[MUSIC] Up until now, we've been considering the functions that you can
get by starting with variables and numbers, and combining them using sums,
products, quotients, and differences. So we can write down, you know, functions
like f(x)=x2+x/(x+1^)^10+x, all of this, -1/x.
But there's more things in heaven and earth that are dreamt of in your rational
functions. For instance, can you imagine a function
f, which is its own derivative? I'm looking for a functions, that if I
differentiate it, I get back itself. Now, if you're thinking cleverly, you
might be able to cook up such a function very quickly.
What if f is just the zero function? Or if I differentiate the zero function,
differentiate a constant function, that's zero.
So this would be an example of function in its own deriviative.
But, that's not a very exciting example. [SOUND] So let's try to think of a
nonzero function, which is its own derivative.
How might we try to find such a function? So to make this concrete, I'm looking for
a function f, so if I differentiate it, I get itself and just make sure that it's
not the zero function. Let's have this function output one if I plug in zero.
Now, how could I rig this function to have the correct derivative at zero?
If the derivative of this function itself, the derivative of this function
at zero should also be one. Can you think of a function whose value
at zero is one and whose derivative at zero is one?
Yes. Here is a function, f(x)=1+x.
This function's value with zero is one, and this function's derivative at zero is
also one. But if the derivative of f is f, then the
derivative of the derivative of f is also the derivative of f, which is also f.
So, the second derivative must be f as well.
So, if this function is its own derivative, the second derivative of f
would also be equal to f. Now, specifically, at the point zero,
that means the second derivative of the function at the point zero would be the
function's value with zero which should be equal to one.
is this function's second derivative at zero equal to one?
No. If I differentiate this function twice, I
just get the zero function, but I can fix this at least to the point
zero. If I add on x^/2, now, this function's derivative at zero
is one and this function's second derivative at zero is one.
Since f is its own derivative, the third derivative of f must also be f.
No worries. If the thid derivative of f is also equal
to f, which is a consequence of the derivative of f being equal to f.
That means the third derivative of f at zero is equal to one, but this thing's
third derivative is just zero. But if I add on x^3/6, now, if I take the
third derivative of this function and plug in zero, I get out one.
The fourth derivative of f must also be f.
Okay, yeah. I gotta deal with the fourth derivative.
I'm out of space here, but no worries, I'll just get more paper.
Here, I've written down a function whose value at zero is one, whose derivative at
zero is one, whose second derivative at zero is one, whose third derivative at
zero is one, whose fourth derivative at zero is one.
And you can see, this is sort of building me closer and closer to a function which
is its own derivative. If I try to differentiate this function,
what do I get? Well, the derivative of one is zero, but the derivative of x is
one, and the derivative of x^2/2 is x, and the derivative of x^3/6, well, that's
x^2/2, and the derivative of x^4/24, well, that's x^3/6.
And yeah, I mean, this function isn't its own
derivative, but things are looking better and better.
But the fifth derivative of f must also be equal to f.
Okay, yeah. The fifth derivative.
I'll just add on another term, x^5/120.
And if you check, take the fifth derivative now of this function,
its value at zero is one. I've written down a function,
so that if I take its fifth derivative at zero,
I get one. The sixth derivative of f must be equal
to f. The sixth derivative I am out of room,
but here we, go. Here is a polynomial whose value first,
second, third, fourth, fifth and sixth derivative at the point zero are all one.
And you can see how this is edging us a little bit closer still to a function
which is its own derivative, because if I differentiate this function,
yeah, the one goes away, but the x gives me the one back, and the x^2/2, when I
differentiate that, gives me the x. X^3/6, when I differentiate that, gives
me x^2/2. X^4/24, when I differentiate that gives
me x^3/6. x^5/120, when I differentiate that, gives
me x^4/24. X^6/720, when I differentiate that, I've got x^5/120. And now, of
course, these aren't the same, but I'm doing better.
The seventh derivative must be equal to f.
To get the seventh derivative at zero to be correct, I'll add on x^7/5040.
The eighth derivative, I'll add on x^8/40,320.
The ninth derivative, I'll add on x^9/362,880.
Okay, okay. This is, isn't working out.
We're not really succeeding in writing down a function which is its own
derivative. Let's introduce a new friend, the number
e to help us. Here is how we're going to get to the
number e. This limit,
the limit of 2^h-1/h as h approaches zero is about 0.69, a little bit more.
On the other hand, this limit, the limit of three to the h minus one
over h as h approaches zero is a little bit more than one,
it's about 1.099. If you think of this as a function that
depends not on two or three, you could define a function g(x), right?
The limit as h approaches zero of x to the h minus one over h. In that case,
this first statement, the statement about the limit of two to the h minus one over
h, that's really saying that g(2) is a bit less than one.
And, this statement over here, and if you think of this as a function g, this
statement is really saying that g(3) is a bit more than 1.
Now, if you're also willing to concede that this function g is continuous, which
is a huge assumption to make, but let's suppose that's the case.
If that's the case, I've got a continuous function, let's say, and if I plug in
two, I get a value that's a little bit less than one,
and if I plug in three, I get a value that's a little bit more than one.
Well, by the intermediate value theorem, that would tell me there must be some
input so that the output is exactly one. I'm going to call that input e.
In other words, e is the number, so that the limit of e^h-1/h as h approaches zero
is equal to one, and this number is about 2.7183 blah,
blah, blah. Now lets consider the function f(x)=e^x.
So let's think about this function f(x)=e^x.
Now, what's the derivative of this function? Well, from the definition,
that's the limit as h approaches zero of f of x plus h minus f of x over h.
Now, in this case f is just e^x, so this is the limit as h approaches zero
of e to the x plus h minus e to the x over h.
And this is e to the x plus h minus e to the x over h,
so I can write this as e to the x times e to the h.
This is the limit then as h goes to zero of e to the x, e to the h minus e to the
x over h, Now I've got a common factor of e to the
x. So I'll pull out that common factor and
I've got the limit as h approaches zero of e to the x times e to the h minus one
over h. Now, as far as h is concerned, e to the x
is a constant, and this is the limit of a constant times
something, so I can pull that constant out.
This is e to the x time the limit as h goes to zero of e to the h minus one over
h. But I picked the number e precisely, so
that this limit was eqal to one. And consequently, this is e to the x
times one, this is just e to the x. Look,
I've got a function whose derivative is the same function.
We've done it. We've found a function which is its own
derivative. The derivative of e^x is e^x.
E^x is honestly different from this polynomials and rational functions.
We couldn't have produced that number e without using a limit.
E^x is the function that only calculus could provide us with.