(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 couple of videos just talking about this
topic. But for now just note that because 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 Leibniz-Madhava formula I'll tell you
more about the theorem, including 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 think about the theorem 4(good - bad)
doesn't this already feel kind of "Leibnizy". 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) formula. For that let's return to the
radius 7 circle. So what we have to 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 which is a nice simple one. What about
this one? Well if there were no integer 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 Fermat wrote about this theorem in a
letter to the mathematician Marin Mersenne 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, Fröhliche 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 Fröhliche Weihnachten.
Kinda weird that he linked 3blue1brown's basel problem video and not the video covering exactly this series for calculating pi.
Anyway, good video. Liked grant's approach a bit more, but I think this is more accessible to someone that's not familiar with complex numbers.
I just learned that the 2019 point occurs before the Feynman point.
Mathologer gave a talk at the Australian maths society conference three weeks ago (he is a maths prof at Monash uni). One very surprising thing he said was that he does not watch YouTube.
Thats just the taylor series of arctan(1)
How can you formalise this wrt Riemann Rearrangement? I don't see how his steps don't work if you do, e.g., 1 + 1/5 - 1/3 + 1/9 + \cdots
Wow, mind blown in love his style.
Missed an opportunity for "naughty" and "nice" numbers!
What happens if you use a triangular grid instead of a square grid?
How does the 4(good - bad) theorem account for numbers like 21 where the result is negative? It seems an important point to bring up (unless I'm missing something which I may very well be).