Worked example: estimating e_ using Lagrange error bound

WEBVTT
Kind: captions
Language: en

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- [Instructor] Estimating
e to the 1.45 using

00:00:02.052 --> 00:00:04.782
a Taylor polynomial about x equals two,

00:00:04.782 --> 00:00:07.164
what is the least degree of the polynomial

00:00:07.164 --> 00:00:10.581
that assures an error smaller than 0.001?

00:00:11.779 --> 00:00:13.957
In general, if you see
a situation like this

00:00:13.957 --> 00:00:16.342
where we're talking about
approximating a function

00:00:16.342 --> 00:00:19.082
with a Taylor polynomial
centered about some value,

00:00:19.082 --> 00:00:21.095
and we wanna know, well,
how many terms do we need,

00:00:21.095 --> 00:00:23.750
what degree do we need to bound the error?

00:00:23.750 --> 00:00:26.424
That's a pretty good clue
that we're going to be using

00:00:26.424 --> 00:00:29.880
the LaGrange error bound or
Taylor's remainder theorem.

00:00:29.880 --> 00:00:31.861
And just as a reminder of that,

00:00:31.861 --> 00:00:35.119
this is a review of
Taylor's remainder theorem,

00:00:35.119 --> 00:00:38.080
and it tells us that the
absolute value of the remainder

00:00:38.080 --> 00:00:40.169
for the nth degree Taylor polynomial,

00:00:40.169 --> 00:00:43.309
it's gonna be less than this
business right over here.

00:00:43.309 --> 00:00:46.527
Now, n is the degree of our polynomial

00:00:46.527 --> 00:00:49.075
that in question, so that's the n.

00:00:49.075 --> 00:00:52.915
The x is the x value at which
we are calculating that error,

00:00:52.915 --> 00:00:55.544
in this case it's going to be this 1.45.

00:00:55.544 --> 00:00:58.343
And c is where our Taylor
polynomial is centered.

00:00:58.343 --> 00:00:59.775
But what about our M?

00:00:59.775 --> 00:01:02.837
Well, our M is an upper
bound on the absolute value

00:01:02.837 --> 00:01:06.470
of the n plus oneth
derivative of our function.

00:01:06.470 --> 00:01:07.779
And that might seem like a mouthful,

00:01:07.779 --> 00:01:09.262
but when we actually
work through the details

00:01:09.262 --> 00:01:12.208
of this example, it'll make
it a little bit more concrete.

00:01:12.208 --> 00:01:14.813
So for this particular thing,
we're trying to estimate,

00:01:14.813 --> 00:01:17.460
we're trying to estimate e to the x.

00:01:17.460 --> 00:01:21.255
So I could write f of x,
let me write this this way.

00:01:21.255 --> 00:01:24.338
So, f of x is equal to e to the x,

00:01:25.631 --> 00:01:29.577
and we're trying to estimate f of 1.45.

00:01:29.577 --> 00:01:31.949
And let's just to get the bound here,

00:01:31.949 --> 00:01:34.279
to figure out what M is,
let's just remind ourselves

00:01:34.279 --> 00:01:36.263
that while the first derivative of this

00:01:36.263 --> 00:01:37.537
is going to be e to the x,

00:01:37.537 --> 00:01:38.934
the second derivative
is gonna be e to the x,

00:01:38.934 --> 00:01:41.355
the nth derivative is gonna be e to the x,

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the n plus oneth derivative
is gonna be e to the x.

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So the n plus oneth
derivative of f is gonna be,

00:01:48.621 --> 00:01:49.538
is gonna be

00:01:50.787 --> 00:01:53.524
e to the x which is convenient.

00:01:53.524 --> 00:01:55.439
These types of problems
are very, very hard

00:01:55.439 --> 00:01:59.910
if it's difficult to bound
the n plus oneth derivative.

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Well this we know, we know that e the x,

00:02:03.968 --> 00:02:05.765
we know that e the x, and I can even say

00:02:05.765 --> 00:02:08.219
the absolute value of this,
but this is gonna be positive,

00:02:08.219 --> 00:02:10.500
is going to be less than or equal to,

00:02:10.500 --> 00:02:15.000
let's say this is gonna be
less than or equal to e squared

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for zero is less than x

00:02:19.610 --> 00:02:22.655
is less than or equal to two.

00:02:22.655 --> 00:02:25.771
e the x isn't bounded over the entire,

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for over its entire domain.

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If x goes to infinity,

00:02:29.533 --> 00:02:31.498
e to the x will also go to infinity.

00:02:31.498 --> 00:02:33.031
But here I set up an interval.

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I've set up an interval that
contains the x we care about.

00:02:35.685 --> 00:02:38.505
Remember, the x we care about is 1.45,

00:02:38.505 --> 00:02:42.541
and it also contains where
our function is centered.

00:02:42.541 --> 00:02:44.190
Our function is centered at two.

00:02:44.190 --> 00:02:46.099
So we know we're bounded by e squared,

00:02:46.099 --> 00:02:50.374
so we can say, we can
use e squared as our M.

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We can use e squared as our M.

00:02:52.667 --> 00:02:55.355
We're able to establish this bound.

00:02:55.355 --> 00:02:57.969
And so doing that, we can now go straight

00:02:57.969 --> 00:02:59.327
to LaGrange error bound.

00:02:59.327 --> 00:03:03.177
We can say, we can say that the remainder

00:03:03.177 --> 00:03:05.255
of our nth degree Taylor polynomial,

00:03:05.255 --> 00:03:06.315
we wanna solve for n.

00:03:06.315 --> 00:03:10.602
We wanna figure out what n
gives us the appropriate bound

00:03:10.602 --> 00:03:12.102
evaluated at 1.45.

