Welcome to Digital Calculus. I'm your synthesized, digitized Professor
Greist. We're about to begin Lecture 47 on the
Discrete Calculus. [SOUND]. In our last two lessons, we have covered
sequences and a discrete form of differentiation
that works on sequences. In this lecture, we're going to go fully digital and build a discrete calculus for
sequences. [SOUND]. The discrete calculus mirrors the differential calculus in the following
ways. The discrete version of a function is a
sequence. The discrete version of a derivative is a
difference. The discrete version of an integral is a
sum. And the discrete version of a differential
equation is a recurrence relation, also called a recursion relation. [SOUND]. Let's start with something big, something
fundamental. The Fundamental Theorem of Integral
Calculus has a digital form. It takes the form of the sum, as n goes
from A to B and the forward difference of a sequence U
is equal to that sequence U.
Evaluated from N equals A to N equals B plus one.
Now what does that mean? That means you evaluate the
anti-derivative that is the anti-difference at the end points, but notice, that there's a, a
little bit of a difference here. We have to add a plus one to the upper
limit. That's in a case of a forward difference, for backward difference, there
is one subtracted from the lower limit. Now what's the proof of the Fundamental
Theorem of discrete calculus. Well, let's move this sum up, and expand
out exactly what it means. Remember, the forward difference says, we
take the next term in the sequence, minus the
current term. So if we add up all of the forward
differences as n goes from A to B.
Then what do we see? We see that most of the terms cancel, because you have a plus, and then a minus
of the same term. The only things that do not cancel are the
first term, negative U sub A and the last term, plus U sub B
plus 1. And that is, indeed, as we have written
it, the sequence evaluated at the limits. Now, sometimes, these kinds of sums are
called Telescoping Sums. But if you ever see reference to that, now you know you're really seeing the
fundamental theorem in disguise. [SOUND]. Here are a few examples of the fundamental
theorem of discrete calculus in action. Since the forward difference of the
sequence 1 over n is equal to, by definition 1 over n plus 1, minus 1 over
n, which when simplified is negative 1 over n squared
plus n, then we can, if you like, integrate both sides.
To obtain, an explicit formula for the sum as n goes from A to B
of one over n squared plus n.
It is the anti-derivative, if you will. Negative one over n evaluated from A to B
plus one. This can be written with a little bit of
algebraic simplification. As B minus A plus 1 over A times quantity
B plus 1. Here's another example. Since the forward difference of the
sequence n factorial is, by definition, n plus one factorial
minus n factorial. Which after factoring out n factorial
gives n factorial times n, then we get an unusual looking formula.
The sum, as n goes from A to B of n factorial times n, is equal to n
factorial, evaluated from A to B plus one.
That's simply B plus one factorial minus A factorial, very, very simple. [SOUND]. Now, since we know a few things about the differences of falling powers, we can say
some stronger things. For example, the sum as n goes from one to
k of n. We've seen that sum before. Here's a way to get it by means of
integration. This is really the sum of n to the falling
one power. What's the anti-difference of that?
Well, it's one half n to the falling two . When we evaluate that from one to k plus
one, we get k plus one times k over two. Now, you've seen this formula before. But I bet you've never seen this simple
derivation. Here's one that's less obvious, the sum as
n goes from one to k of n squared, we can rewrite n squared as n to the
falling two plus n to the falling one. We know how to anti-differentiate these
following powers and so we get one third n to the falling
three. Plus one half end of the following two
evaluated from one to k plus one. I'm going to let you do the algebra here
to obtain the final answer of k times k plus one times two k plus one
over six. [SOUND]. By reversing the product rule for forward
differencing, it's possible to obtain an integration by parts formula.
Here's what it looks like. The sum, as n goes from A to B of U, times
the forward difference of V is equal to U
times V evaluated from A to B plus one.
Minus, the sum from A to B of Ev, the forward shift of v times the
difference, delta u. I'm not going to go through a proof of this. We'll leave that to some homework
problems. Let's just do a simple example. The sum, as n goes from zero to K of n
times 2 to the n, is equal to what?
Well, let's let u be equal to n, because I know what the forward difference
in that is, that's simply one. And, we're going to let the v, and hence v, be the exponential sequence, two to the n.
Now, in this case, what does the integration by
parts formula tell us. It tells us that this sum is equal to U
times v, that is N times 2 to the n, evaluated from zero to K plus one,
minus the sum of the forward shift of v. That's 2 to the n plus 1, times the
forward difference out of n. Well that forward difference is simply
one. And so, we get when we evaluate that
middle term, a plus one times two to the k plus one minus zero.
