Proof of special case of l'Hôpital's rule

WEBVTT
Kind: captions
Language: en

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What I want to go
over in this video

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is a special case
of L'Hopital's Rule.

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And it's a more
constrained version

00:00:10.031 --> 00:00:11.780
of the general case
we've been looking at.

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But it's still very powerful
and very applicable.

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And the reason why we're going
to go over this special case

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is because its proof is
fairly straightforward

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and will give you an intuition
for why L'Hopital's Rule works

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at all.

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So the special case
of L'Hopital's Rule

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is a situation where
f of a is equal to 0.

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f prime of a exists.

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g of a is equal to 0.

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g prime of a exists.

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If these constraints
are met, then the limit,

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as x approaches a of
f of x over g of x,

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is going to be equal to f
prime of a over g prime of a.

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So it's very similar
to the general case.

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It's little bit
more constrained.

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We're assuming that
f prime of a exists.

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We're not just
taking the limit now.

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We're assuming f prime of a and
g prime of a actually exist.

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But notice if we substitute
a right over here we get 0/0.

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But that if the
derivatives exist

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we could just evaluate
the derivatives at a,

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and then we get the limit.

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So this is very close
to the general case

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of L'Hopital's Rule.

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Now let's actually prove it.

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And to prove it, we're going
to start with the right hand

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and then show that if we use
the definition of derivatives,

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we get the left hand
right over here.

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So let me do that.

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So I'll do it right over here.

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So f prime of a
is equal to what,

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by the definition
of derivatives?

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Well, we could view
that as the limit

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as x approaches a of f of x
minus f of a over x minus a.

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So this is literally just
a slope between two points.

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So like, if you have your
function f of x like this,

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this is the point a, f
of a right over here.

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This right over here
is the point x, f of x.

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This expression
right over here is

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the slope between
these two points.

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The change in our y value
is f of x minus f of a.

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The change in our f
value is x minus a.

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So this expression is just
the slope of this line.

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And we're just taking
the-- let me actually

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do that in a different
color-- the line that

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connects these two points,
that's the slope of it.

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I'll do that in white.

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The slope of the line that
connects those two points.

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And we're taking
the limit as x gets

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closer and closer
and closer to a.

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So this is just
another way of writing

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the definition of
the derivative.

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So that's fine.

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Let's do the same
thing for g prime of a.

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So f prime of a
over g prime of a,

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is going to be this
business which is in orange,

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f prime of a over g prime of a.

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Which we can write as
the limit as x approaches

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a of g of x minus g
of a over x minus a.

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Well, in the numerator,
we're taking the limit

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as x approaches a, and
in the denominator,

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we're taking the limit
as x approaches a.

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So we can just rewrite this.

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This we can rewrite as
the limit as x approaches

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a of all this
business in orange.

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f of x minus f of
a, over x minus a,

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over all the business in green.

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g of x minus g of a, all
of that over x minus a.

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Now, to simplify this, we
can multiply the numerator

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and the denominator by x minus
a to get rid of these x minus

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a's.

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So let's do that.

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Let's multiply by x
minus a over x minus a.

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So the numerator, x minus a,
and we're dividing by x minus a.

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Those cancel out.

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And then these two cancel out.

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And we're left with this thing
over here is equal to the limit

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as x approaches a
of, in the numerator

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we have f of x minus f of a.

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And in the denominator, we
have g of x minus g of a.

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And I think you see
where this is going.

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What is f(a) equal to?

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Well, we assumed f
of a is equal to 0.

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That's why we're
using L'Hopital's Rule

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from the get go. f of a is equal
to 0, g of a is equal to 0.

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f of a is equal to 0.
g of a is equal to 0.

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And this simplifies to the
limit as x approaches a of f

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prime of x, sorry of f of
x, we've got to be careful.

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Of f of x over g of x.

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So we just showed that if f of
a equals 0, g of a equals 0,

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and these two derivatives exist,
then the derivatives evaluated

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at a over each other are
going to be equal to the limit

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as x approaches a of
f of x over g of x.

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Or the limit as x
approaches a of f

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of x over g of x is going to
be equal to f prime of a over g

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prime of a.

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So fairly straightforward
proof for the special case--

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the special case, not
the more general case--

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of L'Hopital's Rule.


Proving a sequence converges using the formal definition

WEBVTT
Kind: captions
Language: en

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I made a claim that for
this sequence-- and this

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was in a previous video--
that for this sequence

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right over here that can
be defined explicitly

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in this way, that the
limit of the sequence--

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and so I can write this as
negative 1 to the n plus 1

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over n.

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That's one way of defining
our sequence explicitly--

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the limit of this as
n approaches infinity

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is equal to 0.

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And it seems that way.

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As n gets larger and
larger and larger,

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even though the numerator
oscillates between negative 1

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and 1, it seems like
it will get smaller

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and smaller and smaller.

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But I didn't prove
it, and that's

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what I want to do in this video.

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In order to prove it, this is
going to be true if and only

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if for any epsilon
greater than 0,

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there is a capital
M greater than 0

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such that if lowercase n, if our
index is greater than capital

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M, then the nth
term in our sequence

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is going to be within epsilon of
our limit, within epsilon of 0.

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So what does that say?

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That says, hey, give
me-- our limit is 0.

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Let me do this in a new color.

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So our limit right
over here is 0.

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That's our limit.

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So our limit right
over here is--

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we're saying the sequence
is converging to 0.

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What we're saying is, give
us an epsilon around 0.

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So let's say that this right
over here is 0 plus epsilon.

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That is 0 plus epsilon.

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The way I've drawn it here
looks like epsilon would be 0.5.

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This would be 0 minus epsilon.

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Let me make it a
little bit neater.

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So this would be
zero minus epsilon.

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So this is negative epsilon, 0
minus epsilon, 0 plus epsilon.

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Our limit in this case, or
our claim of a limit, is 0.

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Now, this is saying
for any epsilon,

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we need to find an M such
that if n is greater than M,

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the distance between our
sequence and our limit

00:02:03.486 --> 00:02:04.860
is going to be
less than epsilon.

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So if the distance between
our sequence and our limit

00:02:08.330 --> 00:02:10.780
is less than epsilon,
that means that the value

00:02:10.780 --> 00:02:13.800
of our sequence for
a given n is going

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to be within these two bounds.

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It's got to be in this
range right over here

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that I'm shading
above a certain n.

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So if I pick an n
right over here,

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it looks like anything
larger than that

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is going to be the case that
we're going to be within those

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bounds.

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But how do we prove it?

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Well, let's just
think about what

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needs to happen for
this to be true.

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So what needs to be happen
for a sub n minus 0,

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the absolute value
of a sub n minus 0,

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what needs to be true for
this to be less than epsilon?

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Well, this is
another way of saying

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that the absolute value of a sub
n has to be less than epsilon.

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And a sub n is just this
business right here,

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so it's another way of saying
that the absolute value

00:02:58.510 --> 00:03:02.180
of negative 1 to
the n plus 1 over n

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has to be less than epsilon,
which is another way of saying,

00:03:05.590 --> 00:03:08.020
because this negative
1 to the n plus 1,

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this numerator just swaps
us between a negative

00:03:10.310 --> 00:03:12.322
and a positive
version of 1 over n.

00:03:12.322 --> 00:03:14.030
But if you take the
absolute value of it,

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this is always just
going to be positive.

00:03:16.550 --> 00:03:19.250
So this is the same
thing as 1 over n,

00:03:19.250 --> 00:03:23.570
as the absolute value of 1 over
n has to be less than epsilon.

00:03:23.570 --> 00:03:26.400
Now, n is always
going to be positive.

00:03:26.400 --> 00:03:28.860
n starts at 1 and
goes to infinity.

00:03:28.860 --> 00:03:31.250
So this value is always
going to be positive.

00:03:31.250 --> 00:03:33.510
So this is saying the
same thing that 1 over n

00:03:33.510 --> 00:03:36.380
has to be less than epsilon
in order for this stuff

00:03:36.380 --> 00:03:37.840
to be true.

00:03:37.840 --> 00:03:40.327
And now we can take the
reciprocal of both sides.

00:03:40.327 --> 00:03:42.910
And if you take the reciprocal
of both sides of an inequality,

00:03:42.910 --> 00:03:45.310
you would have that
n-- if you take

00:03:45.310 --> 00:03:49.640
the reciprocal of both
sides of an inequality,

00:03:49.640 --> 00:03:51.260
you swap the inequality.

00:03:51.260 --> 00:03:57.870
So for this to be true, n has to
be greater than 1 over epsilon.

00:03:57.870 --> 00:03:59.780
And we essentially
have proven it now.

00:03:59.780 --> 00:04:02.880
So now we've said, look, for
this particular sequence,

00:04:02.880 --> 00:04:06.950
you give me any
epsilon, and I'm going

00:04:06.950 --> 00:04:12.030
to set M to be 1 over epsilon.

00:04:12.030 --> 00:04:22.760
Because if n is greater than
M, which is 1 over epsilon,

00:04:22.760 --> 00:04:25.880
then we know that this right
over here is going to be true.

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That is going to be true.

00:04:30.680 --> 00:04:34.380
So the limit does
definitely exist.

00:04:34.380 --> 00:04:36.810
And so over here, for
this particular epsilon,

00:04:36.810 --> 00:04:40.200
it looks like we've picked
0.5 or 1/2 as our epsilon.

00:04:40.200 --> 00:04:43.250
So as long as n is greater
than 1 over 1/2, which is 2,

00:04:43.250 --> 00:04:46.630
so in this case we could
say, look, you gave me 1/2.

00:04:46.630 --> 00:04:48.464
My M is going to be a
function of epsilon.

00:04:48.464 --> 00:04:50.630
It's going to be defined
for any epsilon you give me

00:04:50.630 --> 00:04:53.030
greater than 0.

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So here, 1 over 1/2
is right over here.

00:04:55.530 --> 00:04:58.170
I'm going to make my
M right over here.

00:04:58.170 --> 00:05:00.290
And you see it is
indeed the case

00:05:00.290 --> 00:05:02.600
that my sequence
is within the range

00:05:02.600 --> 00:05:05.240
as we passed for any
n greater than 2.

00:05:05.240 --> 00:05:06.970
So for n is equal to
3 it's in the range.

00:05:06.970 --> 00:05:08.553
For n is equal to 4
it's in the range.

00:05:08.553 --> 00:05:10.620
For n equals 5-- and it
keeps going and going.

00:05:10.620 --> 00:05:12.370
And we're not just
taking our word for it.

00:05:12.370 --> 00:05:14.410
We've proven it right over here.

00:05:14.410 --> 00:05:15.710
So we've made the proof.

00:05:15.710 --> 00:05:17.510
You give me any
other any epsilon,

00:05:17.510 --> 00:05:21.000
I said M is equal to
1 over that thing.

00:05:21.000 --> 00:05:24.080
And so for n greater than
that, this is going to be true.

00:05:24.080 --> 00:05:25.990
So this is definitely the case.

00:05:25.990 --> 00:05:30.220
This sequence converges to 0.


Formal definition for limit of a sequence

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:04.820
what i want to do in this video is to provide ourselves with a rigorous definition

00:00:04.820 --> 00:00:09.460
of what it means to take the limit of a sequence as n approaches infinity

00:00:09.844 --> 00:00:15.508
and what we'll see is actually very similar to the definition of any

00:00:15.508 --> 00:00:22.182
function as a limit approaches infinity and this is because the sequences really can be just viewed

00:00:22.182 --> 00:00:25.732
as a function of their indices, so let's say

00:00:25.871 --> 00:00:30.302
let me draw an arbitrary sequence right over here

00:00:30.302 --> 00:00:33.240
so actually let me draw like this just to make it clear

00:00:33.240 --> 00:00:36.504
but the limit is approaching so let me draw a sequence

00:00:36.504 --> 00:00:39.573
let me draw a sequence that is jumping around

00:00:39.573 --> 00:00:43.966
little bit, so lets say when n=1, a(1) is there,

00:00:43.966 --> 00:00:50.802
when n = 2, a(2) is there, when n = 3, a(3) is over there

00:00:50.802 --> 00:00:59.004
when n=4, a(4) is over here, when n=5 a(5) is over here

00:00:59.004 --> 00:01:04.049
and it looks like is n is so this is 1 2 3 4 5

00:01:04.093 --> 00:01:08.071
so it looks like that as n gets bigger and bigger and bigger

00:01:08.071 --> 00:01:13.539
a(n) seems to be approaching, seems to be approaching some value

00:01:14.277 --> 00:01:18.373
it seems to be getting closer and closer, seems to be converging

00:01:18.373 --> 00:01:23.264
to some value L right over here. What we need to do is come up with a definition of

00:01:23.264 --> 00:01:27.363
what is it really mean to converge to L.

