Approximating functions with polynomials (part 3)

WEBVTT
Kind: captions
Language: en

00:00:00.700 --> 00:00:02.590
I had dinner between the
last video and this one.

00:00:02.590 --> 00:00:05.300
So I might have forgotten
what I just did.

00:00:05.300 --> 00:00:08.740
I think I was about to-- if
what I see on my board makes

00:00:08.740 --> 00:00:13.010
sense-- I was about to use the
Taylor series, or in this

00:00:13.010 --> 00:00:15.990
specific example the Maclaurin
series approximation, to figure

00:00:15.990 --> 00:00:21.410
out a polynomial version, a sum
of polynomial terms to

00:00:21.410 --> 00:00:23.480
approximate e to the x.

00:00:23.480 --> 00:00:27.200
And remember let me write
here what the definition of

00:00:27.200 --> 00:00:31.380
the Maclaurin series was.

00:00:31.380 --> 00:00:42.560
That we said that f of x is
equal to the sum from n equals

00:00:42.560 --> 00:00:49.890
0 to infinite of the nth
derivative of f evaluated at

00:00:49.890 --> 00:00:52.460
0-- I don't know if I
remembered to put it evaluated

00:00:52.460 --> 00:00:55.840
at 0 last time I wrote this
down-- times x to the

00:00:55.840 --> 00:00:57.990
n over n factorial.

00:00:57.990 --> 00:00:59.430
And hopefully that
makes sense to you.

00:00:59.430 --> 00:01:01.920
This might seem really
confusing and strange.

00:01:01.920 --> 00:01:04.930
But now that we're going to
apply it to e to the x, this

00:01:04.930 --> 00:01:06.780
should maybe be a little
bit more concrete.

00:01:06.780 --> 00:01:09.650
And I think at the end of the
last video I said that if f of

00:01:09.650 --> 00:01:15.140
x is e to the x, f of 0 is
e to the 0, which is 1.

00:01:15.140 --> 00:01:21.180
And f prime of x, any
derivative of e to the x

00:01:21.180 --> 00:01:23.810
is equal to e to the x.

00:01:23.810 --> 00:01:30.090
So you take any derivative of
at 0, and it equals 1 for e to

00:01:30.090 --> 00:01:32.490
the x for this particular
case of f of x.

00:01:32.490 --> 00:01:33.200
And that's really neat.

00:01:33.200 --> 00:01:37.770
That means that the rate of
change of y with respect to x

00:01:37.770 --> 00:01:43.300
is for every 1 you move in x,
you move 1 in y at e to the 0.

00:01:43.300 --> 00:01:46.610
But that also means that
the rate of change of the

00:01:46.610 --> 00:01:47.710
rate of change is also 1.

00:01:47.710 --> 00:01:49.060
At the rate of change at
the rate of change at the

00:01:49.060 --> 00:01:50.030
rate of change is also 1.

00:01:50.030 --> 00:01:57.470
So at either the 0 or x equals
0 of e to the x, the slope of

00:01:57.470 --> 00:01:59.470
the slope of the slope of the
slope of the slope of the slope

00:01:59.470 --> 00:02:00.710
of the slope, they're all 1.

00:02:00.710 --> 00:02:05.420
Which to me tells me something
mysterious is happening.

00:02:05.420 --> 00:02:08.040
It's another reason why you
should just sit and ponder e.

00:02:08.040 --> 00:02:10.240
But anyway, back to what
we were trying to do.

00:02:10.240 --> 00:02:11.370
So how would we do this?

00:02:11.370 --> 00:02:12.805
How would be write
the approximation?

00:02:16.660 --> 00:02:17.680
Let me write the approximate.

00:02:17.680 --> 00:02:18.800
I'll call that p of x.

00:02:18.800 --> 00:02:19.620
Because it's going
to be a polynomial.

00:02:22.740 --> 00:02:30.040
Well in this particular case,
what's the derivative of any

00:02:30.040 --> 00:02:33.590
derivative evaluated at f of 0?

00:02:33.590 --> 00:02:34.490
Well that term is 1.

00:02:34.490 --> 00:02:37.290
We wrote that down right here.

00:02:37.290 --> 00:02:39.890
f of 0 is 1.

00:02:39.890 --> 00:02:41.490
The first derivative at 0 is 1.

00:02:41.490 --> 00:02:42.830
The second derivative
at 0 is 1.

00:02:42.830 --> 00:02:43.010
Right?

00:02:43.010 --> 00:02:44.410
That's what's special
about e to the x.

00:02:44.410 --> 00:02:48.680
So all of these terms
are going to equal 1.

00:02:48.680 --> 00:02:56.740
So this polynomial simplifies
to the sum from n equals 0 to

00:02:56.740 --> 00:03:02.470
infinite of x to the
n over n factorial.

00:03:02.470 --> 00:03:03.920
That to me is very neat.

00:03:03.920 --> 00:03:06.010
Remember these are
all 1 in every term.

00:03:06.010 --> 00:03:07.960
So that's why I took it out.

00:03:07.960 --> 00:03:08.810
So what does that mean?

00:03:08.810 --> 00:03:11.960
Well that tells us that e to
the x can be approximated.

00:03:11.960 --> 00:03:13.630
And actually, I don't
prove it here.

00:03:13.630 --> 00:03:16.660
But it actually turns out that
we take the infinite sum that

00:03:16.660 --> 00:03:19.030
the Maclaurin series not only
approximates e to the

00:03:19.030 --> 00:03:21.245
x at x equals 0.

00:03:21.245 --> 00:03:23.620
When you take the infinite
series, it actually

00:03:23.620 --> 00:03:25.460
equals e to the x.

00:03:25.460 --> 00:03:30.490
So when you take a Maclaurin
series at 0, and the resulting

00:03:30.490 --> 00:03:33.670
function, the resulting
polynomial actually converges--

00:03:33.670 --> 00:03:35.560
and that's something we'll
learn a little bit more

00:03:35.560 --> 00:03:43.340
rigorously hopefully later when
we start doing analysis-- but

00:03:43.340 --> 00:03:46.090
it can actually converge to the
function at all points.

00:03:46.090 --> 00:03:48.490
And it actually is the
case with e to the x.

00:03:48.490 --> 00:03:51.240
So we can actually say
that e to the x is equal.

00:03:51.240 --> 00:03:52.050
I didn't prove this.

00:03:52.050 --> 00:03:54.330
But you can take
my word for it.

00:03:54.330 --> 00:03:56.100
And you can even test it
out with some numbers.

00:03:56.100 --> 00:03:57.340
It equals this sum.

00:03:57.340 --> 00:03:59.060
Well what is this sum?

00:03:59.060 --> 00:04:08.020
It's x to the 0 over 0
factorial plus x to the 1 over

00:04:08.020 --> 00:04:12.660
1 factorial plus x-squared
over 2 factorial.

00:04:12.660 --> 00:04:13.780
And you keep going.

00:04:13.780 --> 00:04:14.780
Of course that's equal.

00:04:14.780 --> 00:04:17.950
So e to x is equal
to-- x to the 0 is 1.

00:04:17.950 --> 00:04:19.870
0 factorial, I said in
the last video is 1.

00:04:19.870 --> 00:04:26.740
So it's 1 plus this is just x,
plus x-squared over 2 factorial

00:04:26.740 --> 00:04:32.610
plus x to the third over
3 factorial plus x to the

00:04:32.610 --> 00:04:34.810
fourth over 4 factorial.

00:04:34.810 --> 00:04:36.790
And you just keep
going on forever.

00:04:36.790 --> 00:04:38.480
And that's e to the x.

00:04:38.480 --> 00:04:40.490
And to me that is amazing.

00:04:40.490 --> 00:04:44.810
Because this strange number e,
this 2.7 whatever, whatever,

00:04:44.810 --> 00:04:48.130
that we got from compound
interest, it can be written as

00:04:48.130 --> 00:04:51.600
an infinite polynomial, this
polynomial series or this

00:04:51.600 --> 00:04:54.730
Maclaurin series, that actually
has a certain beauty to it.

00:04:54.730 --> 00:04:56.870
This number is kind of ugly
when you write it out,

00:04:56.870 --> 00:04:58.380
2.7 whatever, whatever.

00:04:58.380 --> 00:05:01.560
But when you write it to an
exponent power as an infinite

00:05:01.560 --> 00:05:04.960
sum, it kind of has
a nice rhythm to it.

00:05:04.960 --> 00:05:06.680
It's very patterned.

00:05:06.680 --> 00:05:08.540
Who would have guessed that
you could have written it

00:05:08.540 --> 00:05:10.390
in such a simple form.

00:05:10.390 --> 00:05:15.460
And even more, what happens
when x is equal to 1.

00:05:15.460 --> 00:05:15.770
Right?

00:05:15.770 --> 00:05:16.770
So what's e to the 1?

00:05:16.770 --> 00:05:19.125
Well then we set x equal
to 1 on both sides.

00:05:19.125 --> 00:05:25.820
And I think I have space to
do it here, e to the 1,

00:05:25.820 --> 00:05:27.350
which is equal to e.

00:05:27.350 --> 00:05:28.810
We just set all the x's to 1.

00:05:28.810 --> 00:05:35.570
So we get that's equal to 1
plus 1 plus 1 over 2 factorial

00:05:35.570 --> 00:05:40.340
plus 1 over 3 factorial plus 1
over 4 factorial plus

00:05:40.340 --> 00:05:42.070
1 over 5 factorial.

00:05:42.070 --> 00:05:44.950
That to me, once
again, is amazing.

00:05:44.950 --> 00:05:48.170
That the number e, and we've
just stumbled on another

00:05:48.170 --> 00:05:56.400
definition of e, e is equal to
the sum from n equals 0 to

00:05:56.400 --> 00:06:03.490
infinite of 1 over n factorial.

00:06:03.490 --> 00:06:05.270
That is amazing.

00:06:05.270 --> 00:06:06.790
So now we have two
definitions for e.

00:06:06.790 --> 00:06:08.520
We have this one that we
stumbled on, and of course we

00:06:08.520 --> 00:06:10.320
had the ones from compound
interest that I will

00:06:10.320 --> 00:06:11.750
do in magenta.

00:06:11.750 --> 00:06:17.150
The limit as n approaches
infinite, 1 plus 1

00:06:17.150 --> 00:06:19.690
over n to the n.

00:06:19.690 --> 00:06:22.510
That also is equal to e.

00:06:22.510 --> 00:06:25.280
This is starting to
give me chills.

00:06:25.280 --> 00:06:29.660
Because this very strange,
bizarre number is popping out.

00:06:29.660 --> 00:06:31.210
This might not seem
so natural to you.

00:06:31.210 --> 00:06:32.340
But it's neat.

00:06:32.340 --> 00:06:34.520
And it comes out in compound
interest, and continuously

00:06:34.520 --> 00:06:35.920
compounding interest.

00:06:35.920 --> 00:06:38.170
But this is even simpler.

00:06:38.170 --> 00:06:40.550
I just keep picking one over
the factorial of numbers and

00:06:40.550 --> 00:06:41.420
I add them all together.

00:06:41.420 --> 00:06:45.010
And if I take every number
really in existence, and I

00:06:45.010 --> 00:06:48.320
sum them all up, I get e.

00:06:48.320 --> 00:06:50.170
That to me is amazing.

00:06:50.170 --> 00:06:52.890
1 over n factorial of
essentially every integer

00:06:52.890 --> 00:06:54.200
from 0 to infinite.

00:06:54.200 --> 00:06:57.860
If I sum them up, I
get the number e.

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You hopefully are getting
chills right now.

00:07:01.740 --> 00:07:05.170
Well anyway, let's do the
Maclaurin series for a

00:07:05.170 --> 00:07:05.990
couple more functions.

00:07:05.990 --> 00:07:10.090
And then we'll get to something
that is even more mind-blowing.

00:07:10.090 --> 00:07:11.140
I'll see you in the next video.


Polynomial approximation of functions (part 1)

WEBVTT
Kind: captions
Language: en

00:00:00.670 --> 00:00:02.800
Let's say I have this
bizarre-looking function.

00:00:02.800 --> 00:00:04.600
It's just some
arbitrary function.

00:00:04.600 --> 00:00:06.970
And we'll call that f of x.

00:00:06.970 --> 00:00:09.880
So this function right
there is f of x.

00:00:09.880 --> 00:00:11.780
And what we're going to do in
this video is, it's not an

00:00:11.780 --> 00:00:13.700
experiment, but we're going to
play around a little bit, and

00:00:13.700 --> 00:00:15.920
we're going to try to
approximate this function

00:00:15.920 --> 00:00:18.720
using a polynomial with
some coefficients.

00:00:18.720 --> 00:00:20.840
And this polynomial we're going
to do, we're going to keep

00:00:20.840 --> 00:00:23.670
adding terms to the polynomial,
so that we can better and

00:00:23.670 --> 00:00:25.380
better approximate
this function.

