The Basel Problem Part 1: Euler-Maclaurin Approximation

[Music] welcome to zetamath over this in the next video we'll discuss this sum how euler discovered its surprising value pi squared over 6 and how euler went on to prove the sum has this value let's jump now to the 1600s as mathematicians are beginning to understand the notion of infinite series recall that to say that a series converges means that the series has a precise numerical value so for example this first series here this 1 over 1 times 2 plus 1 over 2 times 3 plus 1 over 3 times 4 and so on if you add all of those numbers together even though it's infinitely many numbers the sum as it happens is exactly the number one and i'll show you why that's true in a minute if you take the series 1 minus a half plus one third minus a fourth plus and so on back and forth this is called the alternating harmonic series and in fact this converges its exact value happens to be the natural log of two on the other hand if you took the series which is the same series we just did but with all plus signs so one plus a half plus a third plus and so on and so forth this series which is called the harmonic series it diverges because if you add up all of these numbers the total value that you get is infinite let's go back to this series 1 over 1 times 2 plus 1 over 2 times 3 plus 1 over 3 times 4 plus and so the kth term of this series is 1 over k times k plus 1. i want to make the following clever algebraic note which is that one over k times k plus one is the same thing as one over k minus one over k plus one and i'll leave it in your capable hands to check that and so what that means is that i can replace the first term one over one times two with one minus a half and the second term one over two times three with a half minus a third and one over three times four with a third minus a fourth and so on forever but if i now go and redo my parentheses then it's pretty easy to see that i now have a minus a half plus a half that's zero a minus a third plus a third that's zero a minus a fourth plus a fourth at zero and so on forever and so in fact everything after that initial one that one at the very very start those things all cancel and so the value of this series in fact is exactly one so now i want to look a little bit at euler's series that is this 1 over 1 plus 1 over 4 plus 1 over 9 plus 1 over 16 and so on what i want to do is i want to line up the terms of euler's series with the series that we just computed namely i'll take this one-fourth and i'll notice that one-fourth that's smaller than a half likewise a ninth is smaller than a sixth and a sixteenth is smaller than a twelfth and so on down the line and so what i'm saying here is that if i add up all the numbers in the top row i get one but each number in the second row is smaller than the corresponding number in the top row and so when i add all of these up i must get some number that's less than one but then the only difference between this and euler's sum is that euler's sum starts with a one so i can add one to the front of this equation and i'll get that euler's sum one plus a fourth plus a nine plus and so on and so forth and that's less than two determining the value of this sum came to be known as the basel problem the name is taken from the city of basel switzerland where the bernoullis popularized this problem and where euler ultimately solved it roughly speaking the process that euler went through to solve this problem had two parts the first was he wanted to approximate the value of the series to enough accuracy that he was able to guess what the number was that the sum was equal to so that he could then try to prove it in fact as we'll go over in just a moment euler was able to calculate the value of this series to 17 decimal places despite the fact that he was centuries before a computer so in this video i'm going to explain to you how he did that and what sort of math he invented along the way just to be able to approximate this series so accurately now after he calculated these 17 decimal places he realized aha this looks just like the decimal expansion of pi squared over six so it must be the case that the thing i'm trying to prove is that this sum is pi squared over 6. so then i just have to figure out how to go about bending and stretching and breaking the limits of math to make that happen in the next video i'll discuss euler's proof that this sum is exactly pi squared over six but for now let's focus on approximating this sum to get started i want to visualize this sum graphically as the area of this cascading arrangement of rectangles drawing in the graph of y equals 1 over x squared we can pull out this area and evaluate it as this integral from one to infinity of one over x squared dx and that integral we can readily calculate to be one if i draw in these diagonal lines i can also capture the area of this bunch of triangles i'm going to leave it to you to show that the area of this bunch of triangles is exactly a half but as a hint i'm going to show you this picture where i slide all the triangles over to live in the same column if you're unsure how to proceed i encourage you to talk about this in the comments this means that the sum as n goes from one to infinity of one over n squared it's one plus a half plus the area of these sort of crescent shaped pieces if i zoom into this unit length interval from i to i plus one then i can calculate the area of this crescent which i'm going to call a sub i and it's going to be the area of the trapezoid minus the integral and so the area of the trapezoid is going to be f of i plus f of i plus 1 over 2 and then minus the integral from i to i plus 1 of f of x thus our sum as n goes from 1 to infinity of 1 over n squared