00:01:27.363 --> 00:01:33.486
So let's say for any, so we're gonna say that you converge to L for any,

00:01:33.486 --> 00:01:37.454
for any ε > 0, for any positive epsilon, you can,

00:01:37.877 --> 00:01:44.819
you can come up, you can get or you can, there is

00:01:44.819 --> 00:01:52.835
let me rewrite it this way, for any positive epsilon there is a positive,

00:01:52.835 --> 00:01:57.989
positive M, capital M, such that,

00:01:57.989 --> 00:02:07.578
such that if, if, lower case n is greater than capital M,

00:02:07.578 --> 00:02:17.269
then the distance between a(n) and our limit, this L right over here

00:02:17.269 --> 00:02:22.286
the distance between those two points is less than epsilon.

00:02:22.286 --> 00:02:25.199
If you can do this for any epsilon, for any epsilon,

00:02:25.199 --> 00:02:30.117
greater than 0, there is a positive M, such that if n is greater than M,

00:02:30.117 --> 00:02:33.669
the distance between a(n) and our limit is less than epsilon

00:02:33.669 --> 00:02:36.288
then we can say, then we can say

00:02:36.288 --> 00:02:43.193
that the limit of a(n) as n approaches infinity is equal to L and

00:02:43.193 --> 00:02:51.286
we can say that a(n) converges, converges, converges to L.

00:02:51.286 --> 00:02:56.500
So let's, let's, let's parse this, so here I was making the claim

00:02:56.500 --> 00:02:58.786
that a(n) is approaching this L right over here,

00:02:58.786 --> 00:03:01.388
I tried to draw it as a horizontal line.

00:03:01.388 --> 00:03:05.794
This definition of the of what it means to converge for sequence to converge

00:03:05.794 --> 00:03:09.045
says look for any epsilon greater than zero.

00:03:09.045 --> 00:03:14.795
So let me pick an epsilon greater than zero, so I am gonna go to L plus epsilon,

00:03:14.795 --> 00:03:17.939
actually let me do it right over here, L, so see this is L plus epsilon

00:03:17.939 --> 00:03:24.353
and let's say this is right here this is L minus epsilon.

00:03:24.353 --> 00:03:28.657
So let me draw those two bounds, right over here.

00:03:28.657 --> 00:03:32.568
And so I picked an epsilon here so for any,

00:03:32.568 --> 00:03:38.831
for any arbitrary epsilon, any arbitrary positive epsilon I pick, we can find a positive

00:03:38.831 --> 00:03:43.277
M, we can find a positive M, so that as long as,

00:03:43.277 --> 00:03:45.755
so let's say that is our M right over there.

00:03:45.755 --> 00:03:50.881
So that as long as our n is greater than our M, then our a(n)

00:03:50.881 --> 00:03:55.932
our a(n) is within epsilon of L, so being within the epsilon,

00:03:55.932 --> 00:03:59.747
being within epsilon of L is essentially being in this range.

00:03:59.747 --> 00:04:04.839
This right over here is just saying, look that the distance between a(n) and L is less than epsilon,

00:04:04.839 --> 00:04:09.618
so that would be any of these, anything that is in this, between L minus epsilon and L plus epsilon.

00:04:09.618 --> 00:04:15.292
The distance between that and our limit is going to be less than epsilon.

00:04:15.292 --> 00:04:16.992
And we see right over here, at least visually,

00:04:16.992 --> 00:04:20.978
if we pick M there and if you can take an n that's larger than that M,

00:04:20.978 --> 00:04:26.019
if you pick an N that's larger than M, if M is equal to 3, a(n) seems to be

00:04:26.019 --> 00:04:30.843
close enough. If M is 4, a(n) is even getting closer. It's within our epsilon.

00:04:30.843 --> 00:04:36.559
So if we can say, if we can say that it is true, for any epsilon that we pick, then we can say,

00:04:36.559 --> 00:04:41.528
we can say that the limit exists, that a(n) converges to L.

00:04:41.528 --> 00:04:45.528
In the next video we will use this definition to actually prove that a sequence converges.


Convergent and divergent sequences

WEBVTT
Kind: captions
Language: en

00:00:00.540 --> 00:00:02.430
Let's say I've got a sequence.

00:00:02.430 --> 00:00:06.210
It starts at 1, then let's
say it goes to negative 1/2.

00:00:06.210 --> 00:00:08.500
Then it goes to positive 1/3.

00:00:08.500 --> 00:00:10.580
Then it goes to negative 1/4.

00:00:10.580 --> 00:00:12.650
Then it goes to positive 1/5.

00:00:12.650 --> 00:00:16.400
And it just keeps going on
and on and on like this.

00:00:16.400 --> 00:00:18.010
And we could graph it.

00:00:18.010 --> 00:00:22.040
Let me draw our vertical axis.

00:00:22.040 --> 00:00:24.005
So I'll graph this
as our y-axis.

00:00:24.005 --> 00:00:27.230
And I'm going to graph
y is equal to a sub n.

00:00:27.230 --> 00:00:31.580
And let's make this our
horizontal axis where I'm going

00:00:31.580 --> 00:00:33.430
to plot our n's.

00:00:33.430 --> 00:00:35.040
So this right over
here is our n's.

00:00:35.040 --> 00:00:38.030
And this is, let's say this
right over here is positive 1.

00:00:38.030 --> 00:00:39.890
This right over
here is negative 1.

00:00:39.890 --> 00:00:41.370
This would be negative 1/2.

00:00:41.370 --> 00:00:43.540
This would be positive 1/2.

00:00:43.540 --> 00:00:46.740
And I'm not drawing the
vertical and horizontal axes

00:00:46.740 --> 00:00:49.530
at the same scale, just so that
we can kind of visualize this

00:00:49.530 --> 00:00:50.420
properly.

00:00:50.420 --> 00:00:57.350
But let's say this
is 1, 2, 3, 4, 5,

00:00:57.350 --> 00:00:59.420
and I could keep going
on and on and on.

00:00:59.420 --> 00:01:02.980
So we see here that
when n is equal to 1,

00:01:02.980 --> 00:01:05.220
a sub n is equal to 1.

00:01:05.220 --> 00:01:06.940
So this is right over there.

00:01:06.940 --> 00:01:09.310
So when n is equal to 1,
a sub n is equal to 1.

00:01:09.310 --> 00:01:12.940
So this is y is
equal to a sub n.

00:01:12.940 --> 00:01:17.720
Now, when n is
equal to 2, we have

00:01:17.720 --> 00:01:19.430
a sub n is equal
to negative 1/2.

00:01:25.420 --> 00:01:30.890
When n is equal to 3, a
sub n is equal to 1/3,

00:01:30.890 --> 00:01:34.150
which is right about there.

00:01:34.150 --> 00:01:39.990
When n is equal to 4, a sub
n is equal to negative 1/4,

00:01:39.990 --> 00:01:43.910
which is right about there.

00:01:43.910 --> 00:01:47.190
And then when n is
equal to 5, a sub n

00:01:47.190 --> 00:01:52.050
is equal to positive 1/5, which
is maybe right over there.

00:01:52.050 --> 00:01:53.790
And we keep going
on and on and on.

00:01:53.790 --> 00:01:55.790
So you see the points,
they kind of jump around,

00:01:55.790 --> 00:01:58.580
but they seem to be getting
closer and closer and closer

00:01:58.580 --> 00:01:59.600
to 0.

00:01:59.600 --> 00:02:02.070
Which would make us ask
a very natural question--

00:02:02.070 --> 00:02:06.560
what happens to a sub n
as n approaches infinity?

00:02:06.560 --> 00:02:10.930
Or another way of saying
that is, what is the limit--

00:02:10.930 --> 00:02:18.730
let me do this in a new color--
of a sub n as n approaches

00:02:18.730 --> 00:02:20.450
infinity?

00:02:20.450 --> 00:02:24.780
Well, let's think about if we
can define a sub n explicitly.

00:02:28.040 --> 00:02:33.590
So we can define this sequence
as a sub n where n starts at 1

00:02:33.590 --> 00:02:39.680
and goes to infinity
with a sub n equaling--

00:02:39.680 --> 00:02:40.730
what does it equal?

00:02:40.730 --> 00:02:43.110
Well, if we ignore
sign for a second,

00:02:43.110 --> 00:02:45.430
it looks like it's
just 1 over n.

00:02:45.430 --> 00:02:48.010
But then we seem like
we oscillate in signs.

00:02:48.010 --> 00:02:51.410
We start with a positive, then
a negative, positive, negative.

00:02:51.410 --> 00:02:54.970
So we could multiply this times
negative 1 to the-- let's see.

00:02:54.970 --> 00:02:57.085
If we multiply it times
negative 1 to the n,

00:02:57.085 --> 00:02:59.544
then this one would be negative
and this would be positive.

00:02:59.544 --> 00:03:00.793
But we don't want it that way.

00:03:00.793 --> 00:03:02.530
We want the first
term to be positive.

00:03:02.530 --> 00:03:05.990
So we say negative 1
to the n plus 1 power.

00:03:05.990 --> 00:03:07.240
And you can verify this works.

00:03:07.240 --> 00:03:10.900
When n is equal to 1, you have
1 times negative 1 squared,

00:03:10.900 --> 00:03:13.040
which is just 1, and it'll
work for all the rest.

00:03:13.040 --> 00:03:16.250
So we could write this
as equaling negative 1

00:03:16.250 --> 00:03:20.260
to the n plus 1 power over n.

00:03:20.260 --> 00:03:22.200
And so asking what
the limit of a sub

00:03:22.200 --> 00:03:25.900
n as n approaches infinity
is equivalent to asking what

00:03:25.900 --> 00:03:29.660
is the limit of
negative 1 to the n

00:03:29.660 --> 00:03:33.420
plus 1 power over
n as n approaches

00:03:33.420 --> 00:03:35.950
infinity is going
to be equal to?

00:03:35.950 --> 00:03:38.370
Remember, a sub n, this
is just a function of n.

00:03:38.370 --> 00:03:40.970
It's a function where we're
limited right over here

00:03:40.970 --> 00:03:43.000
to positive integers
as our domain.

00:03:43.000 --> 00:03:45.690
But this is still just a
limit as something approaches

00:03:45.690 --> 00:03:46.634
infinity.

00:03:46.634 --> 00:03:48.300
And I haven't rigorously
defined it yet,

00:03:48.300 --> 00:03:51.560
but you can conceptualize
what's going on here.

00:03:51.560 --> 00:03:53.470
As n approaches
infinity, the numerator

00:03:53.470 --> 00:03:55.934
is going to oscillate between
positive and negative 1,

00:03:55.934 --> 00:03:57.350
but this denominator
is just going

00:03:57.350 --> 00:03:59.309
to get bigger and bigger
and bigger and bigger.

00:03:59.309 --> 00:04:01.474
So we're going to get really,
really, really, really

00:04:01.474 --> 00:04:02.440
small numbers.

00:04:02.440 --> 00:04:07.800
And so this thing right over
here is going to approach 0.

00:04:07.800 --> 00:04:09.790
Now, I have not proved
this to you yet.

00:04:09.790 --> 00:04:11.690
I'm just claiming
that this is true.