00:00:25.380 --> 00:00:26.890
And that's actually
called a power series.

00:00:26.890 --> 00:00:29.060
And we'll do more about that
later, but we're going to

00:00:29.060 --> 00:00:32.360
specifically try, in this case,
to approximate the function

00:00:32.360 --> 00:00:34.320
around x is equal to 0.

00:00:34.320 --> 00:00:36.490
So around this point.

00:00:36.490 --> 00:00:40.020
So the easiest way to
approximate it is to say, well,

00:00:40.020 --> 00:00:43.000
the simplest polynomial is
just a constant, right?

00:00:43.000 --> 00:00:44.590
Let's say this is my
polynomial, let me call

00:00:44.590 --> 00:00:47.100
my polynomial p of x.

00:00:47.100 --> 00:00:49.320
The simplest polynomial is just
a constant, and it would just

00:00:49.320 --> 00:00:51.280
be a horizontal line someplace.

00:00:51.280 --> 00:00:54.310
So if I just wanted this one
term polynomial, what would be

00:00:54.310 --> 00:00:57.300
my best approximation for this
function, at least

00:00:57.300 --> 00:00:58.290
at this point?

00:00:58.290 --> 00:01:02.850
Well, I could just set p
of x is equal to f of 0.

00:01:02.850 --> 00:01:06.130
And in that case, p of x would
just look like a horizontal

00:01:06.130 --> 00:01:08.600
line going through f of 0.

00:01:08.600 --> 00:01:10.050
It would just look like that.

00:01:10.050 --> 00:01:12.460
I'm going to erase that now,
just so I don't dirty up

00:01:12.460 --> 00:01:13.590
this picture too much.

00:01:13.590 --> 00:01:17.540
But that's, you could say,
a very rough approximation

00:01:17.540 --> 00:01:19.050
of f of x, right?

00:01:23.070 --> 00:01:24.860
So that's a start.

00:01:24.860 --> 00:01:27.450
Well, what would be one way
to approximate it even more?

00:01:27.450 --> 00:01:33.140
Well, what if not only does p
of x equal f of 0 at x is equal

00:01:33.140 --> 00:01:44.900
to 0, so that horizontal line
we did, we got p of 0 is equal

00:01:44.900 --> 00:01:48.450
to f of 0, so we knew that at x
equals 0, at least that

00:01:48.450 --> 00:01:50.430
horizontal line is the same
value of f of x, that's a

00:01:50.430 --> 00:01:51.920
very rough approximation.

00:01:51.920 --> 00:02:00.010
But what if we set it up so
that the derivative of p of 0

00:02:00.010 --> 00:02:06.020
is equal to the derivative
of the function at 0?

00:02:06.020 --> 00:02:08.350
All of a sudden, this could be
a little bit more interesting.

00:02:08.350 --> 00:02:09.790
So how could we set
it up like that?

00:02:09.790 --> 00:02:14.350
Well, what if we set p of x,
and I'm doing it very general,

00:02:14.350 --> 00:02:16.180
and we're going to do specific
examples, and actually, the

00:02:16.180 --> 00:02:19.160
first example we're going to do
is probably the coolest one.

00:02:19.160 --> 00:02:23.460
So what if p of x is equal to,
well, the constant term is f of

00:02:23.460 --> 00:02:31.650
0, and then it's plus the
derivative of this function, so

00:02:31.650 --> 00:02:37.110
the slope of this function at
that point, f prime

00:02:37.110 --> 00:02:39.780
of 0 times x.

00:02:39.780 --> 00:02:42.020
Let's say I'm defining,
so this is a polynomial.

00:02:42.020 --> 00:02:44.040
I just added a first
degree term here.

00:02:44.040 --> 00:02:46.270
I had a constant, and now I'm
adding a first degree term.

00:02:46.270 --> 00:02:48.630
And let me confirm that this
will have the same derivative.

00:02:48.630 --> 00:02:49.990
So let's see.

00:02:49.990 --> 00:02:52.800
First of all, let's confirm
that p of 0 is equal to f of 0.

00:02:52.800 --> 00:03:03.130
Well, p of 0 is equal to f of
0 plus f prime of 0 times 0.

00:03:03.130 --> 00:03:05.980
Well, this last term just
goes to, is nothing, right?

00:03:05.980 --> 00:03:06.560
Times 0.

00:03:06.560 --> 00:03:09.180
So that checks out.

00:03:09.180 --> 00:03:12.500
At x is equal to 0,
the two functions are

00:03:12.500 --> 00:03:13.840
equal to each other.

00:03:13.840 --> 00:03:16.190
Now let me confirm that their
derivative, their first

00:03:16.190 --> 00:03:17.990
derivatives are the same.

00:03:17.990 --> 00:03:21.450
So what's the first
derivative of p?

00:03:21.450 --> 00:03:25.470
p prime of x is equal to,
well, the derivative of a

00:03:25.470 --> 00:03:27.660
constant term is 0, right?

00:03:27.660 --> 00:03:30.850
Plus, and what's the derivative
of a next term, of a

00:03:30.850 --> 00:03:32.170
first degree term?

00:03:32.170 --> 00:03:33.800
Well, it's just f prime of 0.

00:03:36.800 --> 00:03:37.780
So this checks out.

00:03:37.780 --> 00:03:45.290
My new polynomial that I've
defined right here is equal to

00:03:45.290 --> 00:03:49.360
the function f at x is equal to
0, and its derivative is equal

00:03:49.360 --> 00:03:52.380
to the function f at
x is equal to 0.

00:03:52.380 --> 00:03:54.740
So what would it look like?

00:03:54.740 --> 00:03:58.830
Well, it would intersect, at
x is equal to 0, the two

00:03:58.830 --> 00:04:00.170
functions would overlap.

00:04:00.170 --> 00:04:02.460
And also, they would have the
same slope at that point.

00:04:02.460 --> 00:04:04.180
So whatever f of x is
doing, that function's

00:04:04.180 --> 00:04:04.640
going to be doing.

00:04:04.640 --> 00:04:07.780
So it's going to look something
like, I'm going to try my best

00:04:07.780 --> 00:04:11.650
to, it's going to look
something like that.

00:04:14.290 --> 00:04:17.930
And so that is a better
approximation, if we had to use

00:04:17.930 --> 00:04:20.770
a line, that's as good as any,
especially around 0,

00:04:20.770 --> 00:04:23.470
of what f of x is.

00:04:23.470 --> 00:04:27.180
f of x might have been some
really crazy weirdo function,

00:04:27.180 --> 00:04:31.350
but we were able to approximate
it reasonably well with this

00:04:31.350 --> 00:04:33.760
very simple linear equation.

00:04:33.760 --> 00:04:36.750
Well, that's all good, but
let's approximate it with a

00:04:36.750 --> 00:04:40.490
quadratic equation, with adding
another x squared term.

00:04:40.490 --> 00:04:42.910
And we're going to do that way,
but we're going to say, well,

00:04:42.910 --> 00:04:47.050
we said, when at x is equal
to 0, the functions

00:04:47.050 --> 00:04:47.900
equal each other.

00:04:47.900 --> 00:04:49.590
That's what we did here.

00:04:49.590 --> 00:04:51.570
Then we said, the derivatives
equal each other, and

00:04:51.570 --> 00:04:53.270
so we added this term.

00:04:53.270 --> 00:04:56.130
And now I'm going to say, what
happens when their second

00:04:56.130 --> 00:04:58.480
derivatives equal each other?

00:04:58.480 --> 00:05:00.050
So what has to happen for
their second derivatives

00:05:00.050 --> 00:05:02.430
to equal each other?

00:05:02.430 --> 00:05:05.610
Well, let's try out this,
and I think you'll start

00:05:05.610 --> 00:05:06.960
to see the intuition here.

00:05:06.960 --> 00:05:10.860
Let me define my new p of x,
so let me add another term.

00:05:10.860 --> 00:05:14.070
p of x, the first terms
are going to be the same.

00:05:14.070 --> 00:05:14.820
They're going to be f of 0.

00:05:14.820 --> 00:05:17.100
Remember, this is just
a constant term.

00:05:17.100 --> 00:05:21.530
Plus f prime of 0, the
first derivative at 0,

00:05:21.530 --> 00:05:24.050
the slope at 0 times x.

00:05:24.050 --> 00:05:32.790
Plus f prime prime, the second
derivative of the function at

00:05:32.790 --> 00:05:37.610
0, times x squared over 2.

00:05:37.610 --> 00:05:40.500
Now you're probably saying,
why are you multiplying

00:05:40.500 --> 00:05:41.390
it by 1/2 here?

00:05:41.390 --> 00:05:43.300
And you'll see, and maybe
you'll even realize it, when

00:05:43.300 --> 00:05:45.500
you take a second derivative,
what happens, right?

00:05:45.500 --> 00:05:48.460
You multiply the expression by
the exponents so you can have a

00:05:48.460 --> 00:05:50.800
2 come down, it's going to
cancel out with the 1/2.

00:05:50.800 --> 00:05:52.150
And that's why I
put the 1/2 there.

00:05:52.150 --> 00:05:55.390
So that when I take the
derivative, that 2 and the 1/2

00:05:55.390 --> 00:05:59.460
cancel out, and I'm just left
with the second derivative

00:05:59.460 --> 00:06:00.470
of the original function.

00:06:00.470 --> 00:06:02.570
So let me confirm that.

00:06:02.570 --> 00:06:09.670
So p of 0 is equal to f of 0
plus, well when x is equal to

00:06:09.670 --> 00:06:12.290
0, that's 0, this term is 0,
and when x is equal to 0,

00:06:12.290 --> 00:06:14.010
that term is 0, right?

00:06:14.010 --> 00:06:16.380
So that checks out.

00:06:16.380 --> 00:06:20.790
What's the first
derivative of p?

00:06:20.790 --> 00:06:26.630
The first derivative of p is
going to be, up here, this

00:06:26.630 --> 00:06:31.570
was the first derivative
of p at 0, right?

00:06:31.570 --> 00:06:34.700
So what's the first
derivative of p?

00:06:34.700 --> 00:06:38.720
Well, the constant term becomes
0, plus-- oh, actually, no,

00:06:38.720 --> 00:06:40.770
this was actually of x, sorry.

00:06:40.770 --> 00:06:43.130
I shouldn't go back on my work,
I know it best when I'm doing

00:06:43.130 --> 00:06:44.080
it the first time around.

00:06:44.080 --> 00:06:44.970
Anyway.

00:06:44.970 --> 00:06:49.530
The first derivative of p of x,
this is my current p of x, this

00:06:49.530 --> 00:06:51.200
constant term, derivative of 0.

00:06:51.200 --> 00:06:55.780
This x term, the derivative
is f prime of 0.

00:06:55.780 --> 00:06:58.380
And then, what's the
derivative of this term?

00:06:58.380 --> 00:07:00.990
Well, we take the exponent,
multiply it by the expression.

00:07:00.990 --> 00:07:03.650
We have 2 times 1/2,
that cancels out.

00:07:03.650 --> 00:07:09.340
So we're just left with
f prime prime of 0 x.

00:07:09.340 --> 00:07:10.020
Right?

00:07:10.020 --> 00:07:12.550
You take the exponent, multiply
by the whole thing, and then

00:07:12.550 --> 00:07:14.900
decrement the exponent by 1.

00:07:14.900 --> 00:07:18.020
So what is p prime of 0?

00:07:18.020 --> 00:07:20.730
What is the derivative at 0?

00:07:20.730 --> 00:07:22.960
Well, it equals,
this is nothing.

00:07:22.960 --> 00:07:27.030
It equals f prime of 0
plus, and, well, this

00:07:27.030 --> 00:07:28.990
term's going to be 0.

00:07:28.990 --> 00:07:30.930
So that checks out.

00:07:30.930 --> 00:07:33.990
And so what's the
third derivative?

00:07:33.990 --> 00:07:36.380
Let me clean up some of
the stuff on the top.

00:07:36.380 --> 00:07:38.780
Since this is our current f
of x anyway, I can clean

00:07:38.780 --> 00:07:41.360
up all of this stuff.

00:07:41.360 --> 00:07:43.050
Let me clean up all of this.

00:07:50.150 --> 00:07:55.700
So what is the third derivative
of this p that I defined here?

00:07:55.700 --> 00:07:57.540
This is our current p.

00:07:57.540 --> 00:08:02.900
Well, the third derivative is
going to be, so p prime prime

00:08:02.900 --> 00:08:07.540
prime of x, we could have also
written p3 of x, is equal

00:08:07.540 --> 00:08:09.900
to the derivative of this.

00:08:09.900 --> 00:08:11.360
Oh, sorry, we're not on
the third, we're only on

00:08:11.360 --> 00:08:12.650
the second derivative.

00:08:12.650 --> 00:08:16.060
p, and I'll write prime prime,
I was going to write a 2 there.

00:08:16.060 --> 00:08:17.750
p prime prime of x.

