Fermat's Christmas theorem: Visualising the hidden circle in pi/4 = 1-1/3+1/5-1/7+...

(Thanks to Karl for the 2019 Easter egg idea :) Welcome to the 2019 Mathologer Christmas video. In this video we'll investigate
that famous and amazing formula over there PI over 4 is equal to 1 minus 1/3
plus 1/5 minus 1/7 and so on. It's usually called a Leibniz formula after
Gottfried Wilhelm Leibniz one of the genius inventors of calculus. Sadly, like
many other results in mathematics, the formula was not discovered by the
mathematician it's named after, at least not first of all. In this case, Leibniz's
formula was first discovered by the indian mathematician Madhava of
Sangamagrama in the 14th century, more than 200 years before Leibniz. Anyway
this formula is definitely very beautiful. At the same time it's very
mysterious. Think about it, pi is of course a circle thing to do with
conferences and diameters and stuff. On the other hand, our formula is stitched
together from the odd numbers, without a circle in sight anywhere.
However, and hardly anybody knows this, when you look hard enough you can find a
huge circle hiding within this iconic formula. The first time I stumble across
this wonderful connection was over 40 years ago in a book by mathematical
megastar David Hilbert and his colleague Stefan Cohn-Vossen. This book "Anschauliche Geometrie" (German)
or "Geometry and the imagination" in English is a popular
account of modern geometry. If you're not familiar with this book, definitely check
it out. An absolute must-read. In their book Hilbert and Cohn-Vossen show how
the Leibniz Madhava formula follows from the area formula of the circle. And the
key to the ingenious argument is a result known as Fermat's Christmas theorem.
What a great hook for a Christmas video, don't you think? Now before we get into
the details there's a feature of our formula that will be very important and
that I'd like you to keep in the back of your mind. In the Leibniz-Madhava
formula the denominators are one, three, five, seven, etc. that's just the odd
numbers. We can think of these numbers as being split into two classes
corresponding to the negative and the positive terms of the series. In the
following I'll call the green numbers one, five, nine etc. "good" and the
remaining odd numbers three, seven, eleven etc. "bad". You'll see why later. Okay, let's
start with the xy-plane, highlight all the points with integer coordinates to
make a lattice and draw a circle centered at the origin.
Now count the lattice points within the circle. That number will be approximately
the area of our circle. Why? Because each point is the center of a little unit
square and then the total area of the circle is approximately the sum of the
areas of those squares. So pi r squared the area of the circle is approximately
equal to N(r), the number of those lattice points. Does this look familiar?
Most of us would have done something like this in primary school: draw a
squiggly loop on grid paper and estimate the area within the loop by counting
the number of squares inside. Back to the circle on our grid, solving for pi
gives an approximation to our favorite number, there. In the example here the
radius is 7 and the number of points is 149 and 7 squared is 49. So we have pi is
approximately 149 divided by 49 which is equal to three point zero four zero eight
Well not a great approximation but not bad either. At least the leading
three is right there Okay can we do better? I can see you
nodding and yawning and so let's get on with it and zoom out. Now, choosing a
larger circle makes the blue area more circulars and then also results in a
better approximation of pi. Go again ... even better. Now pushing the radius r
to 1,000 gives an approximation correct to those first four famous decimals:
3.1415. And pushing r all the way out to infinity the approximate sign turns
into an equal sign. And that's a challenge for you: find a short proof
that we get equality in the limit. As always give you ideas in the comments.
What comes next? Well we're supposed to be heading for
the Leibniz Madhava formula. So if our circle lattice games are going to help
this has to happen by finding some other way to calculate the numbers N(r).
Okay, let's focus on one of the lattice points that one there yeah. The
coordinates of the lattice points are integers and a distance of the point
from the origin is less than or equal to the radius of the circle, right? And with
the Pythagoras that is staring at us in the diagram we can summarize all this
information like this: five squared plus three squared equal to d squared which
is less or equal to r squared. And in the case of this lattice point five squared
plus three squared equals 34 and of course r squared equals 49. This means
that one way to count the number of points in the circle is to do this: first
we list all the integers from 0 to 49 then for every number in our list we
figure out all the different ways to write this number as a sum of two
integer squares. Finally the total number of all these
different ways is the number we're after, the number of blue lattice points. To get
a feel for how this works, let's figure out the different ways to
write the first few numbers in our list as sums of two integer squares. Okay
0 is first. How many ways are there to write 0 as a sum of two integer squares?
