Welcome to another Mathologer video. A while ago I did this video in which I wanted to explain the mysterious identity e to the i pi is equal to minus 1 to Homer Simpson who encountered it in one of the Treehouse of Horror episodes. So my mission was to explain why this identity is true to someone who knows only basic arithmetic: plus, minus, times, divided. Today I’d like to do the same for the other fundamental facts about the number e. These fundamental facts I’m thinking of include this truly amazing identity here that expresses the exponential function as an infinite sum. I’ll first show Homer and you how you can derive this identity from scratch just using basic arithmetic. Then we can set x equal to 1 and use the resulting sum to calculate an arbitrary number of digits of e. Now here the main point is to figure out when you can stop adding the terms of this infinite sum and be absolutely sure that you’ve calculated the first, say 1 million digits of e or however many digits you are after. Then I’ll show you that e is an irrational number, why e cannot be written as a fraction. And for the finale I have to catch Homer on one of those days that he’s super smart, that he knows a little bit of calculus for example. The finale is about showing why the exponential function is its own derivative, some super famous facts about the natural logarithm, the inverse of the exponential function and a couple of other super cool insights to round things off. Now the main point of the video is to show you WHY all these things are true and not just talk about them like it’s usually done on YouTube and to show this to you in as elementary and accessible way as possible. As usual please let me know in the comments how I did in this respect. In the e to the i pi video I first showed Homer that if he invests one dollar at an interest rate of one hundred percent and compounds the interest N times during one year he’ll have that many dollars at the end of the year. For example, if you compound twice you end up with two dollars and twenty-five cents. Now the more often you compound the interest, that is, the larger the N, the more money you’ll have by the end of the year. But this amount does not grow forever. There’s a limiting value and this limiting value is the number e which is approximately 2.718. Then I show Homer that this actually extends in a straightforward way to the exponential function. So plug in any value for x like for example 5, crank up the N and in the limit you get exactly e to the power of 5. Then we just go for it and plug in i pi for x and see what happens when we crank up N. Translating all this into geometry leads to this neat complex number pokeball there. It looks like a pokeball right? The green point shows what complex number we are dealing with for N=6. Then cranking up N you can see the green point tending towards -1 and so e to the i pi is equal to minus 1. Super neat, check it out 🙂 For today I’ll return to this point of the argument here and branch off. Now let’s try something super natural. I think anybody would try this. Let’s expand the right side for a couple of small N and see whether we can see some sort of pattern. So N equals 1, nothing to expand here. N equal to 2, I’m sure everybody watching can do this. N equal to 3 and equal to 4, and so on. A couple of patterns are emerging, right? In terms of green and blue bits it’s clear how this continues. What about those other numbers? Well, there’s also a pattern and I’m sure many of you will be familiar with it from school. These are binomial coefficients and can be written like this. We’ve got our pattern right there. But before we carry on just a reminder that there is a special way to write and refer to the denominators 1, 1 x 2, 1 x 2 x 3, and so on. For example, 1 x 2 x 3 is written three exclamation mark and it’s called 3 factorial. And there we have 1 and 2 factorial. Now that we’ve established a pattern let’s crank up the N and replace 4 by something big, say 1000. All right, so here we go. Now what I want to do is swap things like that. And what I mean by this is I want to swap 1! by 1000, 2! by a thousand squared and 3! by a thousand cubed, and so on. So let’s just do it. Now let’s have a look at the fractions. Okay this guy here is equal to 1. That one is equal to 1, too. What about that one here? Well it’s close to 1 and as we replace 1000 by larger and larger values this fraction will tend to 1. All the other fractions that you see here also go to 1 and so all the numbers here will go to 1. And so in the limit we get this identity here. And if you are a mathematician you can dot all the i’s and cross all the t’s to prove that this amazing identity is really always true no matter what you plug in here for X. If you know more maths you may have seen this identity before but derived in a very different way and in a more general context. In that general context it’s called the Maclaurin series of the exponential