Euler’s real identity NOT e to the i pi = -1

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Welcome to another Mathologer video. Everybody who watches these videos knows Euler’s identity e to the I pi is equal to minus one except it’s not really Euler’s identity. The mathematician Roger Cotes already wrote about it in 1714 when Euler was only seven years old. I actually find it a bit sad that people associate the math super hero Euler with a result that’s not really by him rather than with one of the zillion’s of his really original amazing discoveries. Of course it’s really sad for Roger Cotes since he doesn’t get mentioned for anything and nobody’s ever heard of him. Anyway I thought I make a video about the real Euler identity, the identity that made Euler into a mathematical celebrity.

It’s this one here. Very curious, right? It says that pi squared over six is equal to the infinite sum of the reciprocals of the squares. This identity is a very surprising answer to the problem of whether the infinite sum on the right adds up to anything nice. A lot of the most famous mathematicians of Euler’s time had tried and failed to answer this problem and so it was a bit of a sensation when in 1734 Euler managed to find the answer. And of course that the answer involves pi with no circle in sight anywhere added really to the appeal of this solution.

The result is not only pretty but also turns out to be important. Up there is the definition of the famous Riemann zeta function which is the function at the heart of the Riemann hypothesis, one of the most important unsolved problems in mathematics. So what Euler managed to do was to find the precise value of the Riemann zeta function at 2. Not only that, he also did so for all even numbers. Now it turns out that all these values involved the circle number pi. For example, this up there is Zeta evaluated at 4 and this is Zeta evaluated at 6. To this day nobody knows whether zeta evaluates to anything nice at the odd values 3, 5, 7, etc. Now Euler was also interested in assigning reasonable values to the exploding sums that you get when you try to evaluate Zeta at negative values: 1+2+3+… you know that one, right? In fact, he was one of the first, if not the first mathematician to really demonstrate that there is a lot more to these sums than meets the eye at first glance. If you dare check out my video on this really mind-boggling topic. Anyway my mission today is to motivate and animate Euler’s proof of his wonderful identities for you. So here we go. To start with it’s not clear at all that this infinite sum will actually add to a finite value and if it does add to a finite value whether it adds to anything nice. If you’ve already looked a bit at infinite sums you may have come across a beautiful closely related infinite sum. It gives a lot of insight in this respect so let’s warm up with that infinite sum. To get to it we first rewrite Euler’s sum like this.

So, for example, we write 1 over 2 squared as 1/2 times 1/2. Make a copy and let’s shift the blue bits one to the right. What results is this other infinite sum which is also very pretty. Let’s compare the two sums term by term. Okay, in the green we have 1 on top and at the bottom. Here the green factors are the same and the bottom blue one is bigger … green the same, bottom blue bigger, green the same bottom blue bigger and so on overall it’s clear that the bottom sum is bigger than Euler’s sum at the top. Now for the really ingenious argument that evaluates the sum at the bottom. Let’s chop off this infinite sum at the fourth term there and calculate this partial sum. Now get ready for a bit of mathematical magic. Here 1 times 1/2 is equal to 1 minus 1/2, neat right? 1/2 times 1/3 is equal to 1/2 minus 1/3, and so on, now we get rid of the brackets and this gives you a strange mathematical telescope of the pirate variety 🙂 Let me explain. This here is one of the segments of the telescope and here’s another one. Okay now let’s collapse the telescope: – 1/2 plus 1/2 is 0, – 1/3 plus 1/3 is 0, and so our partial sum evaluates to 2 minus 1/4. Now we chopped off the infinite sum at this term over there. If instead we would have chopped off at that term here a longer telescope would have given us the partial sum 2 minus 100. Now the further down the sum we cut, the smaller the number we minus gets. At infinity it vanishes and so our telescoping sum adds up to 2, exactly. Okay so these infinite sums can really add to nice values and so Euler & Co. had some hope that the same might be the case for the closly related infinite sum they were chasing. Also since the sum is greater than the Euler sum we know the Euler sum adds to a finite number which is less than 2. And, in fact, pi square divided by 6 is approximately 1.644 which is less than 2. So it’s all looking really good.

Ok warmup is over. Now how did Euler manage to prove his identity. Well he starts with the so called Maclaurin series of the sine function. Most of you who’ve done some calculus are familiar with writing sine as this sort of infinite polynomial. Definitely Euler and all his contemporaries knew this very well. Now to present a complete picture and to motivate where Euler’s idea for his proof comes from, it’s important to know how mathematicians derive this formula and I’ll explain it in a moment but for now let’s have a look at this identity. Obviously, the sine function is derived from the circle and it should therefore not come as a surprise that pi is hiding in sine in a couple of different ways. For example, setting X equal to pi sine becomes 0 and so we get this funky expression for pi here which is almost as nice as Euler’s identity, right? Question for you: How could you use this expression to calculate better and better approximations for pi? If you’ve got an idea tell us in the comments. All right, let’s say you suspect sine of x can be expressed as an infinite polynomial like this. How do you go about finding those coefficients? Well a is easy. Just set x equal to 0 and sine is equal to 0 and so is this expression over there and this means that a has to be equal to 0. Now let’s find the derivatives on both sides of the equation. If you don’t know what a derivative is just go with the flow.

