You might know this series involving pi, but the usual proof involves essentially reproving
this series for arctangent, and then plugging in x = 1, but this typically involves differentiation
or integration. There are alternative proofs, which both 3Blue1Brown and Mathologer have
covered before, but those involve complicated number theoretic arguments. Is it possible
to avoid both number theory, and differentiation or integration at the same time? The answer
is yes! And it starts with a clock, or rather, a clock with an infinite number of hands.
I know that looks dizzying, but let’s see what the infinite hands are about.
[The setup]
We are going to label the hands by an integer n, which means this hand will complete n revolutions
per hour. For example, this is hand 0. It completes 0 revolutions per hour, i.e. it
doesn’t move at all. What about hand number 1? As said before, it will complete one full
revolution in an hour. For convenience afterwards, we will call this the minute hand, because
the usual minute hand will also revolve around the clock in an hour. What about hand number
-1? It is similar, completing 1 revolution in an hour, but goes anticlockwise, so actually,
hands 1 and -1 are mirror images of each other, which is a useful perspective later on. But
for hand number 2, while it means that it completes 2 revolutions in an hour, I want
you to think of it as going twice as fast as the minute hand, like here: rotating twice
as fast as the yellow minute hand. And again, for hand number -2, it is just the mirror
image of hand number 2, and we can keep going.
When we add more hands together, it looks like this dizzying mess. But we can clean
this up by just focusing on the tip of the hands. So altogether, it looks like this instead.
But with this setup, what are we going to do? We want to focus on the average positions
of the tips over a certain period of time, let’s say from quarter past to quarter to,
so the minute hand initially starts here on the right. And after this half-hour period,
it ends up on the left, but we want to know, *on average*, where is the tip? The tip has
travelled, at a constant speed, the lower half of the clock. Because of the symmetry,
we expect the average position lies somewhere along the vertical line, but because the tip
only travels the lower half of the clock, the average is also somewhere in the lower
half, say around here. We *will* need to figure out exactly where this average position is,
but for the time being, let’s just say the distance to the centre is d, which we are
going to calculate later. [The main argument]
What we have done is calculating the average position of hand 1, during the period of quarter
past to quarter to. But we can do the same for hand -1, and so on, for all the infinite
hands on the clock. Then we add these up. But adding the averages should be the same
as first adding the positions of the tips at each moment in time, then take the average.
Now, whenever we add up an infinite number of things, we have to be very careful, especially
the right side, if we want to be rigorous. More specifically, we will need to use something
called Cesàro sums. However, for the time being, we will treat this more informally
and intuitively first, and then use the framework of Cesàro summation to patch things up.
So first, we compute the left side, and let’s start with hand number 0, which is the easiest
to compute the average position of. Hand 0 is stationary - it always points up! Let’s
just say the clock hand is 1 unit long, then the position of the tip is just 1 unit upwards.
Ok, now onto hand 1 and -1. We have established that for hand 1, it is d units down from the
centre. What about hand -1? Well, it is just the mirror image of hand 1, so the average
position is also its mirror image. In this case, the mirror image of the yellow dot is
still that yellow dot, so altogether, these two hands contribute -2d units, or 2d units
downwards from the centre. Next, we go for hands 2 and -2. For this, just note that we
are considering a half-hour period, which means that hand number 1 will go halfway around
the clock, but because hand number 2 goes twice as fast as hand number 1, it just undergoes
a full revolution within that half hour, and so the average position would be just the
centre of the clock. Of course the same goes for hand -2. So altogether, the two hands
contribute 0 to the sum.
In fact, exactly the same logic goes for any even-numbered hands. During the half-hour
period, the hand 2n will go n full revolutions around the clock, so by the same argument,
all the even hands will contribute 0 to the sum. Now let’s go to hands 3 and -3, where
we will start to see a pattern. In this half hour period, initially, where is hand 3? Well,
the minute hand would be on the right, because from the hour, it travelled a quarter of the
way around the clock at quarter past. So for hand number 3, travelling 3 times as fast,
should go around 3 quarters of the clock to end up on the left, so in the half-hour period,
hand 3 *starts* on the left. So during the half-hour period, starting on the left, it
goes through 3 half revolutions around the clock. (wait 4 seconds) But where should be
the average position?
Well, the first half revolution has the average being d units upwards from the centre, then
the second half revolution contributes d units down from the centre, and the final half revolution
contributes d units upwards, so altogether, we just take the average of these three, and
get d/3 unit upwards. And by being the mirror image, hand -3 would also have this average
position. So going back to our sum, we have the total for hands 3 and -3 to be 2d/3 unit
upwards. In fact, all the hands afterwards would fall into either of these two categories.
In general, for any odd-numbered hands n, the hand goes through an odd number of half
revolutions, and we just need to take care of whether the unpaired half is the lower
half like here, or the upper half. This means the average position is going to be either
plus or minus d / n, where the sign just depends on whether you got the upper half as the unpaired
half, or the lower half.
