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PROFESSOR: All right, so
let's begin lecture six.
We're talking today about
exponentials and logarithms.
And these are the last functions
that I need to introduce,
the last standard
functions that we
need to connect with calculus,
that you've learned about.
And they're certainly as
fundamental, if not more so,
than trigonometric functions.
So first of all, we'll
start out with a number,
a, which is positive, which
is usually called a base.
And then we have
these properties that
a to the power 0 is always 1.
That's how we get started.
And a^1 is a.
And of course a^2,
not surprisingly,
is a times a, etc.
And the general rule is that
a^(x_1 + x_2) is a^(x_1) times
a^(x_2).
So this is the basic rule of
exponents, and with these two
initial properties, that defines
the exponential function.
And then there's an
additional property,
which is deduced
from these, which
is the composition of
exponential functions, which
is that you take a to the
x_1 power, to the x_2 power.
Then that turns out to be
a to the x_1 times x_2.
So that's an additional property
that we'll take for granted,
which you learned
in high school.
Now, in order to understand
what all the values of a^x are,
we need to first remember that
if you're taking a rational
power that it's the ratio
of two integers power of a.
That's going to be a^m, and
then we're going to have to take
the nth root of that.
So that's the definition.
And then, when
you're defining a^x,
so a^x is defined for
all x by filling in.
So I'm gonna use that
expression in quotation marks,
"filling in" by continuity.
This is really what
your calculator
does when it gives
you a to the power x,
because you can't even punch
in the square root of x.
It doesn't really exist
on your calculator.
There's some decimal expansion.
So it takes the decimal
expansion to a certain length
and spits out a
number which is pretty
close to the correct answer.
But indeed, in theory,
there is an a to the power
square root of 2, even
though the square root of 2
is irrational.
And there's a to
the pi and so forth.
All right, so that's the
exponential function,
and let's draw a picture of one.
So we'll try, say y = 2^x here.
And I'm not going to draw
such a careful graph,
but let's just plot the
most important point, which
is the point (0,1).
That's 2^0, which is 1.
And then maybe we'll go
back up here to -1 here.
And 2 to the -1 is
this point here.
This is (-1, 1/2),
the reciprocal.
And over here, we have 1, and so
that goes all the way up to 2.
And then exponentials
are remarkably fast.
So it's off the board what
happens next out at 2.
It's already above
my range here,
but the graph looks
something like this.
All right.
Now I've just visually,
at least, graphically
filled in all the
rest of the points.
You have to imagine all these
rational numbers, and so forth.
So this point here
would have been (1, 2).
And so forth.
All right?
So that's not too far along.
So now what's our goal?
Well, obviously we want
to do calculus here.
So our goal, here, for now -
and it's gonna take a while.
We have to think
about it pretty hard.
We have to calculate
what this derivative is.
All right, so we'll get started.
And the way we get
started is simply
by plugging in the
definition of the derivative.
The derivative is
the limit as delta
x goes to 0 of a to the x plus
delta x, minus a to the x,
divided by delta x.
So that's what it is.
And now, the only step that
we can really perform here
to make this is into
something a little bit simpler
is to use this very first
rule that we have here.
That the exponential of
the sum is the product
of the exponentials.
So we have here, a^x .
So what I want to use is just
the property that a^(x + delta
x) = a^x a^(delta x).
And if I do that, I see that I
can factor out a common factor
in the numerator, which is a^x.
So we'll write this as the
limit as delta x goes to 0,
of a to the x times this ratio,
now a to the delta x, minus 1,
divided by delta x.
So far, so good?
We're actually almost to
some serious progress here.
So there's one other
important conceptual step
which we need to understand.
And this is a
relatively simple one.
We actually did this
before, by the way.
We did this with
sines and cosines.
The next thing I want
to point out to you
is that you're used to thinking
of x as being the variable.
And indeed, already
we were discussing
x as being the variable
and a as being fixed.
But for the purposes
of this limit,
there's a different variable
that's moving. x is fixed
and delta x is the
thing that's moving.
So that means that this factor
here, which is a common factor,
is constant.
And we can just factor
it out of the limit.
It doesn't affect
the limit at all.
A constant times a
limit is the same
as whether we multiply before
or after we take the limit.
So I'm just going
to factor that out.
So that's my next step here.
a^x, and then I have the
limit delta x goes to 0
of a to the delta x minus
1, divided by delta x.