00:04:11.690 --> 00:04:16.279
But if this is true-- so
let me write this down.

00:04:16.279 --> 00:04:23.540
If true, if the limit of a sub
n as n approaches infinity is 0,

00:04:23.540 --> 00:04:33.940
then we can say that a
sub n converges to 0.

00:04:33.940 --> 00:04:37.200
That's another way of
saying this right over here.

00:04:37.200 --> 00:04:40.190
If it didn't, if the
limit as n approaches

00:04:40.190 --> 00:04:43.230
infinity didn't go to
some value right here--

00:04:43.230 --> 00:04:45.950
and I haven't rigorously
defined how we define a limit--

00:04:45.950 --> 00:04:49.175
but if this was not true, if
we could not set some limit--

00:04:49.175 --> 00:04:50.550
it doesn't have
to be equal to 0.

00:04:50.550 --> 00:04:54.450
As long as it-- if this was
not equal to some number, then

00:04:54.450 --> 00:04:57.900
we would say that
a sub n diverges.


Geometric sequences

WEBVTT
Kind: captions
Language: en

00:00:00.500 --> 00:00:08.130
Let's talk about
geometric sequences,

00:00:08.130 --> 00:00:11.990
which is a class of sequences
where we start at some number,

00:00:11.990 --> 00:00:15.030
then each successive number
is the previous number

00:00:15.030 --> 00:00:16.830
multiplied by the same thing.

00:00:16.830 --> 00:00:18.120
So what am I talking about?

00:00:18.120 --> 00:00:21.160
Well let's multiply a times r.

00:00:21.160 --> 00:00:22.450
And then I'm going to get ar.

00:00:25.310 --> 00:00:28.290
Let's multiply it times,
but to get the third term,

00:00:28.290 --> 00:00:30.946
let's multiply the
second term times r.

00:00:30.946 --> 00:00:32.320
And then what am
I going to have?

00:00:32.320 --> 00:00:35.810
I'm going to have-- it's a
different shade of yellow-- I'm

00:00:35.810 --> 00:00:38.050
going to have ar squared.

00:00:38.050 --> 00:00:39.450
Multiply by r
again, you're going

00:00:39.450 --> 00:00:41.670
to get ar to the third
power, and you just

00:00:41.670 --> 00:00:43.110
keep on going like that.

00:00:43.110 --> 00:00:45.620
And this is, the way
I've denoted this,

00:00:45.620 --> 00:00:47.380
this is an infinite
geometric sequence.

00:00:47.380 --> 00:00:50.190
We just keep going on
and on and on and on.

00:00:50.190 --> 00:00:52.000
And the different
ways we can denote it,

00:00:52.000 --> 00:00:53.430
we can denote it explicitly.

00:00:53.430 --> 00:00:57.920
We could say that our
sequence is a sub n starting

00:00:57.920 --> 00:01:03.630
with the first term going all
the way to infinity, with a sub

00:01:03.630 --> 00:01:05.710
n equaling-- well,
we see an a here

00:01:05.710 --> 00:01:09.359
for any term-- is
going to be a times r.

00:01:09.359 --> 00:01:11.150
And just to be clear,
this right over here,

00:01:11.150 --> 00:01:15.125
a is the same thing as a times r
to the zeroth power, r to the 0

00:01:15.125 --> 00:01:16.090
is just 1.

00:01:16.090 --> 00:01:19.550
This second term is
ar to the first power.

00:01:19.550 --> 00:01:21.770
The third term is ar
to the third power.

00:01:21.770 --> 00:01:23.790
It looks like the
nth term is going

00:01:23.790 --> 00:01:26.010
to be ar to the n minus 1 power.

00:01:26.010 --> 00:01:29.030
So ar to the n minus 1.

00:01:29.030 --> 00:01:30.200
And you could verify it.

00:01:30.200 --> 00:01:34.180
If you want the second term, you
say a times r to the 2 minus 1,

00:01:34.180 --> 00:01:36.000
a times r to the first power.

00:01:36.000 --> 00:01:37.580
It works out.

00:01:37.580 --> 00:01:39.130
This is defining it explicitly.

00:01:39.130 --> 00:01:40.930
We could also define
it recursively.

00:01:40.930 --> 00:01:45.540
We could say a sub n from
n equals 1 to infinity,

00:01:45.540 --> 00:01:50.190
with a sub 1 being equal to a.

00:01:50.190 --> 00:01:52.480
That's the base case.

00:01:52.480 --> 00:01:56.220
a sub 1 is equal to a,
ar to the 0 is just a.

00:01:56.220 --> 00:02:01.989
Or we could say for n equals
1, and then we could say a--

00:02:01.989 --> 00:02:03.530
and I don't even
have to really write

00:02:03.530 --> 00:02:05.155
that because we're
making it very clear

00:02:05.155 --> 00:02:10.070
that a sub 1 is equal to a--
and then we could say a sub n is

00:02:10.070 --> 00:02:16.680
equal to the previous term, a
sub n minus 1, times r, for n

00:02:16.680 --> 00:02:18.086
is greater than or equal to 2.

00:02:18.086 --> 00:02:19.710
So this is saying,
look, our first term

00:02:19.710 --> 00:02:22.775
is going to be a, that right
over there is a, ar to the 0

00:02:22.775 --> 00:02:26.110
is just a, and then each
successive term is going

00:02:26.110 --> 00:02:27.860
to be the previous
term times r, which

00:02:27.860 --> 00:02:29.980
is exactly what
we did over there.

00:02:29.980 --> 00:02:32.700
So let's look at some
geometric sequences.

00:02:32.700 --> 00:02:35.930
So I could have a geometric
sequence like this.

00:02:35.930 --> 00:02:44.630
I could have a sub n, n is
equal to 1 to infinity with,

00:02:44.630 --> 00:02:47.350
let's say, a sub n is
equal to, let's say

00:02:47.350 --> 00:02:53.350
our first term is, I don't know,
let's say it is equal to 20.

00:02:53.350 --> 00:02:56.260
And then r, the
number that we're

00:02:56.260 --> 00:02:58.930
multiplying to get
each successive term,

00:02:58.930 --> 00:03:00.800
let's say it's equal to 1/2.

00:03:00.800 --> 00:03:03.632
1/2 to the n minus 1.

00:03:03.632 --> 00:03:05.590
So what would this sequence
actually look like?

00:03:05.590 --> 00:03:06.673
Well let's think about it.

00:03:06.673 --> 00:03:08.030
The first term is 20.

00:03:08.030 --> 00:03:10.200
If you say, if n
is 1, this is going

00:03:10.200 --> 00:03:11.760
to be 1/2 to the 0-th power.

00:03:11.760 --> 00:03:14.100
So it's going to be 1 times 20.

00:03:14.100 --> 00:03:16.640
So the first term is 20,
and then each time we're

00:03:16.640 --> 00:03:18.130
multiplying by what?

00:03:18.130 --> 00:03:20.550
Well here each time
we're multiplying by 1/2.

00:03:20.550 --> 00:03:25.930
So this could be 20 times
1/2 is 10, 10 times 1/2 is 5,

00:03:25.930 --> 00:03:28.933
5 times 1/2 is
2.5-- actually let

00:03:28.933 --> 00:03:33.370
me just write that as a
fraction, is 5/2, 5/2 times 1/2

00:03:33.370 --> 00:03:38.170
is 5/4, and you can just
keep going on and on and on.

00:03:38.170 --> 00:03:42.079
This is a geometric sequence.

00:03:42.079 --> 00:03:43.620
Now let me give you
another sequence,

00:03:43.620 --> 00:03:45.670
and tell me if it is geometric.

00:03:45.670 --> 00:03:51.660
So let's say we start at 1,
so then I'm going to go to 2,

00:03:51.660 --> 00:03:54.410
and then I'm going to
go to 6, and then I'm

00:03:54.410 --> 00:03:56.630
going to go to--
let me see what I

00:03:56.630 --> 00:04:00.690
want to do-- I want to go to 24.

00:04:00.690 --> 00:04:05.470
And then I could go to 120,
and I go on and on and on.

00:04:05.470 --> 00:04:08.566
Is this a geometric sequence?

00:04:08.566 --> 00:04:10.190
Well let's think
about what's going on.

00:04:10.190 --> 00:04:14.550
To go from 1 to 2,
I multiplied by 2.

00:04:14.550 --> 00:04:19.000
To go from 2 to 6,
I multiplied by 3.

00:04:19.000 --> 00:04:26.620
To go from 6 to 24,
I multiplied by 4.

00:04:26.620 --> 00:04:29.080
So I'm always multiplying
not by the same amount.

00:04:29.080 --> 00:04:30.530
You have to multiply by
the same amount in order

00:04:30.530 --> 00:04:32.000
for it to be a
geometric sequence.

00:04:32.000 --> 00:04:34.270
Here I'm multiplying it
by a different amount.

00:04:34.270 --> 00:04:39.160
So this sequence that I just
constructed has the form,

00:04:39.160 --> 00:04:42.320
I have my first term,
and then my second term

00:04:42.320 --> 00:04:48.580
is going to be 2 times my first
term, and then my third one

00:04:48.580 --> 00:04:54.300
is going to be 3 times my second
term, so 3 times 2 times a.

00:04:54.300 --> 00:04:59.260
My fourth one is 4 times the
third term, so 4 times 3 times

00:04:59.260 --> 00:05:01.120
2 times a.

00:05:01.120 --> 00:05:04.620
And we go on and on and on.

00:05:04.620 --> 00:05:07.470
So this sequence, which is
not a geometric sequence,

00:05:07.470 --> 00:05:09.170
we can still define
it explicitly.

00:05:09.170 --> 00:05:12.570
We could say that
its set or it's

00:05:12.570 --> 00:05:19.140
the sequence a sub n from n
equals 1 to infinity with a sub

00:05:19.140 --> 00:05:22.040
n being equal to, let's see
the fourth one is essentially

00:05:22.040 --> 00:05:24.080
4 factorial times a.

00:05:24.080 --> 00:05:26.760
Well, actually, if we
look at this particular,

00:05:26.760 --> 00:05:29.120
these particular
numbers our a is 1.

00:05:29.120 --> 00:05:31.650
So this is actually, let
me write this, this is 1,

00:05:31.650 --> 00:05:35.820
this is 2 times 1, this
is 3 times 2 times 1,

00:05:35.820 --> 00:05:40.140
this is 4 times 3
times 2 times 1.

00:05:40.140 --> 00:05:49.370
And so a sub n is just
equal to n factorial.

00:05:49.370 --> 00:05:52.930
This right over here, which
is not a geometric sequence,

00:05:52.930 --> 00:05:56.370
describes exactly this
sequence right over here.

00:05:56.370 --> 00:05:59.277
Just to get some
practice with-- Here

00:05:59.277 --> 00:06:01.110
we've defined it
explicitly, but we can also

00:06:01.110 --> 00:06:02.900
define it recursively.

00:06:02.900 --> 00:06:05.950
We could also say-- do it
in white-- we could also

00:06:05.950 --> 00:06:10.970
say that a sub n takes
us from n equals 1

00:06:10.970 --> 00:06:18.280
to infinity, with a sub 1, or
maybe at a sub 1 is equal to 1.

00:06:18.280 --> 00:06:19.780
That's our first term.

00:06:19.780 --> 00:06:22.220
And then each
successive term is going

00:06:22.220 --> 00:06:26.310
to be equal to the
previous term times n.

00:06:26.310 --> 00:06:29.510
So the second term is equal
to the previous term times 2.

00:06:29.510 --> 00:06:32.860
The nth term is going to be
the previous turn times n

00:06:32.860 --> 00:06:38.770
So this is another valid
way of defining it.


Introduction to arithmetic sequences

WEBVTT
Kind: captions
Language: en

00:00:00.570 --> 00:00:03.420
What I want to do in this
video is familiarize ourselves

00:00:03.420 --> 00:00:06.830
with a very common
class of sequences.

00:00:06.830 --> 00:00:11.630
And this is
arithmetic sequences.