Here the derivative of sine X is cosine X, the derivative of a is 0, the derivative of b x is b, the derivative of c x squared is 2 c x, the derivative of d x cubed is 3 d x squared, etc. Ok, now if we set x equal to 0 again. This gives on the left side 1 and here on the right side 0. And so b is equal to 1. Now just rinse and repeat, find the derivative, set x equal to 0 and conclude that c is equal to 0. And again and again and again and again and that gives you all the coefficients. So if sine can be written as an infinite polynomial that’s what it’s going to be. Anyway that’s where the Maclaurin series for sine comes from. This identity actually works for all x but that it really does of course requires a separate proof. Let me just demonstrate how well this works by showing you how the partial sums of this infinite sum are better and better approximations to sine of x. There we go. Let’s plot sine of x, there we go and first partial sum that’s just a line, second partial sum is this cubic here, and then we keep on going, you can see these guys wrap closer and closer to sine and at infinity they coincide. Anyway, as you’ve seen, the infinite sum on the right is constructed term-by-term based on higher and higher derivatives of sine at zero.

That’s really wonderful and actually gives similar formulas for all the really famous functions, like the exponential function or cosine, etc. At the same time this method of pinning down this infinite polynomial is different from what most of us would try first for finite polynomials, right? For example, for a linear function like this we construct it maybe from two points on its graph. Or for a quadratic function from three different points. And so it’s kind of natural to see whether we can also determine that infinite right side up there by taking a similar approach. And that’s exactly what Euler does. What he does is he constructs the infinite polynomial in a different way from these special points here which, of course, correspond to the zeros of sine. Okay let’s first focus on the middle three zeros. That part of sine looks like a cubic curve and we can straight away write down a cubic polynomial that has those three zeros. Here it is. Pretty obvious it’s got these three zeros. Let’s graph it. Well it does have those three zeros but otherwise it’s not a great fit. Of course this is just one of the infinitely many cubic polynomials that has these zeros. We get the other ones by putting a constant in front and varying it.

So let’s just do that. Okay down, down down, down, back up a bit and so we get a very close fit when the slopes of sine and the cubic coincide at zero. And it’s not hard to see that this happens when the constant is 1/pi squared. Well let’s just run with it. Let’s clean up a bit. first pull those constants into the brackets. Two 1s there and there, put the factors in order and combine the last two factors as you learned in primary school. Hope you paid attention. Okay looks neat and now it’s also easy to check that this cubic just like sine of x has slope 1 at 0. Next let’s repeat the same calculation for the middle 5 zeros. The polynomial we get here is of order 5 and it looks like this. In terms of the algebra you get exactly the same answer as before except you get another factor that features 2pi instead of a pi and which takes care of the outer zeros 2pi and -2pi. Now just repeat over and over adding two zeros at a time. You can see a closer and closer fit and at infinity we’ve got coincidence again. And so Euler now has two infinite polynomials that are both equal to sine x, right? Now since the infinite expressions are equal in a way, by expanding the infinite product we expect to recover the infinite sum. Let’s just have a look at how this pans out numerically. First for the infinite sum. So that’s what it is and here the smallest terms you get when you expand the first couple of factors of the infinite product. As you can see, you get pretty good coincidence in the coefficients which gets better and better the more factors you use. So this really seems to work. Now how do you expand the whole infinite product? Well, just start expanding one factor at a time. So highlight the first two factors and multiply them together and you get this. Take the result and the next factor, multiply and you get this guy. Now let’s also highlight the x cubed term because that’s going to be very important in a second, and focus on how it evolves. Okay, next factor is there. Multiply and you get this one here, and again. And now just have a look at the green. It’s pretty clear what’s going to happen in the end once you’ve finished expanding. The x cubed term will look like this. But remember this x cubed term of the expanded infinite product has to be equal to the x cubed term of the infinite sum and so we get this equation here which simplifies. First multiply by minus 1 then times pi squared and there you have it, Euler’s real identity, pretty ingenious argument isn’t it.

And all of other identities that I mentioned earlier, like this one up there, you also get by comparing coefficients of the infinite sum and the expanded product. For example, you get the identity up there by comparing the coefficients of x to the power of 5 in the Maclaurin series to the one in the expanded infinite product. Maybe give this a shot yourself. It’s really very rewarding if you managed to do this on your own. If you get stuck I’ll also do the nifty calculation in a separate video on Mathologer 2. Eventually of course there’s a lot more I could say about this identity. Most importantly I should stress that what I’ve shown you is not a complete proof. It still takes a bit of work to make this argument completely rigorous but it can be done, However, what I showed you is pretty much all that Euler had when he decided to go public with his result. In fact, believe it or not :)he didn’t have access to Mathematica and so all the numerical evidence to support his argument he had to produce by hand and he actually had a lot less numerical evidence than I showed you. Now it still took Euler a couple of years after he published his identity to come up with a completely rigorous proof. Now to finish let me just mention two more amazing facts about Euler’s sum. For my first amazing fact, let’s have another look at Euler’s way of writing sine as an infinite product up there. If we set x equal to pi divided by 2, sine becomes 1 and when we write the factors on the right as fraction, we get this. Solving for pi divided by 2 gives the so called Wallis product, named after the mathematician John Walli, a really beautiful and useful infinite product featuring squares of all the even numbers in the numerator and squares of all the odd numbers in the denominator, really really nice in it. Amazing fact number two: a lot of you will be familiar with the Leibniz formula, this absolutely wonderful identity up there.

I didn’t know about this myself until recently but Euler actually points out in one of his papers that this identity can be derived in exactly the same way as Euler’s sum, the sum that this video is about, by writing the function 1 – sine x as a product using its zeros so there’s pretty much all you need and then comparing the x coefficient to the Maclaurin series with that of the expanded infinite product not hard once you know so maybe one of you may also be able to supply the details in the comments. Oh, by the way, although this amazing identity is named after Leibniz, it, as well as the Maclaurin series of all the main trigonometric functions were already known to Indian mathematicians at least 300 years before Leibniz and Maclaurin were born and it may very well be due to the mathematician Madhava of Sanggammagrammar. And hardly anybody knows about this or cares how very say it anyway knows about this or cares. How very sad. And this is it for today. As usual let me know how well this video work for you and see you next time.

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