Hmm well of course there's just one such way.
This equation corresponds to the origin the point with coordinates (0, 0). What
about 1? Well there's obviously just these four different ways, corresponding
to four of the lattice points. Next the number 2 can also be written as four
different sums, corresponding to four points. What about 3 how many ways?
Hmmm, actually, none! And then 4. There are also four ways again. For 5 there are
eight different ways, and so on. Figuring out the ways of writing integers as the
sums of integers squares has a long long history and I could actually spend a
of the symmetry inherent in the diagram the number of ways of writing our
integers as a sum of two integer squares is always a multiple of 4: zero ways
4 ways, 8 ways, 12 ways, and so on. The one exception is 0
corresponding to the point in the middle of the diagram, which can be written in
only one way. Now remember the fact that I asked you to keep in the back of your
mind? Remember my way of splitting up the odd numbers into the good ones and the
bad ones? Well that was to prepare you for a
stunningly beautiful theorem. This theorem expresses the number of ways of
writing a positive integer as a sum of two integer squares in terms of ... the
good and the bad odd factors of that integer. For the moment I'll just
introduce and apply this theorem later after we've successfully chased down the
Christmas connection. Okay I'll tell you what the theorem says using the number
18 as an example. The odd factors of 18 are 1, 3 and 9, as you can see up there. 1
and 9 are good and 3 is bad. Now you simply go number of good ones minus
the number of bad ones and then you times the resulting number by 4. Then
this magical theorem says that the number you get this way is the number of
ways to write 18 as the sum of two integer squares. How pretty is that?
So for 18 we have two good factors and one bad factor, so 2 minus 1 that's 1,
times 4 that's 4. So there are exactly four different ways to write 18 as the
sum of two integer squares. Three challenges for you: first what are the 4 ways to
write 18 as a sum of two integer squares. Second how many different ways are there
to write the number 2020 as the sum of integer squares. Third find all those
ways of writing 2020. Anyway what a stunningly slick and beautiful theorem,
don't you agree? A real shame that so few people ever get
to learn about it. Hopefully that will change because of this video. But now
There's a telltale 4 at the top and at the bottom and the bad odd
numbers get subtracted from the good ones in both expressions. The plot is
definitely thickening. So what comes next? Well you've probably already guessed it.
We'll now calculate the number of lattice points using our 4(good - bad)
do is to calculus 4(good - bad) for each integer from 1 to 49, sum all
the numbers we get and then a final plus 1 for the point at the origin. Can we do
this in a systematic manner? Yep, easy peasy :) But to be able to isolate and
really appreciate a trick that will give us our mysterious pi formula, let's be
super systematic. Have a look at these good odds and bad odds vector tables.
The numbers from 1 to 49 are at the top, the good odd numbers are listed here and
the bad odd numbers are listed below. And the dots indicate who is a factor of who.
For example, this dot here indicates that the good 5 is a factor of 10. Pretty
straightforward, right? And now we just tally up. Start with 1 up there. The
number of green dots here minus the number of orange dots here, so that's 1
minus 0 which is 1. For 2 we get again 1 minus 0 equals 1. And for 3 we get 1
minus 1 so 0. And now continue all the way up to 49, adding up all those
differences gives 37. Then multiplying by 4 gives 148 plus 1 for a grand total of
149. And 149 is the number of lattice points we found earlier. Ok, now a
slightly more efficient way of calculating that critical number 37 is
to just first adding up all the green dots and then all the orange dots and
then subtracting orange from green. But now comes the trick. It turns out to be
much, much easier to calculate the green and orange totals by tallying row by row
instead of column by column as we've done so far. Those of you who've made it
through our recent monster Euler-Maclaurin video may remember that we
used a similar trick there.
