1:18

For this reason we're going to reconsider the notion of a sequence.

Now we have seen sequences before, but here is a slightly different perspective.

A sequence is a discrete or digital input function with an analog output.

That is, the inputs are natural numbers.

The outputs are reals.

Now sequences are everywhere, from economic data to digital signals.

But thinking of them as a function with an input, say n,

and an output, a sub n, is going to be the focus this chapter.

The notation that we'll use sometimes is writing out the terms of the sequence.

That is the outputs, a sub 0, a sub 1, a sub 2, etc.

3:09

How would you describe the sequence 1, -1, 1, -1, repeating?

Well seems a little odd at first, but

one could describe this as a trigonometric function, for

example as cosine of n times pi.

Now more interesting sequences are out there.

For example, what if we remove the pi and consider the sequence cosine n?

If we plot the terms of that sequence as a function of n,

we see a curios mixture of wavelike behavior.

But non-repeatability the sequence never repeats and

in that way it's a little bit unlike the trigonometric function cosine.

This phenomenon is rather curious.

You've seen it before if you've ever seen an object that spins so

fast that your visual refresh rate can't keep up with it and

it looks like it's spinning at a different rate or even in a different direction.

This is sometimes called aliasing.

Let's begin our construction of discrete calculus with the notion of a limit.

Now what does it mean to take the limit of a sequence?

Well, I believe our adventure begins with failure because of the discrete input.

It doesn't make sense to talk about getting as close as you want to n = 11.

The one circumstance under which it does makes sense, is to take a limit

as n goes to infinity in this setting.

Then the original definition of the limit as n goes to infinity makes sense.

We say that that limit is L if for every epsilon

we find some value, let's say capital M,

past which the output of the function lies within epsilon of the limit.

This has to continue for any value of epsilon that is desired.

As we change the tolerance on the output we can find a new

tolerance on the input in order to satisfy the condition.

7:14

Solving for the limit as n goes to infinity, follows the exact same method.

We take the log of both sides, pull that log inside the limit, and

use it to remove the exponent n out in front.

Then we use our knowledge of Taylor series to expand that logarithm

as alpha over n +big O of 1 over n squared.

And this is the key point, that our use of Taylor series and

big O works just as well in the discrete setting as in the continuous.

And so we can evaluate the limit, exponentiate, and

get an answer that should be very familiar.

8:04

Another example would be a sub n = 2 n squared-

n square root of quantity 4n squared + 5.

We would attack this by doing some algebraic

manipulation, factoring out a 2 n squared.

What is left is of the form 1- square root of 1 + 5 over 4n squared.

Here we see another Taylor Series lurking, that involving the square root.

Using the binomial series we can obtain,

after a little bit of algebraic simplification,

that the leading order term is negative five-fourths.

Everything else is in big O of (1 over n squared).

And so when we take the limit, we obtain negative five-fourths.

Now not every limit is so transparent.

For example, what would you do if asked to evaluate something of the form square root

of 2 + square root of 2 + square root of 2 + square root of 2, etc., all nested.

Well, the appropriate way to handle something like this is to realize it

as the limit of a sequence where we build things up step by step,

beginning with root 2 and then with root 2 + root 2 etc.

We want to compute the limit of this sequence, a sub n.

9:43

Now in order to do so, there's one difficult and critical step.

That is, you need to determine the recursion relation that the terms satisfy.

In this case, I'll tell you that it is the following.

a sub n = square root of 2 + a sub n-1.

Let's say I give you that recursion relation.

How do you compute the limit?

Let us, as before, denote this limit by L and

then apply the limit to the recursion relation.

On the left, we have the limit, as n goes to infinity, of a sub n.

That's L.

On the right we have an a sub n-1 term.

What's the limit of that as n goes to infinity?

Well of course, it too must be L.

Now we're assuming that we can slip that limit in under the square root,

let's forget about whether that's legal or not, and

see what happens when we try to simplify this algebraically.

Squaring both sides, we obtain L squared = 2 + L.

With a little bit of rearrangement, we get a polynomial

that easily factors into (L-2) and (L+1).

Now, this polynomial has two solutions,

namely negative 1 and positive 2.

We know that the limit of this sequence is not a negative number.

And therefore, if the limit exists, it must be equal to 2.

Indeed the limit does exist and is 2.

Let's look at another limit of this form.

In this case, 1+1 over 1+1 over 1+1,

etc., all nested together.

As before we can write out a sequence that gets closer and closer to this.

a naught is 1, a1 is 1 over 1 + 1, a2 is 1 over 1 +

1 over 1 over 1 or something like that, I don't know.

It keeps going.

The tricky part in this case is to find

the recurrence relation that these terms satisfy.

Once again, I’m going to tell you what it is.

It is that a sub n = 1 + 1 over a sub n-1.

This instruction tells you how to build the next term.

In order to compute the limit of the a sub ns, we follow the same

procedure as before, taking the limit of the recursion relation,

substituting in L for a sub n, and a sub n-1.

And then, with a little bit of algebraic rearrangement, what do we see?

We get L squared- L- 1 = 0.

This has roots 1 plus or minus square root of 5 over 2.

One of these is negative, the other positive.

We'll take that positive square root, and that is our limit.

This particular value is of some independent interest.

It is sometimes called the golden ratio and given the symbol phi.

It equals 1 + root 5 over 2.

Now perhaps you've seen the golden ratio before in

the context of some other sequences.

Perhaps you've seen the Fibonacci sequence,

that begins with 0, 1, 1, 2, 3, 5, 8,

13, 21, 34, 55, 89, etc.

There's a lot of interesting mathematics hiding behind here.

And if you'd like, you might want to take a look at the bonus material for

a hint at what calculus will be able to tell us about the Fibonacci sequence.