Welcome to another Mathologer video :) Most
of you will have heard of mathematical superhero Carl Friedrich Gauss and many
of you will be familiar with a great story of Gauss at the schoolboy. Gauss's
teacher, so the story goes, had asked the students in his class to add up all
integers from 1 to 100. So, a tedious task to keep the little monsters busy. But
Gauss surprised everybody by blurting out the answer pretty much immediately.
And in doing so, Gauss announced to the world that a great thinker was emerging.
The story featured prominently in a recent German movie Die Vermessung der Welt - measuring the world. That's a screenshot over there, check it
out. Tt's a great story and it's even sort of true. But how did Gauss do it?
Well, although hundreds of retellings of the story, including the movie version,
suggest one particular method, it turns out that nobody knows for certain what
Gauss did. In fact there's a number of details of the standard story that are
probably fictional embellishments. Check out the fascinating story behind the
story in the article by Brian Hayes linked in from the description of this
video, great stuff. Gauss's story also has a sequel. Well, I guess a prequel since
it happened before Gauss was born. The prequel involves another famous
mathematician Jakob Bernoulli. In his book Ars Conjectandi Jacob Bernoulli boasted
that he was able to find the exact sum of 1 to the 10th power plus 2 to the
10th plus 3 to the 10th all the way up to 1000 to the 10th power and he was able
to sum all this in just 7 1/2 minutes. Think you could do it? Give it a
try, with or without a calculator. And let us know in the comments how you go.
And if, incredibly, you somehow finish and you want to check, here's the answer. What
a monster and 7 1/2 minutes, by hand !? And, by the way, notice the remarkable
symmetry present in this number. What's the explanation for that. Again,
let us know in the comments if you come up with anything worth sharing with the
world. Two great stories about our two great mathematicians, both in
possession of some magic trick to simplify a painful sum. But Gauss and
Bernoulli are not alone in this. The quest to find tricks and formulas for
sums of powers of integers has been going on for millennia and is still
ongoing. There have been contributions by many
great mathematicians including Archimedes, Euler, Pascal, Bernoulli,
Ramanujan, Riemann and many more. What I'd like to do today is to take you on
another magical Mathological journey. With nicely animated derivations I'll
show you how to discover many spectacular power sum formulas and
computational tricks. My ultimate goal in all this is to chase
down the famous Bernoulli numbers and the infamous Euler-Maclaurin summation
formula. The Bernoulli numbers are a very strange sequence that is at the heart of
ALL the integer power sums. Yes, Bernoulli numbers even lead to that
notorious - 1/12 weirdness. The Euler- Maclaurin summation formula is a
mathematical mega weapon. The formula allows us to go way beyond sums of
powers to magically calculate very general finite sums. One of the most
important, beautiful and applicable formulas in mathematics and a formula
that only the experts tend to understand. Until today,
just wait! :) As you've probably already guessed, this video is another
challenging Mathologer master class. Anyway, the usual rules apply. The video
is split up into chapters, eight chapters in total today. We'll start with some really
easy stuff and then proceed to ever more crazy and deep stuff. And a warning: today
the crazy and deep stuff is even crazier and deeper than usual. But, as usual, give
it your best shot, follow as long as you can and it's fun
and don't worry if you don't get every detail in the first viewing. We'll start with an easy one. So how did
Gauss sum the numbers from 1 to 100. Well as I confessed we don't really know but
there are a few things Gauss might have done. There are a few different ways to
see the answer pretty much at a glance. Now to try to keep everybody entertained
including you know-it-alls who are very familiar with this sum, here is a nice and
natural method that may be new to you. Let's first add 1 plus 2 plus all the
way up to 10. We can represent the sum as a triangle like this 1 plus 2, 3, 4, 5, 6, 7,
8, 9 and 10. So the question is how many little unit squares are in this triangle,
or, what is the same, what's the area of the triangle. And that's easy, right? For
probably the millionth time in our life we go "base times height divided by 2". So
10 squared that's 100, divided by 2 that's 50 and that's our sum, right? Nope,
close but no banana :) Why, what went wrong? Of course, the problem is that our step
triangle is not truly ruly a triangle. That 50 we calculate is actually the
area of the genuine triangle here. We failed to account for these 10 half
squares whose area is, well, 10 halves which is 5. So the answer to our question,
the sum we are after is 50 plus 5 which equals 55. And what about the sum from 1 to 100?