is this 1 plus this one half and then plus this a1 plus this a2 plus this a3 and so on our plan here is that we're going to estimate each of the ais individually and then at the end we're going to combine all of it together so let's go to a zoomed in view of just one of these ais to avoid letting our notation get too cumbersome i'll let our interval run simply from zero to one but as you'll see by the end of this it won't matter what the interval is to emphasize that this works for any function i'm just going to call the function here f x i'm going to call the area of the crescent a and i want to remind you here that a in this case is given by f of zero plus f of one over two minus the integral from zero to one of f of x dx what we're going to do is rewrite this integral from zero to one of f of x using a clever integration by parts okay so here we go we're assuming f is just some differentiable function and i want to rewrite this integral from zero to one of f of x using integration by parts now this might seem a bit strange as you might remember that usually you apply integration by parts to a product of things and even more strange because we don't know anything about f of x but bear with me for a minute and i'll show you how this is going to work out so i want to remind you what integration by part says in this case namely if you take the integral from 0 to 1 of some function u times the derivative of some function v then what you get is the product function u v evaluated at one then minus that same product u v evaluated at zero minus the integral from zero to one of v d u so the whole application of integration by parts comes down to choosing what we want to be our u and what we want to be rv usually what we do when we're integrating by parts is to start by choosing a u and a dv so that is we choose two functions that multiply together to be the function that we're integrating in this case the function we're integrating is f of x so the only real way i can write that as a product is as f of x times one in this case i'm going to choose u to be f of x and dv to be 1. once we've chosen u to be f of x then d u is just the derivative of that so it's f prime of x and we've chosen dv to be 1 and so our natural inclination is to take v to be the anti-derivative of one namely x but i want to remind you that in fact one has lots of anti-derivatives namely not only x but x plus any constant in this case for reasons that we'll see very shortly if we choose the constant minus a half that is if we choose v to be x minus a half then things work out well all right so let's apply the formula so our formula gives us that we have uv which is f of x times x minus a half evaluated from zero to one i'm just going to leave the integral here as the integral from zero to one of vdu because the reason why this x minus 1 half is going to be so nice has everything to do with what's happening here with this term at the front so i want to focus on that for now in particular when we plug in 1 we get f of 1 times 1 minus a half minus and then when we plug in 0 we get f of 0 times 0 minus a half and of course we still have the minus the integral from zero to one of vdu piece now f of one times one minus a half that's just f of one over two similarly f of zero times minus a half is minus f of zero over two but we also have this minus and so together that gives us two minuses which means we get a plus which means the second term here is plus f of zero over two and so these two pieces together give us f of one plus f of zero divided by two and then we have minus the integral from zero to one of and i'll go ahead and write it out this time x minus a half times f prime of x dx i'll bring back this sheet of paper where we were writing down our original formula for a and remind you that what we just did was to rewrite the part underlined in red namely we rewrote the integral from 0 to 1 of f of x dx so i'm going to now plug in what was our result from before so that's going to give us f of 0 plus f of 1 over 2 minus and then f of one plus f of zero over two minus the integral from zero to one of x minus a half times f prime of x dx and here is where the magic happens namely our opening f of zero plus f of one over two is exactly cancelled by our new f of one plus f of zero over two and so at the end of the day we're left with a very simple formula for a namely that a is just the integral from 0 to 1 of x minus a half times f prime of x dx why did this work out so nicely y c equals minus a half to understand that we have to go back to what our formula for a looked like immediately after we integrated by parts before we did anything else so i'm going to write that here as a equals and then we had this part from the trapezoid this f of 0 plus f of one over two then minus this u v term evaluated from x equals zero to x equals one that's the part out of the integration by parts that eventually ends up canceling with this first term and we wanna understand why and then we have plus this integral from 0 to 1 of x minus a half times f prime of x dx to understand why we get so lucky i want to focus on what happens when f of x is linear in this case it's clear from the geometry that a equals zero so we can just cross it out here on the other hand let's look at what happens to the integral when f is linear in this case f prime of x is just a constant so let's call it little c thus we can pull this out of the integral so we're left with just little c times the integral from 0 to 1 of x minus a half dx but this integral is zero thus returning to our formula we can cross out the integral term too but now it follows immediately that the f of zero plus f of one over two term and the u v evaluated from x equals zero to x equals one term