Let's see how this works. How many green dots are in the first row. Well, shall
I make it a challenge? Obviously, 1 is a factor of every positive integer and so
there are 49 dots in the first row. Now our second good number is 5. How many
dots in that row? Well those dots are equally spaced which is nice, right,
corresponding to all the multiples of five below 49. So how many are there? Well
that's simply 49 divided by 5, rounded down. That's called the integer part of 49
divided by 5 and we denote it with square brackets around a fraction. For the next
row we get 49 divided by the next good number so the integer part of 49 divided
by 9, and so on. Time to wrap up the proof.
So the number of lattice points N(7) is four times the total number of
green dots minus the total number of orange dots plus one and with our new
way of tallying the goods and the bads this equation looks like this. But anyway
to make the bit in the brackets look more "Leibnizy", let's alternate the
positive and negative terms. Alright, now remember for this specific example
49 is just a square of the radius which is 7 and so the general
formula looks like this. Now, can you see it coming? I sure hope you do. Remember
how we got our approximation for pi. We simply divided N(r) by r-squared. So
let's divide both sides of the equation above by r-squared.
Now zoom are off to infinity and let's see what happens to all the fractions.
Okay we already know that the fraction on the left will exactly zoom to pi. What
about on the right? Well the first fraction is r squared over r squared
part brackets there then again our square on
top and the one at the bottom would cancel leaving us with 1/3. And, in
fact, as r zooms off to infinity, the limit of this fraction is 1/3. You can
fill in the details in the comments, which shouldn't be too hard of a
challenge. Next fraction. Well if the previous fraction zooms to 1/3 then this
one zooms to ... well what? 1/5 of course. And so on. And that very last
fraction? Well, of course as r gets huge that fraction just saps to zero. And the
final tweak, just divided by 4 and we're done. Tada
and it's my Christmas present for you. Like it? So now you see, the Leibniz-
Madhava formula really is a circle thing. It just comes from the formula for the
area of a circle. Pretty amazing isn't it? Now to make the zooming bit of our proof
completely bulletproof we actually have to worry a little bit more about some
details that I glossed over. For the experts among you think Riemann
rearrangement theorem and how exactly the series we're dealing with here grows
as the radius tends to infinity. Not hard at all,
just a little bit fiddly. Anyway if you're interested in these details, i'll
link to the relevant pages from Hilbert and Cohn-Vossens beautiful book in
the description. Well we're not quite done yet. Of course I still owe you some
details of the 4(good - bad) theorem and I have to explain the Christmas
connection. Right?
Let's see what our 4(good - bad) theorem says about a prime number like
17. Well a prime number has only two factors, the good 1 and the prime
itself. So what if the prime is itself good, like in the case of 17? Then we
have good - bad equals 2 and so our theorem guarantees that every good prime
can be written as a sum of two integer squares. Right? On the other hand, if the
prime is bad like 11 and then good - bad will equal zero. So our theorem says that
bad primes cannot be written as a sum of two integer squares. And that's known
as Fermat's Christmas theorem: good primes can be written as sums of integer squares
and bad primes cannot. That's also why I labeled the odd numbers good and bad
earlier on. Now the Christmas in the name of this theorem is standard, although the
connection with Christmas is pretty flimsy. It solely derives from the fact that
on Christmas Day in the year 1640. Still if you are a desperate Mathologer, like
me, looking for a Christmas hook, you take what you can get. And there's also twist
to the Christmas hook. Yes you guessed it Fermat's Christmas theorem is not Fermat's.
The theorem was actually first stated by the mathematician Albert Girard 15 years
earlier. And that's a picture of Girard there. Well actually it's not Girard,
it's the cartographer Jodocus Hondius which is what Google spits out when you
ask for Girard. In fact Google choosing some designated replacement when it can't
find the correct portrait seems to be just as common as theorems being named
after the wrong person and sadly, as for Madhava, no picture for Girard seems to
have survived. Anyway, neither Fermat nor Girard provided a proof of the theorem and
the first to publish one was Euler. Well it's always Euler, isn't it? Actually while
we're delaying proving the Christmas theorem it's worth mentioning another
reason why the theorem is now so famous. In 1990 the mathematician Don Zagier came
up with an absolutely incredible one-sentence proof of the Christmas
theorem. There it is, but good luck with that sentence. Figure it out and you're probably
ready to begin a PhD in math(s). Now, historically, the Christmas theorem
preceded our 4(good - bad) theorem. The 4(good - bad) is known as Jacobi's two
square theorem and, wonder of wonders, appears to actually have been first
proved by Carl Jacoi and, yes, that's Jacobi there. And now we'll prove the
Christmas theorem and Jacobi's theorem? No, I'm sorry, definitely not today. To
mathologerise these theorems and to make them truly accessible is very
tricky and is still work in progress. But if I can't give you the proofs yet
I'd like to finish today by at least mentioning a couple of easy and
beautiful ideas that will give you a feel for where these theorems come from.