00:00:11.630 --> 00:00:13.890
And they are usually
pretty easy to spot.

00:00:13.890 --> 00:00:17.290
They are sequences where each
term is a fixed number larger

00:00:17.290 --> 00:00:18.920
than the term before it.

00:00:18.920 --> 00:00:21.820
So my goal here is to figure
out which of these sequences

00:00:21.820 --> 00:00:23.259
are arithmetic sequences.

00:00:23.259 --> 00:00:25.050
And then just so that
we have some practice

00:00:25.050 --> 00:00:26.680
with some of the
sequence notation,

00:00:26.680 --> 00:00:30.220
I want to define them
either as explicit functions

00:00:30.220 --> 00:00:32.880
of the term you're looking for,
the index you're looking at,

00:00:32.880 --> 00:00:35.110
or as recursive definitions.

00:00:35.110 --> 00:00:37.500
So first, given that
an arithmetic sequence

00:00:37.500 --> 00:00:40.540
is one where each successive
term is a fixed amount larger

00:00:40.540 --> 00:00:45.790
than the previous one, which of
these are arithmetic sequences?

00:00:45.790 --> 00:00:47.940
Well let's look at this
first one right over here.

00:00:47.940 --> 00:00:52.470
To go from negative 5 to
negative 3, we had to add 2.

00:00:52.470 --> 00:00:56.780
Then to go from negative 3 to
negative 1, you have to add 2.

00:00:56.780 --> 00:01:00.490
Then to go from negative
1 to 1, you had to add 2.

00:01:00.490 --> 00:01:02.810
So this is clearly an
arithmetic sequence.

00:01:02.810 --> 00:01:04.669
We're adding the same
amount every time.

00:01:04.669 --> 00:01:07.210
And there are several ways that
we could define the sequence.

00:01:07.210 --> 00:01:10.022
We could say it's a sub n.

00:01:10.022 --> 00:01:11.480
And you don't always
have to use k.

00:01:11.480 --> 00:01:14.360
This time I'll use n
to denote our index.

00:01:14.360 --> 00:01:19.480
From n equals 1 to infinity
with-- and there's two ways

00:01:19.480 --> 00:01:20.300
we could define it.

00:01:20.300 --> 00:01:21.520
We could either
define it explicitly,

00:01:21.520 --> 00:01:23.070
or we could define
it recursively.

00:01:23.070 --> 00:01:24.910
So if we wanted to
define it explicitly,

00:01:24.910 --> 00:01:29.070
we could write a sub n is equal
to whatever the first term is.

00:01:29.070 --> 00:01:33.290
In this case, our first
term is negative 5.

00:01:33.290 --> 00:01:36.340
It's equal to negative
5 plus-- we're

00:01:36.340 --> 00:01:40.530
going to add 2 one less
times than the term we're at.

00:01:40.530 --> 00:01:42.290
So for the second
term, we add 2 once.

00:01:42.290 --> 00:01:44.620
For the third term,
we add 2 twice.

00:01:44.620 --> 00:01:47.490
For the fourth term,
from our base term,

00:01:47.490 --> 00:01:49.000
we added 2 three times.

00:01:49.000 --> 00:01:51.150
So we're going to add 2.

00:01:51.150 --> 00:01:54.890
We're going to add positive
2 one less than the index

00:01:54.890 --> 00:01:58.380
that we're looking
at-- n minus 1 times.

00:01:58.380 --> 00:02:00.960
So this is an
explicit definition

00:02:00.960 --> 00:02:03.260
of this arithmetic sequence.

00:02:03.260 --> 00:02:04.950
If I wanted to write
it recursively,

00:02:04.950 --> 00:02:10.910
I could say a sub 1 is
equal to negative 5.

00:02:10.910 --> 00:02:14.490
And then each successive term,
for a sub 2 and greater--

00:02:14.490 --> 00:02:20.780
so I could say a sub n is equal
to a sub n minus 1 plus 3.

00:02:20.780 --> 00:02:24.120
Each term is equal to the
previous term-- oh, not 3--

00:02:24.120 --> 00:02:25.790
plus 2.

00:02:25.790 --> 00:02:31.230
So this is for n is
greater than or equal to 2.

00:02:31.230 --> 00:02:34.030
So either of these
are completely

00:02:34.030 --> 00:02:39.420
legitimate ways of defining
the arithmetic sequence

00:02:39.420 --> 00:02:40.230
that we have here.

00:02:40.230 --> 00:02:41.970
We can either define
it explicitly,

00:02:41.970 --> 00:02:44.950
or we could define
it recursively.

00:02:44.950 --> 00:02:46.330
Now let's look at this sequence.

00:02:46.330 --> 00:02:48.950
Is this one arithmetic?

00:02:48.950 --> 00:02:50.480
Well, we're going from 100.

00:02:50.480 --> 00:02:52.252
We add 7.

00:02:52.252 --> 00:02:54.976
107 to 114, we're adding 7.

00:02:54.976 --> 00:02:58.440
114 to 121, we are adding 7.

00:02:58.440 --> 00:03:00.760
So this is indeed an
arithmetic sequence.

00:03:00.760 --> 00:03:03.240
So just to be
clear, this is one,

00:03:03.240 --> 00:03:05.300
and this is one right over here.

00:03:05.300 --> 00:03:09.840
And we could write that this
is the sequence a sub n,

00:03:09.840 --> 00:03:14.800
n going from 1 to infinity
of-- and we could just

00:03:14.800 --> 00:03:17.610
say a sub n, if we want
to define it explicitly,

00:03:17.610 --> 00:03:23.190
is equal to 100 plus
we're adding 7 every time.

00:03:23.190 --> 00:03:26.330
And then each term-- the
second term we added 7 once.

00:03:26.330 --> 00:03:28.280
Third term-- we add 7 twice.

00:03:28.280 --> 00:03:32.430
So for the nth term, we're
going to add 7 n minus 1 times.

00:03:32.430 --> 00:03:34.340
So this is an explicit
definition of it,

00:03:34.340 --> 00:03:35.840
but we could also
do it recursively.

00:03:38.650 --> 00:03:42.270
So just to be clear, this is
one definition where we write it

00:03:42.270 --> 00:03:47.730
like this, or we
could write a sub n,

00:03:47.730 --> 00:03:50.240
from n equals 1 to infinity.

00:03:50.240 --> 00:03:53.836
And in either case
I should write with.

00:03:53.836 --> 00:03:55.460
And if I want to
define it recursively,

00:03:55.460 --> 00:03:58.950
I could say a sub
1 is equal to 100.

00:03:58.950 --> 00:04:04.170
And then, for anything larger
than 1, for any index above 1,

00:04:04.170 --> 00:04:09.560
a sub n is equal to the
previous term plus 7.

00:04:09.560 --> 00:04:10.822
And so we're done.

00:04:10.822 --> 00:04:12.280
This is another
way of defining it.

00:04:12.280 --> 00:04:14.920
So in general, if you
wanted a generalizable way

00:04:14.920 --> 00:04:18.149
to spot or define an
arithmetic sequence,

00:04:18.149 --> 00:04:20.680
you could say an
arithmetic sequence

00:04:20.680 --> 00:04:23.910
is going to be of the form
a sub n-- if we're talking

00:04:23.910 --> 00:04:26.930
about an infinite one--
from n equals 1 to infinity.

00:04:26.930 --> 00:04:28.460
If you want to
define it explicitly,

00:04:28.460 --> 00:04:33.540
you could say a sub n is
equal to some constant, which

00:04:33.540 --> 00:04:35.790
would essentially
the first term.

00:04:35.790 --> 00:04:41.470
It would be some
constant plus some number

00:04:41.470 --> 00:04:43.090
that your incrementing--
or I guess

00:04:43.090 --> 00:04:45.089
this could be a negative
number, or decrementing

00:04:45.089 --> 00:04:47.670
by-- times n minus 1.

00:04:47.670 --> 00:04:51.380
So this is one way to define
an arithmetic sequence.

00:04:51.380 --> 00:04:54.000
In this case, d was 2.

00:04:54.000 --> 00:04:55.160
In this case, d is 7.

00:04:55.160 --> 00:04:57.350
That's how much you're
adding by each time.

00:04:57.350 --> 00:05:02.280
And in this case, k is negative
5, and in this case, k is 100.

00:05:02.280 --> 00:05:04.850
The other way, if you
wanted to the right

00:05:04.850 --> 00:05:08.000
the recursive way of defining an
arithmetic sequence generally,

00:05:08.000 --> 00:05:12.180
you could say a sub
1 is equal to k,

00:05:12.180 --> 00:05:16.270
and then a sub n is
equal to a sub n minus 1.

00:05:16.270 --> 00:05:18.170
A given term is equal
to the previous term

00:05:18.170 --> 00:05:22.250
plus d for n greater
than or equal to 2.

00:05:22.250 --> 00:05:24.110
So once again, this is explicit.

00:05:24.110 --> 00:05:26.110
This is the recursive
way of defining it.

00:05:26.110 --> 00:05:28.140
And we would just
write with there.

00:05:28.140 --> 00:05:30.780
Now the last question I have
is is this one right over here

00:05:30.780 --> 00:05:32.630
an arithmetic sequence?

00:05:32.630 --> 00:05:34.730
Well, let's check it out.

00:05:34.730 --> 00:05:35.860
We start at 1.

00:05:35.860 --> 00:05:37.910
Then we add 2.

00:05:37.910 --> 00:05:38.929
Then we add 3.

00:05:38.929 --> 00:05:40.970
So this is an immediate
giveaway that this is not

00:05:40.970 --> 00:05:41.970
an arithmetic sequence.

00:05:41.970 --> 00:05:43.470
Now we are adding 4.

00:05:43.470 --> 00:05:46.780
We're adding a different
amount every time.

00:05:46.780 --> 00:05:51.130
So this, first of all,
this is not arithmetic.

00:05:51.130 --> 00:05:53.400
This is not an
arithmetic sequence.

00:05:53.400 --> 00:05:56.139
But how could we define
this, since we're

00:05:56.139 --> 00:05:57.430
trying to define our sequences?

00:06:02.100 --> 00:06:05.220
Let's say we wanted to
define it recursively.

00:06:05.220 --> 00:06:10.790
So we could say, this is
equal to a sub n, where

00:06:10.790 --> 00:06:13.830
n is starting at 1 and
it's going to infinity,

00:06:13.830 --> 00:06:20.100
with-- we'll say our base
case-- a sub 1 is equal to 1.

00:06:20.100 --> 00:06:23.580
And then for n is 2
or greater, a sub n

00:06:23.580 --> 00:06:25.400
is going to be equal to what?

00:06:25.400 --> 00:06:29.810
So a sub 2 is the previous
term plus 2. a sub 3

00:06:29.810 --> 00:06:32.170
is the previous
term plus 3. a sub 4

00:06:32.170 --> 00:06:34.200
is the previous term plus 4.

00:06:34.200 --> 00:06:38.980
So it's going to be the
previous term plus whatever

00:06:38.980 --> 00:06:40.210
your index is.

00:06:42.770 --> 00:06:45.030
So this looks close,
but notice here we're

00:06:45.030 --> 00:06:46.640
changing the amount
that we're adding

00:06:46.640 --> 00:06:47.765
based on what our index is.

00:06:47.765 --> 00:06:50.090
We're adding the amount of
index to the previous term.

00:06:50.090 --> 00:06:53.340
And so this is for n is
greater than or equal to 2.

00:06:53.340 --> 00:06:55.000
Well for an arithmetic
sequence, we're

00:06:55.000 --> 00:06:58.110
adding the same
amount regardless

00:06:58.110 --> 00:06:59.900
of what our index is.

00:06:59.900 --> 00:07:01.800
Here we're adding
the index itself.

00:07:01.800 --> 00:07:03.440
So this one is not
arithmetic, but it's

00:07:03.440 --> 00:07:05.970
an interesting
sequence nonetheless.