Of course, it's the same thing, just a bigger triangle with one more zero
everywhere. 100 squared that's 10,000, half that is 5,000, half of 100 is 50 and
so our answer is 5050. And however he got there that would also have been Gauss's
answer. Applause for one very smart little kid. It's now also easy to see
that in general if we sum from 1 to any integer n we get this formula here:
n squared over 2 plus n divided by 2. Very pretty! Just simplifying a bit more,
this formula can also be written like this: n (n + 1)/2.
I suspect many of you have seen this formula before. Anyway, we're just warming
up, right. Okay, what about the sum of the squares? Can we do something similar? Well, just like the sum of the integers corresponds to a stepped 2d triangle, the
sum of the squares corresponds to a step 3d pyramid: 1 squared plus 2 squared plus
3 squared all the way up to n squared. And here is our complete Mayan style
pyramid. Our new sum, the number of unit cubes in the step pyramid is
approximated by the volume of this proper square pyramid, there. Then, dusting
off the mental cobwebs, we remember that the formula for the volume of a pyramid
is 1/3 the area of the base times the height. Then the area of the base, that's
N squared, and the height is also n so the volume of our pyramid turns out to
be n cubed over 3. Again, this number is not exactly the volume we want but it's
a good approximation. Can we now adjust this volume to obtain the exact formula
we're after just like before? Yes, but it's not very Mathological, it's just not
much fun. The method of correction and the resulting solution aren't so pretty
and, as far as elegant and exact solutions go, this step approach is
pretty much at a dead end. Pretty much but not quite. Before giving up
completely, aren't you tempted to take a peek into higher dimensions? First 2d
triangles then 3d pyramids and so what's next? If there's a Mayan god, the sum of
the cubes should be a step 4-dimensional pyramid
with a 3d cubic base of side length n and also a height of n and this stepped 4d pyramid is approximated by a normal :) 4d pyramid of the same basic shape.
And then the volume of this normal 4d pyramid is n to the power of 4 divided
by 4 and you can guess the rest, right? There is indeed a Mayan God and the
pattern is clear, Whatever the exponent appearing in our
sum, that number plus 1 will appear in our approximate formula, here and here. And so
the general formula works out to be something like this. A very pretty and handy approximate
formula and very easy to remember. And does this formulas seem kind of familiar,
does it remind you of something? Well, if you know some calculus it really should.
Remember that one? Quick and dirty what this tells us is that the discrete sum
at the top is approximated very well by its continuous counterpart, an integral.
Interesting, isn't it? It's the harbinger of a very deep and beautiful truth,
it's our first hint of the promised Euler-Maclaurin formula, the culmination of
our video, just wait. I'll now get going properly by showing
you very pretty and animated derivations of the exact formulas for these three
sums. As before, we'll picture these sums in terms of physical objects: squares,
cubes, stepped triangles and stepped pyramids. Okay, here we've got the stepped
triangle again standing for one plus two plus three plus four. There we get second
one, so now we've got two of them and the steps go away, which is really really
good!! :) Now the area of the rectangle is 4 times 4+1. Okay, now if you
had used things up to 5, things would change like this. If you added up to 6,
things would change like this. And, in general, things would look like that. And
now we just have to divide by 2 and that's it! Great stuff. We're not quite finished but
let's pause and admire this amazing identity. The some of the cubes equals
the square of the sum of the integers. How pretty is that! And so, for example, 1
cubed plus 2 cubed plus 3 cubed is equal to 1 plus 2 plus 3 squared. Or, 1 cubed
plus 2 cubed plus 3 cubed plus 4 cubed is 1 plus 2 Plus 3 plus 4 squared, and so
on. And, finally, to obtain a properly compact formula for the sum of those
cubes all we have to do is to plug our formula for the sum of the integers in
this. And so we get this. I'm sure you agree it's all super pretty stuff and if
you'd like to see more of the same head over to Think Twice a very nice
youtube channel dedicated to short animated proofs. What's next? Well, the
next sum of course. But before that let's fill in a hidden gap and ask what
comes before. Weird question, Well, 1 plus 2 plus 3 and so on
is the sum of the first powers. So what about the sum of the zeroth powers? Easy,
each of these integers to the power of 0 is of course 1 and there are n of these,
so the sum totals to n. Not earth-shattering or rocket science but
it's cute and it will help us later on. And now we can get on with what comes
next. What is the sum of the fourth powers? Well, can we spot any patterns
here? Well all the expressions on the right contain the factors n and n+1
and that will turn out to be true for all the higher power formulas as well.