must exactly cancel each other out as we saw above so just to emphasize why x minus a half because that's the only antiderivative of one we could have chosen whose integral from zero to one is 0. so what i'd like to do here is integrate by parts again so here i'm going to set u to be f prime of x so d u is f double prime of x into dv is going to be x minus half so then v is going to be x squared over 2 minus x over 2 and then plus a constant and this time i'm going to just leave this as plus a constant and i'm going to let us figure out what we want that constant to be so take a brief moment to relax while i churn through the integration by parts formula here having done that now we see that we get that a is c times f prime of 1 minus f prime of 0 minus the integral from 0 to 1 of f double prime of x times x squared over 2 minus x over 2 plus c i want to emphasize here that this equality is true for all values of c no matter what constant i choose for c this is a true statement so which c should i pick at first maybe you're thinking oh we can pick c equal to zero because if we pick c equal to zero then the first term goes away but actually the first term is easier to contend with than the second because the first term is just evaluating some information about our function at one and zero namely the derivative but this integral is a pretty complicated integral because it's some complicated combination of polynomials so what i want to look back to is i want to look back to what happened in the previous step where we noted that if f was linear f prime was constant and then this minus a half factor turned out to make the whole integral vanish in the case that f was linear if f is quadratic then f double prime is a constant i'm going to call that constant little d just so we don't confuse it with our big c over here so i can pull the little d out of the integral which will give me the integral from 0 to 1 of x squared over 2 minus x over 2 plus c can you see where this is going just as before i'm going to choose my constant c so that i can make this integral equal to zero in this case the c that makes this happen is c equals a twelfth thus if we choose the constant c to be a twelfth then we'll make sure that this formula for a is this piece at the front with the derivatives but then this integral part just completely disappears i want to go ahead and give a name to this process for a polynomial little p of x we define script a p of x to be the unique anti-derivative big p of x such that the integral from zero to one of big p of x dx is zero i encourage you to work out the details for yourself to convince yourself that for every polynomial little p of x this script ap of x is a well-defined polynomial here are our three formulas for a so far from the first to the second we've pulled the linear part of f out of the integral and when we did that it cancelled with this front term from the second to the third we pulled the quadratic part of f of x out of the integral and that gave us this term we want to understand how we can continue this process so that is how we can pull the cubic and cortic and kintic parts of f of x out of the integral each part of f that i pull out of the integral is going to give me a new term here up front and i'm going to call all of the terms here up front collectively the boundary terms and the reason why i'm calling it the boundary term is because it only depends on what happens on the boundary of the interval from 0 to 1 that we're talking about here namely the points 0 and 1 themselves nothing about what happens between 0 and 1. we're going to call this the zeroth formula for a this the first formula for a this the second formula for a and so we're looking to figure out how we can find the third the fourth and so on formula for a so i want to start keeping track of the things i'm using as the v in the integration by parts i'm going to call the little v in the k-th integration by parts big v sub k so in the formulas you can see this is my v0 this is my v1 and this is my v2 so what's our general kth formula going to look like my kth formula is going to look like a is equal to and then i'm gonna have a sum of boundary terms one for each time we've done this so b t one plus b t two plus up to b t k minus one plus and then some plus or minus one times and then the integral from 0 to 1 of our v sub k times the kth derivative of f dx note our second formula here for a already has this form with b t 1 equal to f prime of 1 minus f prime of 0 all divided by 12. okay so what i want to do now is i want to give sort of an inductive definition of what the kth step here is going to look like so the way i'm going to do that is say well i've already shown you what the first two steps look like and then i'm going to show you if you've already taken some number of steps how do you go one step further here we have our formula for a which is the sum of some boundary terms plus minus 1 to the k plus first power times this integral involving the kth derivative of f and what we'd like to do is figure out how to change this into a formula again with some boundary terms but instead with an integral in terms of the k plus first derivative of f and so not surprisingly what we're going to do here is we're going to integrate by parts so i'm going to pull out this integral and then we're going to do this integral by parts as we've done each time before we're going to take u to be the f part in this case the kth derivative of f and we're going to take dv to be this v part the v sub k so then d u is just the k plus first derivative of f and for v i'm going to take this thing that we called a of v k of x that is we're going to take the anti-derivative of v k of x whose integral from zero to one