Explicit and recursive definitions of sequences

WEBVTT
Kind: captions
Language: en

00:00:00.500 --> 00:00:04.090
What I want to do in this
video is familiarize ourselves

00:00:04.090 --> 00:00:06.880
with the notion of a sequence.

00:00:06.880 --> 00:00:09.410
And all a sequence is is
an ordered list of numbers.

00:00:09.410 --> 00:00:12.172
So for example, I could
have a finite sequence--

00:00:12.172 --> 00:00:14.380
that means I don't have an
infinite number of numbers

00:00:14.380 --> 00:00:17.600
in it-- where, let's say, I
start at 1 and I keep adding 3.

00:00:17.600 --> 00:00:19.246
So 1 plus 3 is 4.

00:00:19.246 --> 00:00:20.366
4 plus 3 is 7.

00:00:20.366 --> 00:00:21.990
7 plus 3 is 10.

00:00:21.990 --> 00:00:25.670
And let's say I only have these
four terms right over here.

00:00:25.670 --> 00:00:29.530
So this one we would
call a finite sequence.

00:00:29.530 --> 00:00:31.560
I could also have an
infinite sequence.

00:00:31.560 --> 00:00:33.360
So an example of an
infinite sequence--

00:00:33.360 --> 00:00:37.940
let's say we start at
3, and we keep adding 4.

00:00:37.940 --> 00:00:43.180
So we go to 3, to 7, to 11, 15.

00:00:43.180 --> 00:00:45.180
And you don't always have
to add the same thing.

00:00:45.180 --> 00:00:46.734
We'll explore fancier sequences.

00:00:46.734 --> 00:00:48.900
The sequences where you
keep adding the same amount,

00:00:48.900 --> 00:00:50.570
we call these
arithmetic sequences,

00:00:50.570 --> 00:00:54.514
which we will also
explore in more detail.

00:00:54.514 --> 00:00:55.930
But to show that
this is infinite,

00:00:55.930 --> 00:00:58.360
to show that we keep this
pattern going on and on and on,

00:00:58.360 --> 00:00:59.920
I'll put three dots.

00:00:59.920 --> 00:01:02.670
This just means we're going to
keep going on and on and on.

00:01:02.670 --> 00:01:07.679
So we could call this
an infinite sequence.

00:01:07.679 --> 00:01:09.470
Now, there's a bunch
of different notations

00:01:09.470 --> 00:01:11.670
that seem fancy for
denoting sequences.

00:01:11.670 --> 00:01:13.090
But this is all they refer to.

00:01:13.090 --> 00:01:15.300
But I want to make us
comfortable with how

00:01:15.300 --> 00:01:20.497
we can denote sequences and
also how we can define them.

00:01:20.497 --> 00:01:22.080
We could say that
this right over here

00:01:22.080 --> 00:01:31.740
is the sequence a sub k
for k is going from 1 to 4,

00:01:31.740 --> 00:01:34.100
is equal to this
right over here.

00:01:34.100 --> 00:01:35.690
So when we look
at it this way, we

00:01:35.690 --> 00:01:38.430
can look at each of these as
the terms in the sequence.

00:01:38.430 --> 00:01:40.790
And this right over here
would be the first term.

00:01:40.790 --> 00:01:42.600
We would call that a sub 1.

00:01:42.600 --> 00:01:44.520
This right over here
would be the second term.

00:01:44.520 --> 00:01:46.070
We'd call it a sub 2.

00:01:46.070 --> 00:01:48.880
I think you get the
picture-- a sub 3.

00:01:48.880 --> 00:01:51.130
This right over here is a sub 4.

00:01:51.130 --> 00:01:54.900
So this just says, all of the
a sub k's from k equals 1,

00:01:54.900 --> 00:01:58.520
from our first term, all
the way to the fourth term.

00:01:58.520 --> 00:02:02.620
Now, I could also define it
by not explicitly writing

00:02:02.620 --> 00:02:03.700
the sequence like this.

00:02:03.700 --> 00:02:06.530
I could essentially do
it defining our sequence

00:02:06.530 --> 00:02:09.740
as explicitly using kind of a
function notation or something

00:02:09.740 --> 00:02:11.540
close to function notation.

00:02:11.540 --> 00:02:14.690
So the same exact
sequence, I could define it

00:02:14.690 --> 00:02:23.880
as a sub k from k equals 1 to
4, with-- instead of explicitly

00:02:23.880 --> 00:02:27.160
writing the numbers
here, I could say a sub k

00:02:27.160 --> 00:02:29.737
is equal to some function of k.

00:02:29.737 --> 00:02:30.820
So let's see what happens.

00:02:30.820 --> 00:02:32.890
When k is 1, we get 1.

00:02:32.890 --> 00:02:36.370
When k is 2, we get 4.

00:02:36.370 --> 00:02:39.509
When k is 3, we get 7.

00:02:39.509 --> 00:02:40.050
So let's see.

00:02:40.050 --> 00:02:43.660
When k is 3, we added 3 twice.

00:02:43.660 --> 00:02:44.800
Let me make it clear.

00:02:44.800 --> 00:02:47.040
So this was a plus 3.

00:02:47.040 --> 00:02:49.890
This right over
here was a plus 3.

00:02:49.890 --> 00:02:52.390
This right over
here is a plus 3.

00:02:52.390 --> 00:02:55.690
So whatever k is,
we started at 1.

00:02:55.690 --> 00:02:59.520
And we added 3 one less
than the k term times.

00:02:59.520 --> 00:03:04.420
So we could say that this
is going to be equal to 1

00:03:04.420 --> 00:03:13.685
plus k minus 1
times 3, or maybe I

00:03:13.685 --> 00:03:15.780
should write 3 times k
minus 1-- same thing.

00:03:19.920 --> 00:03:21.530
And you can verify
that this works.

00:03:24.840 --> 00:03:31.070
If k is equal to 1, you're
going to get 1 minus 1 is 0.

00:03:31.070 --> 00:03:33.660
And so a sub 1 is going to be 1.

00:03:33.660 --> 00:03:37.200
If k is equal to 2, you're going
to have 1 plus 3, which is 4.

00:03:37.200 --> 00:03:41.310
If k is equal to 3, you
get 3 times 2 plus 1 is 7.

00:03:41.310 --> 00:03:42.980
So it works out.

00:03:42.980 --> 00:03:47.270
So this is one way to explicitly
define our sequence with kind

00:03:47.270 --> 00:03:48.360
of this function notation.

00:03:48.360 --> 00:03:50.193
I want to make it clear--
I have essentially

00:03:50.193 --> 00:03:51.200
defined a function here.

00:03:51.200 --> 00:03:53.370
If I wanted a more
traditional function notation,

00:03:53.370 --> 00:03:55.960
I could have written
a of k, where

00:03:55.960 --> 00:03:58.190
k is the term that
I care about. a

00:03:58.190 --> 00:04:02.610
of k is equal to 1
plus 3 times k minus 1.

00:04:02.610 --> 00:04:04.180
This is essentially
a function, where

00:04:04.180 --> 00:04:06.000
an allowable input,
the domain, is

00:04:06.000 --> 00:04:09.340
restricted to positive integers.

00:04:09.340 --> 00:04:12.310
Now, how would I denote this
business right over here?

00:04:12.310 --> 00:04:16.100
Well, I could say that
this is equal to--

00:04:16.100 --> 00:04:17.380
and people tend to use a.

00:04:17.380 --> 00:04:20.310
But I could use the notation
b sub k or anything else.

00:04:20.310 --> 00:04:23.110
But I'll do a again-- a sub k.

00:04:23.110 --> 00:04:26.140
And here, we're going
from our first term--

00:04:26.140 --> 00:04:30.170
so this is a sub 1,
this is a sub 2--

00:04:30.170 --> 00:04:33.180
all the way to infinity.

00:04:33.180 --> 00:04:36.040
Or we could define it-- if we
wanted to define it explicitly

00:04:36.040 --> 00:04:41.780
as a function-- we could write
this sequence as a sub k, where

00:04:41.780 --> 00:04:45.570
k starts at the first
term and goes to infinity,

00:04:45.570 --> 00:04:53.720
with a sub k is equaling--
so we're starting at 3.

00:04:53.720 --> 00:04:58.180
And we are adding
4 one less time.

00:04:58.180 --> 00:04:59.980
For the second term,
we added 4 once.

00:04:59.980 --> 00:05:02.100
For the third term,
we add 4 twice.

00:05:02.100 --> 00:05:05.800
For the fourth term,
we add 4 three times.

00:05:05.800 --> 00:05:09.280
So we're adding 4 one less
than the term that we're at.

00:05:09.280 --> 00:05:15.440
So it's going to be
plus 4 times k minus 1.

00:05:19.150 --> 00:05:20.670
So this is another
way of defining

00:05:20.670 --> 00:05:22.890
this infinite sequence.

00:05:22.890 --> 00:05:24.770
Now, in both of these
cases, I defined it

00:05:24.770 --> 00:05:26.690
as an explicit function.

00:05:26.690 --> 00:05:29.670
So this right over
here is explicit.

00:05:29.670 --> 00:05:31.590
That's not an attractive color.

00:05:31.590 --> 00:05:34.010
Let me write this in.

00:05:34.010 --> 00:05:36.724
This is an explicit function.

00:05:36.724 --> 00:05:38.140
And so you might
say, well, what's

00:05:38.140 --> 00:05:40.390
another way of defining
these functions?

00:05:40.390 --> 00:05:44.080
Well, we can also define
it, especially something

00:05:44.080 --> 00:05:47.140
like an arithmetic sequence, we
can also define it recursively.

00:05:47.140 --> 00:05:49.850
And I want to be clear-- not
every sequence can be defined

00:05:49.850 --> 00:05:51.580
as either an explicit
function like this,

00:05:51.580 --> 00:05:52.950
or as a recursive function.

00:05:52.950 --> 00:05:54.452
But many can,
including this, which

00:05:54.452 --> 00:05:55.910
is an arithmetic
sequence, where we

00:05:55.910 --> 00:05:59.180
keep adding the same
quantity over and over again.

00:05:59.180 --> 00:06:00.610
So how would we do that?

00:06:00.610 --> 00:06:02.730
Well, we could also--
another way of defining

00:06:02.730 --> 00:06:07.140
this first sequence,
we could say a sub k,

00:06:07.140 --> 00:06:11.390
starting at k equals
1 and going to 4 with.

00:06:11.390 --> 00:06:14.900
And when you define a
sequence recursively,

00:06:14.900 --> 00:06:19.870
you want to define what your
first term is, with a sub 1

00:06:19.870 --> 00:06:22.390
equaling 1.

00:06:22.390 --> 00:06:25.370
You can define every other term
in terms of the term before it.

00:06:25.370 --> 00:06:33.660
And so then we
could write a sub k

00:06:33.660 --> 00:06:37.810
is equal to the previous term.

00:06:37.810 --> 00:06:39.700
So this is a sub k minus 1.

00:06:39.700 --> 00:06:41.700
So a given term is equal
to the previous term.

00:06:41.700 --> 00:06:48.530
Let me make it clear-- this is
the previous term, plus-- in

00:06:48.530 --> 00:06:50.350
this case, we're
adding 3 every time.

00:06:52.791 --> 00:06:54.040
Now, how does this make sense?

00:06:54.040 --> 00:06:55.977
Well, we're defining
what a sub 1 is.

00:06:55.977 --> 00:06:58.310
And if someone says, well,
what happens when k equals 2?

00:06:58.310 --> 00:07:01.900
Well, they're saying, well, it's
going to be a sub 2 minus 1.

00:07:01.900 --> 00:07:04.010
So it's going to
be a sub 1 plus 3.

00:07:04.010 --> 00:07:06.300
Well, we know a sub 1 is 1.

00:07:06.300 --> 00:07:08.630
So it's going to be
1 plus 3, which is 4.

00:07:08.630 --> 00:07:10.620
Well, what about a sub 3?

00:07:10.620 --> 00:07:12.920
Well, it's going to be
a sub 2 plus 3. a sub 2,

00:07:12.920 --> 00:07:13.990
we just calculated as 4.

