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And what follows, I going to give you is the language of operators.

Â Where for purposes of this course, an operator is something that takes as it's

Â input function and returns as it's output another function.

Â What are some examples of operators? Well, there's one that we've been looking

Â at rather closely. The derivative, I'm going to give it an

Â operator symbol as a capital D. Then, what does capital D do?

Â It takes a function f and returns the derivative df dx.

Â Now, what's so special about this? Well, operators have an algebra

Â associated with them. For example, we can take D and raise it

Â to certain powers. We could talk about D squared, as meaning

Â we apply D twice. You take the derivative, then you take

Â the derivative again. So, the operator D squared is really

Â taking the second derivative, D cubed is taking the third derivative.

Â What is D to the 0? Well, that too has a name.

Â It is called the identity operator, and is given the special symbol I.

Â The identity operator takes f and returns f.

Â It does nothing. That is why it is called the identity.

Â So, d to the zero; the zeroth derivative of f is of course, f.

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

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

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

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

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

Â the number 1 in the algebra that will follow.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â language for doing algebra with these operators.

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

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

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

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

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

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

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

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

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

Â f. So, let's use our differentiation

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

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

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

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

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

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

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

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

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

Â What is that? Where have I seen something like that

Â before? This really looks like an exponential,

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

Â differentiation operator and applying that to f.

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

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

Â differentiation operator. That is really a deep observation.

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

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

Â The shift is the exponential of the differenciation operator.

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

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

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

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

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

Â We will continue asking this question throughout the semester.

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

Â layers of understanding what we mean, by exponentiation.

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

Â information about a function.

Â