All right?
And so what I have
here, so this is
by definition the derivative.
So here is d/dx of a^x, and it's
equal to this expression here.
Now, I want to stare
at this expression,
and see what it's telling
us, because it's telling us
as much as we can get
so far, without some--
So first let's just
look at what this says.
So what it's saying is that the
derivative of a^x is a^x times
something that we
don't yet know.
And I'm going to call this
something, this mystery number,
M(a).
So I'm gonna make the label,
M(a) is equal to the limit
as delta x goes to 0
of a to the delta x
minus 1 divided by delta x.
All right?
So this is a definition.
So this mystery number M(a)
has a geometric interpretation,
as well.
So let me describe that.
It has a geometric
interpretation,
and it's a very, very
significant number.
So let's work out what that is.
So first of all, let's rewrite
the expression in the box,
using the shorthand
for this number.
So if I just rewrite it, it says
d/dx of a^x is equal to this
factor, which is
M(a), times a^x.
So the derivative of the
exponential is this mystery
number times a^x.
So we've almost solved
the problem of finding
the derivative of a^x.
We just have to figure
out this one number, M(a),
and we get the rest.
So let me point out two more
things about this number, M(a).
So first of all,
if I plug in x = 0,
that's going to be
d/dx of a^x , at x = 0.
According to this formula,
that's M(a) times a^0,
which of course M(a).
So what is M(a)?
M(a) is the derivative
of this function at 0.
So M(a) is the slope of
a^x at x = 0, of the graph.
The graph of a^x at 0.
So again over here, if
you looked at the picture.
I'll draw the one
tangent line in here,
which is this one here.
And this thing has slope,
what we're calling M(2).
So, if I graph the
function y = 2^x,
I'll get a certain slope here.
If I graph it with
a different base,
I might get another slope.
And what we got so far is
the following phenomenon:
if we know this one number, if
we know the slope at this one
place, we will be able to figure
out the formula for the slope
everywhere else.
Now, that's actually
exactly the same thing
that we did for
sines and cosines.
We knew the slope of the
sine and the cosine function
at x = 0.
The sine function had slope 1.
The cosine function had slope 0.
And then from the
sum formulas, well
that's exactly this
kind of thing here,
from the sum formulas.
This sum formula, in fact
is easier than the ones
for sines and cosines.
From the sum formulas,
we worked out
what the slope was everywhere.
So we're following the same
procedure that we did before.
But at this point we're stuck.
We're stuck, because
that time using radians,
this very clever idea
of radians in geometry,
we were able to actually
figure out what the slope is.
Whereas here, we're not so sure,
what M(2) is, for instance.
We just don't know yet.
So, the basic question that
we have to deal with right now
is what is M(a)?
That's what we're left with.
And, the curious fact is
that the clever thing to do
is to beg the question.
So we're going to go through
a very circular route here.
That is circuitous,
not circular.
Circular is a bad word in math.
That means that one
thing depends on another,
and that depends on it,
and maybe both are wrong.
Circuitous means, we're going
to be taking a roundabout route.
And we're going to discover
that even though we refuse
to answer this
question right now,
we'll succeed in
answering it eventually.
All right?
So how are we going
to beg the question?
What we're going
to say instead is
we're going to define a
mystery base, or number e,
as the unique number,
so that M(e) = 1.
That's the trick that
we're going to use.
We don't yet know what e
is, but we're just going
to suppose that we have it.
Now, I'm going to show you a
bunch of consequences of this,
and also I have to persuade you
that it actually does exist.
So first, let me explain what
the first consequence is.
First of all, if M(e)
is 1, then if you
look at this formula over here
and you write it down for e,
you have something which
is a very usable formula.
d/dx of e^x is just e^x.
All right, so that's an
incredibly important formula
which is the fundamental one.
It's the only one you have to
remember from what we've done.
So maybe I should
have highlighted it
in several colors here.
That's a big deal.
Very happy.
And again, let me
just emphasize,
also that this is the one
which at x = 0 has slope 1.
That's the way we
defined it, alright?
So if you plug in x = 0 here on
the right hand side, you got 1.
Slope 1 at x = 0.
So that's great.
Except of course, since
we don't know what e is,
this is a little bit dicey.
So, next even before
explaining what e is...
In fact, we won't
get to what e really
is until the very
end of this lecture.
