PROFESSOR: Welcome
back to recitation. In this video, I want us to
compute the following limit. It's the limit is n goes to
infinity of the sum for i equals 0 to n minus 1 of the
following, 2 over n times the quantity 2i over n
quantity squared minus 1. Now this might look
a little intimidating to try and take a
limit of this, but what I'd like you to do,
as a hint to you, is that you should think about
this as being potentially a Riemann sum of a
certain function. So if you can figure
out the function, and you can figure out
the appropriate interval that you're taking a Riemann
sum over, as n goes to infinity, you should be able to
write this as an integral. We know how to use the
fundamental theorem of calculus to determine that a definite
integral in many cases. Hopefully this is a function for
which we know a way to do that. So that's my hint to you. Think about it, it's a Riemann
sum approximating an integral, and I'll give you a
while to work on it, and then I'll be back. OK, welcome back. Well hopefully it's been fun
for you to look at this problem so far. Let me just remind you
what we were doing. We were trying to
compute a limit as n goes to infinity of the
sum from i equals 0 to n minus 1 of 2 over n times 2i
over n squared minus 1. So I gave you the big hint
that this is probably going to be written as an integral. So let me show you some
pieces of this sum that should help us see what the
integral is, and then I'll make a guess about
what this is, and then I'll try to give an educated
way to check my guess. So the first thing we noticed
is that there is one thing, this is a product of two
functions, and one of them-- well, of n, I guess. But this is a product
of two things. One thing appears
in every single term that you have for i. So the sum has n
terms, and they're all going to be 2 over n times this,
and the i is going to change. But this does not change, this
2 over n does not change, right? In fact, I could even
pull that out if I wanted. But, I don't want to pull
it out of the sum right now. I want us to look
at what's actually going on in this product. So if this thing is appearing
over and over again, and we know this is
probably a Riemann sum, then chances are
this is our delta x. So delta x being
equal to 2 over n, we know delta x equals b minus
a over n, where b and a are our left endpoint-- oh,
sorry, our right endpoint-- and our left endpoint. Right? We integrate from a to b. So b minus a is the
length of the interval. So this is really
dividing up whatever interval we're integrating
over, into n equal subintervals. So that's my first thought,
is that b minus a over n is equal to 2 over n. And now we want to try
and figure out, well, what the heck is this. Well, when we take
a Riemann sum, remember when we take a
Riemann sum what we get. We get the sum of delta
x times f of x sub i, and i is what's varying
from 0 to n minus 1. Let me put a little
curve in here so we see those are
two different things. So this is i equals
0 to n minus 1. I have this delta x here. I'm anticipating this
is some f of x sub i. And so the question
is, what f is it? Right? If I know what f it is, than
I know that this sum will be equal to something,
the integral from a to b of f of x dx, and a
and b will differ by 2. So that's where I'm heading. So now the question is,
what is this a function of? What function is
this, I should say. Now first guess
would be something like, well, I'm taking some
quantity, I'm squaring it, and subtracting 1. So my first guess for this
function is x squared minus 1. I mean, that seems easy to me. Let's see if this would
actually even make sense just by looking at the subscripts,
or sorry, the index, the indices I have here. So what do I have? Well, when I put in
i equals 0-- let's put down some of these values--
when I put in i equals 0, I get 2 times 0 over
n squared minus 1. When I put in i equals
1, I get 2 times 1 over n squared minus 1. And I go all the
way up, to 2 times n minus 1 over n squared minus 1. So it's kind of a long
sum there, but these are, this is what our
sum of these things looks like if I pull
out the 2 over n. So here I get 0 squared minus 1. That looks pretty good. Here I get 2 times 1
over n squared minus 1. So it does look like I'm doing
something, taking something, squaring it, subtracting 1. Does it make sense
that these are the kind of x values
I would expect to get if this were the Riemann
sum of x squared minus 1? It does, and let's
think about why. I'm starting at x equals 0
here, it sure looks like. Let's look at what happens when
I go all the way over here. What happens when n gets
really, really big, is it this ratio approaches 2. So it's 2 times n minus 1
over n. n minus 1 over n, as n gets arbitrarily
large, as n gets really big, then this approaches 2. So this is approaching
2 squared minus 1. So it's giving me more
evidence that this is probably the function x squared minus 1. And now I'm starting to
believe the interval is 0 to 2. I know it's a length 2
interval, and it's looking like the interval is 0 to 2. Let's come back and talk
about one more thing. The one other thing
that you should notice is that how does
this value differ from this value, and the next,
and the next, and the next? They differ by 2 over n. So each time whatever
input I had previously, I'm now adding 2 over
n to the next input. And that should
make sense of what we know about Riemann
sums, because what I do, is I divide my interval
into these subintervals of length 2 over n. I'm evaluating it
at one x-value, that-- I'm starting,
in this case, at 0. Then the next interval
is 2 over n more. Then I evaluate at that x-value. The next one is 2 over n more,