00:08:17.750 --> 00:08:20.000
That equals what?

00:08:20.000 --> 00:08:21.990
That's the derivative of this.

00:08:21.990 --> 00:08:24.440
So there was a 0 here,
that goes to nothing.

00:08:24.440 --> 00:08:27.090
This is now a constant term,
that becomes nothing.

00:08:27.090 --> 00:08:29.250
And then we take the
derivative of this term.

00:08:29.250 --> 00:08:32.050
Well, we just, it's
a constant times x.

00:08:32.050 --> 00:08:34.040
Remember, this might look like
a function or some variable.

00:08:34.040 --> 00:08:34.680
It's just a constant.

00:08:34.680 --> 00:08:39.140
Because we evaluated this
curvy function, it's

00:08:39.140 --> 00:08:41.570
second derivative 0, so we
just got a number here.

00:08:41.570 --> 00:08:43.900
So this derivative is
just this number.

00:08:43.900 --> 00:08:48.650
So it equals f
prime prime of 0.

00:08:48.650 --> 00:08:54.000
And so our current p of x has
the same value when x is equal

00:08:54.000 --> 00:08:58.080
to 0 as f of x, it has the same
first derivative at xis equal

00:08:58.080 --> 00:09:01.050
to zero as f of x, it has
the same second derivative.

00:09:01.050 --> 00:09:05.270
And I don't, this is getting
beyond my visualization

00:09:05.270 --> 00:09:07.360
ability, especially for an
arbitrary function like this,

00:09:07.360 --> 00:09:10.140
but I could guess that maybe
it looks something like this.

00:09:10.140 --> 00:09:10.620
I don't know.

00:09:10.620 --> 00:09:14.940
Maybe it looks, maybe our new
function will curve, and it'll

00:09:14.940 --> 00:09:17.180
approximate it a little
bit better, and then maybe

00:09:17.180 --> 00:09:18.180
it'll come down like that.

00:09:18.180 --> 00:09:18.480
I don't know.

00:09:18.480 --> 00:09:19.710
I'm just guessing.

00:09:19.710 --> 00:09:22.460
But around x is equal to 0,
it's going to be a better

00:09:22.460 --> 00:09:24.560
approximation of f of x.

00:09:24.560 --> 00:09:27.170
Well, we could just keep doing
this, and actually, we will

00:09:27.170 --> 00:09:31.540
keep doing this, and you know,
just saying, well, the zeroth

00:09:31.540 --> 00:09:34.430
derivative, or at the value, is
the same the first derivative

00:09:34.430 --> 00:09:36.600
is the same at 0, the second
derivative is the same at 0,

00:09:36.600 --> 00:09:38.380
we'll say the third derivative,
the fourth derivative, and

00:09:38.380 --> 00:09:39.520
we'll keep doing that.

00:09:39.520 --> 00:09:41.730
And I only have 20 seconds left
in this video, so we will

00:09:41.730 --> 00:09:44.250
continue that in
the next video.


Proof: d/dx(e^x) = e^x

WEBVTT
Kind: captions
Language: en

00:00:04.490 --> 00:00:08.710
Let's prove with the derivative
of e to the x's, and I think

00:00:08.710 --> 00:00:11.400
that this is one of the most
amazing things, depending on

00:00:11.400 --> 00:00:14.060
how you view it about either
calculus or math

00:00:14.060 --> 00:00:14.705
or the universe.

00:00:17.220 --> 00:00:19.470
Well we're essentially going to
prove-- I've already told you

00:00:19.470 --> 00:00:23.235
before that the derivative of e
to the x is equal to e to

00:00:23.235 --> 00:00:24.600
the x, which is amazing.

00:00:24.600 --> 00:00:29.660
The slope at any point of that
line is equal to the x value--

00:00:29.660 --> 00:00:33.050
is equal to the function at
that point, not the x value.

00:00:33.050 --> 00:00:35.010
The slope at any
point is equal to e.

00:00:35.010 --> 00:00:36.980
That is mind boggling.

00:00:36.980 --> 00:00:39.220
And that also means that the
second derivative at any point

00:00:39.220 --> 00:00:42.110
is equal to the function of
that value or the third

00:00:42.110 --> 00:00:46.520
derivative, or the infinite
derivative, and that never

00:00:46.520 --> 00:00:47.380
ceases to amaze me.

00:00:47.380 --> 00:00:49.300
But anyway back to work.

00:00:49.300 --> 00:00:50.220
So how are we going
to prove this?

00:00:50.220 --> 00:00:53.410
Well we already proved-- I
actually just did it right

00:00:53.410 --> 00:00:57.630
before starting this video--
that the derivative-- and some

00:00:57.630 --> 00:00:59.770
people actually call this
the definition of e.

00:00:59.770 --> 00:01:00.610
They go the other way around.

00:01:00.610 --> 00:01:02.520
They say there is some number
for which this is true,

00:01:02.520 --> 00:01:04.140
and we call that number e.

00:01:04.140 --> 00:01:10.310
So it could almost be viewed as
a little bit circular, but be

00:01:10.310 --> 00:01:14.770
we said that e is equal to the
limit as n approaches infinity

00:01:14.770 --> 00:01:18.140
of 1 over 1 plus n to the end.

00:01:18.140 --> 00:01:21.930
And then using this we actually
proved that derivative of

00:01:21.930 --> 00:01:25.710
ln of x is equal to 1/x.

00:01:25.710 --> 00:01:28.630
The derivative of log base
e of x is equal to 1/x.

00:01:28.630 --> 00:01:33.060
So now that we prove this out,
let's use this to prove this.

00:01:38.310 --> 00:01:40.290
Let me keep switching colors
to keep it interesting.

00:01:40.290 --> 00:01:46.820
Let's take the derivative
of ln of e to the x.

00:01:50.280 --> 00:01:51.420
This is almost trivial.

00:01:51.420 --> 00:01:57.010
This is equal to the logarithm
of a to the b is equal to b

00:01:57.010 --> 00:01:59.200
times the logarithm of a,
so this is equal to the

00:01:59.200 --> 00:02:04.600
derivative of x ln of e.

00:02:04.600 --> 00:02:07.640
And this is just saying e to
what power is equal to e.

00:02:07.640 --> 00:02:09.650
Well, to the first
power, right?

00:02:09.650 --> 00:02:13.730
So this just equals the
derivative of x, which we

00:02:13.730 --> 00:02:14.840
have shown as equal to 1.

00:02:14.840 --> 00:02:17.480
I think we have shown it,
hopefully we've shown it.

00:02:17.480 --> 00:02:20.650
If we haven't, that's actually
a very easy one to prove.

00:02:20.650 --> 00:02:21.700
OK fair enough.

00:02:21.700 --> 00:02:23.630
We did that.

00:02:23.630 --> 00:02:25.030
But let's do this another way.

00:02:25.030 --> 00:02:28.280
Let's use the chain rule.

00:02:28.280 --> 00:02:29.420
So what doe the chain rule say?

00:02:29.420 --> 00:02:32.610
If we have f of g of x, where
we have one function embedded

00:02:32.610 --> 00:02:35.660
in another one, the chain rule
say we take the derivative of

00:02:35.660 --> 00:02:50.550
the inside function, so
d/dx of e to the x.

00:02:50.550 --> 00:02:53.320
And then we take the derivative
of the outside function or the

00:02:53.320 --> 00:02:56.340
derivative of the outside
function with respect

00:02:56.340 --> 00:02:57.400
to the inner function.

00:02:57.400 --> 00:03:00.570
You can almost
view it that way.

00:03:00.570 --> 00:03:03.000
So the derivative ln
of e to the x with

00:03:03.000 --> 00:03:04.530
respect to e to the x.

00:03:04.530 --> 00:03:05.530
I know that's a
little confusing.

00:03:05.530 --> 00:03:09.680
You could have written a d e to
the x down over here, but I

00:03:09.680 --> 00:03:11.760
think you know the
chain rule by now.

00:03:11.760 --> 00:03:15.650
That is equal to 1
over e to the x.

00:03:15.650 --> 00:03:18.055
And that just comes from this.

00:03:18.055 --> 00:03:19.940
But instead of an x,
we have e to the x.

00:03:19.940 --> 00:03:23.340
So this is just a chain rule.

00:03:23.340 --> 00:03:24.830
Well what else do we know?

00:03:24.830 --> 00:03:28.090
We know that this is equal to
this, and we also know that

00:03:28.090 --> 00:03:29.840
this is equal to this.

00:03:29.840 --> 00:03:31.910
So this must be equal to this.

00:03:31.910 --> 00:03:34.740
So this must be equal to 1.

00:03:34.740 --> 00:03:37.150
Well let's just multiply
both sides of this

00:03:37.150 --> 00:03:41.160
equation by e to the x.

00:03:41.160 --> 00:03:43.070
We get on the left hand
side, we're just left

00:03:43.070 --> 00:03:44.246
with this expression.

00:03:44.246 --> 00:03:51.450
The derivative of e to the x
times- we're multiplying both

00:03:51.450 --> 00:03:54.090
sides by e to the x, times
e to the x over e to the x.

00:03:54.090 --> 00:03:57.650
I just chose to put the e
to the x on this term,

00:03:57.650 --> 00:03:59.740
is equal to e to the x.

00:03:59.740 --> 00:04:01.320
This is 1.

00:04:01.320 --> 00:04:02.670
Scratch it out.

00:04:02.670 --> 00:04:03.960
We're done.

00:04:03.960 --> 00:04:05.890
That might not have been
completely satisfying

00:04:05.890 --> 00:04:07.640
for you, but it works.

00:04:07.640 --> 00:04:12.910
The derivative of e to the
x is equal to e to the x.

00:04:16.490 --> 00:04:21.260
I think the school or the
nation should take a national

00:04:21.260 --> 00:04:23.250
holiday or something,
and people should just

00:04:23.250 --> 00:04:27.620
ponder this, because it
really is fascinating.

00:04:27.620 --> 00:04:31.730
But then actually this will
lead us to I would say even

00:04:31.730 --> 00:04:36.200
more dramatic results in the
not too far off future.

00:04:36.200 --> 00:04:37.870
Anyway, I'll see in
the next video.


Proof: d/dx(ln x) = 1/x

WEBVTT
Kind: captions
Language: en

00:00:00.720 --> 00:00:03.640
I'm now going to show you what
I think are probably the two

00:00:03.640 --> 00:00:06.490
coolest derivatives
in all of calculus.

00:00:06.490 --> 00:00:08.680
And I'll reserve that.

00:00:08.680 --> 00:00:10.850
None of the other ones have
occurred to me right now.

00:00:10.850 --> 00:00:13.780
But these are definitely to
me some of the neatest.

00:00:13.780 --> 00:00:15.795
So let's figure out
what the derivative of

00:00:15.795 --> 00:00:16.810
the natural log is.

00:00:16.810 --> 00:00:18.530
And just as a review,
what is the natural log?

00:00:18.530 --> 00:00:23.380
Well the natural log of
something is the exact same

00:00:23.380 --> 00:00:27.720
thing as saying logarithm
base e of that something.

00:00:27.720 --> 00:00:28.810
That's just a review.

00:00:28.810 --> 00:00:30.390
So let's take the
derivative of this.

00:00:33.440 --> 00:00:35.455
I think I'm going to need
a lot of space for this.

00:00:35.455 --> 00:00:37.430
I'm going to try to do it
as neatly as possible.

00:00:37.430 --> 00:00:47.090
So the derivative of the
natural log of x equals-- well

00:00:47.090 --> 00:00:49.780
let's just take the definition
of a derivative, right?

00:00:49.780 --> 00:00:53.780
We just take the slope at some
point and find the limit as we

00:00:53.780 --> 00:00:56.720
take the difference between
the two points to 0.

00:00:56.720 --> 00:01:08.240
So let's take the limit as
delta x approaches 0 of

00:01:08.240 --> 00:01:13.340
f of x plus delta x.

00:01:13.340 --> 00:01:15.440
So I'm going to take the
limit of this whole thing.

00:01:15.440 --> 00:01:21.450
The natural log ln of x plus
delta x-- right, that's like

00:01:21.450 --> 00:01:24.660
one point that I'm going to
take evaluate the function--

00:01:24.660 --> 00:01:29.490
minus the ln of x.

00:01:29.490 --> 00:01:31.010
All of that over delta x.

00:01:31.010 --> 00:01:34.250
And if you remember from the
derivative videos, this is just

00:01:34.250 --> 00:01:37.670
the slope, and I'm just taking
the limit as I find the slope

00:01:37.670 --> 00:01:41.040
between a smaller and
a smaller distance.

00:01:41.040 --> 00:01:42.476
Hopefully you remember that.

00:01:42.476 --> 00:01:44.800
So let's see if we can do some
logarithm properties to

00:01:44.800 --> 00:01:46.530
simplify this a little bit.