But I have to persuade
you why e exists.
We have to have some
explanation for why
we know there is such a number.
Okay, so first of all, let
me start with the one that we
supposedly know, which
is the function 2^x.
We'll call it f(x) is 2^x.
All right?
So that's the first thing.
And remember, that the
property that it had,
was that f'(0) was M(2).
That was the derivative of this
function, the slope at x = 0
of the graph.
Of the tangent line, that is.
So now, what we're
going to consider
is any kind of stretching.
We're going to stretch this
function by a factor k.
Any number k.
So what we're going
to consider is f(kx).
If you do that, that's
the same as 2^(kx).
Right?
But now if I use the second law
of exponents that I have over
there, that's the same thing
as 2 to the k to the power x,
which is the same
thing as some base b^x,
where b is equal to-- Let's
write that down over here.
b is 2^k.
Right.
So whatever it is, if I have
a different base which is
expressed in terms of
2, of the form 2^k,
then that new function is
described by this function
f(kx), the stretch.
So what happens when
you stretch a function?
That's the same thing
as shrinking the x axis.
So when k gets larger, this
corresponding point over here
would be over here, and so
this corresponding point
would be over here.
So you shrink this picture,
and the slope here tilts up.
So, as we increase k, the
slope gets steeper and steeper.
Let's see that explicitly,
numerically, here.
Explicitly, numerically, if
I take the derivative here...
So the derivative with
respect to x of b^x,
that's the chain rule, right?
That's the derivative
with respect
to x of f(kx), which is what?
It's k times f'(kx).
And so if we do it at 0,
we're just getting k times
f'(0), which is k
times this M(2).
So how is it exactly that
we cook up the right base b?
So b = e when k =
1 over this number.
In other words, we can pick all
possible slopes that we want.
This just has the effect
of multiplying the slope
by a factor.
And we can shift the slope
at 0 however we want,
and we're going to do it so
that the slope exactly matches
1, the one that we want.
We still don't know what k is.
We still don't know what e is.
But at least we know that
it's there somewhere.
Yes?
Student: How do you
know it's f(kx)?
PROFESSOR: How do I know?
Well, f(x) is 2^x.
If f(x) is 2^x, then the
formula for f(kx) is this.
I've decided what f(x)
is, so therefore there's
a formula for f(kx).
And furthermore,
by the chain rule,
there's a formula
for the derivative.
And it's k times
the derivative of f.
So again, scaling does this.
By the way, we did
exactly the same thing
with the sine and
cosine function.
If you think of
the sine function
here, let me just
remind you here,
what happens with
the chain rule,
you get k times cosine k t here.
So the fact that we set things
up beautifully with radians
that this thing is, but we could
change the scale to anything,
such as degrees, by the
appropriate factor k.
And then there would be
this scale factor shift
of the derivative formulas.
Of course, the one with
radians is the easy one,
because the factor is 1.
The one with
degrees is horrible,
because the factor is some
crazy number like 180 over pi,
or something like that.
Okay, so there's
something going on here
which is exactly the same
as that kind of re-scaling.
So, so far we've got only one
formula which is a keeper here.
This one.
We have a preliminary
formula that we still
haven't completely
explained which
has a little wavy line there.
And we have to fit all
these things together.
Okay, so now to
fit them together,
I need to introduce
the natural log.
So the natural log is
denoted this way, ln(x).
So maybe I'll call
it a new letter name,
we'll call it w = ln x here.
But if we were reversing
things, if we started out with
a function y = e^x , the
property that it would have is
that it's the inverse
function of e^x.
So it has the property that
the log of y is equal to x.
Right?
So this defines the log.
Now the logarithm has
a bunch of properties
and they come from the
exponential properties
in principle.
You remember these.
And I'm just going to
remind you of them.
So the main one that I
just want to remind you of
is that the logarithm
of x_1 * x_2
is equal to the logarithm of
x_1 plus the logarithm of x_2.
And maybe a few more are
worth reminding you of.
One is that the
logarithm of 1 is 0.
A second is that the
logarithm of e is 1.
All right?
So these correspond to the
inverse relationships here.
If I plug in here,
x = 0 and x = 1.
If I plug in x = 0 and x = 1,
I get the corresponding numbers
here: y = 1 and y = e.
And maybe it would be worth
it to plot the picture once
to reinforce this.
So here I'll put them
on the same chart.