00:03:13.211 --> 00:03:17.447
When x is 1.45 is going to
be less than or equal to

00:03:17.447 --> 00:03:21.192
the absolute value, our M is e squared,

00:03:21.192 --> 00:03:22.442
e squared over,

00:03:24.437 --> 00:03:27.020
over n plus one factorial times

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1.45, that's our x that we care about,

00:03:32.609 --> 00:03:33.926
that's where we're calculating the error,

00:03:33.926 --> 00:03:35.287
we're trying to bound the error,

00:03:35.287 --> 00:03:37.016
minus where we're centered,

00:03:37.016 --> 00:03:40.616
minus two to the n plus oneth power.

00:03:40.616 --> 00:03:42.199
Now 1.45 minus two,

00:03:44.225 --> 00:03:46.214
that is negative 0.55.

00:03:46.214 --> 00:03:47.815
So let me just write that.

00:03:47.815 --> 00:03:48.732
So this is,

00:03:49.658 --> 00:03:51.408
this is negative 0.55

00:03:53.572 --> 00:03:56.095
to the n plus oneth power.

00:03:56.095 --> 00:03:59.929
And we wanna figure out for
what n is all of this business,

00:03:59.929 --> 00:04:04.012
is all of this business
gonna be less than 0.001.

00:04:05.012 --> 00:04:07.941
Well let's do a little bit of
algebraic manipulation here.

00:04:07.941 --> 00:04:11.054
This term is positive,
this is gonna be positive.

00:04:11.054 --> 00:04:13.155
This right over here, or this part of it,

00:04:13.155 --> 00:04:14.675
it's not an independent term,

00:04:14.675 --> 00:04:16.643
but the e to the squared
is gonna be positive,

00:04:16.643 --> 00:04:18.173
n plus one factorial is gonna be positive,

00:04:18.173 --> 00:04:20.621
the negative 0.55 to some power,

00:04:20.621 --> 00:04:23.225
that's gonna flip between
being positive or negative.

00:04:23.225 --> 00:04:25.035
But since we're taking the absolute value,

00:04:25.035 --> 00:04:26.530
we could write it this way.

00:04:26.530 --> 00:04:28.613
We could write e squared,

00:04:30.385 --> 00:04:34.983
e squared, since we're taking
the absolute value times 0.55

00:04:34.983 --> 00:04:38.566
to the n plus one over
n plus one factorial

00:04:40.837 --> 00:04:44.754
has to be less than, has
to be less than 0.001.

00:04:46.277 --> 00:04:49.027
Or, since we want to solve for n,

00:04:50.146 --> 00:04:52.790
let's divide both sides by e squared.

00:04:52.790 --> 00:04:55.707
So we could write, we could write,

00:04:55.707 --> 00:04:57.957
let's find the n where 0.55

00:05:00.073 --> 00:05:02.950
to the n plus oneth power

00:05:02.950 --> 00:05:05.033
over n plus one factorial

00:05:07.072 --> 00:05:08.155
is less than,

00:05:09.164 --> 00:05:10.664
is less than 0.001

00:05:13.020 --> 00:05:14.270
over e squared.

00:05:16.904 --> 00:05:20.309
Now to play with this, we're
gonna have to use a calculator.

00:05:20.309 --> 00:05:23.772
From this point we're just
gonna try larger and larger ns

00:05:23.772 --> 00:05:26.178
until we get an n that makes this true.

00:05:26.178 --> 00:05:28.876
And we wanna find the smallest possible n

00:05:28.876 --> 00:05:30.306
that makes this true.

00:05:30.306 --> 00:05:32.905
But let's get out our calculators
so that we can actually,

00:05:32.905 --> 00:05:35.371
so that we can actually do this.

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So, first I'm just gonna figure out

00:05:37.112 --> 00:05:40.454
what is 1/1000 divided by e squared.

00:05:40.454 --> 00:05:42.835
So make sure it's cleared out.

00:05:42.835 --> 00:05:44.895
So let's take e squared,

00:05:44.895 --> 00:05:46.931
I'm gonna take its reciprocal,

00:05:46.931 --> 00:05:49.215
and then I'm gonna
multiply that times 1/1000.

00:05:49.215 --> 00:05:51.382
So times .001 is equal to,

00:05:54.526 --> 00:05:55.693
so it's about,

00:05:56.907 --> 00:05:58.242
so I'll say,

00:05:58.242 --> 00:06:01.056
so it's three zeroes, this is a 10/1000,

00:06:01.056 --> 00:06:02.239
and then three five.

00:06:02.239 --> 00:06:05.166
So it's three zeroes, so
I'll say one three six.

00:06:05.166 --> 00:06:09.333
So this needs to be less than
zero point one two three,

00:06:11.131 --> 00:06:13.216
then I'll say one three six.

00:06:13.216 --> 00:06:16.693
If I can find an n that is less than this,

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then I'm in, then I am in good shape.

00:06:18.567 --> 00:06:19.443
Actually let me say this.

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Less than one three five, I
wanna be less than that value,

00:06:22.454 --> 00:06:25.120
then I can be, then I
will be in good shape.

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This is a little bit
more than one three five,

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but if I can find an n where
that is less than this,

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then I'm in good shape.

00:06:33.063 --> 00:06:36.646
So let me write this
0.55 to the n plus one

00:06:37.976 --> 00:06:40.143
over n plus one factorial.

00:06:41.555 --> 00:06:43.116
So let's try out some ns.

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And I'm gonna have to
get my calculator out.

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So, let's see, did I do that right?

00:06:48.064 --> 00:06:49.231
Yeah, .000135.

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If we get something below
this, then we're in good shape,

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because this is even less than that.

