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Certain algebraic properties are manifest and well known to us.

From your prior exposure to calculus, you should have seen some differential and

integral properties. E to the X is that remarkable function

that is its own derivative. And thus, it is its own integral, up to

the constant of integration. There's one other fact that we need to

move forward, and that is Euler's formula that tells us something about

exponentiating a complex input, namely e to the ix is cosine of x plus i times

sine of x. None of these properties however, tell us

what e to the x really means. We therefore move to the definition for

the purposes of this course. This is what e to the x is and means.

We define e to the x as 1 plus x, plus one half x squared, plus one sixth x

cubed, plus one 24th x to the fourth, plus one over 120 times x to the fifth

and this keeps on going and going. Where do these numbers come from?

What do they mean? Well, another way to write this is using

factorial notation. That e to the x is 1 + x + x squared over

2 factorial + x cubed over 3 factorial, etcetera.

All the way down the line, one never stops this sum; keeps going forever and

ever. Now recall that k factorial for a

positive integer k is defined to be, k times k minus 1 times k minus 2, and all

the way down until you get to 3 times 2 times 1.

That gives the sequence of numbers that we saw at the beginning.

Recall also that by convention and for very good reasons, 0 factorial is defined

to be 1. Thus we could write our definition for e

to the x using summation notation. As the sum k goes from 0 to infinity of x

to the k divided by k factorial. This is a formula that you are going to

want to memorize as we will be using it throughout this course.

Well, definitions may be nice, but what do we do with it?

How do we deal with this statement? How do we even make sense of this

infinite sum. Well, certainly for specific values of x,

say x equal to 1, we can try to compute what e to the one would be, as plus one

plus one half plus one sixth plus one 24th, etcetera.

It seems as though this converges to the familiar decimal expansion for e that we

know. In general, the principle that you should

follow in trying to understand statements such as the definition of e to the x is

to pretend that this is a long polynomial; a polynomial of unbounded

degree. Now, polynomials are wonderful objects to

work with, very simple from the point of view of differential and integral

calculus. Recall that when it comes to

differentiation, the derivative of x to the k is k times x to the k minus one.

Likewise the integral of x to the k is x to the k plus one over k plus one.

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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.

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Our conclusion from this rather simplistic manipulation is, that we now

have alternate expressions for certain trigonometric functions.

The cosine of x is 1 minus x squared over 2 factorial plus x to the fourth over 4

factorial minus x to the sixth over 6 factorial, etcetera.

In summation notation, we can use a wonderful little trick to express this

compactly, as the sum k goes from 0 to infinity of negative 1 to the k times x

to the 2k over quantity 2k factorial. That builds in the alternating signs and

the even powers. Likewise, for sine of x, we can write

this in a summation notation, with a similar idea as the sum k goes from 0 to

infinity of negative 1 to the k times x to the 2k plus 1 over quantity 2k plus 1

factorial. This gives us the odd powers of x.

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Now, you may recall that the trigonometric functions have some very

nice properties, with respect to calculus.

For example, you may remember something about the derivative sign of x.

Let's see what happens, when we take our newly derived expression and

differentiate it, as if it were the long polynomial.

The derivative of x is 1. The derivative of x cubed is 3x squared,

we must divide this by 3 factorial. The derivative of x to the fifth and x to

the seventh follow the familiar pattern with a little bit of cancellation of the

coefficents, what do we see? Well, we get 1 minus x squared over 2

factorial, plus x to the fourth over 4 factorial, minus x to the sixth over 6

factorial, etcetera. This is an expression that we have very

recently seen. this is out derived expression for the

cosine of x. And you may recall that the derivative of

sine is cosine. But without any complicated proof, we've

derived this expression very simply, by pretending that everything in sight is a

long polynomial. You could write this out in terms of

summation notation. It looks a bit complicated, but it's a

wonderfully convenient and compact way to perform this derivation.