and I evaluate at that x-value. So this is looking like--
I'm going to write it here, this is my guess-- integral from
0 to 2 of x squared minus 1 dx. And now to make myself
feel good about this-- I'm pretty sure it's that. To make you feel
good about this, I'm going to divide this
into four subintervals, and I'm going to show you
what the Riemann sum with four intervals looks
like, and then we can talk about how it relates
to this one over here. OK, so let me draw a graph. Actually, I'll use
just white chalk again. Let me draw a graph of x
squared minus 1 from 0 to 2. So 0, 1, 2, minus 1. OK, so at 0, x squared
minus 1 is negative 1. At x equals 1, x
squared minus 1 is 0. And at 2, x squared
minus 1 is 3. So hopefully, this is all
going to go into the video, and-- in the video
screen, I mean. And there we go,
something like that. So this is, you know,
it continues over here, but I'm really only
interested in this part. So now let's look at what
the subintervals are. And now I'm going to
get some colored chalk. So what are the subintervals? I'm taking 1 over 4, OK? And so delta x, in
this case, is 2 over 4, which is equal to 1/2. Right? And so what are my,
what are-- so what is my sum going to look like? Well, I am going to tell
you that I'm also going to use left-handed endpoints. And I mentioned earlier
why that is, I believe. Maybe I didn't. But, I started
off at i equals 0, and my first input value was 0. My last input value had an n
minus 1 in it instead of an n. So I was doing, somehow,
the second-to-last place that I was interested in here. So it's definitely more of
a left-hand endpoint thing. So I'm going to do this
with left-hand endpoints. And I'm not going
to simplify as I go. I'm going to write it out
in sort of an expanded form. OK, so let's write it
out in expanded form. So the Riemann sum-- this is
y equals x squared minus 1. The Riemann sum is, the
first term is 1/2 times what? It's the value,
this x-value, which is 0, evaluated on this
curve, so 0 squared minus 1. The next term-- I'll just write
them right below each other-- is 1/2. 'Cause again, let's
draw a picture of what the first one is, sorry. It's this rectangle. Right? It's evaluated--
It's length 1/2, and it's the function
evaluated at 0. The next one is
length 1/2, and it's going to be the
function evaluated at whatever this left-hand
endpoint is here. So it's going to be this area. So it's going to be
length 1/2, and then the height is going
to be at x equals 1/2, so 1/2 quantity squared minus 1. The next one is going
to be this interval. Well, there's no rectangle
to draw because it's just, the output is zero at
the left endpoint here. So it's just going
to be-- it's going to have output equal to 0,
at length 1/2 and height 0. But we'll write it out anyway. It's going to be 1/2
times the quantity-- now, I went up 1/2 more, so
it's going to be two 1/2's, two times 1/2 squared minus 1. Let me just show
you why I did this. OK, if we look at
the picture, here I'd gone up 1/2 from
my initial value. Here I'd gone up another
1/2 from my initial value. So that's two 1/2's from
my initial value of 0. The next one is going to
be three 1/2's, so this is three 1/2's away, or
commonly known as 3/2. OK? So that one is going to be--
1/2 is the base length again, times the quantity 3
times 1/2 squared minus 1. And that is in the
picture, this rectangle. Great. So what are we see here? If we look at this, these
four pieces, what do we have? We have a 1/2 in
front each time. Which, what was the 1/2? It was b minus a over n. So b minus a was 2, n was 4. So maybe I should have
kept that as 2 over 4. But, it's a little
easier to write it as 1/2 because of what I'm doing next. I square something,
I subtract 1. I go up by the value that this
is from the initial one here. And so now I'm taking the
output of what was in here. I now take the output at
1/2 more than what was here. Now I take it at two 1/2's more
than what was here, or 1/2 more than what was there,
and then three 1/2's more than what was here, or
one more than what was there. That's kind of confusing, but
let's go back to the picture and see what it is. My delta x was 1/2, right? So I evaluate at
the first place, and I evaluate one more up, and
then I evaluate one more up, and I evaluate one more up,
which gives me outputs here, there, there, and there. Right? So really if you go
back and you look at the formulation of the
sum, this was 2 over n times quantity 2i over
n squared minus 1, you can see the 2
over n is my 1/2, and then this is maybe
the hardest part to see, but that's the 2 over
n is my 1/2 again, and the i is this thing
that's coming in as 1, 2, 3. And so that i was going
from 0 to n minus 1-- so I should have said 0, 1, 2, 3. Right? So that i is the 0 to n minus
1, and then I'm evaluating that, and then I add them all up. So when I take the sum, I get,
for n equals 4, I get this. So in fact, this
is just a guess, still maybe you
should, maybe you should convince yourself more. I'm actually convinced
at this point, based on not just this
evidence, but the evidence I understood before about
how the function works. Maybe you want to
compare it when n equals 6, or something like that. You may need a
little more evidence to make you understand
this particular piece. But, hopefully that
makes sense to you that this is really just
i times delta x, and then evaluated somewhere. That's the main idea
of this component. OK, well hopefully this
is informative to you. If you want to know
the exact answer of how to compute the sum, obviously
you just take the integral. So I know you can do that. So that's where I'll stop.
