Welcome to calculus. 
I'm Professor Ghrist. We're about to begin lecture two on the 
exponential. In this lesson, we'll introduce the 
principle hero of our story: the exponential function, and dive deeper 
into its definition and the implications thereof. 
We'll also see how Euler's formula entwines the exponential function with 
our principle supporting actors, the trigonometric functions. 
We begin this lecture, and indeed the course with the question, what is e to 
the x, the exponential function. Well, of course it's a function. 
Hence we can plot the graph of this function and consider what the output e 
to the x looks like for various inputs x. But how do we compute some of these 
outputs? Let's say for an irrational input, like 
pi. How would we even make sense of 
exponentiating an imaginary or a complex number. 
Is it possible to give meaning or sense to exponentiating an operator like the 
derivative, or more unusual objects that you may have seen such as matrices. 
Well, we won't answer all of those questions today, but let's recall a few 
facts about the exponential function. 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. Don't forget the arbitrary constant, and 
don't forget that something unusual happens when k is equal to negative one. 
Both of these properties, should be familiar from your previous exposure to 
calculus. 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. Likewise, the sin of x must be the 
imaginary portion of this expression. That is, x minus x cubed over 3 
factorial, plus x to the fifth over 5 factorial, etcetera. 
With odd powers and alternating signs. 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. 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. I'll leave it to you to check, that you 
can do this same thing with the derivative of cosine of x, obtaining 
minus sine of x. We now have new interpretations for the 
exponential function as well as for sine and cosine. 
Don't panic if these feel unfamiliar, take a look at the bonus lecture to see a 
more concrete perspective and to hear a few words of comfort. 
Be sure to take a look at that before moving on to our next lesson, where we 
introduce a series perspective for a broader array of functions.