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 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 if top of 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.
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