Note that the first half revolution is always going to be the extra one, so the sign actually
depends on where the hand starts. If you are like here, starting on the *right* ending
on the left, the sign is negative, because you are travelling the lower half. If, however,
you start on the left ending on the right, the sign would be positive, so starting left
means positive, and starting right means negative. Alright, so in general, for hand 4k+1, *at
quarter past*, it should have gone 4k+1 quarters around the clock, which means k full revolutions
and an extra quarter, so at quarter past, the hand is on the right, and so this hand
will contribute negatively to the sum. As for hand 4k+3, similarly, at quarter past,
it has gone through 4k+3 quarters of the clock, which means k full revolutions and 3 extra
quarters, i.e. starting on the left side. And this means it contributes positively to
the sum.
Ok, so let’s go back to the infinite sum we have. As said before, we can cancel the
even-numbered hands, except for hand 0, so we ignore these, and then the contribution
from hands 5 and -5 would be altogether -2d/5, because this hand is of the form 4k+1, so
we have a negative contribution, and we have both hand 5, and its mirror image, explaining
the factor of 2 in front. Exactly the same logic shows the next term would be +2d/7,
and so on. So altogether, tidying a little bit, the sum looks like this, and there we
go, we have the series. Now, there are just two more things to do: what is d? And what
does this whole thing equal to? Let’s first tackle what d is. As a reminder, d is the
average position of the lower half of the clock.
At any point in time, the velocity of the tip is always perpendicular to the position
vector of the tip. The position vector is 1 unit long, and for the minute hand, the
velocity is 1 revolution per hour, but the circumference of the clock is 2 pi, and so
we say it is 2 pi units per hour. And so, to obtain the position vector, we can just
use the velocity vector. More precisely, we can rotate it anticlockwise by a right angle,
then divide the value by 2 pi, to get the position vector, because while the velocity
is 2 pi, the position vector is only 1 unit long. But the same thing happens for the average
position.
We can do these to the average velocity vector to get the average position vector. And calculating
the average velocity of the tip is not difficult: the tip has displaced 2 units to the left,
doing so in half an hour, which means that the average velocity is 4 units per hour to
the left. Next, we need to rotate this average velocity anticlockwise by a right angle. Then
divide by 2 pi, which in this case is 4 divided by 2 pi units downwards, or in other words,
the d that we are looking for is 2/pi. So we plug this value of d back into our sum.
In this case, 2d is 4/pi, and we finally have this complete sum. The really final thing
is to see what this series equals. From the very beginning, we had this plan, and what
we have done is that the left side can be represented by this series, and we now need
to focus on the right side, i.e. summing the hands at each moment, then take the average.
Let’s say this is how the tips are distributed at a moment in time. It seems random, but
it has some pattern in it. If the minute hand has travelled this far around the clock, say
the angular displacement is theta, then hand number 2, travelling twice as fast, will travel
exactly one more theta. And obviously the same goes for hand number 3: the same rotation
could get us from hand 2 to hand 3. It might be also worth noting that if we start from
hand -1, then applying the same rotation would also get us to hand 0. What this means is
that this rotation sends hand -1 to hand 0, hand 0 to hand 1 and so on, kind of like the
Hilbert hotel. So each of these positions will still be filled after a rotation by theta.
Like here, we almost have a complete match, except for these two places, but if we have
an infinite number of points, the before and after would be indistinguishable.
So *if* the sum of the positions is like this white dot, because a rotation by theta gives
exactly the same picture, the sum would need to still be at the original position. The
only resolution to this discrepancy is if the sum was originally at the centre, which
wouldn’t be changed by a rotation. So because the sum would always be at the centre, i.e.
0 in this half-hour period, the average necessarily has to be 0. So finally, we see that this
series needs to equal to 0, which implies that this series has to be pi/4. This proof,
I think, is arguably the simplest because it doesn’t require any knowledge of calculus,
or number theory - but we cheated a bit here, especially when it comes to the right side
of the equation.
[Making this rigorous]
This is only going to be a bit of a taster of how we can make this argument rigorous.
Normally, when we sum a series, what we do is to sum the first term, the first two terms,
first three terms and so on. These numbers we obtained by summing the first few terms
are called the partial sums of the series. In this case, we can plot the partial sums,
and while the graph is oscillating, it oscillates less and less, and we would expect that it
converges to a certain value, which in this case, turns out to be pi/4. This is what we
mean when we write down the sum equals pi/4. But this is not what we mean when we say the
sum of the positions is 0. This is the partial sums graph, which doesn’t converge to anything.
And this phenomenon occurs at all times on the clock! Look - none of these partial sums
graphs actually converge, so why do we say the sum is 0? Well, if you are forced to make
up a value to this series, what would you pick? It makes sense to pick 0 in this case,
because it is like the “middle ground” of all these partial sums. This middle ground
is called the Cesàro sum of the series.
More precisely, we don’t just consider the partial sums themselves, but rather, we take
the average of the first few partial sums, and then see whether the averages themselves
converge or not. If it does, it will converge to that “middle ground”. But in this series,
because the partial sums already converge to pi/4, this will also be the “middle ground”,
so pi/4 here acts as both the regular sum, and the Cesàro sum of the series. But in
summing the positions of the tips, if we actually take the average of the first few partial
sums, we will instead get this graph, where we can see directly that the average does
go to 0. This is what we really mean when we say the sum is 0 - the Cesàro sum is 0.
So hopefully, this is a more complete justification of how this argument really works. And if
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