1:30

Certain algebraic properties are manifest and well known to us.

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

Â integral properties. E to the X is that remarkable function

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

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

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

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

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

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

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

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

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

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

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

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

Â 2 factorial + x cubed over 3 factorial, etcetera.

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

Â ever. Now recall that k factorial for a

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â know. In general, the principle that you should

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

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

Â degree. Now, polynomials are wonderful objects to

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

Â calculus. Recall that when it comes to

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

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

Â 5:46

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

Â the x. For example, if we tried to differentiate

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

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

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

Â one? That's clearing zero.

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

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

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

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

Â Etcetera. Following the constants as we go.

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

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

Â gives us simply x squared over 2 factorial.

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

Â one factorial. And what do we observe?

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

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

Â well? Let's see.

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

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

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

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

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

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

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

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

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

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

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

Â front. We could absorb that negative 1 into the

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

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

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

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

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

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

Â squared. That is, i squared times x squared

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â plus x to the fourth over 4 factorial, etcetera.

Â With the signs alternating and with even powers of x.

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

Â 11:20

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

Â have alternate expressions for certain trigonometric functions.

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

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

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

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

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

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

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

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

Â factorial. This gives us the odd powers of x.

Â 12:34

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

Â nice properties, with respect to calculus.

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

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

Â differentiate it, as if it were the long polynomial.

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

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

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

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

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

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

Â recently seen. this is out derived expression for the

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

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

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

Â long polynomial. You could write this out in terms of

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

Â wonderfully convenient and compact way to perform this derivation.

Â