If you have here e^x over here.
It looks like this.
Then the logarithm which I'll
maybe put in a different color.
So this crosses at this
all-important point
here, (0,1).
And now in order to figure out
what the inverse function is,
I have to take the flip
across the diagonal x = y.
So that's this shape here,
going down like this.
And here's the point (1, 0).
So (1, 0) corresponds
to this identity here.
But the log of 1 is 0.
And notice, so this is
ln x, the graph of ln x.
And notice it's only
defined for x positive,
which corresponds to the fact
that e^x is always positive.
So in other words, this white
curve is only above this axis,
and the orange one
is to the right here.
It's only defined
for x positive.
Oh, one other thing I should
mention is the slope here is 1.
And so the slope there
is also going to be 1.
Now, what we're allowed to do
relatively easily, because we
have the tools to do it, is
to compute the derivative
of the logarithm.
So to find the
derivative of a log,
we're going to use
implicit differentiation.
This is how we
find the derivative
of any inverse function.
So remember the
way that works is
if you know the derivative
of the function,
you can find the derivative
of the inverse function.
And the mechanism
is the following:
you write down here w = ln x.
Here's the function.
We're trying to find
the derivative of w.
But now we don't know how to
differentiate this equation,
but if we exponentiate it, so
that's the same thing as e^w =
x.
Because let's just
stick this in here.
e^(ln x) = x.
Now we can differentiate this.
So let's do the
differentiation here.
We have d/dx e^w is equal
to d/dx x, which is 1.
And then this, by
the chain rule,
is d/dw of e^w times dw/dx.
The product of
these two factors.
That's equal to 1.
And now this guy,
the one little guy
that we actually know and can
use, that's this guy here.
So this is e^w times
dw/dx, which is 1.
And so finally,
dw/dx = 1 / e^w .
But what is that?
It's x.
So this is 1/x.
So what we discovered
is, and now I
get to put another
green guy around here,
is that this is equal to 1/x.
So alright, now we have two
companion formulas here.
The rate of change
of ln x is 1/x.
And the rate of change
of e^x is itself, is e^x.
And it's time to
return to the problem
that we were having a
little bit of trouble with,
which is somewhat not explicit,
which is this M(a) times x.
We want to now differentiate
a^x in general, not just e^x .
So let's work that
out, and I want
to explain it in
a couple of ways,
so you're going to
have to remember this,
because I'm going to erase it.
But what I'd like you
to do is, so now I
want to teach you
how to differentiate
basically any exponential.
So now to differentiate
any exponential.
There are two methods.
They're practically
the same method.
They have the same
amount of arithmetic.
You'll see both of them, and
they're equally valuable.
So we're going to
just describe them.
Method one I'm going to
illustrate on the function a^x.
So we're interested
in differentiating
this thing, exactly this problem
that I still didn't solve yet.
Okay?
So here it is.
And here's the procedure.
The procedure is to write, so
the method is to use base e,
or convert to base e.
So how do you convert to base e?
Well, you write a^x
as e to some power.
So what power is it?
It's e to the power
ln a, to the power x.
And that is just e^(x ln a).
So we've made our
conversion now to base e.
The exponential of something.
So now I'm going to carry
out the differentiation.
So d/dx of a^x is equal
to d/dx of e^(x ln a).
And now, this is a step which
causes great confusion when
you first see it.
And you must get used to it,
because it's easy, not hard.
Okay?
The rate of change of
this with respect to x is,
let me do it by analogy here.
Because say I had e^(3x) and
I were differentiating it.
The chain rule would
say that this is just 3,
the rate of change of 3x with
respect to x times e^(3x).
The rate of change of e to
the u with respect to u.
So this is the
ordinary chain rule.
And what we're doing up here
is exactly the same thing,
because ln a, as
frightening as it
looks, with all three letters
there, is just a fixed number.
It's not moving.
It's a constant.
So the constant just
accelerates the rate of change
by that factor, which is
what the chain rule is doing.
So this is equal to
ln a times e^(x ln a).
Same business here
with ln a replacing 3.
So this is something you've
got to get used to in time
for the exam, for instance,
because you're going
to be doing a million of these.
So do get used to it.
So here's the formula.
On the other hand, this
expression here was the same
as a^x.
So another way of writing this,
and I'll put this into a box,
but actually I never
remember this particularly.