Hmm what else? Well let's expand all the formulas. Ok
the terms of highest degree are exactly what we expected from our previous pyramid
games. And so we expect to also see this in our
next sum, right? What else? Well, those one-half coefficients of the
next highest term are clamouring for attention. It's a pretty good guess that
this will also continue. Can you see anything else? No? No, neither can I :) The pattern guessing is pretty much
stuck at this point, at least for now. We could now go back to our building blocks
to try to produce higher-dimensional animated derivations. But how are you
with 4d blocks? it's actually a lot of fun and well worth exploring and I
actually went wild with 4d blocks a couple of years ago :) But to do this now
I'd have to introduce it to visualizing four dimensions and then there's the
fifth dimension, and so on. Fascinating stuff, but too far afield for us today.
So, instead, we'll do what modern mathematicians do, we'll give up on the
fun but crazy geometry and we look for a purely algebraic way of proceeding.
Alright it's algebra time but first let's give
our sums short names, so that we can fit more of me on the screen. That's a plus,
right? What shall we call the first sum? Let's see, let's go for something super
original like ... S1 :) Then the second sum is S2, and so on. And, remember, there was also
that zeroth sum which of course we'll call, yeah you got it,
S0. Now here's the trick. Remember that the sum of cubes S3 could be expressed
in terms of the sum of the integers S1. There, that one. Similarly, it turns out
that we can express any of our sums in terms of the lower sums which means we
can bootstrap our way up to get any sum we like. Now how do we link to sums? What
we need is an identity that connects different powers. And what comes to mind,
what famous formulas relate different powers?
Well, the binomial identities of course. But after some trial and error it
turns out that instead of the pluses here using minuses is the way to go.
You'll see in a second why. Let me now show you how you can use the fifth power
identity at the bottom to derive the sum of fourth powers S4. That seems a little
weird. Why look at a fifths power formula if you're interested in the sum of
fourth powers. But wait and watch. Those fifth powers will vanish with a really
really cool trick. Okay, we'll begin by moving those weird fifth powers to the
right and everything else to the left. Substitute 1 for x. Ok now also
substitute 2, 3 and 4 for x. Evaluate the simple subtractions on the right. That's
0, that's 1, that's 2 ,and that's, yep, 3. Still a jumble of coefficients and
powers but here's the trick. We'll add all these equations. First on the right,
and there you can see it all telescopes and cancels except for one number. So we
sum everything on the right. Ok, so these two guys cancel out, these two guys
cancel out, these two guys cancel out and the only thing that's left over when we
add up everything there is 4 to the power of 5. Now there's nothing special
about stopping at the fourth equation so let's make it general by summing n
equations. Looks like this. Now what about summing the left sides? Well, let's do
this column by column. Adding the terms in this column we get 5 times 1 to the
power 4 plus 2 to the power 4 plus 3 to the power 4, all the way to n to the
power 4, and of course that is 5 times S4. Adding up the second column we get
-10 times S3, and so on. Just about there. Remember, we're chasing
the formula for S4 but we already know the formulas for everything else here. So
we just have to solve for S4 for plug in the formulas for S0 and S1, and so on,
simplify a little, and we'll be done. There we go, Fantastic!! Great stuff, isn't it? Now, if
you feel bored, by all means double check my calculation. And what then about a
formula for S5. Well, we just rinse and repeat. The sixth power binomial identity
will generate a formula for S5. Then we can generate the formula for S6, and so
on, forever :) It's all easy peasy, even if it's a slow step by step process. But
there's also a super magical way to recast this generating process in a way
that gives all the formulas in one go. It also makes a supercool connection
with another mathematical all-time favorite, Pascal's triangle. Ready for
that, too? Well ready or not, I don't care, let's go :) First, just a quick recap to make sure we
are all up to speed. So what we've just done is to begin with all the summation
formulas up to a certain point and we want to derive the next summation
formula in line. To do this we translate a binomial identity which connects the
powers of x into that other identity that connects our sums. Of course we
don't do this just once but over and over to generate all our formulas.