And then we have to subtract the sum. From zero to K of, two to the n plus one. That sum is not so bad, we know how to do
that. We get, after a little bit of algebra, the
final answer of k minus 1 times 2 to the k plus 1, plus 2. This is a highly non-obvious result, and a
good example of integration by parts. [SOUND]. Let's turn to differential equations. These are are recurrence relations, or
sometimes recursion relations. The shift operator E acts something like a
derivative in discrete calculus. Let's look at a linear first order
recurrence relation. This is something of the form, u sub n
plus 2 equals lambda times u sub n, where lambda
is a constant. Now we can rewrite this using the language
of operators. As Eu equals lamda times u, or moving the lamda u over
to the other side and factoring out the u. We have have the operator E minus lambda I
applied to u equals 0. Now, it's obvious what the solution to
this recurrence relation is, u has to be some constant times lambda to the n.
Why is that? Just look at the recursion relation, un
plus 1, the next term, equals lambda plus un the previous term.
What is that constant big C? Oh, of course, that is the initial
condtion for the quote differential equation.
It is U not A constant. So at each step you are simply multiplying
by lambda, that is the simplest differential
equation in discrete calculus. [SOUND]. Now if we change things just a little bit and look at a linear first order
difference equation. This is something of the form delta u
equals lambda u. Now using what we know about operators, we
can say well delta is really E the shift minus
I the identity. And so what we're really saying is that E
minus I applied to u is lambda u.
Can we solve this? Moving things over to the other side, we
see that this difference relation, reduces to
the recurrence relation. E minus quantity lamda plus one is applied
to u equals zero. We already know the solution to this, it
is constant times quantity, lambda plus one to the n.
Now this is extremely significant. If we compare this to the differential
equation, x prime equals lambda x. We know what the solution is. It is the constant, the initial condition,
C, times E to the lambda t. Now, comapre what's happening here.
This tells you why the exponential sequence two to the n is really the
discrete exponential function. It is the solution to the difference
equation delta u equals u. In other words, it's the solution to this difference relation when lambda
equals one. But notice, according to our solution, it's lambda
plus one to the n and since one plus one equals two. We see that two to the n is the proper version of e to the x in discrete
calculus. [SOUND] .
Let's go back to the Fibonacci sequence. This comes from something like a second
order differential equation. Really, a second order recurrence. Recall, the Fibonacci sequence satisfies.
F sub n plus two equals F sub n plus one plus F sub n.
If we move everything over to the left hand side of the equation and write out
using operators what do we get? E squared minus E minus I applied to F, equals zero.
If we consider that as a second-order polynomial and factor this, just like E was a variable and I
were a variable, then what would we get?
We would get E minus phi I. Times E minus CI, applied to F equals
zero, where fee and C are those two constants
related to the golden ratio that we see whenever we find
Fibonacci. The solutions to this recurrence relation
are built from solutions to the individual first order
pieces. That is, E minus phi I applied to a
sequence F1 equals 0. Or E minus CI applied to a sequence F2
equals 0. You don't quite know why this is true yet. You'll learn that in multivariable
calculus. But lets just follow the conclusion.
Namely that the first type of solution is of the form constant.
C one time phi to the n, the second type of solution is of the form, constant C2
times C to the n. [SOUND]. The general solution as I have claimed, is
really a combination of these primal solutions, these peieces.
The Fibonacci sequence, F sub n, is of the form constant
C1 times phi to the n plus constant C2 times C to the
n. What are these constants?
Well we know what Fibonacci sequence is when n equals zero.
The 0th term is of course 0 applying n equals 0 to our general
solution gives C1 plus C2. We know that the next term in the
Fibonacci sequence F one equals one. That is C one times phi plus C two times
C. This is a system with two equations and
two unknowns. We can solve that for C one and C two.
C one is equal to one over root five. C two is negative one over root five.
Applying that to our general solution for the Fibonacci sequence of Fs
of N we get an exact answer. Fs of N equals one over root five times
phi to the n minus one over root five times psi to the n.
That's a pretty amazing result. One that comes from a little bit of
discrete calculus. [SOUND]. Well, that's the end of our introduction
to discrete calculus. But that's by no means the end of this
subject. There's a lot more to it, but it's not
usually covered by calculus class. [SOUND]. I hope you've enjoyed this foray of the
world of discrete calculus. I hope you didn't find it too confusing. I hope it was a little bit inspiring, but
if it was confusing don't worry. In our next lesson we're going to move on
to some more concrete applications. [SOUND]