I just re-derive it every time,
is that the derivative of a^x
is equal to (ln a) a^x .
Now I'm going to get rid
of what's underneath it.
So this is another formula.
So there's the formula I've
essentially finished here.
And notice, what is
the magic number?
The magic number is
the natural log of a.
That's what it was.
We didn't know what
it was in advance.
This is what it is.
It's the natural log of a.
Let me emphasize to you
again, something about what's
going on here, which has
to do with scale change.
So, for example, the derivative
with respect to x of 2^x is (ln
2) 2^x.
The derivative
with respect to x,
these are the two most obvious
bases that you might want
to use, is ln 10 times 10^x .
So one of the things that's
natural about the natural
logarithm is that
even if we insisted
that we must use base 2, or
that we must use base 10,
we'd still be stuck
with natural logarithms.
They come up naturally.
They're the ones
which are independent
of our human construct
of base 2 and base 10.
The natural logarithm
is the one that
comes up without reference.
And we'll be mentioning
a few other ways
in which it's natural later.
So I told you about
this first method,
now I want to tell you
about a second method here.
So the second is called
logarithmic differentiation.
So how does this work?
Well, sometimes
you're having trouble
differentiating a
function, and it's easier
to differentiate its logarithm.
That may seem peculiar,
but actually we'll
give several examples where
this is clearly the case,
that the logarithm is
easier to differentiate
than the function.
So it could be that this is an
easier quantity to understand.
So we want to relate it
back to the function u.
So I'm going to write it
a slightly different way.
Let's write it in
terms of primes here.
So the basic identity
is the chain rule again,
and the derivative
of the logarithm,
well maybe I'll write
it out this way first.
So this would be d ln
u / du, times d/dx u.
These are the two factors.
And that's the same
thing, so remember
what the derivative
of the logarithm is.
This is 1/u.
So here I have a 1/u,
and here I have a du/dx.
So I'm going to encode this
on the next board here,
which is sort of the main
formula you always need
to remember, which is
that (ln u)' = u' / u.
That's the one to remember here.
STUDENT: [INAUDIBLE].
PROFESSOR: The question is
how did I get this step here?
So this is the chain rule.
The rate of change of
ln u with respect to x
is the rate of change
of ln u with respect u,
times the rate of change
of u with respect to x.
That's the chain rule.
So now I've worked out
this identity here,
and now let's show how it
handles this case, d/dx a^x.
Let's do this one.
So in order to get that
one, I would take u = a^x .
And now let's just take a look
at what ln u is. ln u = x ln a.
Now I claim that this is
pretty easy to differentiate.
Again, it may seem hard, but
it's actually quite easy.
So maybe somebody
can hazard a guess.
What's the derivative of x ln a?
It's just ln a.
So this is the same thing that I
was talking about before, which
is if you've got 3x,
and you're taking
its derivative with respect
to x here, that's just 3.
That's the kind
of thing you have.
Again, don't be put off by this
massive piece of junk here.
It's a constant.
So again, keep that in mind.
It comes up regularly in
this kind of question.
So there's our formula, that the
logarithmic derivative is this.
But let's just rewrite that.
That's the same thing as u' / u,
which is (ln u)' = ln a, right?
So this is our
differentiation formula.
So here we have u'.
u' is equal to u times ln a, if
I just multiply through by u.
And that's what we wanted.
That's d/dx a^x is equal to
ln a (I'll reverse the order
of the two, which is
customary) times a^x.
So this is the way that
logarithmic differentiation
works.
It's the same arithmetic
as the previous method,
but we don't have to
convert to base e.
We're just keeping
track of the exponents
and doing differentiation
on the exponents,
and multiplying
through at the end.
Okay, so I'm going to do
two trickier examples, which
illustrate logarithmic
differentiation.
Again, these could be done
equally well by using base e,
but I won't do it.
Method one and method
two always both work.
So here's a second
example: again this
is a problem when you
have moving exponents.
But this time, we're going
to complicate matters
by having both a moving
exponent and a moving base.
So we have a function u, which
is, well maybe I'll call it v,
since we already had a
function u, which is x^x.
A really complicated
looking function here.
So again you can handle
this by converting
to base e, method one.
But we'll do the logarithmic
differentiation version,
alright?
So I take the logs
of both sides.
And now I differentiate it.
And now when I
differentiate this here,
I have to use the product rule.
This time, instead of
having ln a, a constant,
I have a variable here.