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All right, so let's do it.

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Let's see what this is equal to

00:07:01.316 --> 00:07:03.576
when, I don't know,
when n is equal to two.

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I can start at n equals one,
n equals two, n equals three,

00:07:08.764 --> 00:07:10.682
but if n equals two is good enough,

00:07:10.682 --> 00:07:12.520
then I might try n equals one.

00:07:12.520 --> 00:07:13.982
But if n equals two isn't good enough,

00:07:13.982 --> 00:07:16.247
then I'm gonna go to n equals
three or n equals four.

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So let's start with,

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actually let's just start
with n equals three.

00:07:19.233 --> 00:07:22.573
So if n equals three, it's gonna
be 0.55 to the fourth power

00:07:22.573 --> 00:07:24.112
divided by four factorial.

00:07:24.112 --> 00:07:25.816
So let's use that, let's do that.

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So 0.55

00:07:28.945 --> 00:07:31.945
to the fourth power is equal to that

00:07:34.350 --> 00:07:37.079
divided by four factorial.

00:07:37.079 --> 00:07:39.996
So divided by four factorial is 24.

00:07:41.951 --> 00:07:44.784
So that's nowhere near low enough.

00:07:45.878 --> 00:07:47.993
So let's try n equals four.

00:07:47.993 --> 00:07:50.917
If n equals four, then it's
gonna be this to the fifth power

00:07:50.917 --> 00:07:52.942
divided by five factorial.

00:07:52.942 --> 00:07:53.775
So 0.55

00:07:56.638 --> 00:07:59.862
to the fifth power is equal to,

00:07:59.862 --> 00:08:02.584
and then divided by five factorial is,

00:08:02.584 --> 00:08:04.963
five factorial is 120.

00:08:04.963 --> 00:08:07.630
Divided by 120 is equal to that.

00:08:08.744 --> 00:08:11.924
We're almost there with n equals four.

00:08:11.924 --> 00:08:14.603
I'm guessing that n equals
five will do the trick.

00:08:14.603 --> 00:08:18.227
So for n equals five,
let's clear this out.

00:08:18.227 --> 00:08:21.331
So for n equals five, we're
gonna raise to the sixth power

00:08:21.331 --> 00:08:23.498
and divide by six factorial.

00:08:23.498 --> 00:08:24.797
And so let's just remind ourselves

00:08:24.797 --> 00:08:27.631
what six factorial is, 720.

00:08:27.631 --> 00:08:29.286
In fact, I could've done that in my head.

00:08:29.286 --> 00:08:30.230
But anyway.

00:08:30.230 --> 00:08:31.238
All right, so let's see.

00:08:31.238 --> 00:08:35.729
We're gonna go 0.55 to
the, remember, n is five,

00:08:35.729 --> 00:08:37.596
so we're gonna raise to the sixth power,

00:08:37.596 --> 00:08:39.601
to the sixth power,

00:08:39.601 --> 00:08:42.084
and then we're gonna divide by 720,

00:08:42.084 --> 00:08:44.334
divided by 720 is equal to,

00:08:45.605 --> 00:08:48.175
and this number for sure is less than

00:08:48.175 --> 00:08:49.831
this number right over here.

00:08:49.831 --> 00:08:53.325
We've got four zeroes before
this three after the decimal,

00:08:53.325 --> 00:08:54.996
here we only have three.

00:08:54.996 --> 00:08:59.205
So when n equals five, it got
us sufficiently low enough,

00:08:59.205 --> 00:09:02.242
this remainder is going
to be sufficiently low,

00:09:02.242 --> 00:09:04.942
it's gonna be less than
this value right over here.

00:09:04.942 --> 00:09:07.934
So, what is the least
degree of the polynomial

00:09:07.934 --> 00:09:12.255
that assures an error smaller than 1/1000?

00:09:12.255 --> 00:09:16.239
The answer is five, our n, if n is five,

00:09:16.239 --> 00:09:18.877
we're definitely gonna be under this.


Worked example: estimating sin(0.4) using Lagrange error bound

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:01.988
- [Instructor] Estimating sine of 0.4

00:00:01.988 --> 00:00:05.114
using a Maclaurin polynominal,
what is the least degree

00:00:05.114 --> 00:00:09.281
of the polynominal that assures
an error smaller than 0.001?

00:00:11.371 --> 00:00:13.551
So what are we talking about here?

00:00:13.551 --> 00:00:16.255
Well, we could take some function

00:00:16.255 --> 00:00:19.701
and estimate it with an nth
degree Maclaurin polynomial,

00:00:19.701 --> 00:00:21.543
in fact, we could talk more generally

00:00:21.543 --> 00:00:23.230
about a Taylor polynominal,
but let's just say

00:00:23.230 --> 00:00:25.950
this is an nth degree
Maclaurin polynomial,

00:00:25.950 --> 00:00:28.970
but this isn't going to be
a perfect approximation,

00:00:28.970 --> 00:00:31.556
there's going to be some
error or some remainder.

00:00:31.556 --> 00:00:33.978
And so we could call this the remainder

00:00:33.978 --> 00:00:37.263
of that nth degree Maclaurin polynomial,

00:00:37.263 --> 00:00:41.032
and it's going to be
dependent for any given x.

00:00:41.032 --> 00:00:43.010
Now if we want to use
the specifics of this

00:00:43.010 --> 00:00:45.180
exact problem, we could
phrase it this way.

