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# Power sum MASTER CLASS: How to sum quadrillions of powers ... by hand! (Euler-Maclaurin formula)

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 :)