00:01:46.530 --> 00:01:49.820
Hopefully you remember-- and if
you don't, review the logarithm

00:01:49.820 --> 00:01:57.480
properties-- but remember that
log of a minus log of b is

00:01:57.480 --> 00:02:02.390
equal to log of a over b, and
that comes out of the fact that

00:02:02.390 --> 00:02:04.930
logarithm expressions are
essentially exponents, so they

00:02:04.930 --> 00:02:06.710
follow the exponent rules.

00:02:06.710 --> 00:02:08.180
And if that doesn't make
sense to you, you should

00:02:08.180 --> 00:02:09.580
review those as well.

00:02:09.580 --> 00:02:13.090
But let's apply this logarithm
property to this equation.

00:02:13.090 --> 00:02:15.110
So let me rewrite the whole
thing, and I'm going to keep

00:02:15.110 --> 00:02:17.580
switching colors to keep it
from getting monotonous.

00:02:17.580 --> 00:02:26.780
So we have the limit as delta x
approaches 0 of this big thing.

00:02:26.780 --> 00:02:27.230
Let's see.

00:02:27.230 --> 00:02:32.060
So log of a minus b equals log
a over b, so this top, the

00:02:32.060 --> 00:02:38.120
numerator, will equal the
natural log of x plus

00:02:38.120 --> 00:02:42.270
delta x over x.

00:02:42.270 --> 00:02:48.000
Right? a b a/b, all of
that over delta x.

00:02:51.350 --> 00:02:58.090
And so that equals the limit as
delta x approaches 0-- I think

00:02:58.090 --> 00:03:04.240
it's time to switch colors
again-- delta x approaches 0

00:03:04.240 --> 00:03:07.025
of-- well let me just write
this 1 over delta

00:03:07.025 --> 00:03:08.730
x out in front.

00:03:08.730 --> 00:03:13.010
So this is 1 over delta x,
and we're going to take

00:03:13.010 --> 00:03:15.000
the limit of everything.

00:03:15.000 --> 00:03:28.270
ln x divided by x is 1
plus delta x over x.

00:03:28.270 --> 00:03:29.580
Fair enough.

00:03:29.580 --> 00:03:31.770
Now I'm going to throw out
another logarithm property, and

00:03:31.770 --> 00:03:35.130
hopefully you remember that--
and let me put the properties

00:03:35.130 --> 00:03:40.440
separate so you know it's not
part of the proof-- that a log

00:03:40.440 --> 00:03:46.290
b is equal to log
of b to the a.

00:03:46.290 --> 00:03:48.890
And that comes from when you
take something to an exponent,

00:03:48.890 --> 00:03:51.300
and then to another exponent
you just have to multiply

00:03:51.300 --> 00:03:52.670
those two exponents.

00:03:52.670 --> 00:03:53.890
I don't want to confuse
you, but hopefully you

00:03:53.890 --> 00:03:55.370
should remember this.

00:03:55.370 --> 00:03:57.710
So how does apply here?

00:03:57.710 --> 00:04:00.330
Well this would be a log b.

00:04:00.330 --> 00:04:09.020
So this expression is the
same thing as the limit.

00:04:09.020 --> 00:04:23.740
The limit as delta x approaches
0 of the natural log of 1 plus

00:04:23.740 --> 00:04:31.800
delta x over x to the
1 over delta x power.

00:04:31.800 --> 00:04:34.820
And remember all this
is the natural log of

00:04:34.820 --> 00:04:35.890
this entire thing.

00:04:35.890 --> 00:04:36.730
And then we're going
to take the limit as

00:04:36.730 --> 00:04:39.110
delta x approaches 0.

00:04:39.110 --> 00:04:42.270
If you've watched the compound
interest problems and you know

00:04:42.270 --> 00:04:44.480
the definition of e, I think
this will start to

00:04:44.480 --> 00:04:45.190
look familiar.

00:04:45.190 --> 00:04:48.030
But let me make a substitution
that might clean things

00:04:48.030 --> 00:04:49.410
up a little bit.

00:04:49.410 --> 00:04:59.190
Let me make the substitution,
let me call it n-- no, no, no,

00:04:59.190 --> 00:05:05.430
let me call u-- is equal
to delta x over x.

00:05:05.430 --> 00:05:07.860
And then if that's true
then we can multiply both

00:05:07.860 --> 00:05:12.450
sides by x and we get
xu is equal to delta x.

00:05:12.450 --> 00:05:16.650
Or we would also know
that 1 over delta x is

00:05:16.650 --> 00:05:20.780
equal to 1 over xu.

00:05:20.780 --> 00:05:22.150
These are all equivalent.

00:05:22.150 --> 00:05:23.720
So let's make the substitution.

00:05:23.720 --> 00:05:28.030
So if we're taking the limit is
delta x approaches 0, in this

00:05:28.030 --> 00:05:29.930
expression if delta x
approaches 0, what

00:05:29.930 --> 00:05:31.110
does u approach?

00:05:31.110 --> 00:05:32.320
u approaches 0.

00:05:32.320 --> 00:05:35.860
So delta x approaching 0 is the
same exact thing as taking

00:05:35.860 --> 00:05:39.230
the limit as u approaches 0.

00:05:39.230 --> 00:05:46.960
So we can write this as the
limit as u approaches 0 of the

00:05:46.960 --> 00:05:53.200
natural log of 1 plus-- well we
did the substitution, delta x

00:05:53.200 --> 00:05:58.870
over x is now u-- to the 1
over delta x, and that same

00:05:58.870 --> 00:06:00.960
substitution told us that's the
same thing as one over xu.

00:06:05.270 --> 00:06:08.360
Remember we're taking the
natural log of everything.

00:06:08.360 --> 00:06:11.550
And we know this is an exponent
property, which I'll now

00:06:11.550 --> 00:06:13.340
do in a different color.

00:06:13.340 --> 00:06:19.530
We know that a to the bc
is equal to a to the

00:06:19.530 --> 00:06:23.620
b to the c power.

00:06:23.620 --> 00:06:35.200
So that tells us that this me
is equal to the limit as u

00:06:35.200 --> 00:06:46.810
approaches 0 of the natural log
of 1 plus u to the 1/u, because

00:06:46.810 --> 00:06:49.960
this is one over xu, right?

00:06:49.960 --> 00:06:55.270
1/u, and then all of
that to the 1/x.

00:06:55.270 --> 00:06:56.110
And how did I do that?

00:06:56.110 --> 00:06:57.950
Just from this exponent
property, right?

00:06:57.950 --> 00:07:00.810
If I were to simplify this, I
would have 1/x times 1/u, and

00:07:00.810 --> 00:07:04.740
that's where I get
this 1 over xu.

00:07:04.740 --> 00:07:07.780
Well then we can just do this
logarithm property in reverse.

00:07:07.780 --> 00:07:10.370
If I have b to the a I can
put that a out front.

00:07:10.370 --> 00:07:12.670
So I could take this 1/x
and put it in front

00:07:12.670 --> 00:07:14.160
of the natural log.

00:07:14.160 --> 00:07:16.250
So now what do I have?

00:07:16.250 --> 00:07:18.550
We're almost there.

00:07:18.550 --> 00:07:25.780
We have the limit
as u approaches 0.

00:07:25.780 --> 00:07:28.950
Take that 1/x, put it in front
of the natural log sign.

00:07:28.950 --> 00:07:40.490
1/x times the natural log
of 1 plus u to the 1/u.

00:07:40.490 --> 00:07:41.330
Fair enough.

00:07:41.330 --> 00:07:45.080
When we're taking the limit as
u approaches 0, x, this term

00:07:45.080 --> 00:07:47.800
doesn't involve it at all.

00:07:47.800 --> 00:07:50.030
So we could take this out in
front, because the limit

00:07:50.030 --> 00:07:51.430
doesn't affect this term.

00:07:51.430 --> 00:07:55.890
And then we're essentially
saying what happens to this

00:07:55.890 --> 00:07:59.440
expression as the limit
as u approaches 0.

00:07:59.440 --> 00:08:10.220
So this thing is equivalent to
1/x times the natural log of

00:08:10.220 --> 00:08:23.480
the limit as u approaches
0 of 1 plus u to the 1/u.

00:08:23.480 --> 00:08:26.320
And by now hopefully you
would recognize that

00:08:26.320 --> 00:08:27.880
this is the definition.

00:08:27.880 --> 00:08:31.230
This limit comes to e, if
you remember anything

00:08:31.230 --> 00:08:32.260
from compound interest.

00:08:32.260 --> 00:08:38.140
You might remember it as the
limit-- as n approaches

00:08:38.140 --> 00:08:42.430
infinity of 1 plus
1 over n to the n.

00:08:42.430 --> 00:08:43.780
But these things
are equivalent.

00:08:43.780 --> 00:08:46.360
If you just took the
substitution u is equal to

00:08:46.360 --> 00:08:48.330
1/n, you would get this.

00:08:48.330 --> 00:08:49.180
You would just get this.

00:08:49.180 --> 00:08:54.500
So this expression right here
is e That expression is e.

00:08:54.500 --> 00:08:55.670
So we're getting close.

00:08:59.570 --> 00:09:05.760
So this whole thing is
equivalent to 1/x times

00:09:05.760 --> 00:09:09.050
the natural log, and this
we know, this is one of

00:09:09.050 --> 00:09:10.740
the ways to get to e.

00:09:10.740 --> 00:09:13.280
So the limit as u approaches
0 of 1 plus u to the 1/u.

00:09:13.280 --> 00:09:15.000
That is e.

00:09:15.000 --> 00:09:16.030
And what is the natural log?

00:09:16.030 --> 00:09:17.530
Well it's the log base e.

00:09:17.530 --> 00:09:23.000
So you know this is equal to
1/x times the log base e of e.

00:09:23.000 --> 00:09:25.270
So that's saying e
to what power is e.

00:09:25.270 --> 00:09:26.870
Well e to the first
power is e, right?

00:09:26.870 --> 00:09:29.400
This is equal to 1.

00:09:29.400 --> 00:09:33.010
So 1 times 1/x is equal to 1/x.

00:09:33.010 --> 00:09:34.280
There we have it.

00:09:34.280 --> 00:09:39.190
The derivative of the natural
log of x is equal to 1/x, which

00:09:39.190 --> 00:09:42.260
I find kind of neat, because
all of the other exponents

00:09:42.260 --> 00:09:43.280
lead to another exponent.

00:09:43.280 --> 00:09:45.580
But all of a sudden in the mix
here you have the natural log

00:09:45.580 --> 00:09:47.440
and the derivative of that
is equal to x to the

00:09:47.440 --> 00:09:49.420
negative 1 or 1/x.

00:09:49.420 --> 00:09:49.740
Fascinating.


Sequences and series (part 2)

WEBVTT
Kind: captions
Language: en

00:00:00.410 --> 00:00:01.040
Welcome back.

00:00:01.040 --> 00:00:03.460
So where we left off in the
last video, I'd shown you

00:00:03.460 --> 00:00:06.020
this thing called the
geometric series.

00:00:06.020 --> 00:00:07.810
And, you know, we could
have some base a.

00:00:07.810 --> 00:00:08.620
It could be any number.

00:00:08.620 --> 00:00:12.800
It could be 1/2,
it could be 10.

00:00:12.800 --> 00:00:13.870
But that's just--
but some number.

00:00:13.870 --> 00:00:15.770
And we keep taking it to
increasing exponents, and we

00:00:15.770 --> 00:00:18.740
sum them up, and this is
called a geometric series.

00:00:18.740 --> 00:00:23.810
And so I want to figure out the
sum of a geometric series of,

00:00:23.810 --> 00:00:26.940
you know, when I have some base
a, and I go up to some

00:00:26.940 --> 00:00:29.440
number a to the n.

00:00:29.440 --> 00:00:31.520
What-- is this a to
the-- why did I write

00:00:31.520 --> 00:00:32.690
a to n minus 2 there?

00:00:32.690 --> 00:00:37.390
That should be a to the big N.

00:00:37.390 --> 00:00:39.510
My brain must have been
malfunctioning in

00:00:39.510 --> 00:00:40.930
the previous video.

00:00:40.930 --> 00:00:42.520
That always happens when I
start running out of time.

00:00:42.520 --> 00:00:43.740
But anyway.

00:00:43.740 --> 00:00:44.490
Let's go back to this.

00:00:44.490 --> 00:00:48.170
So I defined s as
this geometric sum.

00:00:48.170 --> 00:00:50.160
Now I'm going to
define another sum.

00:00:50.160 --> 00:00:53.170
And that sum I'm going
to define as a times s.

00:00:57.970 --> 00:01:00.970
And that equals-- well, that's
just going to be a times

00:01:00.970 --> 00:01:04.110
this exact sum, right?

00:01:04.110 --> 00:01:06.540
And that's the same
a as this a, right?

00:01:06.540 --> 00:01:08.400
That a is the same as this a.

00:01:08.400 --> 00:01:11.620
So what's a times
this whole thing?