00:00:45.180 --> 00:00:49.989
We want to say, look, if
we're taking the sine of 0.4

00:00:49.989 --> 00:00:53.033
this is going to be
equal to our Maclaurin,

00:00:53.033 --> 00:00:57.579
our nth degree Maclaurin
polynomial evaluated at 0.4

00:00:57.579 --> 00:00:59.701
plus whatever the remainder is

00:00:59.701 --> 00:01:03.887
for that nth degree Maclaurin
polynomial evaluated at 0.4,

00:01:03.887 --> 00:01:07.445
and what we really want to
do is figure out for what n,

00:01:07.445 --> 00:01:09.760
what is the least degree
of the polynomial?

00:01:09.760 --> 00:01:10.677
So what is,

00:01:13.027 --> 00:01:15.289
let me do this in a different color.

00:01:15.289 --> 00:01:20.078
So we want to figure out
what is the smallest n,

00:01:20.078 --> 00:01:21.911
what is the smallest n

00:01:25.578 --> 00:01:27.495
such that the remainder

00:01:31.640 --> 00:01:36.636
of our nth degree Maclaurin
polynomial evaluated at 0.4

00:01:36.636 --> 00:01:40.386
is less than this number,
is less than 0.001?

00:01:43.040 --> 00:01:46.643
So this is just another way
of rephrasing the problem.

00:01:46.643 --> 00:01:49.619
And the way that we
can do it is we can use

00:01:49.619 --> 00:01:51.907
something called the Lagrange error round

00:01:51.907 --> 00:01:53.716
and we have other videos that prove it,

00:01:53.716 --> 00:01:57.616
this is often also called
Taylor's Remainder Theorem.

00:01:57.616 --> 00:01:59.447
And I'll first write it out and I'll try

00:01:59.447 --> 00:02:01.114
to explain it while I write
it out, but it'll actually

00:02:01.114 --> 00:02:03.931
become a lot more concrete
when we work it out.

00:02:03.931 --> 00:02:06.826
So Taylor's Remainder Theorem tells us,

00:02:06.826 --> 00:02:09.360
or Legrange error bound tells us that

00:02:09.360 --> 00:02:13.715
if the n plus oneth
derivative of our function,

00:02:13.715 --> 00:02:17.650
so f, so this is our n
plus oneth derivative

00:02:17.650 --> 00:02:20.349
of our function, if the
absolute value of that

00:02:20.349 --> 00:02:23.016
is less than or equal to some M

00:02:23.016 --> 00:02:24.766
for an open interval,

00:02:28.286 --> 00:02:30.369
open interval, containing

00:02:33.240 --> 00:02:35.990
where our polynomial is centered,

00:02:37.417 --> 00:02:41.094
in this case it's zero, we're
gonna use the Maclaurin case,

00:02:41.094 --> 00:02:44.844
so it's containing zero, and the x,

00:02:44.844 --> 00:02:47.903
or zero and x, the x that we care about

00:02:47.903 --> 00:02:49.907
in this particular video is 0.4,

00:02:49.907 --> 00:02:51.907
but I'll say generally for any x,

00:02:51.907 --> 00:02:54.760
so if this is true, if our
n plus oneth derivative

00:02:54.760 --> 00:02:56.701
of our function, if the
absolute value of it

00:02:56.701 --> 00:02:59.420
is less than or equal to
M, over an open interval

00:02:59.420 --> 00:03:01.399
containing where we're centered,

00:03:01.399 --> 00:03:03.934
this would be C if we're
talking about the general case,

00:03:03.934 --> 00:03:07.515
and x, so this x right over here, then,

00:03:07.515 --> 00:03:10.535
and this is the Legrange, this
is the part that's useful,

00:03:10.535 --> 00:03:13.040
we can say that the remainder is bounded,

00:03:13.040 --> 00:03:15.413
the remainder for that
nth degree polynomial.

00:03:15.413 --> 00:03:17.220
So this is the n plus oneth derivative,

00:03:17.220 --> 00:03:19.275
that's bounded, then we
can say the remainder

00:03:19.275 --> 00:03:23.361
for the nth degree polynomial
that approximates our function

00:03:23.361 --> 00:03:27.173
is going to be less
than or equal to that M

00:03:27.173 --> 00:03:29.590
times x to the n plus one,

00:03:34.359 --> 00:03:36.526
over n plus one factorial.

00:03:39.169 --> 00:03:42.860
So how do we apply that to
this particular problem?

00:03:42.860 --> 00:03:45.645
Well think about the derivatives of sine,

00:03:45.645 --> 00:03:47.833
we know that the absolute value of sine

00:03:47.833 --> 00:03:51.339
is less than or equal to one,
its derivative is cosine of x,

00:03:51.339 --> 00:03:53.992
the absolute value of that
is going to be bounded,

00:03:53.992 --> 00:03:56.312
is going to be less than or equal to one,

00:03:56.312 --> 00:03:58.680
so no matter how many times
we take the derivative

00:03:58.680 --> 00:04:01.196
of sine of x, the absolute
value of that derivative

00:04:01.196 --> 00:04:03.836
is going to be less than or equal to one.

00:04:03.836 --> 00:04:08.777
So we could write generally
that for this particular f of x,

00:04:08.777 --> 00:04:12.223
so let's say, for this particular
f of x right over here,

00:04:12.223 --> 00:04:15.249
we could say that the absolute value

00:04:15.249 --> 00:04:19.249
of the n plus oneth
derivative, evaluated any x,

00:04:20.755 --> 00:04:24.172
is going to be less than or equal to one.

00:04:25.078 --> 00:04:28.167
And this is the case for where f is sine,

00:04:28.167 --> 00:04:30.391
where f is sine of x, and this is actually

00:04:30.391 --> 00:04:32.482
going to be true over any interval,

00:04:32.482 --> 00:04:34.234
it doesn't even have to be over some type

00:04:34.234 --> 00:04:36.374
of a restricted interval
where we can do this.