function and there it is derived using calculus. Setting X equal to 1 we get this identity here and it allows us to approximate e by adding more and more of those simple terms. Now, as promised, and i hope Hormer is still with us, there’s only basic arithmetic and some glossing over details involved. Before we calculated e by cranking up the N in this expression here. In terms of using this expression to actually calculate a good approximation for e the problem is that every time we up the N we have to start afresh and discard everything that we’ve done so far. More importantly, although we know that we get e in the limit, to start with we have no idea how large an N we have to choose to be able to guarantee that we found e correct to the desired number of digits. On the other hand, the infinite sum is much much more user-friendly in this respect. As we add more and more of those simple terms our approximation gets better and better and it’s also very easy to estimate how well an approximation a chopped-off partial sum is. So here we’ve chopped off the infinite sum at the 8th term. How close does this get us to e? Well, the error or the difference between e and this approximation is just the sum of the remaining terms. Let’s estimate how large this error is. The 8! at the bottom is just 1 x 2 x and so on to 7 which is 7! times 8. 9! is 7! times 8 times 9, and so on. Let’s pull out the 1/7! Now the bracket is still very complicated so let’s do something drastic, let’s replace the 8, 9, 10, etc. in the denominators by 2s. This gives the new simpler expression. Is this new simpler expression greater or smaller than the error? Well we’re going down in all the yellow denominators, so that means that the new expression has to be greater, right? But now it’s really easy to see that in the brackets we have 1/2 plus 1/4 plus 1/8, and so on which you all know is equal to 1. And so to summarize, the error we make by chopping the sum off at the 1/N! th term is equal to 1/N! which is very good because 1/N! gets very small very quickly. So let’s just put down all the numbers in sight here. If we focus on the zeros in the estimate for the error it seems clear that our approximation will be correct in the first four digits, right, and it is. So this means that if we wanted to have some fun calculating the first 1 million digits of e from scratch, to pinpoint how many terms of the infinite sum we’d have to add we simply have to figure out which number we have to substitute 7 by to get a million zeros at the beginning of the estimate at the bottom. Can someone discuss in the comments why, just going with what I’ve shown you so far, one might want to aim for a million and one zeros instead of just a million. Anyway, in this way you can figure out that you have to add a little more than the first 205,000 terms of the infinite sum to be able to guarantee a million digits of e, very neat, right? And maybe surprisingly the sort of thinking that went into all this is also very applicable. In fact, you use infinite sums and estimations like this all the time to come up with approximations of complicated numbers that are accurate enough for practical purposes. It also turns out, and this really is very surprising I think, that this estimate for the error also gives a straightforward way to prove that e is an irrational number, that no fraction is equal to e. Are you ready for some real magic? Okay, so let’s summarize what we’ve done so far like this. On the left side we have the difference between the true value and the approximation, that’s the error, right? And we’ve seen that this error is less than 1/7! Now let’s warm up by using this inequality to show that one particular fraction, 19/7 is not equal to e. Why 19/7 ? Well, it does not really matter what fraction we start with. The only reason for us using 19/7 is that it has a 7 in the denominator which will mesh in nicely with the other 7s that are already floating around here. Can e be equal to 19/7 ? Well, let’s assume it is. Then we’ve got this inequality up there. 19/7 minus the junk in the brackets is less than 1/7! Okay, now all the denominators on the left divide 7! , right? Just think about it for a moment. Yes they divide 7! which means that we can combine the left side into a fraction with denominator 7! where the ? stands for 1 or 2 or 3 or some other positive integer. So IF 19/7 was really equal to e, THEN we just showed this fraction, the error would be 1/7! or larger. But, and this is really the punchline, we already know that the error is definitely smaller than 1/7! . So our assumption that 19/7 is equal to e implies a statement that is obviously false. So this means that our assumption, that 19/7 is equal to e must have been false to start with. Tada! Okay you probably didn’t see that coming, right? To show that e is not equal to any fraction a/b we simply have to adjust our argument like this: first replace 19/7 by a/b. Then, instead