00:01:11.620 --> 00:01:15.350
Well, it's the a times a
to the zero is-- let me

00:01:15.350 --> 00:01:16.370
write it down for you.

00:01:16.370 --> 00:01:19.610
So this'll be a because I just
distribute the a, right? a

00:01:19.610 --> 00:01:25.460
times a to the zero, plus a
times a to the 1, plus a times

00:01:25.460 --> 00:01:30.880
a squared, plus all the way a
times a to the n minus one,

00:01:30.880 --> 00:01:34.260
plus a times a to the n.

00:01:34.260 --> 00:01:36.260
I just took an a and I
distributed it along

00:01:36.260 --> 00:01:39.540
this whole sum.

00:01:39.540 --> 00:01:41.200
But what is this equal to?

00:01:41.200 --> 00:01:43.810
Well, this is equal to
a times a to the zero.

00:01:43.810 --> 00:01:51.220
That's a one-- a to the first
power-- plus a squared, plus a

00:01:51.220 --> 00:01:55.420
cubed, plus a to the n, right?

00:01:55.420 --> 00:01:57.770
Because you just add the
exponents, a to the n.

00:01:57.770 --> 00:02:00.600
Plus a to the n plus 1.

00:02:00.600 --> 00:02:03.800
So this is as.

00:02:03.800 --> 00:02:07.450
And we saw before that s
is just our original sum.

00:02:10.530 --> 00:02:16.950
That is just a to the zero,
plus a to the 1, plus a

00:02:16.950 --> 00:02:19.380
squared, plus up, up, up, up.

00:02:19.380 --> 00:02:26.120
All the way to plus
a to the n, right?

00:02:26.120 --> 00:02:29.680
So let me ask you a question.

00:02:29.680 --> 00:02:33.980
What happens if I
subtract this from that?

00:02:33.980 --> 00:02:36.120
What happens?

00:02:36.120 --> 00:02:41.840
If I say, as minus s.

00:02:41.840 --> 00:02:45.980
Well, I subtracted this from
here, on the left hand side.

00:02:45.980 --> 00:02:48.032
What happens on the
right hand side?

00:02:48.032 --> 00:02:50.120
Well, all of these
become negative, right?

00:02:50.120 --> 00:02:51.770
Let me do it in a bold color.

00:02:51.770 --> 00:02:54.240
This becomes-- because I'm
subtracting-- negative,

00:02:54.240 --> 00:02:55.740
negative, these are
all negatives.

00:02:55.740 --> 00:02:56.250
Negative.

00:02:56.250 --> 00:02:57.480
Negative.

00:02:57.480 --> 00:03:00.270
Well, a to the first,
minus a to the first.

00:03:00.270 --> 00:03:02.735
That crosses out. a squared
minus a squared crosses

00:03:02.735 --> 00:03:04.530
out. a to the third,
it'll all cross out.

00:03:04.530 --> 00:03:06.970
All the way up to a
to the n, right?

00:03:06.970 --> 00:03:07.920
So what are we left with?

00:03:07.920 --> 00:03:11.920
We're just left with minus
a to the zero, right?

00:03:11.920 --> 00:03:13.600
We're just left with that term.

00:03:13.600 --> 00:03:15.400
And we're just left
with that term.

00:03:15.400 --> 00:03:19.225
Plus a to the n plus 1.

00:03:19.225 --> 00:03:20.800
And of course, what's
a to the zero?

00:03:20.800 --> 00:03:22.240
That's just 1.

00:03:22.240 --> 00:03:32.240
So we have a times s
minus s is equal to a to

00:03:32.240 --> 00:03:36.110
the n plus 1 minus 1.

00:03:36.110 --> 00:03:37.850
And now let's
distribute the s out.

00:03:37.850 --> 00:03:44.180
So we get s times a minus
1 is equal to a to the n

00:03:44.180 --> 00:03:48.950
plus 1 minus 1, right?

00:03:48.950 --> 00:03:50.500
And then what do we get?

00:03:50.500 --> 00:03:52.600
Well, we can just divide
both sides by a minus 1.

00:03:52.600 --> 00:03:55.820
Let me erase some of
this stuff on top.

00:03:58.890 --> 00:04:05.700
I think I can safely erase
all of this, really.

00:04:05.700 --> 00:04:07.870
Well, I don't want
to erase that much.

00:04:07.870 --> 00:04:09.240
I want to erase this stuff.

00:04:13.830 --> 00:04:15.360
That's good enough.

00:04:15.360 --> 00:04:16.890
OK.

00:04:16.890 --> 00:04:24.150
So I have just-- dividing both
sides of this equation by a

00:04:24.150 --> 00:04:31.520
minus 1, I get s is equal to a
to the n plus 1 minus

00:04:31.520 --> 00:04:37.560
1 over a minus 1.

00:04:37.560 --> 00:04:41.280
So where did that get us?

00:04:41.280 --> 00:04:46.270
We defined the geometric
series as equal to the sum.

00:04:46.270 --> 00:04:51.700
From k is equal to 0,
to n of a to the k.

00:04:51.700 --> 00:04:54.170
And now we've just derived
a formula for what that

00:04:54.170 --> 00:04:55.770
sum ends up being.

00:04:55.770 --> 00:05:03.990
Equals a to the n plus 1
minus 1 over a minus 1.

00:05:03.990 --> 00:05:05.600
And why is this useful?

00:05:05.600 --> 00:05:10.940
We now know, if I were to say,
well, what is-- let me clean

00:05:10.940 --> 00:05:12.490
up all of this, as well.

00:05:12.490 --> 00:05:17.400
Let me clean up all of
this and we can-- OK.

00:05:17.400 --> 00:05:22.260
So if I said, you figure out
the sum of, I don't know, the

00:05:22.260 --> 00:05:28.277
powers of 3 up to 3 to the, I
don't know, 3 to

00:05:28.277 --> 00:05:29.550
the tenth power.

00:05:29.550 --> 00:05:32.760
So, you know, 3.

00:05:32.760 --> 00:05:37.170
So 3 to the zero, plus 3 to the
one, plus 3 squared, plus all

00:05:37.170 --> 00:05:39.280
the way to 3 to the tenth.

00:05:39.280 --> 00:05:43.800
So this is the same thing as
the sum of k equals zero

00:05:43.800 --> 00:05:48.660
to 10, of 3 to the k.

00:05:48.660 --> 00:05:49.520
Right?

00:05:49.520 --> 00:05:54.080
So this formula we just figured
out, a is 3 and n is 10.

00:05:54.080 --> 00:05:58.620
So this sum is just going to be
equal to 3 to the eleventh

00:05:58.620 --> 00:06:05.480
power minus 1 over 3 minus 1.

00:06:05.480 --> 00:06:08.140
Which equals-- well, I
don't know what 3 to

00:06:08.140 --> 00:06:09.420
the eleventh power is.

00:06:09.420 --> 00:06:11.350
Minus 1 over 2.

00:06:11.350 --> 00:06:13.130
So that's kind of useful.

00:06:13.130 --> 00:06:15.050
That is a number.

00:06:15.050 --> 00:06:17.460
Although you'd have to memorize
your exponent tables to the

00:06:17.460 --> 00:06:18.550
eleventh power to do that.

00:06:18.550 --> 00:06:19.880
But I think you get the idea.

00:06:19.880 --> 00:06:23.710
This is especially useful if we
were dealing with-- well, if

00:06:23.710 --> 00:06:27.000
the base was a power of ten,
it would be very, very easy.

00:06:27.000 --> 00:06:31.260
But what I actually want to do
now is I want to take this and

00:06:31.260 --> 00:06:34.510
say, well, what happens
if n goes to infinity?

00:06:34.510 --> 00:06:36.270
Let me show you.

00:06:36.270 --> 00:06:37.440
So what happens?

00:06:37.440 --> 00:06:41.350
So there's two types of series
that we can take-- that's

00:06:41.350 --> 00:06:42.750
not what I wanted to do.

00:06:42.750 --> 00:06:45.250
There are two types of series
that we can take that we

00:06:45.250 --> 00:06:47.450
can find the sums of.

00:06:47.450 --> 00:06:51.220
There's finite series,
and infinite series.

00:06:51.220 --> 00:06:56.310
And in order for an infinite
series to come up to a sum

00:06:56.310 --> 00:06:58.950
that's not infinity, they
need to-- what we say--

00:06:58.950 --> 00:07:00.150
they need to converge.

00:07:00.150 --> 00:07:02.870
And if you think about what has
to happen for them to converge,

00:07:02.870 --> 00:07:06.520
every next digit has to
essentially get smaller and

00:07:06.520 --> 00:07:09.260
smaller and smaller, as
we go towards infinity.

00:07:09.260 --> 00:07:12.910
So let's say that
a is a fraction.

00:07:12.910 --> 00:07:13.700
a is 1/2.

00:07:13.700 --> 00:07:17.870
So how does a geometric series
look like if we have 1/2 there?

00:07:17.870 --> 00:07:22.650
So let's say that we're taking
the geometric series from k

00:07:22.650 --> 00:07:25.910
is equal to 0 to infinity.

00:07:25.910 --> 00:07:27.160
So this is neat.

00:07:27.160 --> 00:07:29.890
We're going to take an infinite
sum, an infinite number of

00:07:29.890 --> 00:07:34.380
terms, and let's see if we can
actually get an actual number.

00:07:34.380 --> 00:07:35.950
You know, we take an infinite
thing, add it up, and it

00:07:35.950 --> 00:07:38.200
actually adds up to
a finite thing.

00:07:38.200 --> 00:07:39.990
This has always amazed me.

00:07:39.990 --> 00:07:43.130
And the base now is
going to be 1/2.

00:07:43.130 --> 00:07:45.410
It's 1/2 and it's going to
be 1/2 to the k power.

00:07:45.410 --> 00:07:45.770
So this is going to be what?

00:07:45.770 --> 00:07:50.220
1/2 to the zero, plus 1/2,
plus-- what's 1/2 squared?

00:07:50.220 --> 00:07:54.860
Plus 1/4, plus 1/8, plus 1/16.

00:07:54.860 --> 00:07:59.680
So as you see, each term is
getting a lot, lot smaller.

00:07:59.680 --> 00:08:03.900
It's getting half of
the previous term.

00:08:03.900 --> 00:08:06.800
Well, let's say, what happens
if this wasn't infinity?

00:08:06.800 --> 00:08:09.880
What happens if this was n?

00:08:09.880 --> 00:08:13.480
Well, then we'd get plus 1
over 2 to the n, right?

00:08:13.480 --> 00:08:16.140
1/2 to the n is the same
thing as 1 over 2 to the n.

00:08:16.140 --> 00:08:19.080
And if we look at the formula
we figured out, we would say,

00:08:19.080 --> 00:08:26.340
well, that is just equal to
1/2 to the n plus 1, minus

00:08:26.340 --> 00:08:32.960
1, over 1/2 minus one.

00:08:32.960 --> 00:08:34.850
And that would be our answer.

00:08:34.850 --> 00:08:36.550
We'd have to know what n is.

00:08:36.550 --> 00:08:39.280
But now we want to know what
happens if we go to infinity.

00:08:39.280 --> 00:08:41.140
So this is essentially
a limit problem.

00:08:41.140 --> 00:08:45.740
What happens-- what's the
limit, as n goes to infinity,

00:08:45.740 --> 00:08:52.980
of 1/2 to the n plus one
minus 1 over 1/2 minus 1?

00:08:52.980 --> 00:08:55.330
Well, all of these are constant
terms, so nothing happens.

00:08:55.330 --> 00:08:58.140
So what happens as this term,
right here, goes to infinity?

00:08:58.140 --> 00:09:01.220
What's 1/2 to the
infinity power?

00:09:01.220 --> 00:09:03.000
Well, that's zero.

00:09:03.000 --> 00:09:04.590
That's an unbelievably
small number.

00:09:04.590 --> 00:09:08.790
Take 1/2 to arbitrarily large
exponents, this just goes to 0.

00:09:08.790 --> 00:09:10.270
And so what are we left with?

00:09:10.270 --> 00:09:16.060
We're just left with this
equals minus 1 over 1/2 minus

00:09:16.060 --> 00:09:19.275
1, or we could multiply the top
and the bottom by negative 1.

00:09:19.275 --> 00:09:22.212
And we get 1 over 1 minus 1/2.

00:09:22.212 --> 00:09:27.320
Which equals 1 over 1/2,
which is equal to 2.

00:09:27.320 --> 00:09:28.810
I find that amazing.

00:09:28.810 --> 00:09:34.090
If I add 0 plus 1/2 plus 1/4
plus 1/8 plus 1/16 and I never

00:09:34.090 --> 00:09:37.410
stop-- I go to infinity-- and
not infinity, but I go to 1

00:09:37.410 --> 00:09:41.330
over essentially 2 to the
infinity-- I end up with

00:09:41.330 --> 00:09:43.080
this neat and clean number.