00:04:36.374 --> 00:04:38.874
So we know that this is our M,

00:04:41.549 --> 00:04:44.169
that sine and its
derivatives are all bounded,

00:04:44.169 --> 00:04:47.527
or their absolute values
are bounded by one.

00:04:47.527 --> 00:04:50.170
And so them we have our M and we can apply

00:04:50.170 --> 00:04:53.258
the Legrange error bound, so we can say

00:04:53.258 --> 00:04:57.425
that the remainder of our nth
degree Maclaurin approximation

00:05:00.066 --> 00:05:04.233
at 0.4, so our x in this
particular case is 0.4,

00:05:05.422 --> 00:05:08.361
we don't have to do it
generally for any x here,

00:05:08.361 --> 00:05:11.096
is going to be less than or equal to,

00:05:11.096 --> 00:05:13.529
our M is one, so I won't
even write that down,

00:05:13.529 --> 00:05:16.779
our x is 0.4, 0.4 to the n plus one,

00:05:20.245 --> 00:05:23.995
to the n plus one, over
n plus one factorial.

00:05:25.544 --> 00:05:27.921
And we're taking the absolute
value of this whole thing.

00:05:27.921 --> 00:05:30.237
This is Legrange error bound,
and we want to figure out,

00:05:30.237 --> 00:05:32.219
if we can figure out
a situation where this

00:05:32.219 --> 00:05:36.213
is less than 0.001, then this for sure

00:05:36.213 --> 00:05:40.435
is going to be less than
0.001, because the remainder

00:05:40.435 --> 00:05:42.427
is less than this, or less than or equal

00:05:42.427 --> 00:05:44.341
to this, which is less than that.

00:05:44.341 --> 00:05:46.221
So how do we do that, how do we figure out

00:05:46.221 --> 00:05:48.823
the smallest n where
this is going to be true?

00:05:48.823 --> 00:05:51.919
Well, we can just try out
some ns and keep increasing

00:05:51.919 --> 00:05:55.415
until this thing actually
becomes smaller than that thing.

00:05:55.415 --> 00:05:59.598
So let's do it, all right,
I'm gonna set up a table here.

00:05:59.598 --> 00:06:02.515
So let me do it relatively cleanly.

00:06:04.445 --> 00:06:07.578
So let's do, this is going to be our n,

00:06:07.578 --> 00:06:09.995
and then this is going to be,

00:06:10.974 --> 00:06:14.474
this is going to be 0.4 to the n plus one,

00:06:15.704 --> 00:06:17.871
over n plus one factorial,

00:06:19.643 --> 00:06:22.511
so let's try it when n is equal to one.

00:06:22.511 --> 00:06:24.784
Well then this is going
to be 0.4 to the two,

00:06:24.784 --> 00:06:28.034
so it's 0.4 squared over two factorial,

00:06:29.341 --> 00:06:31.174
this is 0.16 over two,

00:06:33.889 --> 00:06:36.889
which is equal to 0.08, that's definitely

00:06:36.889 --> 00:06:39.746
not less than one thousandth here.

00:06:39.746 --> 00:06:42.565
So let's try n equals
two, when n equals two,

00:06:42.565 --> 00:06:45.418
it's gonna be 0.4 to the third power,

00:06:45.418 --> 00:06:48.857
over three factorial, and that's equal to,

00:06:48.857 --> 00:06:51.010
what is that, zero point,

00:06:51.010 --> 00:06:52.703
I'm gonna need three
digits behind the decimal,

00:06:52.703 --> 00:06:56.620
0.064 over six, well,
this is a little bit more

00:06:57.496 --> 00:07:01.079
than 0.01, so our n
isn't large enough yet.

00:07:02.933 --> 00:07:06.494
So let's try three, so
this is going to be 0.04

00:07:06.494 --> 00:07:10.123
to the three plus one,
so that's going to be

00:07:10.123 --> 00:07:13.540
to the fourth power, over four factorial,

00:07:14.757 --> 00:07:18.400
and let's see, that is
going to be equal to,

00:07:18.400 --> 00:07:21.733
this is going to be, let's
see, we're gonna have

00:07:21.733 --> 00:07:25.150
four digits behind the decimal, so 0.0256

00:07:26.587 --> 00:07:29.670
over 24, this is, we're almost there,

00:07:31.596 --> 00:07:35.628
this is a little bit, this is
going to be a little bit more

00:07:35.628 --> 00:07:39.628
than 0.001, so that doesn't
do the trick for us.

00:07:40.801 --> 00:07:44.093
So I'm guessing already that n equals four

00:07:44.093 --> 00:07:46.702
is gonna do the trick,
but let's verify that.

00:07:46.702 --> 00:07:50.869
So this is going to be
0.04, or 0.4, I should say,

00:07:52.044 --> 00:07:55.377
to the fifth power, over five factorial,

00:07:56.294 --> 00:07:58.064
and what is this equal to?

00:07:58.064 --> 00:08:00.454
Let's see, four to the fifth is 1024,

00:08:00.454 --> 00:08:03.318
I'm gonna have five
numbers behind the decimal

00:08:03.318 --> 00:08:07.308
and I'm gonna divide it by
five factorial, which is 120.

00:08:07.308 --> 00:08:10.309
And let's see, this right over here, yes,

00:08:10.309 --> 00:08:13.059
this for sure is less than 0.001,

00:08:14.798 --> 00:08:19.153
this is definitely less than
a thousandth right over here.

00:08:19.153 --> 00:08:22.328
So we see that when n is equal to four,

00:08:22.328 --> 00:08:26.747
so we can say that the
remainder for our fourth degree

00:08:26.747 --> 00:08:31.570
polynomial, fourth degree
Maclaurin polynomial, evaluated

00:08:31.570 --> 00:08:35.737
at x equals 0.4 is for sure
going to be less than 0.001.