00:07:13.990 --> 00:07:14.490
You add 3.

00:07:14.490 --> 00:07:15.410
It's going to be 7.

00:07:15.410 --> 00:07:16.910
This is essentially
what we mentally

00:07:16.910 --> 00:07:19.326
did when I first wrote out the
sequence, when I said, hey,

00:07:19.326 --> 00:07:22.480
I'm just going to start with 1.

00:07:22.480 --> 00:07:25.930
And I'm just going to add 3
for every successive term.

00:07:25.930 --> 00:07:27.870
So how would we do this one?

00:07:27.870 --> 00:07:32.320
Well, once again, we could
write this as a sub k.

00:07:32.320 --> 00:07:34.510
Starting at k, the
first term, going

00:07:34.510 --> 00:07:38.790
to infinity with-- our
first term, a sub 1,

00:07:38.790 --> 00:07:40.450
is going to be 3, now.

00:07:40.450 --> 00:07:42.900
And every successive
term, a sub k,

00:07:42.900 --> 00:07:50.240
is going to be the previous
term, a sub k minus 1, plus 4.

00:07:50.240 --> 00:07:52.599
And once again, you start at 3.

00:07:52.599 --> 00:07:54.140
And then if you want
the second term,

00:07:54.140 --> 00:07:55.580
it's going to be the
first term plus 4.

00:07:55.580 --> 00:07:56.780
It's going to be 3 plus 4.

00:07:56.780 --> 00:07:57.420
You get to 7.

00:07:57.420 --> 00:07:58.950
And you keep adding 4.

00:07:58.950 --> 00:08:02.090
So both of these,
this right over here

00:08:02.090 --> 00:08:06.280
is a recursive definition.

00:08:06.280 --> 00:08:09.020
We started with
kind of a base case.

00:08:09.020 --> 00:08:11.780
And then every term is
defined in terms of the term

00:08:11.780 --> 00:08:13.750
before it or in terms
of the function itself,

00:08:13.750 --> 00:08:17.056
but the function for
a different term.


Divergent improper integral

WEBVTT
Kind: captions
Language: en

00:00:00.680 --> 00:00:03.860
Right here I have graphed
part of the graph of y

00:00:03.860 --> 00:00:05.360
is equal to 1/x.

00:00:05.360 --> 00:00:08.890
And what I'm curious about
is the area under this curve

00:00:08.890 --> 00:00:13.890
and above the x-axis between
x equals 1 and infinity.

00:00:13.890 --> 00:00:19.080
So I want to figure out
this area right over here.

00:00:19.080 --> 00:00:20.810
So let's try to do it.

00:00:20.810 --> 00:00:25.180
So we could set this up as an
improper integral going from 1

00:00:25.180 --> 00:00:32.530
to infinity of 1/x dx.

00:00:32.530 --> 00:00:38.110
Well once again-- actually, let
me do that same yellow color.

00:00:38.110 --> 00:00:43.270
I like that more-- we can
view this as the limit as n

00:00:43.270 --> 00:00:47.830
approaches infinity of
the integral from 1 to n

00:00:47.830 --> 00:00:54.470
of 1/x dx, which we can
write as the limit as n

00:00:54.470 --> 00:00:57.870
approaches infinity of
the antiderivative of 1/x,

00:00:57.870 --> 00:01:01.980
which is the natural log
of the absolute value of x.

00:01:01.980 --> 00:01:04.500
So this is going to
be the natural log

00:01:04.500 --> 00:01:06.930
of the absolute value of x.

00:01:06.930 --> 00:01:09.340
And the absolute value of x
won't really matter so much.

00:01:09.340 --> 00:01:10.714
We could just say
x because we're

00:01:10.714 --> 00:01:12.210
dealing with
positive values of x.

00:01:12.210 --> 00:01:14.126
But I'll just write it
down as the natural log

00:01:14.126 --> 00:01:19.300
of the absolute value of x
between x is 1 and x is n.

00:01:19.300 --> 00:01:22.920
And so this is going to
be equal to the limit

00:01:22.920 --> 00:01:27.260
as n approaches infinity
of-- you evaluate this at n,

00:01:27.260 --> 00:01:29.307
so you're going to
get the natural log--

00:01:29.307 --> 00:01:30.890
I could write the
absolute value of n,

00:01:30.890 --> 00:01:32.681
but we know that n is
going to be positive.

00:01:32.681 --> 00:01:34.570
So we could just write
the natural log of n

00:01:34.570 --> 00:01:37.820
minus the natural log of
the absolute value of 1,

00:01:37.820 --> 00:01:40.170
or the natural log of 1.

00:01:40.170 --> 00:01:42.810
Natural log of 1 is just a 0.

00:01:42.810 --> 00:01:45.100
e to the 0-th power
is equal to 1.

00:01:45.100 --> 00:01:48.610
So this boils down
to the limit as n

00:01:48.610 --> 00:01:53.430
approaches infinity of
the natural log of n.

00:01:53.430 --> 00:01:55.000
Now this is interesting.

00:01:55.000 --> 00:01:57.010
Natural log function
just keeps getting

00:01:57.010 --> 00:01:58.900
larger and larger and larger.

00:01:58.900 --> 00:02:03.610
The natural log function keeps
growing and growing and growing

00:02:03.610 --> 00:02:07.030
like this, albeit at a
slower and slower pace,

00:02:07.030 --> 00:02:08.330
but it keeps growing.

00:02:08.330 --> 00:02:10.870
The limit as n approaches
infinity of the natural log

00:02:10.870 --> 00:02:14.940
of n is just equal to infinity.

00:02:14.940 --> 00:02:17.670
So here we do not
have a finite area.

00:02:17.670 --> 00:02:19.370
This is an infinite.

00:02:19.370 --> 00:02:21.040
This is an infinite area.

00:02:21.040 --> 00:02:22.090
It's interesting.

00:02:22.090 --> 00:02:24.220
When this function
decreased faster--

00:02:24.220 --> 00:02:26.960
when it was 1 over x squared--
we had a finite area.

00:02:26.960 --> 00:02:29.080
Now we have an infinite area.

00:02:29.080 --> 00:02:33.990
And so we would say that this
integral right over here,

00:02:33.990 --> 00:02:36.433
this improper
integral, is divergent.

00:02:39.700 --> 00:02:41.520
And we're done.


Integration by parts: Âºln(x)dx

WEBVTT
Kind: captions
Language: en

00:00:00.660 --> 00:00:02.680
The goal of this video
is to try to figure out

00:00:02.680 --> 00:00:05.600
the antiderivative of
the natural log of x.

00:00:05.600 --> 00:00:09.010
And it's not completely obvious
how to approach this at first,

00:00:09.010 --> 00:00:11.360
even if I were to tell you
to use integration by parts,

00:00:11.360 --> 00:00:12.950
you'll say,
integration by parts,

00:00:12.950 --> 00:00:15.190
you're looking for the
antiderivative of something

00:00:15.190 --> 00:00:17.590
that can be expressed as the
product of two functions.

00:00:17.590 --> 00:00:19.173
It looks like I only
have one function

00:00:19.173 --> 00:00:20.950
right over here, the
natural log of x.

00:00:20.950 --> 00:00:22.990
But it might become a
little bit more obvious

00:00:22.990 --> 00:00:25.750
if I were to rewrite
this as the integral

00:00:25.750 --> 00:00:30.740
of the natural log
of x times 1dx.

00:00:30.740 --> 00:00:33.040
Now, you do have the
product of two functions.

00:00:33.040 --> 00:00:35.910
One is a function,
a function of x.

00:00:35.910 --> 00:00:38.410
It's not actually dependent on
x, it's always going to be 1,

00:00:38.410 --> 00:00:40.520
but you could have f
of x is equal to 1.

00:00:40.520 --> 00:00:42.020
And now it might
become a little bit

00:00:42.020 --> 00:00:44.350
more obvious to use
integration by parts.

00:00:44.350 --> 00:00:46.800
Integration by parts
tells us that if we

00:00:46.800 --> 00:00:50.980
have an integral that can be
viewed as the product of one

00:00:50.980 --> 00:00:57.150
function, and the derivative
of another function,

00:00:57.150 --> 00:00:59.370
and this is really just
the reverse product rule,

00:00:59.370 --> 00:01:01.203
and we've shown that
multiple times already.

00:01:01.203 --> 00:01:04.420
This is going to be equal to
the product of both functions,

00:01:04.420 --> 00:01:13.470
f of x times g of x minus
the antiderivative of,

00:01:13.470 --> 00:01:15.270
instead of having f
and g prime, you're

00:01:15.270 --> 00:01:17.230
going to have f prime and g.

00:01:17.230 --> 00:01:27.570
So f prime of x times g of x dx.

00:01:27.570 --> 00:01:29.060
And we've seen this
multiple times.

00:01:29.060 --> 00:01:30.476
So when you figure
out what should

00:01:30.476 --> 00:01:32.077
be f and what should
be g, for f you

00:01:32.077 --> 00:01:33.660
want to figure out
something that it's

00:01:33.660 --> 00:01:36.740
easy to take the derivative
of and it simplifies things,

00:01:36.740 --> 00:01:38.950
possibly if you're taking
the derivative of it.

00:01:38.950 --> 00:01:41.370
And for g prime of x, you
want to find something

00:01:41.370 --> 00:01:44.100
where it's easy to take
the antiderivative of it.

00:01:44.100 --> 00:01:46.522
So good candidate for f
of x is natural log of x.

00:01:46.522 --> 00:01:48.230
If you were to take
the derivative of it,

00:01:48.230 --> 00:01:49.000
it's 1 over x.

00:01:49.000 --> 00:01:50.440
Let me write this down.

00:01:50.440 --> 00:01:56.140
So let's say that f of x is
equal to the natural log of x.

00:01:56.140 --> 00:02:00.230
Then f prime of x is
equal to 1 over x.

00:02:00.230 --> 00:02:03.810
And let's set g prime
of x is equal to 1.

00:02:03.810 --> 00:02:07.050
So g prime of x is equal to 1.

00:02:07.050 --> 00:02:12.340
That means that g of
x could be equal to x.

00:02:12.340 --> 00:02:14.500
And so let's go back
right over here.

00:02:14.500 --> 00:02:19.790
So this is going to be equal
to f of x times g of x.

00:02:19.790 --> 00:02:23.920
Well, f of x times g of
x is x natural log of x.

00:02:23.920 --> 00:02:28.354
So g of x is x, and f of
x is the natural log of x,

00:02:28.354 --> 00:02:29.770
I just like writing
the x in front

00:02:29.770 --> 00:02:32.370
of the natural log of
x to avoid ambiguity.

00:02:32.370 --> 00:02:37.490
So this is x natural log of x
minus the antiderivative of f

00:02:37.490 --> 00:02:41.800
prime of x, which
is 1 over x times g

00:02:41.800 --> 00:02:45.028
of x, which is x, which is xdx.

00:02:48.865 --> 00:02:50.490
Well, what's this
going to be equal to?

00:02:50.490 --> 00:02:53.080
Well, what we have inside
the integrand, this is just 1

00:02:53.080 --> 00:02:57.450
over x times x, which
is just equal to 1.

00:02:57.450 --> 00:03:00.210
So this simplifies quite nicely.

00:03:00.210 --> 00:03:13.530
This is going to end up
equaling x natural log of x

00:03:13.530 --> 00:03:16.530
minus the antiderivative
of, just dx,

00:03:16.530 --> 00:03:20.870
or the antiderivative of
1dx, or the integral of 1dx,

00:03:20.870 --> 00:03:24.009
or the antiderivative
of 1 is just minus x.

00:03:24.009 --> 00:03:25.800
And this is just an
antiderivative of this.

00:03:25.800 --> 00:03:28.091
If we want to write the entire
class of antiderivatives

00:03:28.091 --> 00:03:33.910
we just have to add a plus
c here, and we are done.

