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Hi.

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Welcome back to recitation.

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You've been learning about
infinite series of constants.

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Adding up, you know,
series of numbers.

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And one thing that
we've talked about

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are divergent series
and convergent series,

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and rates of divergence
a little bit.

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So suppose you wanted
to run an experiment.

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So suppose you started with
the series log n over n,

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and you started summing--
I picked n equals 3.

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It doesn't actually
matter that much--

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started summing at n equals
3, and going to infinity.

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And you set your
computer to sort of keep

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track of what the partial
sums are as it does this.

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So you want it to produce
a list of partial sums.

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And you know, maybe
to get a good feel

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for this rate of divergence,
what you want to do

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is let it run for a
long time, collect

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a whole bunch of values.

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Say, you know, stop when it gets
to some fairly large number.

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So suppose we run
this, and we ask

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it to stop when it gets
to a partial sum that's

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greater than or equal to 5,000.

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You know, whatever.

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A moderately large
integer or number.

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Doesn't have to be an integer.

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So it's going to stop as soon as
it reaches a partial sum that's

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bigger than 5,000.

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So what I want to know, is how
long is that going to take?

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Now of course, that
depends on your computer.

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So let's just say, you
know, for sake of argument,

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that your computer-- this is
generous to your computer.

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At least my computer
doesn't go this fast--

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that it takes 10 to
the minus 20 seconds

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to add an additional
term to the partial sums.

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So suppose that your
computer takes 10

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to the minus 20 seconds for
each additional summand.

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How long, how many
terms of the series,

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how long are you going to wait
around, waiting for this sum

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to get to 5,000?

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So that's the question.

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Why don't you pause the
video, take some time,

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try and work this out.

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This is probably not
a quite good question

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for experimentation.

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I would advise against
actually running it

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while you wait to
continue with the video.

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But of course, you're
welcome to try.

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But you can, you know,
work this out, come back,

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and we'll talk
about it together.

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Welcome back.

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Hopefully you had some
fun with this problem.

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Let's let's talk about it.

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So what we want to do, is we
want to look at this series

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and say, you know,
about how long

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it's going to take
to get to 5,000.

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What we want-- which
partial sum of this series

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is about equal to 5,000
for the first time?

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So let's give a notation there.

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So let's call S_N--
big N now, S big N

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is equal to N-th
partial sum-- well, OK.

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Let me not make it
the N-th partial sum.

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Let me just make it
the sum when the top

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is big N. So going from 3 to big
N of log little n divided by n.

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So first of all, the
answer could be-- A priori,

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the answer could be forever.

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It might be that
series converges

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to some number less than 5,000.

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So if that were to
happen, then, you know,

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no matter how long you waited
around, you wouldn't get there.

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But we can check that
that's not the case.

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So we can check that
this series diverges.

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So how can we check that?

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So the series diverges
by the integral test,

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because the integral of
log x over x dx from, say,

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3 to infinity, though of course
the bounds aren't-- the lower

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bound, 3, is not
crucially important.

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This is something we
know how to integrate.

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This is equal to 1/2 log
of x quantity squared,

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taken between, well,
3 and infinity.

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And anyway, you see that as
x goes to infinity, log of x

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goes to infinity, and so log
of x squared goes to infinity.

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So this diverges.

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So the series does diverge
to positive infinity.

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And as a result
of its divergence

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to positive infinity, we
know that eventually, we

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do get to 5,000.

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OK, so this is-- the answer is
not an infinite amount of time.

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The answer is some
finite amount of time.

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So now the question
is, how much.

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Now, the thing to remember
about the integral test

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is not only can it tell you
whether something converges

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or diverges, but it gives
you a really good estimate

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of how quickly it
converges or diverges.

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So I have over here
a little picture.

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This is the function y
equals log of x over x.

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So at x equals 1,
this function is 0,

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and then it reaches a
maximum, and then tapers down.

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And as x gets larger-- so
it's a little hard to tell

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in this picture--
at x equals 6, this

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is still-- well, so the peak
here is a little bit less

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than 0.4.

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At x equals 6, this,
the function value

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is still bigger than a quarter.

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But it does eventually go down
to 0, because x grows faster

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than log x.

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And of course if
you were interested,

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an interesting exercise might
be really precise, you know,

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where, to figure out precisely,
what is this peak, and what's

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the x-value, and what's the
y-value of that peak point

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before it starts turning down.

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If you wanted something
fun to work out.

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And OK.

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But that doesn't
actually have anything

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to do with our
problem, which has

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to do with this integral test.

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So remember that the
integral in question

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is the area under this
curve from 3 onwards,

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and a partial sum
is what you get

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when you take a bunch of these
left endpoint rectangles.

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And because this
function is decreasing,

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the rectangles completely
cover the region.