00:08:39.723 --> 00:08:42.806
So there you go, that is the
least degree of the polynomial

00:08:42.806 --> 00:08:45.858
that assures an error
smaller than one thousandth.


Worked example: limit comparison test

WEBVTT
Kind: captions
Language: en

00:00:00.216 --> 00:00:01.465
- [Instructor] So we're
given a series here,

00:00:01.465 --> 00:00:03.026
and they say, "What series should we use

00:00:03.026 --> 00:00:04.994
"in the limit comparison test?"

00:00:04.994 --> 00:00:07.379
Let me underline that,
"The limit comparison test,

00:00:07.379 --> 00:00:10.559
"in order to determine
whether S converges?"

00:00:10.559 --> 00:00:12.088
So let's just remind ourselves

00:00:12.088 --> 00:00:13.979
about the limit comparison test.

00:00:13.979 --> 00:00:18.490
If we say, if we say
that we have two series,

00:00:18.490 --> 00:00:21.429
and I'll just use this notation, a sub-n,

00:00:21.429 --> 00:00:24.179
and then another series, b sub-n.

00:00:25.107 --> 00:00:29.953
And we know that a sub-n
and b sub-n are greater than

00:00:29.953 --> 00:00:34.120
or equal to zero for all n,
for all n, if we know this,

00:00:36.985 --> 00:00:41.152
then if, then if the limit
as n approaches infinity of

00:00:45.154 --> 00:00:49.321
a sub-n over b sub-n is equal
to some positive constant.

00:00:51.119 --> 00:00:55.110
So zero is less than that
constant is less than infinity,

00:00:55.110 --> 00:00:58.943
then either both converge,
then both converge,

00:01:01.263 --> 00:01:03.680
both converge, both converge,

00:01:06.411 --> 00:01:08.911
or both diverge, both diverge.

00:01:11.497 --> 00:01:13.316
And it really makes a lot of
sense, because it's saying,

00:01:13.316 --> 00:01:16.555
look, as we get into our
really large values of n,

00:01:16.555 --> 00:01:19.230
as we go really far out
there in terms of the terms,

00:01:19.230 --> 00:01:22.759
if our behavior starts to look
the same, then it makes sense

00:01:22.759 --> 00:01:25.310
that both these series
would converge or diverge,

00:01:25.310 --> 00:01:27.262
and we have an introductory
vizio, (chuckles)

00:01:27.262 --> 00:01:30.515
we have an introductory video
on this in another video.

00:01:30.515 --> 00:01:33.081
So let's think about what,
if we say that this is our

00:01:33.081 --> 00:01:37.582
a sub-n, if we say that this
is a sub-n right over here,

00:01:37.582 --> 00:01:40.604
what is a series that we
can really compare to?

00:01:40.604 --> 00:01:45.119
That seems to have the same
behavior as n gets really large?

00:01:45.119 --> 00:01:47.603
Well this one just
seems to gets unbounded.

00:01:47.603 --> 00:01:49.581
This one doesn't look that similar.

00:01:49.581 --> 00:01:51.814
Has a three to the n minus
one in the denominator,

00:01:51.814 --> 00:01:54.716
but the numerator doesn't behave the same.

00:01:54.716 --> 00:01:56.541
This one over here is interesting

00:01:56.541 --> 00:01:58.316
because we could write this,

00:01:58.316 --> 00:02:02.433
the is the same as a sum
n equals one to infinity.

00:02:02.433 --> 00:02:06.426
We could write this is two
to the n over three to the n,

00:02:06.426 --> 00:02:07.689
and these are very similar.

00:02:07.689 --> 00:02:10.368
The only difference between
this and this is that

00:02:10.368 --> 00:02:12.705
in the denominator here, or
in the denominator up here

00:02:12.705 --> 00:02:14.841
we have a minus one, and down here

00:02:14.841 --> 00:02:16.738
we don't have that minus one,

00:02:16.738 --> 00:02:19.054
and so it makes sense given
that that's just a constant,

00:02:19.054 --> 00:02:23.063
that as n gets very large, that
these might behave the same.

00:02:23.063 --> 00:02:24.202
So let's try it out.

00:02:24.202 --> 00:02:25.952
Let's find the limit,

00:02:27.984 --> 00:02:30.705
and we also know that the a
sub-n's and the b sub-n's,

00:02:30.705 --> 00:02:33.275
if we say that this over here is b sub-n,

00:02:33.275 --> 00:02:37.932
we say that's b sub-n, that
this is going to be positive,

00:02:37.932 --> 00:02:39.997
or this is going to be
greater than or equal to zero

00:02:39.997 --> 00:02:42.779
for n equals one, two,
three, so for any values

00:02:42.779 --> 00:02:46.176
this is going to be greater
than or equal to zero,

00:02:46.176 --> 00:02:48.454
and the same thing right over here.

00:02:48.454 --> 00:02:50.119
It's gonna be greater
than or equal to zero

00:02:50.119 --> 00:02:51.999
for all of the n's that we care about.

00:02:51.999 --> 00:02:54.907
So we meet these first
constraints, and so let's find

00:02:54.907 --> 00:02:58.435
the limit as n approaches
infinity of a sub-n,

00:02:58.435 --> 00:03:00.654
which is, I'll write in that red color,

00:03:00.654 --> 00:03:04.821
which is two to the nth power
over three to the n minus one

00:03:07.178 --> 00:03:11.345
over b sub-n, over two to the
nth over three to the nth.

00:03:14.070 --> 00:03:15.855
So let me actually do a little

00:03:15.855 --> 00:03:17.539
algebraic manipulation right over here.

