided by seven, on and on, all of
those factors are less than one. So the result keeps getting smaller and smaller. And this time, the series turns out to approach zero. But what if we mix the two? If you looked at two over one times two over three times four over three times four
over five, 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 𝜋 divided by two. And to understand the connection between this product, apparently unrelated to
circles, and 𝜋, we’re gonna 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 set-up 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 two over one times two over three times four over
three times four over five and so on, we’re gonna 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 gonna 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 gonna stay the same. But the angle it makes with the horizontal will double. That’s how squaring complex numbers works. Similarly, cubing this number is gonna 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 𝑙 zero, 𝑙 one, 𝑙 two, and so on. And they’re rotated in such a way that 𝑙 zero is sitting at the number one on that
right-hand side.

So because the angle that each one of these makes with the horizontal is an integer
multiple of one-seventh 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 𝑙 zero times 𝑥 minus 𝑙 one, on
and on and on, up to 𝑥 minus 𝑙 six. I mean, you plug in any one of these numbers and that product will have to equal
zero. 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 gonna 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 seven here. If you have 𝑛 points evenly spaced around a circle like this, they are the roots of
𝑥 to the 𝑛 minus one equals zero.

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 gonna 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 one 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 happens to be about
1.73.

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 𝜋, which is two times the sine of 𝑓𝜋. 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 one-half. This is the distance between two points on the opposite ends of a unit circle, which
is two. 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 two. 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 zero. The distance-zero 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 one, which
has a root at all of these 𝑛 roots of unity, we’re looking at the polynomial
observer to the 𝑛 minus one divided by observer minus one, which has a root at all
of the roots of unity except for the number one itself.

And a little algebra shows that this fraction is the same thing as one plus observer
plus observer squared, on and on and on, up to observer to the 𝑛 minus one. And so, if you plug in observer equals one, 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 set-up
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 true
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 𝜋.

Take this set-up with 𝑛 lighthouses evenly spaced around a unit circle and imagine
two separate observers, what I’ll call 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 gonna compute this ratio in two separate ways. From the first key fact, we know that the total distance product for the sailor is
two. 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 sailor’s 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 keeper’s
distance product and the sailor’s 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 keeper’s distance product,
meaning just its distance to the keeper, divided by its contribution to these
sailor’s 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-sailor ratio of exactly four-thirds, 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 three, is gonna contribute a fraction that gets closer
and closer to six-fifths as 𝑛 is approaching infinity. Now for this proof, we’re going to wanna consider all the lighthouses on the bottom
of the circle a little bit differently, which is why I’ve enumerated them negative
one, negative two, negative three, 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 two-thirds as 𝑛 approaches
infinity. And then, the second lighthouse before it, 𝑙 negative two here, contributes a ratio
that gets closer and closer to four-fifths. And the third lighthouse, 𝑙 negative three, contributes a fraction closer and closer
to six-sevenths, and so on. Combining this over all of the lighthouses, we get the product two over one times two
over three times four over three times four over five times six over five times six
over seven, 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 keeper’s
distance product to the sailor’s distance product, which we know at every step has
to equal 𝑛 times the distance between the observers divided by two. So what does that value approach as 𝑛 approaches infinity? Well, the distance between the observers is half of one over 𝑛 of a full turn around
the circle. And since this is a unit circle, its total circumference is two 𝜋. So the distance between the observers approaches 𝜋 divided by 𝑛. And therefore, 𝑛 times this distance divided by two approaches 𝜋 divided by
two. So there you have it! Our product, two over one times two over three times four over three times four over
five, on and on and on, must approach 𝜋 divided by two.

This is a truly marvelous result. And it’s known as the Wallis product, named after 17th-century mathematician John
Wallis, who first discovered this fact in a way more convoluted way. And also, 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 indeed 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 infinitary 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 seven, the limit of the products is also
seven. But look at what happens if you take the limits first. If you look at each column, the limit of a given column is gonna be one since at some
point, it’s nothing but ones.

But then, if you’re taking the product of those limits, you’re just taking the
product of a bunch of ones. So you can 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, Sridhar 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, four
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 indeed, we’ll 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 can
get a product that converges to 𝜋 halves. This is something that falls out of the way the dominated convergence justifies us in
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, it would be a shame 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. You 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 sailor’s distance product is now 𝑛 times the
distance between them divided by chord of 𝑓, which approaches 𝑓 times two 𝜋
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 𝑘th lighthouse after the keeper will
contribute a factor of 𝑘 divided by 𝑘 minus 𝑓 to 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 gonna equal 𝑓 times two 𝜋 divided by
chord of 𝑓. Put another way, since chord of 𝑓 is two times the sine of 𝑓𝜋, this product is the
same as 𝑓 times two 𝜋 divided by two times sine of 𝑓𝜋, which is 𝑓𝜋 over sine
of 𝑓𝜋.

Now, rewriting this just a little bit more, what you get is a pretty interesting
fact. Sine of 𝑓 times 𝜋 is equal to 𝑓𝜋 times this really big product, the product of
one 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 Wallis product, which is incredible 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 𝜋 to circles is one thing, but connecting them
to each other is another thing entirely.