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And so the integral
test tells us

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that the integral from 3
to big N of our function,

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log x over x dx, is
strictly less than-- well,

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we want the partial
sum that goes up to,

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so whose right interval
here ends at big N.

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So that's actually the N
minus first partial sum.

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So maybe what I'll do, is
I'll add a plus 1 here,

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and I'll make this the
big Nth partial sum is

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bigger than the sum from 3
to N plus 1 of the function.

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And also, if you
turn this around,

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and you took right endpoint
rectangles instead,

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you see that if you chop
off the first rectangle,

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then what you get
is the-- you get

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an area less than the integral.

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So we also have an
upper bound here.

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This is less than or equal
to the very first term, which

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in our case is ln 3 over
3 plus that same integral.

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So this is this formula that
Professor Jerison showed you

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in class that bounds the partial
sums in between the integral.

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So here, where the
integral, instead

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of being an improper integral,
just goes up to N plus 1.

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So the integral test
gives us, it tells us

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that this diverges if
and only if this does,

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but it also gives
us explicit bounds

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for those two expressions,
how they relate to each other.

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And so it says that the
rate of divergence of S_N

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is very, very closely tied
to the rate of divergence

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of this integral.

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Well, so what?

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So OK.

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So we said, we know, you know,
about how big this integral is.

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How big is it?

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Well, we saw that it's about log
x squared between the bounds.

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So this is equal to,
well, it's about,

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it's 1/2 log of n plus
1 squared minus 1/2 log

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3 squared is less than S_N,
which is less than log 3

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over 3 plus the same thing.

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So this is log N plus 1
quantity squared minus 1/2 log

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3 squared.

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And here this is log N
plus 1 quantity squared.

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OK.

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So this right-hand side is
equal to this right-hand side.

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OK.

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So we have that S-- and
this is a big N. So we have

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that S_N is about bounded by
these-- sorry-- is definitely

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bounded by these
two expressions.

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Another thing to note is
that the constants here

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are pretty small.

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Right?

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This 1/2 log 3 squared and log 3
over 3 times 1/2 log 3 squared,

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those are pretty small.

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From this point on
in this problem,

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I'm going to stop worrying about
precise constants like that,

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and I'm going to start
using an approximate sign.

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So another way to
write this is to say

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that S_N is approximately-- so
this is like an equals sign,

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but both of them are curvy.

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So what I mean by this, it
doesn't mean anything formally.

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So it's not a precise meaning.

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In this case, it's going to
mean always about the same size.

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For purposes of
this argument, it's

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going to be about the same.

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So this is about equal to 1/2
ln of N plus 1 quantity squared.

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OK.

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And what we want--

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So, OK.

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So, good.

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So we have a good approximation
for about how big the Nth

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partial sum is.

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Well, all right.

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So why?

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Well, because I know
that what I want to do is

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I want to keep going until
I reach a partial sum that

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has size about 5,000.

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So I want to know, how
many terms do I need.

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Right?

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00:10:09,790 --> 00:10:11,873
That's going to tell me
how much time it requires,

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and I need to know how
many terms there are.

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So I need to know, when S_N
is about 5,000, how big is N?

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So coming back over here, so
I need to take this equation,

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and I need to solve it
for N in terms of S_N.

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Because I know that S_N is
going to be about 5,000,

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so I want to know, how
big is N going to be?

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So OK.

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00:10:29,790 --> 00:10:30,498
So let's do that.

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00:10:30,498 --> 00:10:33,210
So we can multiply by 2, take a
square root, and exponentiate.

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00:10:33,210 --> 00:10:37,370
And what we get-- so I'm
going to move up here.

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So from this expression, what
I get is that N plus 1 is

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approximately equal to-- so
I'm going to multiply by 2,

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00:10:50,179 --> 00:10:51,720
take a square root,
and exponentiate,

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00:10:51,720 --> 00:10:58,860
so that's e to the
square root of 2*s_N.

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00:11:03,060 --> 00:11:05,180
And you know what, I don't
care about this plus 1.

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00:11:05,180 --> 00:11:06,388
I'm going to forget about it.

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00:11:06,388 --> 00:11:10,290
So N is approximately equal to
e to the 2 square root of S_N.

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00:11:10,290 --> 00:11:14,420
Now in our case, we want to know
particularly when S_N is equal

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to about 5,000.

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So when S_N is approximately
equal to 5,000,

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this means that N is
approximately-- well,

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2 times 5,000 is 10,000,
square root of 10,000 is 100,

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so N is approximately
equal to e to the 100.

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Now e is between
2 and 3-- so OK,

224
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so you know, I
don't, I can't really

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give a very precise estimate
of this off the top of my head.

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But luckily I used a
computer ahead of time

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to approximate it.