00:09:43.080 --> 00:09:43.530
2.

00:09:43.530 --> 00:09:45.350
And this might be a little
project for you, to actually

00:09:45.350 --> 00:09:48.070
draw it out into like maybe a
pie and see what happens as

00:09:48.070 --> 00:09:51.310
you keep adding smaller and
smaller pieces to the pie.

00:09:51.310 --> 00:09:54.550
But it never ceases to amaze
me, that I added an infinite

00:09:54.550 --> 00:09:55.520
number of terms, right?

00:09:55.520 --> 00:09:56.390
This was infinity.

00:09:56.390 --> 00:09:58.530
And I got a finite number.

00:09:58.530 --> 00:09:59.930
I got a finite number.

00:09:59.930 --> 00:10:01.630
Anyway, we ran out of time.

00:10:01.630 --> 00:10:03.132
See you soon.


Sequences and Series (part 1)

WEBVTT
Kind: captions
Language: en

00:00:00.720 --> 00:00:03.240
Let's learn a little bit
about sequences and series.

00:00:03.240 --> 00:00:04.020
So what's a sequence?

00:00:04.020 --> 00:00:06.980
Well, a sequence is just a
bunch of numbers in some order.

00:00:06.980 --> 00:00:11.630
You know, the most difficult
sequence is 1, 2, 3, 4.

00:00:11.630 --> 00:00:13.370
You get the point.

00:00:13.370 --> 00:00:14.480
And what's a series?

00:00:14.480 --> 00:00:17.550
Well, it's often represented--
it's just a sum of sequences,

00:00:17.550 --> 00:00:18.980
a sum of a sequence.

00:00:18.980 --> 00:00:23.350
So, for example, the arithmetic
sequence-- sorry the arithmetic

00:00:23.350 --> 00:00:26.920
series is just the sum of
the arithmetic sequence.

00:00:26.920 --> 00:00:31.630
So 1 plus 2 plus 3 plus--
we could keep going

00:00:31.630 --> 00:00:33.440
until maybe some number.

00:00:33.440 --> 00:00:36.345
This is called the
arithmetic series.

00:00:42.750 --> 00:00:45.370
Nothing too fancy here.

00:00:45.370 --> 00:00:48.860
But before we move forward,
let's get a notation for how we

00:00:48.860 --> 00:00:51.630
can represent these sums
without necessarily having to

00:00:51.630 --> 00:00:54.430
write out all of the digits or
having to keep doing this dot,

00:00:54.430 --> 00:00:57.140
dot, dot, plus notation.

00:00:57.140 --> 00:00:59.475
And that notation
is Sigma notation.

00:01:03.320 --> 00:01:06.740
That's an upper case Sigma.

00:01:06.740 --> 00:01:08.510
And how do you use
Sigma notation?

00:01:08.510 --> 00:01:09.490
Well, let's say I wanted
to represent this

00:01:09.490 --> 00:01:11.020
arithmetic series.

00:01:11.020 --> 00:01:13.590
So I would say, well,
let's add up a bunch of--

00:01:13.590 --> 00:01:14.940
let's call them k's.

00:01:14.940 --> 00:01:16.810
This is an arbitrary variable.

00:01:16.810 --> 00:01:19.480
And we'll start at k equals 1.

00:01:19.480 --> 00:01:22.240
We'll start at k equals
1, and we'll go to k

00:01:22.240 --> 00:01:24.850
is equal to big N.

00:01:24.850 --> 00:01:27.795
And this is the exact same
thing, so we first make k equal

00:01:27.795 --> 00:01:32.100
to 1, and then we add it to k
is equal to 2 plus 3, and we go

00:01:32.100 --> 00:01:38.550
all the way until N minus
1 and then plus N.

00:01:38.550 --> 00:01:45.770
So this is the Sigma notation
for the arithmetic series.

00:01:45.770 --> 00:01:48.110
Before I move on, I think this
is a good time just to like

00:01:48.110 --> 00:01:50.730
learn a little bit more about
the arithmetic series.

00:01:50.730 --> 00:01:54.010
We'll actually focus on this
one and the geometric series

00:01:54.010 --> 00:01:56.560
because those are the two
that you'll see most often.

00:01:56.560 --> 00:01:59.010
And then once you learn
calculus, I'll show you the

00:01:59.010 --> 00:02:02.040
power and Taylor series, which
is exact-- the Taylor series is

00:02:02.040 --> 00:02:05.250
a specific version
of a power series.

00:02:05.250 --> 00:02:08.690
But let's play around with this
arith-- I keep wanting to say

00:02:08.690 --> 00:02:11.175
arith-MET-ic, but a-RITH-metic,
either way-- series.

00:02:14.770 --> 00:02:17.170
So let's call the sum S.

00:02:17.170 --> 00:02:23.360
Let's say that this is equal to
the sum from k is equal to 1 to

00:02:23.360 --> 00:02:30.880
N of k, which is equal to, just
like we said, 1 plus 2 plus 3.

00:02:30.880 --> 00:02:33.800
And we'll just keep adding
them, dot, dot, dot, to a bunch

00:02:33.800 --> 00:02:39.200
of numbers, to big N minus
1 plus big N, right?

00:02:39.200 --> 00:02:41.140
Fair enough.

00:02:41.140 --> 00:02:42.940
Now, bear with me a second.

00:02:42.940 --> 00:02:47.430
I'm just going to write that
same exact sum again, but

00:02:47.430 --> 00:02:48.640
I'm just going to write
it in reverse order.

00:02:48.640 --> 00:02:52.520
And I think it's intuitive to
you that it doesn't matter what

00:02:52.520 --> 00:02:54.170
order I add up numbers in.

00:02:54.170 --> 00:02:55.580
They'll add up to
the same number.

00:02:55.580 --> 00:02:58.290
2 plus 1 is the same thing
as 1 plus 2, right?

00:02:58.290 --> 00:03:00.892
So let me write this exact
same sum, but I'll write

00:03:00.892 --> 00:03:01.800
it in reverse order.

00:03:01.800 --> 00:03:09.040
So that's the same thing as N
plus N minus 1, plus N minus

00:03:09.040 --> 00:03:13.320
2, plus-- and the pluses
keep going-- plus 2 plus 1.

00:03:13.320 --> 00:03:16.190
This is the exact sum, just
in the reverse order.

00:03:16.190 --> 00:03:19.960
And I did that for a reason
because now I'm going to add

00:03:19.960 --> 00:03:21.250
both sides of this equation.

00:03:21.250 --> 00:03:23.430
I'm going to take-- S plus S.

00:03:23.430 --> 00:03:25.450
Well, that's just 2S.

00:03:25.450 --> 00:03:30.870
And that's going to equal
this sum plus this sum.

00:03:30.870 --> 00:03:34.340
I wrote this so that
the sum becomes clean.

00:03:34.340 --> 00:03:35.490
And why do I say that?

00:03:35.490 --> 00:03:38.000
Well, let's add up
corresponding terms.

00:03:38.000 --> 00:03:40.600
We could have added up any
terms, but-- so since they all

00:03:40.600 --> 00:03:43.345
have to add up, let's just add
the 1 plus the N, then we'll

00:03:43.345 --> 00:03:47.000
add the 2 plus the N minus 1,
then we'll add the 3 plus the

00:03:47.000 --> 00:03:48.820
N minus 2, and so forth.

00:03:48.820 --> 00:03:50.560
And I think you'll see in a
second, or maybe you already

00:03:50.560 --> 00:03:51.790
realize why I'm doing this.

00:03:51.790 --> 00:03:56.490
One plus the N, the 2 plus the
N minus 1, the 3 and the N

00:03:56.490 --> 00:04:00.160
minus 2, all the way to the N
minus 1 and the 2,

00:04:00.160 --> 00:04:02.310
the N and the 1.

00:04:02.310 --> 00:04:03.830
What's 1 plus N?

00:04:03.830 --> 00:04:07.500
Well, that's just
N plus 1, right?

00:04:07.500 --> 00:04:10.410
What's 2 plus N minus 1?

00:04:10.410 --> 00:04:14.060
Well, that's also
N plus 1, right?

00:04:14.060 --> 00:04:16.670
What's 3 plus N minus 2?

00:04:16.670 --> 00:04:17.705
I think you could guess.

00:04:17.705 --> 00:04:19.910
It's N minus 1.

00:04:19.910 --> 00:04:21.140
And we just keep doing that.

00:04:21.140 --> 00:04:23.890
And what's N minus 2 plus 2?

00:04:23.890 --> 00:04:25.600
Sorry, this is a plus.

00:04:25.600 --> 00:04:27.780
N plus 1.

00:04:27.780 --> 00:04:29.350
And what's N plus 1?

00:04:29.350 --> 00:04:32.310
Well, that's just N
plus 1, of course.

00:04:32.310 --> 00:04:34.190
So my question to you
is how many of these

00:04:34.190 --> 00:04:36.020
N plus 1's are there?

00:04:36.020 --> 00:04:37.410
Well, there are N
of them, right?

00:04:37.410 --> 00:04:40.550
Each N plus 1 corresponds
to each of these terms,

00:04:40.550 --> 00:04:41.530
so there are N of these.

00:04:41.530 --> 00:04:45.000
So instead of just adding N
plus 1 N times, we could say

00:04:45.000 --> 00:04:51.580
that this is just
N times N plus 1.

00:04:51.580 --> 00:04:55.485
So we have 2 times the sum is
equal to N times N plus 1, and

00:04:55.485 --> 00:04:58.310
we could divide both sides by
2, and we get the sum is equal

00:04:58.310 --> 00:05:04.570
to N times N plus 1 over 2.

00:05:04.570 --> 00:05:07.800
Now, why is this neat, or
why is this cool at all?

00:05:07.800 --> 00:05:11.210
Well, first of all, we found
out a way to sum this

00:05:11.210 --> 00:05:12.330
Sigma notation up.

00:05:12.330 --> 00:05:16.670
We got kind of a
well-defined formula.

00:05:16.670 --> 00:05:19.350
And what makes this especially
cool is you can use this

00:05:19.350 --> 00:05:23.650
for low-end parlor tricks.

00:05:23.650 --> 00:05:24.530
What do I mean by that?

00:05:24.530 --> 00:05:28.760
Well, you can go up to someone
and you can say, well, how

00:05:28.760 --> 00:05:32.490
quickly do you think I can add
up the numbers between 1 and--

00:05:32.490 --> 00:05:35.000
what am I doing-- oh,
between 1 and 100?

00:05:35.000 --> 00:05:36.480
And, you know, people will say,
oh, it will take you a little

00:05:36.480 --> 00:05:37.780
time: 1 plus 2 plus 3.

00:05:37.780 --> 00:05:40.720
And you say, well, it takes
me no time at all because

00:05:40.720 --> 00:05:42.630
this is what I can do.

00:05:42.630 --> 00:05:47.340
So the sum-- and I just want
to show you that you can use

00:05:47.340 --> 00:05:50.470
different variables from B
equals 1, we're taking the

00:05:50.470 --> 00:05:55.100
variable B, to 100, right?

00:05:55.100 --> 00:05:56.720
That's the sum from 1 to 100.

00:05:56.720 --> 00:05:59.160
And we figured out
what that formula is.

00:05:59.160 --> 00:06:07.230
It's going to be 100
times 101 over 2.

00:06:07.230 --> 00:06:09.340
Well, what's 100 times 101?

00:06:09.340 --> 00:06:13.250
It's just going to be 101
with two zeroes, right?

00:06:13.250 --> 00:06:20.030
10,100 over 2, and
that equals 5,050.

00:06:20.030 --> 00:06:20.730
That's pretty neat.

00:06:20.730 --> 00:06:24.400
Instead of having to say 1 plus
2 plus 3 plus blah, blah, blah,

00:06:24.400 --> 00:06:29.395
blah, blah, blah, blah, blah,
plus 98 plus 99 plus 100, this

00:06:29.395 --> 00:06:31.590
would take you some time, and
there's a very good chance you

00:06:31.590 --> 00:06:32.860
would make a careless mistake.

00:06:32.860 --> 00:06:35.290
We could just plug into this
formula, which we proved and

00:06:35.290 --> 00:06:39.680
hopefully you understood,
and say that equals 5,050.

00:06:39.680 --> 00:06:41.560
You could do even something
more impressive: the

00:06:41.560 --> 00:06:43.940
sum from 1 to 1,000.

00:06:43.940 --> 00:06:45.150
What's the sum from 1 to 1,000?

00:06:45.150 --> 00:06:48.900
Well, our formula, remember,
was N times N plus 1 over 2.

00:06:48.900 --> 00:06:53.715
So if N is equal to 1,000,
then what's our sum?

00:06:53.715 --> 00:07:02.225
It's 1,000 times 1,001 over 2,
which is equal to-- well, we'll

00:07:02.225 --> 00:07:09.290
just add three zeroes to
this: 1,001, one, two, three.