is equal to zero and i'm going to call that thing v sub k plus one so what does that give us it's going to give us the kth derivative of f times v sub k plus one evaluated at one then minus that evaluated at zero and then minus the integral from zero to one of v sub k plus one times the k plus first derivative of f and now we're going to plug this formula that we just got out of the integration by parts back into our original formula for a and so here after we notice that i can pull this minus into the minus 1 to the k plus 2. so we've returned to the form we wanted but now our integral involves the k plus first derivative of f instead of the kth derivative of f and one nice thing about having written it this way is that now we know what the k-th boundary term is namely it must be that this piece i'm underlining in blue is the k-th boundary term and so we can record that formula for the future using this formula for the boundary terms we have the following expression for our kth formula in terms of only the kth derivative of f which is part of an integral and the values of the derivatives of f up to the kth derivative of f at 0 and 1. to summarize our kth formula has the following two key properties first the boundary terms here depend only on f and its derivatives evaluated at zero and one nowhere in between second the integral term vanishes if f is a polynomial of degree less than or equal to k be sure you understand what both of these points are saying and why they're true before we proceed now euler did all of this calculation by hand so let's do at least a little bit of this to show you that it is possible so i've got again v zero is one v one is x minus a half v two is x squared over two minus x over two plus a twelfth so what am i gonna do to find v three well i have to calculate remember this a of v2 so i'm going to write down this anti-derivative so that's going to be x cubed over 6 minus x squared over 4 plus x over 12. and now plus a constant and i need to solve for this constant because remember i need to make it so that the integral from 0 to 1 of this thing that i have left is 0. so let's go ahead and calculate that integral so the integral from 0 to 1 of this polynomial i just wrote down is x to the 4th over 24 minus x cubed over 12 plus x squared over 24 plus whatever our constant is times x but i'm evaluating this from 0 to 1 so i get to plug in 1 and then subtract off what i get when i plug in 0. note conveniently when i plug in 0 everything just disappears so i just get 1 24 minus a 12 plus a 24 plus c which is to say that i get just zero plus c and so that means the constant i'm looking for here is actually zero and so i get my formula for my v three it's x cubed over six minus x squared over four plus x over 12. if i want to think of this graphically we're going to take a look at the graph of the quote unquote obvious antiderivative of x squared over 2 minus x over 2 plus a 12 namely x cubed over 6 minus x squared over 4 plus x over 12. as before i'll go ahead and shade the part above the axis and the part below the axis and calculate the integral and before we moved it up and down but in a minor miracle here the integral is already zero so i don't have to touch anything so in other words i can choose as my constant of integration c here c equals zero that's what we just saw algebraically but in the graph i'll let you watch me continue to do this one k at a time each time i'm taking the antiderivative of the previous v sub k and then i'm integrating that anti-derivative from 0 to 1. so if you look at this page after page what you'll notice is basically every time i have to integrate twice in order to solve this but after i integrate the anti-derivative from 0 to 1 then i solve for the constant to just make that integral 0. and so my hope is that you can see that even if this looks a bit messy you can see that this is definitely something very concrete that you can do speaking of messy as this keeps going by this is getting really messy and there's a good reason for that namely if you look at this first term here this x to the k term it always has a denominator of k factorial so the denominators involved in our v sub ks are growing very quickly and so that's making this polynomial look a little bit more complicated than it has to be because it has this ugly denominator floating around from the beginning because of this we tend to normalize these polynomials by multiplying through by k factorial which also makes them way easier to work with so to be super precise by what i mean i'm going to define a new list of polynomials i'm going to call these b sub k and my b sub k is just going to be literally equal to k factorial times the v sub k these things are known as the bernoulli polynomials and we'll discuss later exactly how bernoulli happened upon them but as you can see these things are much more palatable than the polynomials we were calculating before i should make a quick note about how we go about calculating the bernoulli polynomials without thinking about the v sub ks so in other words i have to explain how to get from b sub k to b sub k plus 1. so to do that i'm going to write down this system of equations basically my b sub k plus 1 it's k plus 1 factorial times v sub k plus 1 that's just the definition my b sub k is k factorial times v sub k it's also just the definition and my v sub k plus 1 that's this a operator of v sub k it's also just our definition i'm going to substitute that last equation into the first one to get that b sub k plus 1 it's k plus 1 factorial times this a operator of v sub k now remember my goal is to not mention the v sub ks so i can rewrite v sub k as being b sub k over k factorial that's by the second equation over here and so then