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So e to the 100 is
about equal to 2 times

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00:12:01,520 --> 00:12:04,880
10 to the forty-third power.

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So that's the number, so
that's about how big N

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is, order of magnitude.

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00:12:10,380 --> 00:12:15,970
So we need to do about 2 times
10 to the forty-third terms.

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So now that's the
number of terms.

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00:12:18,289 --> 00:12:20,080
Now, if you remember,
back in our question,

235
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we were asked, on the assumption
that each term takes 10

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to the minus 20 seconds, how
long is this going to take.

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00:12:28,070 --> 00:12:47,460
So if each term requires
10 to the minus 20 seconds,

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00:12:47,460 --> 00:12:52,850
this means we're dealing with
2 times 10 to the 23 seconds.

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00:12:52,850 --> 00:12:54,710
That's how long
you'll have to wait.

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Now, this is a big number.

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Right?

242
00:12:57,710 --> 00:13:01,150
So let's think about how
big a number this is.

243
00:13:01,150 --> 00:13:03,160
Because there are
big numbers, and then

244
00:13:03,160 --> 00:13:05,220
there are ridiculously
big numbers.

245
00:13:05,220 --> 00:13:07,350
So, a big number
would be a number

246
00:13:07,350 --> 00:13:11,181
where like, maybe this 10 to
the minus 20 is a bad estimate,

247
00:13:11,181 --> 00:13:11,680
right?

248
00:13:11,680 --> 00:13:15,130
I mean, maybe you are
emperor of the world,

249
00:13:15,130 --> 00:13:17,430
and you can harness
all the computing

250
00:13:17,430 --> 00:13:20,470
power in the entire world
to work on this problem.

251
00:13:20,470 --> 00:13:22,590
And so then, in
that case, maybe you

252
00:13:22,590 --> 00:13:26,730
can do, say, 10 to
the 25 computations

253
00:13:26,730 --> 00:13:29,940
per second, something.

254
00:13:29,940 --> 00:13:31,550
Or whatever.

255
00:13:31,550 --> 00:13:35,480
So how much of a dent would
that make in this number?

256
00:13:35,480 --> 00:13:37,260
Like 2 times 10 to 23.

257
00:13:37,260 --> 00:13:37,760
OK.

258
00:13:37,760 --> 00:13:39,551
It's not going to be
done before lunchtime,

259
00:13:39,551 --> 00:13:40,700
but how long will it take?

260
00:13:40,700 --> 00:13:43,300
So just to-- let's see.

261
00:13:43,300 --> 00:13:45,830
I have this written down here.

262
00:13:45,830 --> 00:13:46,330
OK.

263
00:13:46,330 --> 00:13:58,160
So for comparison, this is about
equal to 50,000 times the age

264
00:13:58,160 --> 00:14:04,010
of the universe.

265
00:14:04,010 --> 00:14:07,680
So if you had, if you
harnessed all the computing

266
00:14:07,680 --> 00:14:10,960
power on earth,
you could probably

267
00:14:10,960 --> 00:14:15,120
get this done in a
few billion years.

268
00:14:15,120 --> 00:14:18,010
So OK, so just a
sense of just how

269
00:14:18,010 --> 00:14:19,410
slowly this thing is getting.

270
00:14:19,410 --> 00:14:22,740
If you wait, and you have all
the computer power on earth

271
00:14:22,740 --> 00:14:27,540
adding terms of the series, in,
say, a few billion years you

272
00:14:27,540 --> 00:14:32,870
will be able to figure
out-- you will sum up

273
00:14:32,870 --> 00:14:35,470
terms that pass 5,000 in value.

274
00:14:35,470 --> 00:14:37,390
So OK.

275
00:14:37,390 --> 00:14:40,360
So I don't know.

276
00:14:40,360 --> 00:14:42,790
I find that amusing.

277
00:14:42,790 --> 00:14:46,090
And remember that the key
here to this whole problem

278
00:14:46,090 --> 00:14:49,860
was just that we were using
this integral test here.

279
00:14:49,860 --> 00:14:52,240
And specifically, we
were using it in the form

280
00:14:52,240 --> 00:14:54,000
that I have over
here, this form that

281
00:14:54,000 --> 00:14:57,390
lets you bound the
partial sums between two

282
00:14:57,390 --> 00:14:58,530
values of the integral.

283
00:14:58,530 --> 00:15:00,600
And because we have
those very tight bounds,

284
00:15:00,600 --> 00:15:03,300
these approximations
are all valid,

285
00:15:03,300 --> 00:15:06,360
and so we can really say,
yes, this problem that I've

286
00:15:06,360 --> 00:15:10,160
described is really something
you can't do in an afternoon

287
00:15:10,160 --> 00:15:11,990
or whatever.

288
00:15:11,990 --> 00:15:13,815
So I'll stop there.