This means we translate all the binomial identities into identities that connect
our sums. Like that. Now, starting from the beginning, we can again get S0 from the
first equation. There we go S0 is equal to n. Yes, works! Then we get S1 by
subbing this into the second equation yes sub in and solve, right? Yep, works
too. Now we sub the formulas for S1 and S0 into the third equation and then
solve for S2, and so on. Looked at from another angle what we are doing here is
solving a system of five linear equations in the five unknowns S0 to S4. However, there are lots of different ways to solve a system of linear
equations. One of the niftiest one is to interpret the system as one matrix
equation. If you've never heard of this just run with it as best as you can. It's
pretty visual so it shouldn't be a problem. Ok here we go.
Here is this matrix equation. Now solve using the inverse of the
square matrix in the middle. The numbers you see there are exactly the
coefficients in our previous formulas and we now get the formulas simply by
performing the multiplication on the right side. Magic, the first five
summation formulas just like that and of course by using more of our
binomial identities to start with we can generate as many of these formulas as we
wish. Magic, definitely, but this very nifty
method of generating all our summation formulas gets even more appealing and
unforgettable once you've looked at it in terms of Pascal's triangle that is an
integral part of our binomial identities of course. Here we go. Suppose you're
interested in generating our first five sum formulas. Then you begin with the
first 5 plus 1 equals 6 rows of Pascal's triangle, there. We all know that, right?
Strip off the row of ones on the right, rearrange into a matrix, decorate every
second diagonal with minus signs and calculate the inverse matrix and the
numbers you obtain are exactly the coefficients in our sum formulas. So, in a
way, our summation formulas are some sort of weird inverse of Pascal's triangle.
I'm excited, are you? And if you want more formulas you just start with a larger
tip of Pascal's triangle. Well, anyway, I think we can all agree that this is
super cool and super easy to remember. Okay now that we know how to generate
any number of these formulas, let's get back to pattern spotting. Over there are
the first eight formulas arranged by descending powers of n. We already
spotted the reciprocals of the integers in the first column, there those, and we
decided that the second column is most likely to be all one halves. What about
the third column? That looks like a complete mess. So let's focus in a bit
more, there. Hmm, well that still looks pretty random but two entries that stick
out are, at least for me, are 5/12 next to the 5th sum and 7/12 next to the 7th
sum. That suggests that the number next to the kth sum could be k/12. Let's see whether this works. Two divided by twelve that's one sixth.
Three divided by twelve that's one forth. Yeah all works out and,
in fact, you can keep on going and it always works out, it's great. So looks
like we've got our pattern here. Okay so we've discovered simple patterns in the
first, second, third columns. Fourth column? Hmm, well, I'll leave it to you to figure
that one out :) And the same for every second column from then on. Looks like
we're getting there. But the master pattern spotter, the first person to
really see the pattern was our friend Jakob Bernoulli, the seven and a half
minute guy. After a lot of playing with the first 15 or so formulas Bernoulli
discovered something amazing. Bernoulli noticed that the numbers in the
different columns always appear to depend in a simple way on the numbers at
the tops of these columns. Now this sequence of numbers 1, 1/2, 1/6, 0,
-1/30, 0, and so on, a named, naturally enough, the Bernoulli numbers. And
subsequently this strange sequence was recognized to be one of the most
important sequences in mathematics. But hardly anybody knows about it, outside
mathematics. Okay here then is the amazing pattern that Bernoulli
discovered. As before we begin with Pascal's triangle. Again strip off the row
of ones, rearrange into a matrix, except this time we don't need the zeros,
decorate all entries and columns with the corresponding Bernoulli numbers,
juxtapose the formulas for the first five sums, strip away all the
coefficients except those in the first column, there. And now magically merge.