of summing up to 1/7! we sum up to 1/b! . The estimate then also changes to 1/b! and this statement down here is still impossible which implies that no fraction can be equal to e or, in other words, that e is irrational. Now you may have to watch this part of the video again to really get what is going on here. What makes this proof work is the fact that the close approximation to here on the left is super close in the sense that it can be written as a fraction that differs from e in less than 1 over its denominator. Similar super close approximations to pi and other irrational numbers also play a crucial role in proving the irrationality and transcendence of these numbers. Actually a video on transcendental numbers is next on my to-do list, so stay tuned. The Simpsons singing nonsensical stuff… (announcer) Tonight’s Simpsons episode was brought to you by the symbol Umlaut and the number e, not the letter e but the number whose exponential function is the derivative of itself. (Mathologer) Hmm, the exponential function is the derivative of itself. Before I deal with this congratulation to Homer and you for making it this far. For this last part of the video I’ll assume that you do know a little bit of calculus. In particular I’ll assume that you know what the derivative of a function is. If you need an introduction or a refresher check out this video up there and then return here or just sit back and admire the mathematical magic doing its thing. Okay we want to convince ourselves that the exponential function is the derivative of itself. Because of this identity the derivative of the exponential function is just the derivative of 1 plus the derivative of this term plus the derivative of that term, and so on so. What’s the derivative of x cubed divided by 3! for example? Let’s see: 3! is 1 times 2 that’s 2! times 3. The derivative of x cubed is 3 x squared. The 3s cancel and of course what remains is equal to this term here. So the derivative of x cubed divided by 3! is simply the previous term and you can check that the derivative of that previous term is the term before that. And the derivative of that term is 1 and the derivative of 1 is 0, and so on. And so you can see that the derivative of all the junk on the right side is again the right side which means the exponential function is its own derivative. Fantastic, right? So, if you made it up to here there is no reason to stop. Let’s use this insight for something fun. The natural logarithm log X is the inverse of the exponential function. This means that e to the log X is equal to X. Let’s find the derivative on both sides. The derivative of X is 1 and the derivative on the other side, well we need that exponential function its own derivative and the chain rule for that but then we get this. Now we have a close look and you see two bits that are the same which means we can also write X down here. Now divide and we’ve got one of those super famous facts from calculus, that the derivative of log X is 1/X. Wonderful! And now I’m just going to go for it. So integrate and we get this guy here. We substitute e for X and that gives this guy here. Now on the right side we’ve got log e and obviously that’s equal to 1 since our logarithm is base e. Almost there ! The standard geometric interpretation of this definite integral here is that the area under the graph of 1/X between 1 and e is exactly equal to 1. Now many people are under the impression that, unlike pi the number e does not have a nice geometric interpretation. Well here we’ve got one, right, in terms of the hyperbola which is both the graph of 1/X and a basic geometric shape closely related to pi’s circle. This is the way to remember it: open up this yellow curtain here under the hyperbola until it’s area is exactly one and you’ve found e. All right, now to wrap things up I’ll quickly sketch how you can also use our infinite sum identity to derive e to the i pi is equal -1 and that’s the standard way this is done in calculus. There are also ways to write the sine and cosine functions as infinite sums like this. After substituting i X for X in the top identity and multiplying the sine identity in the middle by i we’ve got exactly the same stuff in the purple and the blue boxes. Now since we’re dealing with identities, the same has to be true on the left. So the contents of the purple box is equal to the sum of what’s in the blue box, and this is Euler’s formula, super famous, right? Now I go over all this slowly in this other Simpson based video up there. So if you, you know, want it a bit slower, check it out. After plugging in pi for X the right side becomes -1 which then brings us full circle back to Euler’s identity. If you’ve made it this far you’ve really earned yourself the Mathologer’s extra huge seal of approval. So, as usual, I’m very interested in finding out how well all these explanations work for you. So please let me know in the comments. And that’s all for today.