00:07:09.290 --> 00:07:11.390
Sorry, I think that was
my first burp ever on

00:07:11.390 --> 00:07:13.030
one of these videos.

00:07:13.030 --> 00:07:17.300
I should re-record it, but
I'm going to move forward.

00:07:17.300 --> 00:07:19.280
That kind of disconcerted
me a little bit.

00:07:19.280 --> 00:07:20.620
I'd eaten too much.

00:07:20.620 --> 00:07:24.820
Anyway, divided by 2,
and what is that?

00:07:24.820 --> 00:07:28.810
Let's see, it'll be 500--
let's see, this is a million.

00:07:28.810 --> 00:07:32.830
Half of a million is 500,000.

00:07:32.830 --> 00:07:37.510
500,500.

00:07:37.510 --> 00:07:41.190
And that would have taken
you forever to do manually.

00:07:41.190 --> 00:07:43.810
But based on this formula we
just got, you know how to

00:07:43.810 --> 00:07:46.450
do it very, very quickly.

00:07:46.450 --> 00:07:48.620
So that's the
arithmetic series.

00:07:48.620 --> 00:07:49.620
But let's do another one.

00:07:49.620 --> 00:07:54.520
This is another typical
series that you might see.

00:07:54.520 --> 00:07:56.910
Actually, this one you'll see a
lot in your life, especially if

00:07:56.910 --> 00:08:03.730
you go into finance or really a
whole series of scientific--

00:08:03.730 --> 00:08:06.890
this shows up a lot, and this
is called the geometric series.

00:08:06.890 --> 00:08:10.530
And the geometric series is--
essentially you take x.

00:08:10.530 --> 00:08:14.440
And I'll do it generally where
I just take a variable x,

00:08:14.440 --> 00:08:15.880
and I say-- well, no, no.

00:08:15.880 --> 00:08:16.740
Let me just not take an x.

00:08:16.740 --> 00:08:19.020
Let me just take some number.

00:08:19.020 --> 00:08:29.300
So let's say some number a to
the k from-- I don't know.

00:08:29.300 --> 00:08:33.940
Let's say from k is equal
to 0 to k is equal to N.

00:08:33.940 --> 00:08:34.660
What does that mean?

00:08:34.660 --> 00:08:40.950
Well, that means a to the 0,
right, k is 0, plus a 1 plus a

00:08:40.950 --> 00:08:45.780
squared plus a to the third
plus-- and you could keep

00:08:45.780 --> 00:08:53.570
going-- plus a to the N minus
1 plus a to the N minus 2.

00:08:53.570 --> 00:08:55.560
This is called the
geometric series.

00:08:55.560 --> 00:08:58.210
And it might not be obvious to
you, but this type of growth,

00:08:58.210 --> 00:09:00.460
where you keep increasing the
exponent, this is called

00:09:00.460 --> 00:09:02.392
geometric growth.

00:09:02.392 --> 00:09:04.360
So how do you take
the sum of this?

00:09:04.360 --> 00:09:06.300
Well, let's see if we can do a
similar trick, although this

00:09:06.300 --> 00:09:09.010
trick will involve
one more step.

00:09:09.010 --> 00:09:13.640
So let's call the sum S.

00:09:13.640 --> 00:09:23.060
Let's call it the sum from k
equals 0 to N, a to the k.

00:09:23.060 --> 00:09:24.950
And that, of course, is
equal to what I just wrote.

00:09:24.950 --> 00:09:27.570
I probably didn't have
to do it like this.

00:09:27.570 --> 00:09:32.310
a squared plus bup, bup, bup,
bup, plus a to the N minus

00:09:32.310 --> 00:09:37.250
1, plus a to the N minus 2.

00:09:37.250 --> 00:09:42.040
Now let's define another sum,
and I'm going to call that aS.

00:09:44.580 --> 00:09:46.120
Actually, I'm about to run out
of time, so I'll continue

00:09:46.120 --> 00:09:48.430
this in the next video.


Solid of Revolution (part 8)

WEBVTT
Kind: captions
Language: en

00:00:00.820 --> 00:00:03.600
Where I left off, we were just
essentially chugging through

00:00:03.600 --> 00:00:08.110
this fairly hairy derivative--
this definite integral--

00:00:08.110 --> 00:00:08.910
this antiderivative.

00:00:08.910 --> 00:00:10.500
It takes my brain a
little while to come

00:00:10.500 --> 00:00:11.150
up with the next term.

00:00:11.150 --> 00:00:14.300
So we evaluated at 2 and
now we have to subtract

00:00:14.300 --> 00:00:16.320
this evaluated at 1.

00:00:16.320 --> 00:00:24.120
So minus 16 pi minus 4
pi over 3-- oh, sorry.

00:00:24.120 --> 00:00:24.650
Plus.

00:00:24.650 --> 00:00:26.390
The minus is going to
be on everything.

00:00:26.390 --> 00:00:26.910
Oh no, sorry.

00:00:26.910 --> 00:00:28.270
That is a minus.

00:00:28.270 --> 00:00:29.290
That's a minus.

00:00:29.290 --> 00:00:31.870
Plus 4 pi.

00:00:31.870 --> 00:00:32.100
Right?

00:00:32.100 --> 00:00:33.380
Because x is 1.

00:00:33.380 --> 00:00:36.220
Minus pi over 2.

00:00:36.220 --> 00:00:38.440
And now we can simplify it.

00:00:38.440 --> 00:00:40.140
Let's see what we can do.

00:00:40.140 --> 00:00:41.340
This is really a hairy problem.

00:00:44.750 --> 00:00:51.680
The 16 pi minus 8 pi,
that equals 8 pi.

00:00:51.680 --> 00:00:53.030
And then that's a plus.

00:00:53.030 --> 00:00:55.860
The 32 plus 8 pi.

00:00:55.860 --> 00:00:59.480
32 pi plus 8 pi,
that equals 40 pi.

00:00:59.480 --> 00:01:06.110
So let's see, I've simplified
it to 40 pi, and what's minus

00:01:06.110 --> 00:01:08.740
8 times 4 is 32 pi over 3.

00:01:13.830 --> 00:01:16.030
And then all of that, let's
see if I can simplify this.

00:01:16.030 --> 00:01:18.990
Let's see, 16 pi plus
4 pi, that's 20 pi.

00:01:22.560 --> 00:01:23.930
And then, let's see.

00:01:23.930 --> 00:01:28.060
Minus 4 pi over 3
minus pi over 2.

00:01:28.060 --> 00:01:30.740
So let's get a common
denominator for this

00:01:30.740 --> 00:01:31.480
part right here.

00:01:31.480 --> 00:01:36.140
So if I put everything over 6--
20 pi over 6 is the same thing

00:01:36.140 --> 00:01:45.015
is 120 pi over 6, and then
minus-- 4 pi over 3, if I put

00:01:45.015 --> 00:01:48.890
it over 6 it becomes
8 pi over 6, right?

00:01:48.890 --> 00:01:53.060
And then pi over 2, if I put it
over 6 it becomes 3 pi over 6.

00:01:53.060 --> 00:01:56.000
So minus 3 pi.

00:01:56.000 --> 00:01:58.240
So this whole term that we're
going to have to subtract from

00:01:58.240 --> 00:02:04.330
this term is equal to
120 minus 11, right?

00:02:04.330 --> 00:02:07.260
So that's 109 pi over 6.

00:02:07.260 --> 00:02:09.080
This equals 1-- is that right?

00:02:09.080 --> 00:02:11.310
Yeah, 109 pi over 6.

00:02:11.310 --> 00:02:12.980
See what happens when you
make up problems on the fly?

00:02:12.980 --> 00:02:14.500
You get ugly numbers.

00:02:14.500 --> 00:02:15.890
109 pi over 6.

00:02:15.890 --> 00:02:17.970
And what does that
top part translate?

00:02:17.970 --> 00:02:19.000
So this is what we're
going to subtract.

00:02:19.000 --> 00:02:22.955
This is when we evaluated
the antiderivative at 1.

00:02:22.955 --> 00:02:24.510
And let's simplify this one.

00:02:24.510 --> 00:02:29.310
So that one, if we put 3 common
denominator, that's 120 pi

00:02:29.310 --> 00:02:33.190
over 3 minus 32 pi over 3.

00:02:33.190 --> 00:02:38.610
120 minus 32, let's
see, we get 90-- 88?

00:02:38.610 --> 00:02:39.290
Right.

00:02:39.290 --> 00:02:40.680
88 pi over 3.

00:02:40.680 --> 00:02:44.990
So that equals 88 pi over
3, and remember, that's

00:02:44.990 --> 00:02:46.320
just the top part.

00:02:46.320 --> 00:02:48.640
And then-- this is more of a
review of fractions than

00:02:48.640 --> 00:02:50.820
anything else-- and then if
I want to put it over

00:02:50.820 --> 00:02:52.830
6, I just double it.

00:02:52.830 --> 00:02:54.590
So I think we're almost done.

00:02:54.590 --> 00:02:57.230
Let me switch colors.

00:02:57.230 --> 00:02:59.790
So if you go over a
denominator of 6-- 88

00:02:59.790 --> 00:03:01.100
pi over 3-- let's see.

00:03:01.100 --> 00:03:06.670
If I double that, I
get 160-- 176, right?

00:03:06.670 --> 00:03:11.780
176 pi minus this.

00:03:11.780 --> 00:03:13.910
Minus 109 pi.

00:03:13.910 --> 00:03:16.210
I'm sure I made a
careless mistake.

00:03:16.210 --> 00:03:19.100
These are my least favorite
type of things to do.

00:03:19.100 --> 00:03:21.050
Hairy fractions.

00:03:21.050 --> 00:03:23.430
So 176 minus 109.

00:03:23.430 --> 00:03:26.000
That's the same thing
as 76 minus 9.

00:03:26.000 --> 00:03:28.730
And that is 65.

00:03:28.730 --> 00:03:32.890
So our final answer
is 65 pi over 6.

00:03:32.890 --> 00:03:36.070
Which isn't as hairy
as it seemed when we

00:03:36.070 --> 00:03:36.690
started the problem.

00:03:36.690 --> 00:03:37.410
But that's pretty neat.

00:03:37.410 --> 00:03:39.790
This is kind of a
strange shape.

00:03:39.790 --> 00:03:45.500
It's kind of a wide ring that
has-- the upper part of the

00:03:45.500 --> 00:03:48.160
ring and the inner part of
the ring are hard edges.

00:03:48.160 --> 00:03:50.090
But then it's curved
on the outside.

00:03:50.090 --> 00:03:52.830
And we were able to figure
out the volume of that.

00:03:52.830 --> 00:03:54.900
Especially-- and what was weird
about this, is when it was

00:03:54.900 --> 00:04:00.350
rotated around the line
y is equal to minus 2.

00:04:00.350 --> 00:04:01.740
Hopefully I didn't confuse you.

00:04:01.740 --> 00:04:06.120
These are about as difficult
as these volume of

00:04:06.120 --> 00:04:09.060
revolution problems get.

00:04:09.060 --> 00:04:11.530
If you want more,
just let me know.

00:04:11.530 --> 00:04:14.200
I will see you in
the next video.


Solid of Revolution (part 7)

WEBVTT
Kind: captions
Language: en

00:00:00.850 --> 00:00:04.220
Let's do a couple more
rotational volume problems, and

00:00:04.220 --> 00:00:05.710
I'm going to make these a
little bit more difficult.

00:00:05.710 --> 00:00:09.290
And hopefully after these if
you've understood everything

00:00:09.290 --> 00:00:12.370
we've done up to now and the
ones I'm about to do, I think

00:00:12.370 --> 00:00:15.300
you're pretty set for most
of what you should face

00:00:15.300 --> 00:00:17.220
in most math classes.

00:00:17.220 --> 00:00:19.460
And definitely I think you'll
be set for the AP exam,

00:00:19.460 --> 00:00:22.690
and either ab or bc
on this concept.

00:00:22.690 --> 00:00:26.050
So let's do another example.

00:00:26.050 --> 00:00:26.990
OK.

00:00:26.990 --> 00:00:29.510
So let's say that I want
to-- actually, let me

00:00:29.510 --> 00:00:31.005
do something different.

00:00:31.005 --> 00:00:33.780
Let me draw here.

00:00:33.780 --> 00:00:36.480
So that's my y-axis.

00:00:36.480 --> 00:00:38.260
This is my x-axis.

00:00:38.260 --> 00:00:42.100
Now let me draw the function
y is equal to x squared.

00:00:46.560 --> 00:00:50.370
And we know that that could be
written as y is equal to x

00:00:50.370 --> 00:00:53.570
squared, or we could write that
as x is equal to square root of

00:00:53.570 --> 00:00:56.540
y, depending on what we want
to be a function of what.

00:00:59.150 --> 00:01:01.530
This is the y-axis.

00:01:01.530 --> 00:01:03.570
This is the x-axis.

00:01:03.570 --> 00:01:12.950
Let's say I also have
the line y equals 2.