Whoa this is Bernoulli's celebrated pattern. Really quite something. Who would
have thought that calculating the 2d triangular infinite array of
coefficients corresponding to our sums can be reduced to calculating a
one-dimensional infinite sequence, the Bernoulli numbers. Pretty amazing, isn't
it? Again what this means is that once you know the Bernoulli numbers, you can
straightaway write down any of our sums. You can check for yourselves that this
pattern really incorporates all the patterns that we spotted earlier. Now
here's a page from Bernoulli's book where he talks about his discovery. Up
the top are the formulas for S1 to S10. He probably used the last one to
calculate that humongous 32 digit sum that we mentioned earlier.
Now even given that formula, to accomplish it in 7 1/2 minutes is
pretty damn amazing. I wouldn't try this. Down there's Bernoulli's version of the
formula for Sn. At the very bottom we spot the Bernoulli numbers B2, B3, B4 and
B5. And how did Bernoulli prove that his super formula always works. Well he
didn't, he just trusted the pattern to continue. As far as I know the first to
nail it all down with a proof was Leonard Euler. Anyway, definitely another
great way to generate all the formulas for our sums. Of course,
what you really need to use this method is a way to calculate all the Bernoulli
numbers, but that's pretty easy. Now before we go there a challenge here.
There's a mistake in Bernoulli's formulas as they're written over there.
Can you spot it? Anyway, here's one way to calculate the Bernoulli numbers. The
formulas down there work for all n, so in particular they work for n is equal to 1.
Ok for n is equal to 1 all the sums have just one term namely a power of 1 and
therefore, and since all powers of 1 are 1, all these Sn are equal to what?
Well, 1 of course. And now we can solve the system of equations, for example, by
using matrices, just as we did earlier. Very pretty! Very pretty and also very
similar to our original matrix solution for the Sn s. Right? That one. In fact, using the top formula to find the first five Bernoulli
numbers is exactly as much effort as finding the Sn s using the second
formula then to get the Sn s from the Bernoulli numbers takes an additional
step and so for the initial goal we set ourselves, to find a formula for the
Sn s, using the second identity which avoids the extra step is definitely the
way to go. So, then why bother with the Bernoulli
numbers at all, seems way too complicated if all we want is our power sum
formulas, right? Right, except, as I mentioned earlier, the Bernoulli numbers
pop up everywhere! They are not only found inside our finite power sums they
also play a central role in hugely important infinite power sums and
things like the Riemann zeta function. Ready to be dazzled? So far this has all been about these
sums of positive powers. How about negative powers? Unfortunately no one knows any simple
formulas for these negative power sums. However, extremely interesting things
happen if we don't stop, if we add all infinitely many terms. The first sum,
known as the harmonic series, explodes to infinity despite its term getting
infinitely small. I've already shown a couple of different proofs of the
surprising explosion in previous Mathologer videos. I've also talked about
the second series of reciprocal squares in one of my videos. I showed an
ingenious proof by Euler that the sum of the series is pi squared over 6. In fact,
Euler used this method to calculate all the sums of even powers. For example, the
bottom series with fourth powers adds up to pi to the power 4 divided by
90. And after pondering all these even power sums for a number of years Euler
also eventually came up with this very stunning general identity. And, yep, those
are the Bernoulli numbers there, at the heart of Euler's formula. I'll have to
make another video about all this at some point so I can also talk about
those mysterious unmentioned odd power sums. But it's definitely not all as far
as the Bernoulli numbers are concerned. Let's return to the positive powers,
ignoring all common sense and sum to infinity. What do we get? Most of you will
have seen at least a couple of the nonsensical pseudo identities over there,