00:03:33.910 --> 00:03:36.550
We figured out
the antiderivative

00:03:36.550 --> 00:03:38.142
of the natural log of x.

00:03:38.142 --> 00:03:40.100
I encourage you to take
the derivative of this.

00:03:40.100 --> 00:03:42.380
For this part, you're going
to use the product rule

00:03:42.380 --> 00:03:46.910
and verify that you do indeed
get natural log of x when

00:03:46.910 --> 00:03:49.360
you take the derivative of this.


Improper integral with two infinite bounds

WEBVTT
Kind: captions
Language: en

00:00:00.640 --> 00:00:05.330
Right here we have the graph
of y is equal to 250 over 25

00:00:05.330 --> 00:00:06.730
plus x squared.

00:00:06.730 --> 00:00:08.680
And what I'm curious
about in this video

00:00:08.680 --> 00:00:11.820
is the total area under this
curve and above the x-axis.

00:00:11.820 --> 00:00:16.920
So I'm talking about everything
that I'm shading in white here,

00:00:16.920 --> 00:00:19.970
including what we can't see,
as we keep moving to the right

00:00:19.970 --> 00:00:21.690
and we keep moving to the left.

00:00:21.690 --> 00:00:25.190
So I'm talking about from
x at negative infinity all

00:00:25.190 --> 00:00:27.700
the way to x at infinity.

00:00:27.700 --> 00:00:30.035
So first, how would we
actually denote this?

00:00:30.035 --> 00:00:31.660
Well, it would be an
improper integral.

00:00:31.660 --> 00:00:34.630
We would denote this area
as the indefinite integral

00:00:34.630 --> 00:00:37.230
from x is equal to
negative infinity

00:00:37.230 --> 00:00:40.800
to x is equal to
infinity of our function,

00:00:40.800 --> 00:00:49.610
250 over 25 plus x squared, dx.

00:00:49.610 --> 00:00:52.380
Now, we've already
seen improper integrals

00:00:52.380 --> 00:00:54.104
where one of our
boundaries was infinity.

00:00:54.104 --> 00:00:55.520
But how do you do
it when you have

00:00:55.520 --> 00:00:57.007
one boundary at
positive infinity

00:00:57.007 --> 00:00:58.590
and one boundary at
negative infinity?

00:00:58.590 --> 00:01:00.814
You can't take a limit
to two different things.

00:01:00.814 --> 00:01:02.730
And so the way that we're
going to tackle this

00:01:02.730 --> 00:01:04.170
is to actually
break up this area

00:01:04.170 --> 00:01:07.340
into two different
improper integrals, one

00:01:07.340 --> 00:01:10.760
improper integral that describes
this area right over here

00:01:10.760 --> 00:01:13.810
in blue from negative
infinity to 0.

00:01:13.810 --> 00:01:16.840
So we'll say that this is equal
to the improper integral that

00:01:16.840 --> 00:01:23.260
goes from negative infinity
to 0 of 250 over 25

00:01:23.260 --> 00:01:27.490
plus x squared dx, plus
the improper integral that

00:01:27.490 --> 00:01:31.090
goes from 0 to
positive infinity.

00:01:31.090 --> 00:01:33.840
So plus the improper,
or the definite,

00:01:33.840 --> 00:01:41.220
integral from 0 to positive
infinity of 250 over 25

00:01:41.220 --> 00:01:45.830
plus x squared dx.

00:01:45.830 --> 00:01:47.920
And now we can start
to make sense of this.

00:01:47.920 --> 00:01:51.720
So what we have in
blue can be rewritten.

00:01:51.720 --> 00:01:57.060
This is equal to the limit as
n approaches negative infinity

00:01:57.060 --> 00:02:03.480
of the definite integral
from n to 0 of 250 over 25

00:02:03.480 --> 00:02:06.500
plus x squared dx.

00:02:06.500 --> 00:02:09.830
Plus-- and I'm running
out of real estate

00:02:09.830 --> 00:02:14.110
here-- the limit as--
since I already used n,

00:02:14.110 --> 00:02:16.980
let me use m now-- the
limit as m approaches

00:02:16.980 --> 00:02:20.560
positive infinity of the
definite integral from 0

00:02:20.560 --> 00:02:29.460
to m of 250 over 25
plus x squared dx.

00:02:29.460 --> 00:02:31.500
So now all we have
to do is evaluate

00:02:31.500 --> 00:02:32.910
these definite integrals.

00:02:32.910 --> 00:02:34.660
And to do that, we
just have to figure out

00:02:34.660 --> 00:02:39.050
an antiderivative of 250
over 25 plus x squared.

00:02:39.050 --> 00:02:41.280
So let's try to figure
out what that is.

00:02:41.280 --> 00:02:43.360
I'll do it over
here on the left.

00:02:43.360 --> 00:02:51.550
So we need to figure out the
antiderivative of 250 over 25

00:02:51.550 --> 00:02:53.140
plus x squared.

00:02:53.140 --> 00:02:54.640
And it might already
jump out at you

00:02:54.640 --> 00:02:57.390
that trig substitution
might be a good thing to do.

00:02:57.390 --> 00:03:00.490
You see this pattern of
a squared plus x squared,

00:03:00.490 --> 00:03:02.400
where in this
case, a would be 5.

00:03:02.400 --> 00:03:05.020
So we can make the
substitution that x

00:03:05.020 --> 00:03:09.830
is equal to a tangent
theta, 5 tangent theta.

00:03:09.830 --> 00:03:12.650
And since we're going to have
to reverse substitute later on,

00:03:12.650 --> 00:03:14.630
we can also put in
the constraint-- well,

00:03:14.630 --> 00:03:18.610
we'd say x/5 is equal to tangent
of theta, which is completely

00:03:18.610 --> 00:03:20.410
consistent with this
first statement.

00:03:20.410 --> 00:03:22.880
And so if we wanted to
have theta expressed

00:03:22.880 --> 00:03:25.360
as a function of x, we
can put the constraint

00:03:25.360 --> 00:03:31.210
that theta is equal
to arctangent of x/5.

00:03:31.210 --> 00:03:33.420
So once again, this is
completely consistent

00:03:33.420 --> 00:03:36.490
with this over here. x
can be 5 tangent of theta,

00:03:36.490 --> 00:03:41.020
and theta can be equal
to arctangent of x/5.

00:03:41.020 --> 00:03:42.760
So now let's do
the substitution.

00:03:42.760 --> 00:03:44.218
Actually, before
that, we also have

00:03:44.218 --> 00:03:45.690
to figure out what
dx is equal to.

00:03:45.690 --> 00:03:51.287
So dx is equal to-- I'll do
it right over here-- well,

00:03:51.287 --> 00:03:53.120
the derivative of this
with respect to theta

00:03:53.120 --> 00:03:58.200
is 5 secant squared
theta d theta.

00:03:58.200 --> 00:04:00.570
So now we're ready to
substitute back in.

00:04:00.570 --> 00:04:07.180
So all of this business is going
to be equal to 250 times dx.

00:04:07.180 --> 00:04:18.170
Well, 250 times dx is 250 times
5 secant squared theta d theta.

00:04:18.170 --> 00:04:21.089
So that's 250 dx is
this business up here.

00:04:21.089 --> 00:04:25.605
All of that over
25 plus x squared.

00:04:25.605 --> 00:04:33.470
Well, x squared is going to be
25 tangent squared of theta.

00:04:33.470 --> 00:04:35.930
And now we can try to
simplify all of this business.

00:04:35.930 --> 00:04:42.370
This is equal to
250 times 5 secant

00:04:42.370 --> 00:04:47.740
squared theta over
25 times 1 plus

00:04:47.740 --> 00:04:51.960
tangent squared theta d theta.

00:04:51.960 --> 00:04:55.790
So 25 goes into 250 10 times.

00:04:55.790 --> 00:04:59.780
And then 1 plus tangent squared
theta is secant squared theta.

00:04:59.780 --> 00:05:03.090
You can prove it for yourself
if you write tan squared theta

00:05:03.090 --> 00:05:05.407
as sine squared theta
over cosine squared theta.

00:05:05.407 --> 00:05:06.990
Add that to 1, which
is the same thing

00:05:06.990 --> 00:05:09.340
as cosine squared theta
over cosine squared theta.

00:05:09.340 --> 00:05:12.350
And then you can use some
basic trig identities

00:05:12.350 --> 00:05:18.170
to realize that this is equal
to secant squared theta, which

00:05:18.170 --> 00:05:20.230
simplifies our
expression a good bit.

00:05:20.230 --> 00:05:22.780
You get secant squared theta
over secant squared theta,

00:05:22.780 --> 00:05:23.990
which is just 1.

00:05:23.990 --> 00:05:27.800
And so you're just left with
the 10 times the 5, 50 d theta.

00:05:27.800 --> 00:05:30.010
So this is equal
to 50-- I'll take

00:05:30.010 --> 00:05:33.460
the 50 outside the
integral-- 50 d theta, which

00:05:33.460 --> 00:05:37.529
is equal to 50 theta.

00:05:37.529 --> 00:05:39.070
We can write the
plus C, just to show

00:05:39.070 --> 00:05:41.460
that these are all of
the antiderivatives.

00:05:41.460 --> 00:05:44.090
But we only need the
most basic antiderivative

00:05:44.090 --> 00:05:46.560
to evaluate these
definite integrals.

00:05:46.560 --> 00:05:49.020
But we right now only
have it in terms of theta.

00:05:49.020 --> 00:05:50.760
Let's write this in terms of x.

00:05:50.760 --> 00:05:54.890
We set the constraint that theta
is equal to arctangent of x/5.

00:05:54.890 --> 00:06:02.800
So this is equal to 50
arctangent of x/5 plus C.

00:06:02.800 --> 00:06:04.400
These are all the
antiderivatives.

00:06:04.400 --> 00:06:06.620
We can pick C is
equal to 0 to find

00:06:06.620 --> 00:06:08.310
an antiderivative
of these things

00:06:08.310 --> 00:06:10.410
to evaluate the
definite integrals.

00:06:10.410 --> 00:06:12.030
So let's do that.

00:06:12.030 --> 00:06:17.900
So what we have in blue, we
can rewrite as the limit as n

00:06:17.900 --> 00:06:23.190
approaches negative infinity
of the antiderivative of this,

00:06:23.190 --> 00:06:25.640
or an antiderivative of
this, which we could say

00:06:25.640 --> 00:06:32.400
is 50 arctangent of x/5.

00:06:32.400 --> 00:06:35.970
And we're going to
evaluate it at 0 and n.

00:06:35.970 --> 00:06:41.370
And to this, we're going to
add the limit as m approaches

00:06:41.370 --> 00:06:46.000
positive infinity of just
the antiderivative of this,

00:06:46.000 --> 00:06:55.360
50 arctangent-- not "actangent,"
arctangent-- arctangent of x/5

00:06:55.360 --> 00:07:00.050
evaluated from 0 to m.

00:07:00.050 --> 00:07:02.450
And let me put a little
parentheses right

00:07:02.450 --> 00:07:04.485
around this x/5.

00:07:04.485 --> 00:07:05.610
So what's this going to be?

00:07:05.610 --> 00:07:08.150
This is going to be 50--
so let me write this.

00:07:08.150 --> 00:07:11.100
This is going to be the
limit as n approaches

00:07:11.100 --> 00:07:16.960
negative infinity of
50 arctangent of 0/5

00:07:16.960 --> 00:07:24.810
/ minus 50 arctangent of n/5.

00:07:27.660 --> 00:07:32.595
And then to that, we're going
to add-- let me give myself

00:07:32.595 --> 00:07:38.280
even a little bit more space--
the limit as m approaches

00:07:38.280 --> 00:07:50.980
infinity of 50 arctan of m/5
minus 50 arctan-- I think

00:07:50.980 --> 00:07:55.670
you see where all this is
going-- minus 50 arctan of 0/5.

00:07:55.670 --> 00:07:57.520
So now we can
evaluate these things.

00:07:57.520 --> 00:07:59.540
And to help us
evaluate them, let's

00:07:59.540 --> 00:08:01.074
think about a unit circle.