i get that b sub k plus 1 is now k plus 1 factorial times the a operator of b sub k over k factorial but the over k factorial is just a constant so i can pull that out of the whole a operator business and so what i'm just left with is that b sub k plus 1 it is k plus 1 times the a operator of b sub k in other words the process for generating the b sub ks is almost identical to the one we used to generate the v sub ks but with this small extra step where i have to multiply by k plus one at the very end so from our calculations above the sixth bernoulli polynomial is this so my script a of the sixth bernoulli polynomial is going to be this and i still need to solve for this plus c to do that i'm going to integrate this script a of b6 from zero to one i set that equal to zero and then i solve it and so the c i'm looking for turns out to actually be zero and so by what we said before b7 is seven times the script a of b6 we just calculated and so we get this for b7 you can watch me do the same thing for calculating b8 albeit at a much faster pace and you can imagine continuing on through this process and calculating progressively more of these polynomials let's bring back our formula for the area a from earlier and now this still has the v sub k plus ones in it i don't want to think about those anymore so i'm going to rewrite all of them in terms of the b sub ks which is going to leave us with this now i want to look at something and note that b sub k it's a polynomial but if i ignore the integral part at the end i don't actually care about the polynomial i only care about the value of the polynomial at 0 and 1. in other words to say that again the boundary terms don't care about the polynomial the boundary terms only care about the value of the polynomial at 0 and 1. so pulling up this table of the bernoulli polynomials that we've generated i'm going to go ahead and fill in the value of them at 0 and at 1. now almost as soon as i do that there's a couple patterns that really leap out at you so the first pattern you probably notice is that the value at 0 and the value at 1 seems to always be the same to prove this we're going to evaluate the integral from 0 to 1 of b sub k minus 1 of x in two different ways before we do this i just want to recall on screen here that our definition of the bernoulli polynomials is that b sub k plus 1 is equal to k plus 1 times the script a of b sub k thus we can go ahead and rewrite the b sub k minus 1 in the integral as k minus 1 times script a of b sub k minus 2. upon doing that k minus 1 is a constant so we can pull it out of the integral but now we're left with the integral from 0 to 1 of the script a of b sub k minus 2 of x dx but remember by our definition of script a this is exactly zero and so this integral we started with it's zero on the other hand i could notice that b sub k of x that's k times script a of b sub k minus 1 of x and in particular what that means is that b sub k of x divided by k that is an anti-derivative of b sub k minus one and so we can evaluate this integral with the fundamental theorem of calculus and what do we get we get that the integral from zero to one of b sub k minus one is b sub k of one over k minus b sub k of 0 over k but on the other hand we already know this integral is 0 and the only way that can happen is if b sub k of 1 and b sub k of 0 are equal which is what we hope to show so there's no reason to have two columns in this table so i'm just going to merge the columns together and in fact the kth entry in this row that is the kth bernoulli polynomial evaluated at 0 or 1 that's usually called the kth bernoulli number and it's usually denoted by a lowercase b sub k so we're going to do that as well with this definition if we look back at our k formula for a now we can rewrite big b sub k of 1 and big b sub k of 0 as just little b sub k everywhere they appear and once we do that we can factor the little b sub k's out of each term and then we're left with this but almost certainly that's not the only pattern that you notice looking at this table if i remove the polynomials and just list the first 25 bernoulli numbers you're probably going to be pretty convinced that b sub k is zero whenever k is odd as long as k isn't one and you might already have noticed there's all kinds of weirdness with the first bernoulli number anyway the proof of this result is actually pretty neat i'm going to leave this to you to discover feel free to discuss it in the comments but i'm just going to take for granted that you've done this exercise and then i'm going to remove all of the odd terms from our formula for a i want to point out that that actually makes things a lot nicer because it gets rid of all of these alternating signs that i've been generously keeping track of for you throughout this video but now all of them disappear and so we're just left with this cleaner looking formula now if you're following the thread here you'll remember that my stated goal was to approximate this sum which makes it a little weird perhaps that i've done all of this work and there's still an equal sign here this statement here is exactly correct and i want to note this first term here it's easy to calculate this next term here it's easy to calculate and so on in particular the function f that i'm actually interested in applying this to is 1 over x squared so i'm happy to take any number of derivatives of one over x squared that you want a hundred derivatives no problem the part of this that's not easy to evaluate is this ugly looking integral piece here at the end and so the way that i'm going to make this an approximation is that i'm just going to agree that i'm never going to evaluate this integral and so i'm going to throw this away and replace