especially that first notorious - 1/12 sum. Of course, anybody who knows anything realizes that, as written, these identities are ridiculous since all of
these series explore to infinity. Nonetheless something super interesting
and important is hiding behind these identities. First, the numbers on the
right are really just the Bernoulli numbers in disguise. Secondly, the identities
become meaningful when the meaningless left sides are replaced by the
corresponding and meaningful values of the famous Riemann Zeta-function. If this
last remark makes no sense to you, you've got another crazy Mathologer master class
video waiting for you where all this is explained. Still here. Whoa :) Need another
coffee? Maybe pause and get one anyway. I gotta warn you, we're about to enter the
deep part of the deep end. We're now heading into territory that even many of
my mathematician colleagues will know very little of. As always I've attempted
to make the presentation as visual and accessible as possible but we're pushing
the Mathologer boundaries here. Okay, last chance, last chance to bail out and go
watch another cat video. That's a no to the cat video? Great. Mathematical
seatbelts and crash helmets on and let's go. Now that we have sums of simple
powers under control, it's natural to also look for summation formulas for
functions that are composed of these powers, functions such as polynomials or
functions that can be expressed as power series like sine and the exponential
function. Now before we launch into summation formulas, let's have a quick
calculus refresher. Let me remind you how to differentiate and integrate these
power series functions. That's actually really, really easy, exactly as for
polynomials. So to differentiate such a function you simply differentiate term by
term, no matter whether the sum is finite or infinite. There to differentiate the
derivative of the aqua is zero, the derivative of the green is c1, and so on. For the second and higher derivatives
just go again. Integration is just the same in
reverse and also works term wise. So there, term wise,
that one, that one, and so on. Whoa the plus C is annoying here but we can make
it zero by selecting one particular definite integral, this one here. Maybe
pause and ponder this for a moment if this is not immediately clear to you.
Anyway with these preliminaries out of the way, let's do some serious function
summing. Let's begin by choosing f(x) to be our good old friend x squared. Then,
remember, we wrote a summation formula for the squares this way. Now we can
write the left side in terms of the function f(x): the 1 squared is just f(1) the 2 squared is f(2) and in total we get this. Okay, now if we switch
from x squared to x, then S2 on top changes to S1, of course. If we then
change to, well whatever, x to, say, 7x, then the right side changes to 7 times S1. If we change f(x) to this randomly chosen polynomial, then the right side
changes accordingly. All pretty straightforward, right?
So let's call our new sum S. Very original again :) Now, let's cross our
fingers, go completely general and head off to infinity. There. And then the sum
should be this, right? All ok so far? So to get the summation formula for a finite
or infinite polynomial you just replace the powers of x by the corresponding Ss,
right? Well, maybe. MAYBE. That infinite sum of Ss is on the right is definitely a
little bit of a worry. But don't worry, be happy and let's leave the demons for
later. Now the idea is to express this general summation formula in terms of
the Bernoulli numbers. Okay, so first here again are the Bernoulli formulas for the
Ss. Let's write them top to bottom. Okay now connect the Ss into the sum we're
interested in. So the constants have to go in.
Okay, expand out. Okay, so finding our sum amounts to summing all the entries in
all the boxes and the clever trick is to do this row by row rather than column by
column. We started with columns, right? Can you see why this is a great idea? No? Well,
let's do it together. First, row one. Okay, take out the common
factor. Nice! Can you see why? Well the expression
the brackets is exactly the integral from my calculus refresher intro from a
moment ago, evaluated at n, so this guy. Now, on to row two. Take out the common factor.