00:03:17.539 --> 00:03:19.735
This is going to to be the same thing as

00:03:19.735 --> 00:03:23.485
two to the nth over
three to the n minus one,

00:03:24.416 --> 00:03:27.499
times three to the n, over two the n.

00:03:29.910 --> 00:03:31.646
Divide the numerator and the denominators

00:03:31.646 --> 00:03:34.034
by two to the n, those cancel out,

00:03:34.034 --> 00:03:38.438
and so this will give
us, this will give us

00:03:38.438 --> 00:03:42.506
three to the n over
three to the n minus one.

00:03:42.506 --> 00:03:44.366
We can divide the numerator
and the denominator

00:03:44.366 --> 00:03:46.167
by three to the n, and that'll give us

00:03:46.167 --> 00:03:49.750
one over one minus one
over three to the n.

00:03:50.951 --> 00:03:53.896
So we could say this is
the same thing as the limit

00:03:53.896 --> 00:03:58.063
as n approaches infinity
of one over one minus one,

00:03:59.824 --> 00:04:01.537
over three to the n.

00:04:01.537 --> 00:04:03.360
Well what's this going to be equal to?

00:04:03.360 --> 00:04:05.149
Well as that approaches
infinity, this thing,

00:04:05.149 --> 00:04:07.416
one over three to the n,
that's just gonna go to zero.

00:04:07.416 --> 00:04:10.994
So this is, this whole thing
is just going to approach one.

00:04:10.994 --> 00:04:14.330
And one is clearly
between zero and infinity,

00:04:14.330 --> 00:04:17.003
so the destinies of these
two series are tied.

00:04:17.003 --> 00:04:21.286
They either both converge
or they both diverge.

00:04:21.286 --> 00:04:25.315
So this is a good one to use
a limit comparison test with,

00:04:25.315 --> 00:04:26.177
and so let's think about it.

00:04:26.177 --> 00:04:29.809
Do they either both converge,
or do they both diverge?

00:04:29.809 --> 00:04:33.541
Well this is a geometric
series, our common ratio here

00:04:33.541 --> 00:04:37.336
is less than one, so this
is going to converge,

00:04:37.336 --> 00:04:38.750
this is going to converge.

00:04:38.750 --> 00:04:40.217
And because this one converges, by the

00:04:40.217 --> 00:04:45.031
limit comparison test, our
original series S converges.

00:04:45.031 --> 00:04:47.281
Converges, and we are done.


Finding function from power series by integrating

WEBVTT
Kind: captions
Language: en

00:00:00.266 --> 00:00:02.532
- [Instructor] We know that
for x in the open interval

00:00:02.532 --> 00:00:04.561
from negative 1/2 to 1/2,

00:00:04.561 --> 00:00:06.449
that negative two over one minus two x

00:00:06.449 --> 00:00:08.630
is equal to this series.

00:00:08.630 --> 00:00:11.374
And it says using this
fact, find the function

00:00:11.374 --> 00:00:14.261
that corresponds to the following series.

00:00:14.261 --> 00:00:15.471
And like always, pause this video

00:00:15.471 --> 00:00:17.514
and see if you can work through it.

00:00:17.514 --> 00:00:19.532
All right, so the first
thing you might say is well

00:00:19.532 --> 00:00:23.288
how do we know that this
expression is equal to this series?

00:00:23.288 --> 00:00:26.504
And you might recognize this
series as a geometric series

00:00:26.504 --> 00:00:28.945
where the first term is negative two

00:00:28.945 --> 00:00:31.867
and then the common ratio
to get each successive term,

00:00:31.867 --> 00:00:36.034
we're multiplying by two x,
we're multiplying by two x.

00:00:37.771 --> 00:00:39.316
And so for a geometric series like this,

00:00:39.316 --> 00:00:41.506
the sum is gonna be the first term,

00:00:41.506 --> 00:00:44.716
negative two over one
minus the common ratio,

00:00:44.716 --> 00:00:46.884
which is exactly what we have over there.

00:00:46.884 --> 00:00:48.269
So that's why we know that.

00:00:48.269 --> 00:00:49.252
And this right over here,

00:00:49.252 --> 00:00:51.172
this gives us our radius of convergence,

00:00:51.172 --> 00:00:55.788
the x values for which this
thing will actually converge.

00:00:55.788 --> 00:00:57.716
But now that we feel good
about this first statement,

00:00:57.716 --> 00:01:00.709
let's try to answer their actual question.

00:01:00.709 --> 00:01:02.753
So we wanna find the function

00:01:02.753 --> 00:01:05.001
that corresponds to the following series.

00:01:05.001 --> 00:01:08.858
So my instinct is to say, well
how does this series relate

00:01:08.858 --> 00:01:10.645
to the series they gave us?

00:01:10.645 --> 00:01:12.921
Let's see, this first
term is a negative two,

00:01:12.921 --> 00:01:15.402
this first term is a negative two x,

00:01:15.402 --> 00:01:18.562
this is a negative four x, this
is a negative two x squared.

00:01:18.562 --> 00:01:22.343
So the thing that might
jump out at you is that

00:01:22.343 --> 00:01:27.220
this second series they gave
us is the antiderivative

00:01:27.220 --> 00:01:28.066
of this first one.

00:01:28.066 --> 00:01:30.077
Or we could say this first
series is a derivative

00:01:30.077 --> 00:01:31.015
of the second one.

00:01:31.015 --> 00:01:33.645
What's the derivative of
negative two x with respect to x?

00:01:33.645 --> 00:01:35.061
Well, it's negative two.