this nice equals with a squiggly equals now my hope is that because at this point i've pulled out all the stuff of degree 2 3 4 5 up to whatever 2k i happen to have chosen from the integral that hopefully whatever is left in the integral is small enough that this approximation is good and so then the integral will be small and we can just forget about it so far we've been focused exclusively on the interval from zero to one and our conclusion here is that this a for the interval from zero to one that's f of 0 plus f of 1 over 2 minus the integral from 0 to 1 of f x dx that's well approximated by this formula which involves the bernoulli numbers and the derivatives of f at 0 and 1. now if you rewind back to way way earlier when we talked about the situation we actually had these a sub i's representing the area of the same region but over the interval from i to i plus 1. so the key here is that i want to not have to find a new formula i want to be able to basically apply the formula we already have to work on the intervals from i to i plus 1. and so here the key to us doing that is to remember i didn't assume anything specific about the function f in doing this calculation other than i guess that i could take this many derivatives of it so what i'm going to do here amounts to a change of variables just so that i can move this interval over to the interval from 0 to 1. namely i'm going to define this function f sub plus i of x and the value of this function at x is just going to be the function f evaluated at x plus i if you think back to your rules about transformations of graphs the graph of f sub plus i of x is just the graph of f of x shifted to the left by i units in particular this shifts the graph so that the interval in f from i to i plus one is now squarely in the interval from zero to zero plus one to see this in equations i'm going to rewrite this formula for a i that we have as f sub plus i of 0 it's just a different way of writing f of i a weirder way plus f sub plus i of 1 divided by 2 minus the integral from 0 to 1 of f sub plus i of x dx now this is exactly the kind of thing that our formula tells us how to approximate and so i'm going to apply that approximation here giving us that a i is approximately the second bernoulli number times f sub plus i prime of one minus f sub plus i prime of zero over two factorial plus and then we've got this term with the fourth bernoulli number and so on but now it's not so hard to see that f sub plus i prime of one is the same thing as f prime of i plus one use the chain rule if you're unsure where this is coming from and so now i'm just going to clean up all the evidence that i ever mentioned the sub plus i business and i'm going to replace everything back so that i get that a sub i it's approximately b 2 times f prime of i plus 1 minus f prime of i over 2 factorial plus b 4 times f triple prime of i plus 1 minus f prime of i over 4 factorial and so on let's return to euler's context that is let's set f of x equal to one over x squared and so remember that what euler is trying to do here is to approximate the sum as n goes from one to infinity of f n which as we showed before is this one plus this a half plus a1 plus a2 plus dot dot and so on now i'm going to break out each ai here into the approximation that we gave just now you'll see that at first this looks pretty messy however if you squint closely at this you'll notice that it's not nearly as bad as it looks so for example if you look at this b2 times f prime of 2 over 2 factorial here it's exactly cancelled by this b2 times minus f prime of 2 over 2 factorial here likewise this b2 times f prime of 3 over 2 factorial is cancelled by this minus b 2 times f prime of 3 over 2 factorial and so on and this happens in every column and at the end of the day all that's left is the terms that specifically involve the derivatives of f at one everything else washes away and so after all of that has washed away here's what we're left with we're now left with the sum as n goes from one to infinity of one over n squared it's approximately one plus a half minus b two times f prime of one over two factorial minus b four times f triple prime of one over 4 factorial and so on i'll go ahead and show you a few derivatives of 1 over x squared so if f of x is 1 over x squared f prime of x is minus 2 over x cubed f double prime of x is 6 over x to the fourth and it's not so hard to convince yourself that indeed the kth derivative of this f is always minus 1 to the kth power times k plus 1 factorial divided by x to the k plus two now remember that we're going to be evaluating this derivative at one and it's going to be appearing in our formula here however if you evaluate this at 1 then what you find is that the kth derivative of f at 1 is minus 1 to the k times k plus 1 factorial now in general k plus 1 factorial is going to be a pretty big number which means that our approximation is not going to settle down quickly and we would like of course especially if we're euler doing this by hand to use as few terms as we possibly can here so the first thing i'm going to do is i'm going to break my sum as n goes from one to infinity of one over n squared into two pieces one piece where n goes from one to nine plus one piece where n goes from 10 to infinity now the first part here the sum is n goes from 1 to 9 that is just a sum with 9 terms so i can write these all out and all of their decimal values and then we can add them together and so here to 20 decimal places is the sum as n goes from 1 to 9 of 1 over n squared so now let's go back and let's look at the sum as n goes from 10 to infinity of 1 over n squared so we pull up this graph again where the area of the bars in aggregate is the sum as n goes from 10 