We can simplify this a bit. 1/2 times 2 that's 1, 1/3 times 3 is 1, and so on,
everything goes away, right? Okay, and now the sum in the bracket should be
familiar. All right so there we've got a bit of a coincidence. Yep, the sum in the
bracket is exactly f(n) except for the missing c0 which means we can write the
bracket like this. Almost done with row 2. Just one more nice insight worth
incorporating. Evaluate f(n) at 0 that wipes out pretty much everything on the
right. Ok so c0 is just f(0). We substitute that
above and Row 2 is done. On to Row 3. Well, I mean you know the drill,
take out the common factor and looking carefully again at the bit in the
bracket gives this. hmm :) Consider filling in the details as your homework
assignment. Involves fiddling a bit with binomial coefficients but shouldn't be
much of a challenge for you calculusi guys out there. Anyway the pattern that
emerges at this point continues to infinity. And that that's the infamous
Euler Maclaurin summation formula. Great, isn't it. We've turned an everyday finite
sum into a truly Infernal infinite sum. Hmm not obviously so great
but don't worry, I'll conjure up some mathematical magic
that will convince you it's a great formula. But before we apply the formula,
let me fiddle with it a bit. This will highlight the hidden parts of the
underlying pattern and will present the formula in its most useful form. To begin
it often makes sense to replace lots of derivative dashes by numbers in
brackets. So those two dashes get replaced by two in a bracket, this one
here gets replaced by one in a bracket But the function values in the B1 term
involved zero derivatives and so the pattern fits this line as well. There, and
why stop there? The integral in the B0 term is an antiderivative which is
exactly a -1 derivative. So we can write the first term to fit the same
pattern. Very pretty and slick, don't you think? Ok
very pretty and very slick. However, when we actually apply the formula it turns out
to make more sense to use a version of this formula in which this pattern is
only partly visible. So let's start again with our original Euler-Maclaurin
formula and fiddle with it slightly differently.
Ok, B0 that's just 1 and B1 that's 1/2. ok right like that
leave B2 for now because the next one's easy, B3 and all the other odd Bs are
equal to 0 and so we can simply zap all the corresponding terms. Now come a few
more cosmetic twists and turns that morph our formula into this form. Can you
spot the differences. Well I'll give you a moment. Ok,
count down: 3 2 1. You got them all? Well, here they are.
All the zeros here turn into ones over there and the minus sign here turns into
a plus sign. Maybe try to justify this final morph for yourself. It's actually a
little trickier than it looks. Also, just in case you get stuck and
desperate, I'll indicate the argument in the description of this video.
Fantastic, but what is this fancy formula good for? Again, it seems pretty crazy to
turn a finite sum into an infinite sum plus an integral. Well, the Euler-Maclaurin formula can be used to evaluate or approximate tricky integrals
in terms of sums and the other way around.
Very technical mathematics but incredibly powerful and beautiful. To do
justice to this magical formula I should now spend an hour or so showing the
formula in action: applications to the Riemann zeta function like justifying
the -1/12 bit, Sterling formula for approximating factorials, the Euler-Mascheroni constant, and so on. But another hour may be pushing my welcome :) and so I'll save the full Euler-Maclaurin extravaganza for future videos. However,
let me finish this video with just one illustration, an amazing power summation
story. Following little Gauss and not-so-little Bernoulli, it's now Euler's
turn. One of Euler's amazing and most famous
achievements was the solution of the so-called Basel problem, the evaluation
of the sum over there of the reciprocals of squares in terms of pi. While working
on the problem Euler first numerically calculated the sum and many related sums
to 16 plus decimals. Here are two pages from one of Euler's books where some of
his approximations are shown. So, given this approximation, Euler first seems to
have guessed the exact value to be pi squared over 6. Yep Euler was that kind
of guy! So you look at this and you guess this is pi squared over 6. I wouldn't
have guessed it. pi squared over 6 that's the bit in the green box. But how does
one calculate a good approximation of this infinite sum before knowing the sum
is equal to Pi squared over 6. Well easy, right, just add 1 plus 1/4 plus
1/9 and keep going until you get sick of it. Right? Wrong! The trouble is that this
infinite sum converges very, very slowly. For example, summing the first nine
terms gives this. Not even close! In fact, to get those 16 correct decimals that
Euler found you'd have to sum around 10 quadrillion, that's 10 to the 16 terms.
And although Euler was a very stubborn man, he wasn't that stubborn :) This is
simply an impossible task to do by hand and might be even really really hard
with a computer. To do the impossible anyway Euler unleashed the Euler-Maclaurin formula on his problem. Okay, now, if someone tells you that this
formula is the solution to our approximation problem, what function
would you choose? Well, we're adding the reciprocal squares and so f(x) equals
1 over x squared is of course the natural choice.
And, luckily, 1 over x squared is a nice polynomial power series thingy and so uh oh...