So I have two factors.
I have ln x when I
differentiate with respect to x.
When I differentiate with
respect to this factor here,
I get that x times the
derivative of that,
which is 1/x.
So, here's my formula.
Almost finished.
So I have here v' / v. I'm going
to multiply these two things
together.
I'll put it on the other side,
because I don't want to get it
mixed up with
ln(x+1), the quantity.
And now I'm almost done.
I have v' = v (1 + ln x), and
that's just d/dx x^x = x^x (1 +
ln x).
That's it.
So these two methods always
work for moving exponents.
So the next thing
that I'd like to do
is another fairly
tricky example.
And this one is not strictly
speaking within calculus.
Although we're going to use the
tools that we just described
to carry it out, in fact
it will use some calculus
in the very end.
And what I'm going to do is
I'm going to evaluate the limit
as n goes to infinity
of (1 + 1/n)^n.
So now, the reason why I want
to discuss this is, is it
turns out to have a
very interesting answer.
And it's a problem that
you can approach exactly
by this method.
And the reason is that
it has a moving exponent.
The exponent n here is changing.
And so if you want to keep track
of that, a good way to do that
is to use logarithms.
So in order to figure
out this limit,
we're going to
take the log of it
and figure out what
the limit of the log
is, instead of the
log of the limit.
Those will be the same thing.
So we're going to take the
natural log of this quantity
here, and that's n ln(1 + 1/n).
And now I'm going
to rewrite this
in a form which will make
it more recognizable,
so what I'd like to do
is I'm going to write n,
or maybe I should say it this
way: delta x is equal to 1/n.
So if n is going to
infinity, then this delta x
is going to be going to 0.
So this is more familiar
territory for us in this class,
anyway.
So let's rewrite it.
So here, we have 1 over delta x.
And then that is multiplied
by ln(1 + delta x).
So n is the
reciprocal of delta x.
Now I want to change this
in a very, very minor way.
I'm going to subtract 0 from it.
So that's the same thing.
So what I'm going to do is I'm
going to subtract ln 1 from it.
That's just equal to 0.
So this is not a
problem, and I'll
put some parentheses around it.
Now you're supposed to
recognize, all of a sudden,
what pattern this fits into.
This is the thing which we
need to calculate in order
to calculate the derivative
of the log function.
So this is, in
the limit as delta
x goes to 0, equal to
the derivative of ln x.
Where?
Well the base point is x=1.
That's where we're
evaluating it.
That's the x_0.
That's the base value.
So this is the
difference quotient.
That's exactly what it is.
And so this by definition
tends to the limit here.
But we know what the derivative
of the log function is.
The derivative of the
log function is 1/x.
So this limit is 1.
So we got it.
We got the limit.
And now we just have
to work backwards
to figure out what this limit
that we've got over here is.
So let's do that.
So let's see here.
The log approached 1.
So the limit as n goes to
infinity of (1 + 1/n)^n.
So sorry, the log of this.
Yeah, let's write it this way.
It's the same thing, as
well, the thing that we know
is the log of this.
1 plus 1 over n to the n.
And goes to infinity.
That's the one that
we just figured out.
But now this thing is
the exponential of that.
So it's really e
to this power here.
So this guy is the
same as the limit
of the log of the limit of the
thing, which is the same as log
of the limit.
The limit of the log and the
log of the limit are the same.
log lim equals lim log.
Okay, so I take
the logarithm, then
I'm going to take
the exponential.
That just undoes
what I did before.
And so this limit is
just 1, so this is e^1.
And so the limit that we
want here is equal to e.
So I claim that with this
step, we've actually closed
the loop, finally.
Because we have an honest
numerical way to calculate e.
The first.
There are many such.
But this one is a perfectly
honest numerical way
to calculate e.
We had this thing.
We didn't know
exactly what it was.
It was this M(e), there was
M(a), the logarithm, and so on.
We have all that stuff.
But we really need to nail
down what this number e is.
And this is telling
us, if you take
for example 1 plus 1 over 100
to the 100th power, that's
going to be a very good,
perfectly decent anyway,
approximation to e.
So this is a numerical
approximation,
which is all we can
ever do with just
this kind of irrational number.
And so that closes
the loop, and we now
have a coherent
family of functions,
which are actually well
defined and for which we have
practical methods to calculate.
Okay, see you next time.