to infinity of 1 over n squared and so again we break up that area into the part that's under the curve which will be the integral from 10 to infinity of 1 over x squared which is a 10th or if we're doing stuff in decimal now 0.1 we've also got all of these triangles floating around and now you can check the total area of all of these triangles it's now 1 200th or 0.005 now the slivers that is the a10 the a11 the a12 and so on those are almost invisible in this picture and remember we're trying to get a really really accurate approximation so even though they're almost invisible in the picture we still need to take this into account so we need to approximate a10 plus a11 plus a12 plus and so on and just like we did earlier if we replace each of these ai's with our approximation from before and sum them all together then just as before everything telescopes that is everything cancels except for the stuff that involves derivatives of f at 10. and so we get that the sum of the ai's starting at a 10 it's approximately minus b2 times f prime of 10 over 2 factorial minus b4 times f triple prime of 10 over 4 factorial and so on remember that before we showed that the kth derivative of f is minus 1 to the k times k plus 1 factorial divided by x to the k plus 2. and so the kth derivative of f at 10 is minus 1 to the k times k plus 1 factorial divided by 10 to the k plus 2. now if we plug this into the formula above what we see is that in fact we have two factorials in every term one in the numerator and one in the denominator and so those all wipe away and cancel and so we're actually now left with that if i sum the ai's starting at a 10 what i get is approximately just b two times one over ten cubed plus b four times one over ten to the fifth plus b six times one over ten to the seventh plus and so on if you were wondering before why euler chose to pull out nine terms rather than say 10 terms or some other number of terms you can see here that actually pulling out nine terms means that these things are going to look a lot nicer when we write them in decimal which if you're about to do all of this by hand as euler was is very important and so here it is euler's approximation of the ai starting from a10 forward using k equals nine so we have here the first term b2 times 1 over 10 cubed all the way running up to b18 times 1 over 10 to the 19th if i go ahead and fill in for you the values of those bernoulli numbers up till then we get this and you can see here that the bernoulli numbers actually get a little less nice than they appear to be at first and i want to be honest with you i can in no way imagine calculating this sum by hand to 20 decimal places but euler did and upon doing that he gets this number now remember we need to combine this number with our running tally from before and upon doing that we get this which is euler's final approximation of the sum of the reciprocals of the squares as n goes from 1 up to infinity this approximation is remarkable it is accurate to 17 decimal places imagine that you actually calculated this sum in the naive way that is you added the first term the second term the third term imagine that you were really patient like really really patient so you somehow meticulously calculated the sum of the first 10 trillion terms of the series if you did that your answer would still only be accurate to 10 decimal places even with 10 trillion terms whereas euler by adding together just these nine numbers was able to get it accurate to 17 decimal places and that shows you just how powerful this method is and it convinced euler that the sum had to be pi squared over 6 because it agreed with pi squared over 6 to this many decimal places the method we've described here is now known as euler maclaurin approximation it was actually independently discovered by both euler and maclaurin at around the same time and what it does here is to explain the relationship between this sum and this integral this method shows up a lot in math in particular where either the sum or the integral is easy to compute in our case the integral was the one that was easy to compute but you're interested in the other one that is to say in other cases you might know a lot about the sum and then be interested in information about the integral so you can use everything we just did but backwards i should also note that euler maclaurin isn't just an approximation method remember that every formula we wrote down until the very end when we kicked off that integral and decided not to evaluate it all of those formulas were exact and so it's worth noting that euler maclaurin and the ideas behind it appear as part of a lot of proofs in analytic number theory let's talk just a little bit about the error that we introduce when we drop the integral piece at the end in other words what i'm really asking is how big is that integral at the end going to be because remember with the integral the formula was exact so what i've shown you here is a graph that tells you how many digits of accuracy you get if you use k terms of this sum above so for example euler used k equals nine terms so you can see this point here shows you that euler got 17 digits of accuracy in this graph you can see as we saw before that you really quickly reach the 17 digits of accuracy but then sort of mysteriously improvement just seems to stop and things go flat but if we broaden our view we'll see it's actually much worse than that because by the time we use k equals 100 terms rather than having the approximation be astronomically accurate which is probably what you would expect instead we see that we're now getting a negative number of digits of accuracy and you might pause for a second and ask what does that even mean but that means the digits