Well, of course, 1 over x squared is 100% not the kind of nice function
that we had in mind when we derived the formula. But no matter, let's cross our
fingers pray to our Mayan god, go on autopilot and see what happens.
So writing out the formula with f(x) is equal to one over x squared, well that's
just baby calculus and here's the result. Ignoring the dodgy equal sign what have
you got? Well, there's the sum of constants here, plus this sum of terms
that depend on n. It's easy to get the yellow sum under control, at least in
the cavalier manner that we're using here. Just let n dash off to infinity
then the finite green sum on top turns into the infinite sum we're interested
in and all the tiny terms in the aqua vanish like, for example 1 over n,
if you let n go to infinity that goes to 0, right? Cool! So the yellow is apparently
equal to the infinite sum of reciprocal squares above. Back to the full equation,
there. So we can replace the yellow sum by the infinite sum of reciprocal
squares and rearrange the equation like this. Solve for the infinite yellow sum we are chasing. Ok. Now chop off the infinite sum in the
aqua after a few terms. This should give an approximation of the yellow
infinite sum. Now fingers and toes and everything else that we haven't crossed
yet crossed :) Let's see what happens when we compute this approximation for a
small number like n equal to 9. Remember the green sum alone gives a pretty
terrible approximation to the full yellow sum correct to like zero decimal
places. And what if we include the aqua? Well, believe it or not the new
approximation gives the correct answer to a whopping
eight decimal places. What if you wanted Euler's 16 decimal places. Well it turns
out n is equal to nine will do it. With just six more of the aqua terms. Real
mathematical magic. Instead of having to add quadrillions of
terms of the series Euler gets away with just nine terms of the series plus a few
correction terms conjured up by his formula. I glossed over a ton of very
interesting and tricky technicalities, appealed again again to the gods, and
some really crazy stuff that makes all this even more amazing.
Anyway, I'll be back (Schwarzenegger style) to give you more Euler-Maclaurin magic, I promise. Let me
just mention one final super-crazy fact. Having seen Euler's magic you
might expect that by adding more and more of those aqua correction terms, we
could approximate the infinite sum as well as we like to any desired accuracy.
However, this is not the case. In fact, for fixed n like our n is equal to 9, the
aqua sequence actually diverges. To start with you get better and better
approximations but then, from some point, adding in more of the aqua makes the
approximations worse and worse. How weird is that? And why does this happen? Well, as
I said this won't be the last video featuring the Euler Maclaurin formula.
There's just so much good stuff here. If you'd like to play a bit with the Euler-Maclaurin formula you could try to replicate some of Euler's 16-digit
approximations or you could unleash the formula on x squared or some other
positive power of x to see how the Euler Maclaurin formula turns into the
familiar sums. Or make yourself famous and discover a nice pi based way to
express the sum of the reciprocal cubes. That second entry in his list even
defeated Euler and is still defeating the greatest mathematical minds today.
To finish, here's an animation of a second glorious proof of the sum of the
cubes that I ended up putting together while working on this video. You've well
and truly earned it (plus at least a cat video or two) and the
world needs to see this one as well. Oh and can everybody who made it all the
way to the end please give some feedback as to what did and didn't work for you
and what your background in maths is like. This would really be helpful to
know in my quest to get away with ever more insane videos. Okay that's it, bye
for now and enjoy the movie :)
Euler-Maclaurin formula, yes! I wish this video was out 10 years ago when I first tried to wrap my head around this theorem. So, so much cleaner and natural than I remember it.
I love his silly sense of humor. "Last chance to bail out and go watch another cat video."
The pacing for this video was fucking fantastic - it felt so much smoother to watch than normal lectures, even though it contained way more than a usual lecture.
This got me wondering if perhaps ancient pyramid architects had perhaps figured out the squared formula when they were figuring out how much material they would need, or if they just simply made the design and decided to keep getting material until it was done.
The man is a genius teacher (and probably researcher as well)
16:40 — should be 1 at the top, not x
What is this music fragment?
Dude I never really knew what Bernoulli-numbers are but them popping up in the zeta-function blew my mind wtf
Is there a reference or a book that contain all of these and other similar formulas? like a quick guide.