The first thing to note is that the bad half of Fermat's Christmas theorem is
really, really easy. In fact, it's easy to show that not only the bad primes but in
fact all bad odd numbers cannot be written as the sum of two integer
squares. None of these guys up there can be written as the sum of two integer
squares. Since it's so easy to prove, let's do it. First, notice that every bad
odd number is of the form 4k+3 right 4 times 0 plus 3 is 3, 4 times 1
plus 3 is 7, and so on. On the other hand, the good odd numbers are of the form
4k+1. In other words, the good odd numbers are the integers that leave a
remainder of 1 when you divide them by 4 and the bad odd numbers leave a
remainder of 3. What other remainders are there on
division by 4? Well, of course 0 and 2 corresponding to even numbers. So every
integer is of one of these four types. Now let's see what types we get when we
square integers. Obviously the square of a type zero number
gives a type zero back again. And it's easy to see that squaring a type one
number also gives back a type one. Just expand, right? See the pattern?
So, now squaring a type two numbers gives back,... no not a type two :) Did I trick you?
Actually, squaring a type two number gives back a type zero, as you can also
easily check by expanding. And, finally, squaring a type three number gives back
a type one. So, in summary, an integer squared either gives type zero or type
one but then what are the possible types of a number that is a sum of two squares?
Well, effectively, you're adding a couple of zeros or ones, so the sum of two
squares might be of type 0, 1, or two, but there's no way to get to that 3.
In other words, no bad or integer can possibly written as a sum of two integer
squares. And that's the easy half of the famous Christmas theorem. Pretty easy,
right? What else is there easy to say about the proofs of our two theorems. Well,
not a lot, but one aspect worth highlighting is the identity up there.
That identity which was already known to the ancient Greeks is the glue that
holds the two theorems together. What this identity tells us is that if
we have two integers that are both the sum of two integer squares, then their
product is also a sum of two integer squares. You get the sense of how this
might work? Since all positive integers are products
of primes once we know exactly how the primes can
be written as sums of squares, there's some hope that this identity will allow
us to extend the prime number results to all integers and actually also count the
number of ways to represent integers as sums of two squares which is what
Jacobi's theorem is about. And this is indeed what happens. Of course, as usual the
devil is very much in the details. But I'll leave those devilish details
for a time in the hopefully not too distant future. The big devil killer that
I will want to use is the law of quadratic reciprocity. Some of you will
be aware of what a challenge that will be to mathologerise. Okay, and that's just
about it for today. Just one more thing. If you liked what I did today there's
also a really nice video by 3blue1brown in which he animates a circle
based proof of the famous solution of the Basel problem, that infinite pi
series over there. And while you're there maybe also check out Euler's original
solution which I cover in the video at the bottom. Okay and that's really all
for today and all for this year. See you in the new year, FrÃ¶hliche Weihnachten.
Actually, actually, one more final final thing promise. We recently hit 500 000
subscribers which i think is pretty amazing for a hardcore mathematics
channel. Anyway I think it's pretty cool and I would
like to thank you all for your interest and your support over the years.
I'm not at all money minded and so I've always avoided even thinking about
monetizing these videos. However, maybe next year is a good time to take
Mathologer to the next level and hire someone to assist with editing the
videos, preparing subtitles, etc. In preparation for this, I recently
monetized the videos by switching on the least annoying ads on YouTube
I also just put up a Patreon page. If you enjoy these videos and you can afford it
please consider taking out one of the memberships or making a one-time
donation via PayPal the links are in the description of the video.
And now once again, for real, bye for now and FrÃ¶hliche Weihnachten.