00:08:01.074 --> 00:08:02.490
So let's think
about a unit circle

00:08:02.490 --> 00:08:05.900
so we can really visualize
the arctangent function.

00:08:05.900 --> 00:08:08.920
So one way to think
about the tangent

00:08:08.920 --> 00:08:12.940
is the slope of the line
that is on the angle

00:08:12.940 --> 00:08:16.160
or that helps define the
terminal side of an angle.

00:08:16.160 --> 00:08:19.330
So for example, if we have an
angle that looks like this--

00:08:19.330 --> 00:08:21.570
this is an angle between
the positive x-axis

00:08:21.570 --> 00:08:23.110
and this line right over here.

00:08:23.110 --> 00:08:26.540
If we have an angle just like
that, the tangent of that angle

00:08:26.540 --> 00:08:29.661
is going to be the
slope of this line.

00:08:29.661 --> 00:08:31.160
And so one way to
think about it is,

00:08:31.160 --> 00:08:36.380
OK, if I want the arctangent
of 0, it's essentially saying,

00:08:36.380 --> 00:08:42.250
OK, let's get an angle where
the terminal side has slope 0.

00:08:42.250 --> 00:08:45.330
Well, an angle where the
terminal side has slope 0

00:08:45.330 --> 00:08:46.890
is an angle of 0.

00:08:46.890 --> 00:08:50.360
So arctangent of 0 is 0 radians.

00:08:50.360 --> 00:08:51.620
So it's 50 times 0.

00:08:51.620 --> 00:08:53.065
That's just going to be 0.

00:08:53.065 --> 00:08:54.690
And we could also
write that over here.

00:08:54.690 --> 00:08:57.540
This is just going to be 0.

00:08:57.540 --> 00:09:03.740
So what we're left
with is the limit

00:09:03.740 --> 00:09:06.440
as n approaches
negative infinity

00:09:06.440 --> 00:09:19.850
of negative 50 arctangent of n/5
plus the limit as m approaches

00:09:19.850 --> 00:09:28.400
positive infinity of
50 arctangent of m/5.

00:09:28.400 --> 00:09:30.900
So let's think about what
these are going to be.

00:09:30.900 --> 00:09:33.386
So it's the limit as n
approaches negative infinity.

00:09:33.386 --> 00:09:34.760
One way to view
that is the limit

00:09:34.760 --> 00:09:37.120
as the slope of
this terminal side

00:09:37.120 --> 00:09:38.464
approaches negative infinity.

00:09:38.464 --> 00:09:40.380
So that's getting more
and more negative, more

00:09:40.380 --> 00:09:41.544
and more negative.

00:09:41.544 --> 00:09:42.960
It's at negative
infinity, or it's

00:09:42.960 --> 00:09:48.080
approaching negative infinity
when our angle right over here

00:09:48.080 --> 00:09:50.480
is negative pi/2 radians.

00:09:55.520 --> 00:09:59.850
Or you could say the
limit of arctan of n/5

00:09:59.850 --> 00:10:02.270
as n approaches negative
infinity, this part

00:10:02.270 --> 00:10:04.590
right over here, is
going to approach,

00:10:04.590 --> 00:10:06.360
as n approaches
negative infinity,

00:10:06.360 --> 00:10:08.069
it's going to be negative pi/2.

00:10:08.069 --> 00:10:09.610
And then we're just
going to multiply

00:10:09.610 --> 00:10:11.730
that times the negative 50.

00:10:11.730 --> 00:10:18.740
So this is going to be negative
50 times negative pi/2, which

00:10:18.740 --> 00:10:21.020
is going to be positive 25 pi.

00:10:21.020 --> 00:10:24.460
So this is going to be 25 pi.

00:10:24.460 --> 00:10:28.400
And similarly,
arctan of m/5 as m

00:10:28.400 --> 00:10:31.279
approaches infinity-- well,
as m approaches infinity,

00:10:31.279 --> 00:10:33.070
the slope of the terminal
side of the angle

00:10:33.070 --> 00:10:34.100
approaches infinity.

00:10:34.100 --> 00:10:36.240
So the slope is
getting higher, higher.

00:10:36.240 --> 00:10:38.950
It approaches infinity as
it gets closer and closer

00:10:38.950 --> 00:10:40.010
to vertical.

00:10:40.010 --> 00:10:43.080
So the arctan of
m/5 as m approaches

00:10:43.080 --> 00:10:45.870
infinity is going
to be positive pi/2.

00:10:45.870 --> 00:10:49.324
This is an angle
of positive pi/2.

00:10:49.324 --> 00:10:50.990
So what we see in
orange right over here

00:10:50.990 --> 00:10:54.810
is going to be
positive 50 times pi/2,

00:10:54.810 --> 00:10:58.220
which is going to be 25 pi.

00:10:58.220 --> 00:11:00.770
So the area that
we have in blue,

00:11:00.770 --> 00:11:06.280
going back to our original
problem, is 25 pi.

00:11:06.280 --> 00:11:10.144
The area that we have
in orange is 25 pi.

00:11:10.144 --> 00:11:12.310
And so if we wanted to
answer our original question,

00:11:12.310 --> 00:11:14.226
what's the total area
under this curve-- which

00:11:14.226 --> 00:11:17.110
is kind of a cool
question to try

00:11:17.110 --> 00:11:24.680
to answer-- we get it's 25 pi
plus 25 pi, which is 50 pi.

00:11:24.680 --> 00:11:26.550
And we're done.


Introduction to improper integrals

WEBVTT
Kind: captions
Language: en

00:00:00.550 --> 00:00:02.970
What I want to figure
out in this video is

00:00:02.970 --> 00:00:06.600
the area under the curve
y equals 1 over x squared,

00:00:06.600 --> 00:00:10.270
with x equals 1 as
our lower boundary

00:00:10.270 --> 00:00:12.190
and have no upper
boundary, just keep

00:00:12.190 --> 00:00:14.110
on going forever and forever.

00:00:14.110 --> 00:00:16.960
It really is essentially
as x approaches infinity.

00:00:16.960 --> 00:00:22.140
So I want to figure out
what this entire area is.

00:00:22.140 --> 00:00:23.990
And one way that
we can denote that

00:00:23.990 --> 00:00:26.540
is with an improper
definite integral,

00:00:26.540 --> 00:00:28.010
or an improper integral.

00:00:28.010 --> 00:00:30.520
And we would denote it as
1 is our lower boundary,

00:00:30.520 --> 00:00:32.414
but we're just going to
keep on going forever

00:00:32.414 --> 00:00:33.330
as our upper boundary.

00:00:33.330 --> 00:00:36.120
So our upper
boundary is infinity.

00:00:36.120 --> 00:00:41.960
And we're taking the integral
of 1 over x squared dx.

00:00:41.960 --> 00:00:43.170
And so let me be very clear.

00:00:43.170 --> 00:00:51.170
This right over here is
an improper integral.

00:00:51.170 --> 00:00:53.190
Now how do we actually
deal with this?

00:00:53.190 --> 00:00:56.210
Well, by definition
this is the same thing

00:00:56.210 --> 00:01:01.460
as the limit as n
approaches infinity

00:01:01.460 --> 00:01:09.380
of the integral from 1 to
n of 1 over x squared dx.

00:01:09.380 --> 00:01:12.270
And this is nice, because we
know how to evaluate this.

00:01:12.270 --> 00:01:15.430
This is just a definite integral
where the upper boundary is n.

00:01:15.430 --> 00:01:17.050
And then we know
how to take limits.

00:01:17.050 --> 00:01:20.750
We can figure out what the limit
is as n approaches infinity.

00:01:20.750 --> 00:01:23.680
So let's figure out if we can
actually evaluate this thing.

00:01:23.680 --> 00:01:26.186
So the second fundamental
theorem of calculus,

00:01:26.186 --> 00:01:28.060
or the second part of
the fundamental theorem

00:01:28.060 --> 00:01:30.685
of calculus, tells us that
this piece right over here--

00:01:30.685 --> 00:01:32.060
just let me write
the limit part.

00:01:32.060 --> 00:01:33.730
So this part I'll just rewrite.

00:01:33.730 --> 00:01:37.762
The limit as n
approaches infinity of--

00:01:37.762 --> 00:01:39.970
and we're going to use the
second fundamental theorem

00:01:39.970 --> 00:01:40.480
of calculus.

00:01:40.480 --> 00:01:44.490
We're going to evaluate
the antiderivative of 1

00:01:44.490 --> 00:01:47.170
over x squared or x
to the negative 2.

00:01:47.170 --> 00:01:49.920
So the antiderivative
of x to the negative 2

00:01:49.920 --> 00:01:52.700
is negative x to the negative 1.

00:01:52.700 --> 00:01:58.040
So negative x to the negative
1 or negative 1 over x.

00:01:58.040 --> 00:02:00.885
So negative 1/x is
the antiderivative.

00:02:00.885 --> 00:02:04.950
And we're going to evaluate
at n and evaluate it at 1.

00:02:04.950 --> 00:02:12.030
So this is going to be equal
to the limit as n approaches

00:02:12.030 --> 00:02:14.210
infinity.

00:02:14.210 --> 00:02:17.890
Let's see, if we evaluate this
thing at n, we get negative 1

00:02:17.890 --> 00:02:18.550
over n.

00:02:21.910 --> 00:02:23.810
And from that we're
going to subtract

00:02:23.810 --> 00:02:24.965
this thing evaluated at 1.

00:02:24.965 --> 00:02:27.710
So it's negative 1 over
1, or it's negative 1.

00:02:27.710 --> 00:02:29.930
So this right over
here is negative 1.

00:02:29.930 --> 00:02:33.160
And so we're going to find the
limit as n approaches infinity

00:02:33.160 --> 00:02:34.270
of this business.

00:02:34.270 --> 00:02:36.500
This stuff right here is
just the stuff right here.

00:02:36.500 --> 00:02:38.250
I haven't found the limit yet.

00:02:38.250 --> 00:02:42.680
So this is going to be equal
to the limit as n approaches

00:02:42.680 --> 00:02:47.220
infinity of-- let's see,
this is positive 1--

00:02:47.220 --> 00:02:53.340
and we can even write that minus
1 over n-- of 1 minus 1 over n.

00:02:53.340 --> 00:02:56.770
And lucky for us, this
limit actually exists.

00:02:56.770 --> 00:03:00.010
Limit as n approaches infinity,
this term right over here

00:03:00.010 --> 00:03:02.670
is going to get closer and
closer and closer to 0.

00:03:02.670 --> 00:03:06.240
1 over infinity you can
essentially view as 0.

00:03:06.240 --> 00:03:08.720
So this right over
here is going to be

00:03:08.720 --> 00:03:10.900
equal to 1, which
is pretty neat.

00:03:10.900 --> 00:03:13.790
We have this area that
has no right boundary.

00:03:13.790 --> 00:03:15.570
It just keeps on going forever.

00:03:15.570 --> 00:03:18.960
But we still have a
finite area, and the area

00:03:18.960 --> 00:03:23.780
is actually exactly equal to 1.

00:03:23.780 --> 00:03:25.710
So in this case we had
an improper integral.

00:03:25.710 --> 00:03:28.950
And because we were actually
able to evaluate it and come up

00:03:28.950 --> 00:03:31.500
with the number that this
limit actually existed,

00:03:31.500 --> 00:03:34.270
we say that this improper
integral right over here

00:03:34.270 --> 00:03:37.500
is convergent.

00:03:37.500 --> 00:03:39.345
If for whatever reason
this was unbounded

00:03:39.345 --> 00:03:41.860
and we couldn't come up with
some type of a finite number

00:03:41.860 --> 00:03:43.750
here, if the area
was infinite, we

00:03:43.750 --> 00:03:45.930
would say that it is divergent.

00:03:45.930 --> 00:03:48.620
So right over here we figured
out a kind of neat thing.

00:03:48.620 --> 00:03:51.141
This area is exactly 1.