before the decimal place aren't even right one of the really weird properties of euler maclaurin approximation is that this almost always happens namely any infinite sum we want to approximate we tend to dive in really really close to the actual value but then hang there for a little bit and then diverge to wrap up this video i want to discuss how these things came to be known as bernoulli numbers and bernoulli polynomials because as these names suggest euler and maclaurin were not the first people to discuss these particular polynomials and the numbers associated with them indeed bernoulli in switzerland and sekitakakazu in japan happened upon these numbers and polynomials independently when they were answering a particular natural question namely they were trying to answer the question if i sum the kth powers of the first n integers what do i get so for example you probably know that if you sum one plus two plus three plus up to n that would be the sum of the first powers of the first n integers then you get n times n plus one over two and likewise you might know that if you sum the squares of the first n integers then you get n times n plus one times two n plus one over six now you might even know that if you sum the first n cubes you get exactly this which weirdly as it happens is exactly the square of the sum of the first n integers so what did bernoulli and seki want to know they wanted to know okay if i sum the first and k powers what do i get now i'm not going to go into much depth here but if you want to know a lot more about this i really recommend checking out this mythologer video where he goes into this in much more depth the key observation here is that the bernoulli polynomials have this unexpected property that if you integrate the kth bernoulli polynomial from x to x plus one then you always get exactly x to the kth power now i'm going to leave verifying that this identity follows from our definition of the bernoulli polynomials as an exercise for you but once you do that little exercise then i can rewrite the sum 1 to the k plus 2 to the k plus 3 to the k plus dot dot plus n to the k as the integral from 0 to 1 of the kth bernoulli polynomial plus the integral from one to two of the kth bernoulli polynomial plus the integral from two to three of the k bernoulli polynomial and so on up until i get to the integral from n to n plus one of the k bernoulli polynomial but of course once you see this giant list of integrals like this we can zip them all together and so it turns out that this sum on the left is just given by the single integral from zero to n plus one of the kth bernoulli polynomial but even more impressive here is that i know an antiderivative of the kth bernoulli polynomial namely the k plus first bernoulli polynomial divided by k plus one and so i can use the fundamental theorem of calculus here to evaluate the integral from zero to n plus one of the kth bernoulli polynomial and what i get is the k plus first bernoulli polynomial evaluated at n plus one divided by k plus one minus the k plus first bernoulli polynomial evaluated at zero divided by k plus one now of course you might notice the k plus first bernoulli polynomial evaluated at zero that's just exactly what we define the k plus first bernoulli number to be or if you want to think about it a different way it's just the constant term of the k plus first brilliant polynomial so really what this is saying is that in order to sum the kth powers of the first n integers all you have to do is start with a k plus first bernoulli polynomial throw away the constant term because that's the k plus first bernoulli number plug in n plus one that is one more than the number you wanted to evaluate and then divide by k plus one and voila that will give you an exact formula for the sum of the kth powers so just to check this we'll bring back what's often known as gauss's formula for the sum of the first n integers even though gauss was 2000 years too late to be the first person to write this formula down so according to our formula here we're supposed to start with the second bernoulli polynomial which is x squared minus x plus a sixth but then we're supposed to subtract off the constant term so that's just x squared minus x and now we're supposed to plug in n plus one which if you do then we'll get n squared plus two n plus one minus n minus one which is n squared plus n and of course if you divide n squared plus n by two then you get n times n plus one over two which is exactly our usual formula here i'll leave it for you to check that this gives you the correct formula for two and three and however far down this list you want to go that's all for this episode join us next time where we'll pick up the story with euler again where euler now knows that he's trying to prove that this sum is pi squared over 6. and we want to investigate carefully exactly what that proof looks like and then it turns out by somewhat remarkable coincidence that euler's proof of the value of this sum turns out to be connected to a lot of important ideas in particular it is very closely related with the contents of riemann's eight page paper in which he outlines and originally states the riemann hypothesis and so we'll discuss that and dreyman's explicit formula and all of that next time thank you for watching our theme music was composed by friend of the channel keith wilker so thanks keith please leave a thumbs up and a comment if you liked the video and subscribe so that you'll be notified when we post videos like this in the future but for now until next time be well you

Channel: zetamath

Views: 7,844

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Id: nxJI4Uk4i00

Length: 59min 6sec (3546 seconds)

Published: Thu Jul 09 2020