Given these facts about polynomials, let's see what we can observe about e to

the x. For example, if we tried to differentiate

e to the x by using our definition, then what would we obtain?

Well, thinking of u to the x as a long polynomial in x, allows us to apply what

we already know. For example, what is the derivative of

one? That's clearing zero.

The derivative of x is clearly 1. What is the derivative of 1 over 2

factoral times x squared. Well, it's 1 over 2 factorial times the

derivative of x squared, which is 2x. We can continue on down the line, taking

the derivative of x cubed, to be 3x squared.

Etcetera. Following the constants as we go.

Now a little bit of simplification tells us that the 2x divided by 2 factorial

gives us simply x. The 3x squared divided by 3 factorial

gives us simply x squared over 2 factorial.

This pattern continues since k divided by k factorial is one over quantity k minus

one factorial. And what do we observe?

We observe that we obtain the definition of e to the x by simply following what

seemed to be the obvious thing to do. Will that work if we try to integrate as

well? Let's see.

If we try to integrate our definition, v to the x.

1 plus x plus x squared over 2, etcetera. What will we get?

While the integral of 1 gives us x, the integral of x gives us one half x

squared. If we have a 1 over 2 factorial times the

integral of x squared, that's 1 3rd x cubed.

Now, I'll let you follow this pattern all the way down the line, and see that with

a little bit of simplification, we wind up getting, not quite e to the x.

It appears as though, we're missing the first term.

We're missing the 1 out in front. So now, we've obtained e to the x minus

1, that's not quite the way I remember the integral of e to the x going.

However, we have forgotten as one often does, the arbitrary constant out in

front. We could absorb that negative 1 into the

arbitrary constant, and what we've obtained is up to a constant e to the x.

That bodes well. Let's see what else we can do.

We'll recall Euler's formula that tells us something about exponentiating i times

x in terms of cosines and sines. What happens if we apply our definition

of the exponential in this case. If we want to take e to the i times x.

Well, this is 1 plus i times x, plus 1 over 2 factorial times quantity ix

squared. That is, i squared times x squared

etcetera, etcetera. There are a lot of terms here.

Then it appears as though there are some simplifications that we can do.

Recall that by definition, i squared and the square root of negative 1 squared,

must be negative 1. Therefore, if we look at i cubed, we have

to get negative i. And i to the fourth, being i squared,

squared, must be equal to 1. Therefore, we have a sick-lick pattern in

our powers of i that allows us to simplify this expression as 1 plus ix

minus x squared over 2 factorial, minus ix cubed over 3 factorial plus x to the

fourth over 4 factorial etcetera. You can see the pluses and the minuses

coming in alternating pairs, and the real versus imaginary terms alternating with

each term. Now, if we were to do what we do when we

work with complex numbers and collect all of the real terms into one part, and all

of the imaginary terms into the other then what would we obtain?

Well, the real portion of this expression is, 1 minus x squared over 2 factorial,

plus x to the fourth over 4 factorial, etcetera.

With the signs alternating and with even powers of x.

From Euler's formula, that must be the cosine of x.