What happens when we take powers of I? Well, I squared means do nothing I, and

then nothing again. So, I squared is equal to I.

This might remind you of a number with which you're somewhat familiar, that is

the number 1 under multiplication. 1 times 1 is 1.

I squared is equal to I. This identity operator plays the role of

the number 1 in the algebra that will follow.

There are a few other interesting operators that we're going to take a look

at. One, let's call it E, capital E.

This corresponds to a shift of f. E of f is going to be the function which

when evaluated at x, gives you f at x plus 1.

This has the effect of shifting the graph of f to the left by one unit.

Now, what happens when we take E squared? That means shift to the left by one, then

shift to the left by one again. We can talk about E cubed, E to the

fourth, etcetera. We're continually shifting f to the left.

Now, can you talk about E raised to a non integer power?

E to the H. Well, surely that makes sense, we can

simply evaluate f not at x, but at x plus h.

And this means of course that we can talk about E to the negative 1, which we would

call a right shift. Instead of evaluating f at x plus 1, it

evaluates at x minus 1. But note, that this right shift, E

inverse really is the inverse of E. If I multiply E to the negative 1 times

E, I get E to the 0 or the identity operator.

And of course, this works whether I multiply E inverse by E, or E by E

inverse. And this means that I now have a a nice

language for doing algebra with these operators.

Let's see what happens when we think in this language for a little bit.

What is E to the h of f evaluated at x. This is a shift by h.

I need to take f and evaluate it at x plus h.

Now, let's use our intuition from Taylor series.

If I take the Taylor expansion of this, it is the sum k goes from 0 to infinity

of 1 over k factorial, times the kth derivative of f evaluated at x times h to

the k. Now for the moment, let's not worry about

where we evaluate this so much. Let's try to rewrite this in the language

of operators. Now notice, there's a kth derivative of

f. So, let's use our differentiation

operator D to the k for taking that kth derivative of f.

And now, we see that there are two terms that are raised to the kth power.

D, the differentiation operator and h, this perturbation to the input.

So, if I collect those two terms together and say, I've got the sum k goes from 0

to infinity of 1 over k factorial times quantity hD to the k, I apply that to f

and evaluate it at x. That is what this shift is.

Now, step back for a moment and take a look at what we've done.

We've really said that this operator, E to the h of f, is equal to the sum k goes

from 0 to infinity of something to the k over k factorial applied to f.

What is that? Where have I seen something like that

before? This really looks like an exponential,

but what am I exponentiating? I am exponentiating h times D, the

differentiation operator and applying that to f.

If we remove the dependents on f, then what we see is that this shift operator,

capital E, to the h, is precisely the exponentiation of h times the

differentiation operator. That is really a deep observation.

That the shift E, just moving to the left by one unit is I plus D plus 1 half D

squared plus 1 over 3 factorial, D cubed, et cetera, et cetera.

The shift is the exponential of the differenciation operator.

Or, if you like, you can think of the differentiation operator D as the natural

logarithm of the shift operator. Now, this probably, is causing a little

bit of surprise or confusion. What's going on?

Well, don't worry too much. But think back to how this course began.

We asked the question, what is e to the x?

We will continue asking this question throughout the semester.

What we have seen is that there are many responses to this question, and many

layers of understanding what we mean, by exponentiation.

What we mean by differentiation and how derivatives conspire to give you deep

information about a function.