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.