00:01:12.950 --> 00:01:15.270
Goes over what I just wrote.

00:01:15.270 --> 00:01:18.640
y equals 2.

00:01:18.640 --> 00:01:22.170
Now this problem is going to
be slightly different then

00:01:22.170 --> 00:01:25.310
what we've done so far.

00:01:25.310 --> 00:01:28.570
I'm going to take a rotation,
but instead of taking a

00:01:28.570 --> 00:01:31.370
rotation around the y or the
x-axis, I'm going to take a

00:01:31.370 --> 00:01:33.610
rotation around another line.

00:01:33.610 --> 00:01:40.330
So let's say I want to
take a rotation between

00:01:40.330 --> 00:01:41.510
x is equal to 0.

00:01:44.680 --> 00:01:46.130
Actually let me do
something arbitrary.

00:01:46.130 --> 00:01:51.370
Let me say between x is equal
to 1, so that's that point,

00:01:51.370 --> 00:01:52.590
and x is equal to 2.

00:01:52.590 --> 00:01:55.230
That's where they intersect.

00:01:55.230 --> 00:01:56.710
This is the point right
here, this is 2,2.

00:02:00.050 --> 00:02:01.990
I'm sorry, no, 2,4.

00:02:01.990 --> 00:02:03.230
Because y is equal
to x squared.

00:02:03.230 --> 00:02:04.610
So this is 2,4.

00:02:04.610 --> 00:02:07.190
This is the point 4.

00:02:07.190 --> 00:02:10.100
So what point is this.

00:02:10.100 --> 00:02:13.670
This is the point 1, right?

00:02:13.670 --> 00:02:16.980
So our y values go from 4 to 1,
our x values go from 1 to 2.

00:02:16.980 --> 00:02:19.170
And that makes sense,
because y is x squared.

00:02:19.170 --> 00:02:21.560
And so if we were to kind of
take the area that we're going

00:02:21.560 --> 00:02:23.230
to rotate, and I haven't told
you what we're going to rotate

00:02:23.230 --> 00:02:27.230
it around yet, and this might
prove to be shocking to you.

00:02:27.230 --> 00:02:31.005
So this is the area
we're going to rotate.

00:02:34.320 --> 00:02:37.580
Instead of rotating it around
the y-axis, I want to rotate

00:02:37.580 --> 00:02:44.010
it around the line y
is equal to minus 2.

00:02:44.010 --> 00:02:49.340
So if that's 2, y equals minus
2 should be roughly here.

00:02:49.340 --> 00:02:55.440
So I'm going to rotate this
area around this line.

00:02:59.390 --> 00:03:00.450
So what's it going
to look like?

00:03:00.450 --> 00:03:03.780
It's going to be a
fairly big ring.

00:03:03.780 --> 00:03:07.300
Like if I were to try to draw
it-- let me see if I can

00:03:07.300 --> 00:03:08.870
even make an attempt.

00:03:08.870 --> 00:03:11.370
Once again this is always the
hardest part, just drawing

00:03:11.370 --> 00:03:13.180
what I'm trying to rotate.

00:03:13.180 --> 00:03:15.450
I'll try to do it from
an upward perspective.

00:03:15.450 --> 00:03:18.040
So that's kind of the inner
loop, and then there

00:03:18.040 --> 00:03:18.880
will be an outer loop.

00:03:18.880 --> 00:03:22.150
The top is flat right, because
it's defined by y is equal

00:03:22.150 --> 00:03:23.320
to 4, so that's the top.

00:03:28.850 --> 00:03:31.010
And the inside is also
going to be a hard edge.

00:03:36.760 --> 00:03:39.180
But then the outside is
going to curve inward.

00:03:39.180 --> 00:03:42.300
I don't know if you see what
I'm saying, because this is the

00:03:42.300 --> 00:03:44.150
outside, it's curving inward.

00:03:44.150 --> 00:03:48.490
So it's going to be a big ring.

00:03:48.490 --> 00:03:52.180
So if I were to draw the axis,
this would be the y-axis access

00:03:52.180 --> 00:03:53.570
coming in-- no, no, sorry.

00:03:57.150 --> 00:03:57.400
Whoops.

00:04:00.630 --> 00:04:05.320
The y-axis is actually going to
be closer to this hand side.

00:04:05.320 --> 00:04:10.160
The y-axis is going to be in
the middle of kind of-- so

00:04:10.160 --> 00:04:12.860
this is going to be the
y-axis coming up here.

00:04:17.240 --> 00:04:19.140
And then the x-axis is
going to come below that.

00:04:19.140 --> 00:04:21.780
I'm drawing everything at
an angle as best as I can.

00:04:21.780 --> 00:04:23.290
The x-axis is going to
come a below that.

00:04:23.290 --> 00:04:26.960
And then this line we're
rotating it around,

00:04:26.960 --> 00:04:28.690
that's going to be
someplace over here.

00:04:33.670 --> 00:04:36.880
That's going to be
something like that.

00:04:36.880 --> 00:04:40.440
It's going to go behind there
and come back over there.

00:04:40.440 --> 00:04:41.310
Hopefully that makes sense.

00:04:41.310 --> 00:04:43.010
We're just getting a big ring.

00:04:43.010 --> 00:04:46.780
So how are we going to do this?

00:04:46.780 --> 00:04:48.810
Well actually there's a
couple of things we can do.

00:04:48.810 --> 00:04:50.700
First we could just
use the shell method,

00:04:50.700 --> 00:04:52.650
using the x value.

00:04:52.650 --> 00:04:54.362
So how do we do that?

00:04:54.362 --> 00:04:56.650
The important thing is
to always visualize

00:04:56.650 --> 00:04:57.710
the shell or the disk.

00:04:57.710 --> 00:05:04.450
So the shell method we're going
to take slivers like this,

00:05:04.450 --> 00:05:06.220
where the width of
that sliver is dx.

00:05:06.220 --> 00:05:08.650
I could draw it really big.

00:05:08.650 --> 00:05:11.510
So that's our direct
angle, is going to be dx.

00:05:11.510 --> 00:05:13.530
What's the height going
to be of the sliver?

00:05:13.530 --> 00:05:15.590
Well it's going to be
the top function minus

00:05:15.590 --> 00:05:16.360
the bottom function.

00:05:16.360 --> 00:05:21.760
It's going to be y equals
4 minus x squared.

00:05:21.760 --> 00:05:25.650
So this is going to be 4 minus
x squared, the height at

00:05:25.650 --> 00:05:27.510
any point right here.

00:05:27.510 --> 00:05:30.880
And then if I were to do
the shell just like we did

00:05:30.880 --> 00:05:36.300
before-- let me see if I
can draw a decent shell.

00:05:40.650 --> 00:05:43.390
I think I'm getting
better at this.

00:05:43.390 --> 00:05:48.560
This is one edge of the shell,
that's the other shell.

00:05:48.560 --> 00:05:52.020
We already figured out that
the width of the shell is dx.

00:05:52.020 --> 00:05:55.410
The height is 4 minus x
squared, the top function minus

00:05:55.410 --> 00:05:58.070
the bottom function because of
the distance between the two.

00:05:58.070 --> 00:06:01.200
And then what's the
radius going to be?

00:06:01.200 --> 00:06:05.040
What's the radius of
that shell going to be?

00:06:05.040 --> 00:06:07.570
Well, is it going to be just x?

00:06:07.570 --> 00:06:08.220
Is it just going to be x value?

00:06:08.220 --> 00:06:12.580
No, the x value will tell you
the distance from the y-axis to

00:06:12.580 --> 00:06:16.750
that shell, and it's going to
be from minus 2 to the e value.

00:06:16.750 --> 00:06:22.630
So it's going to be
essentially 2 plus x.

00:06:22.630 --> 00:06:25.220
That's going to be the
radius at any point.

00:06:25.220 --> 00:06:28.030
And this is where we diverge
from what we've done before.

00:06:28.030 --> 00:06:31.630
Before the radius was just
x, but now it's 2 plus x.

00:06:31.630 --> 00:06:35.280
So what's the circumference
of each shell going to be?

00:06:35.280 --> 00:06:39.460
Well circumference is
equal to 2 pi r, and

00:06:39.460 --> 00:06:40.890
our radius is 2 plus x.

00:06:40.890 --> 00:06:48.690
So it's 2 pi times 2 plus
x, which equals 4 pi

00:06:48.690 --> 00:06:54.310
plus 4 pi plus 2 pi x.

00:06:54.310 --> 00:06:55.840
That's the circumference.

00:06:55.840 --> 00:06:57.940
And then what's the
surface area of this?

00:06:57.940 --> 00:07:00.960
Well it's going to be the
circumference times the height.

00:07:00.960 --> 00:07:04.630
So surface area is
equal to that.

00:07:04.630 --> 00:07:12.180
The circumference 4 pi plus 2
pi x, all of that times the

00:07:12.180 --> 00:07:17.220
height-- times 4
minus x squared.

00:07:17.220 --> 00:07:19.540
And let's see if we
can foil this out or

00:07:19.540 --> 00:07:20.870
distribute this out.

00:07:20.870 --> 00:07:24.810
So 4 pi times 4 is 16 pi.

00:07:24.810 --> 00:07:31.580
4 pi minus x squared
minus 4 pi x squared.

00:07:31.580 --> 00:07:38.040
2 pi x times 4 plus 8 pi x, and
then 2 pi x times minus x

00:07:38.040 --> 00:07:43.030
squared, so that's minus
2 pi x to the third.

00:07:43.030 --> 00:07:44.980
So that's the surface
area of each ring.

00:07:44.980 --> 00:07:48.820
And then if we want the volume
of each shell, essentially

00:07:48.820 --> 00:07:50.890
we multiply it times
the width, the dx.

00:07:56.390 --> 00:07:58.220
So that's the volume of
each shell, and if we want

00:07:58.220 --> 00:08:03.180
the volume of all the
shells, we sum them up.

00:08:03.180 --> 00:08:04.800
So we take the integral, that's
an integral sign, and I'm

00:08:04.800 --> 00:08:06.720
running out of space
like I always do.

00:08:06.720 --> 00:08:08.015
And where did I take
the integral from?

00:08:08.015 --> 00:08:11.550
I take the integral from x is
equal to 1 to x is equal to 2.

00:08:11.550 --> 00:08:13.340
From 1 to 2.

00:08:13.340 --> 00:08:16.980
That's probably too
small for you to read.

00:08:16.980 --> 00:08:20.710
So let's see if we can take
the antiderivative of this.

00:08:20.710 --> 00:08:23.610
Let me make some space
free just so I don't

00:08:23.610 --> 00:08:26.220
have to write so small.

00:08:26.220 --> 00:08:28.360
So I'll keep this down here,
because that's the set

00:08:28.360 --> 00:08:28.965
up of the problem.

00:08:31.600 --> 00:08:35.360
I think I can get rid
of a lot of this.

00:08:35.360 --> 00:08:36.995
I think that is pretty good.

00:08:41.460 --> 00:08:42.360
OK.

00:08:42.360 --> 00:08:43.876
And let me switch
to another color.

00:08:43.876 --> 00:08:45.280
And we're going to take
the antiderivative.

00:08:45.280 --> 00:08:47.550
So what's the
antiderivative of this?

00:08:47.550 --> 00:08:52.405
So the antiderivative
of 16 pi is 16 pi x.

00:09:01.250 --> 00:09:02.570
And then what's the
antiderivative of

00:09:02.570 --> 00:09:03.530
4 pi x squared?

00:09:03.530 --> 00:09:07.370
It's going to be x to the third
over 3, so it'll be minus 4

00:09:07.370 --> 00:09:14.370
pi over 3-- sorry-- 4 pi
over 3 x to the third.

00:09:17.350 --> 00:09:19.100
Well this will be x squared
over 2, so it'll be

00:09:19.100 --> 00:09:23.400
plus 4 pi x squared.

00:09:23.400 --> 00:09:25.200
And then minus, this will
be x to the fourth over

00:09:25.200 --> 00:09:28.390
4, so minus pi over 2.

00:09:28.390 --> 00:09:30.550
Just divided by 4.

00:09:30.550 --> 00:09:32.170
x to the fourth.

00:09:32.170 --> 00:09:34.320
I'm going to evaluate that.

00:09:34.320 --> 00:09:36.840
This is a much hairrier problem
then what we've been doing.

00:09:36.840 --> 00:09:37.940
2 and at 1.

00:09:37.940 --> 00:09:40.770
So what is it evaluated at 2?

00:09:40.770 --> 00:09:50.720
It is 32 pi minus 4 pi over 3
times 8 plus 4 pi times 4 plus

00:09:50.720 --> 00:10:00.640
16 pi minus 2 to the fourth is
16 divided by 2 minus 8 pi.

00:10:00.640 --> 00:10:02.920
And then I just realized I'm
running out time, so I will

00:10:02.920 --> 00:10:04.130
continue this in
the next video.