00:01:35.061 --> 00:01:36.983
What's the derivative of
negative two x squared

00:01:36.983 --> 00:01:38.310
with respect to x?

00:01:38.310 --> 00:01:41.941
Well, it's negative four
x and so on and so forth.

00:01:41.941 --> 00:01:46.072
And so if we were to
call this, let me call,

00:01:46.072 --> 00:01:48.597
let me call this thing right over here,

00:01:48.597 --> 00:01:53.439
actually, let me just say
that this is equal to g of x.

00:01:53.439 --> 00:01:55.454
Well, the way I can take
this right hand side

00:01:55.454 --> 00:01:58.682
and get g of x is by
taking the antiderivative,

00:01:58.682 --> 00:02:00.256
taking the indefinite integral.

00:02:00.256 --> 00:02:02.559
And so what we can do, we can take

00:02:02.559 --> 00:02:06.059
the indefinite integral of both sides, dx.

00:02:07.375 --> 00:02:08.208
So dx.

00:02:09.267 --> 00:02:13.582
On the right-hand side
I'm gonna get g of x.

00:02:13.582 --> 00:02:17.221
And on the left-hand
side, I am going to get,

00:02:17.221 --> 00:02:20.383
actually let me write it this way.

00:02:20.383 --> 00:02:22.113
So on the left-hand side,
I'll just rewrite it,

00:02:22.113 --> 00:02:24.563
I'm gonna get the indefinite integral of,

00:02:24.563 --> 00:02:27.230
I'll write it as negative two dx

00:02:28.341 --> 00:02:30.091
over one minus two x.

00:02:31.478 --> 00:02:32.764
And that's just another way of writing

00:02:32.764 --> 00:02:34.585
what I have on the left-hand side here.

00:02:34.585 --> 00:02:37.531
This is equal to g of x, right?

00:02:37.531 --> 00:02:39.062
If I take the antiderivative of this

00:02:39.062 --> 00:02:40.601
or an antiderivative of this,

00:02:40.601 --> 00:02:44.629
is this g of x, where the
constant would've been zero.

00:02:44.629 --> 00:02:47.212
So I'll say is equal to g of x.

00:02:49.330 --> 00:02:51.457
And so the key here is well,
what's the indefinite integral

00:02:51.457 --> 00:02:54.927
of this stuff, and you
might immediately recognize

00:02:54.927 --> 00:02:57.430
what I have here in the bottom,

00:02:57.430 --> 00:02:59.742
I have its derivative here on the top.

00:02:59.742 --> 00:03:01.905
If I consider this to be u.

00:03:01.905 --> 00:03:04.745
If I say u is equal to one minus two x,

00:03:04.745 --> 00:03:06.346
this is just u substitution,

00:03:06.346 --> 00:03:08.952
then du is equal to the derivative of this

00:03:08.952 --> 00:03:12.641
with respect to x which
is negative two dx.

00:03:12.641 --> 00:03:15.352
And so I have the du right over here.

00:03:15.352 --> 00:03:17.096
So let me rewrite all of this.

00:03:17.096 --> 00:03:21.013
I can rewrite this as
the integral of du over u

00:03:23.990 --> 00:03:25.573
is equal to g of x,

00:03:27.050 --> 00:03:30.550
and this we can rewrite as the natural log

00:03:31.999 --> 00:03:34.749
of the absolute value of u plus c

00:03:35.969 --> 00:03:37.552
is equal to g of x.

00:03:38.670 --> 00:03:41.591
And then we can undo our u substitution,

00:03:41.591 --> 00:03:45.033
and so for u I'll undo,
I will substitute back

00:03:45.033 --> 00:03:46.722
the one minus two x.

00:03:46.722 --> 00:03:49.424
So I can write the natural log of

00:03:49.424 --> 00:03:53.591
the absolute value of one
minus two x is equal to

00:03:56.212 --> 00:03:58.369
plus c, I don't wanna for get that,

00:03:58.369 --> 00:04:00.536
plus c is equal to g of x.

00:04:03.275 --> 00:04:06.107
And so the next thing we wanna
do is well, what is our c?

00:04:06.107 --> 00:04:07.974
And the easiest way to figure out c is

00:04:07.974 --> 00:04:10.307
let's substitute zero for x.

00:04:11.260 --> 00:04:13.666
And so let's think
about this a little bit.

00:04:13.666 --> 00:04:17.565
So if I put a zero
here, if I put x equals,

00:04:17.565 --> 00:04:18.666
actually let me just write it again.

00:04:18.666 --> 00:04:21.805
So I have the natural
log, if I say x is zero,

00:04:21.805 --> 00:04:23.411
this is gonna be the natural log

00:04:23.411 --> 00:04:26.284
of the absolute value of one plus c

00:04:26.284 --> 00:04:28.496
is equal to g of zero.

00:04:28.496 --> 00:04:30.158
So what's g of zero?

00:04:30.158 --> 00:04:32.872
G of zero, everyone of these
terms is equal to zero.

00:04:32.872 --> 00:04:35.157
G of zero is equal to zero.

00:04:35.157 --> 00:04:36.931
So is equal to zero.

00:04:36.931 --> 00:04:39.879
Well, natural log of one is just zero.

00:04:39.879 --> 00:04:43.474
So zero plus c is equal to
zero, c is equal to zero.

00:04:43.474 --> 00:04:44.853
So there you have it.

00:04:44.853 --> 00:04:47.032
I just took the
antiderivative of both sides

00:04:47.032 --> 00:04:48.911
of this equation right over here.

00:04:48.911 --> 00:04:51.890
I figured out, when I
substituted zero for x,

00:04:51.890 --> 00:04:55.097
then okay, my constant
here is going to be zero,

00:04:55.097 --> 00:04:58.350
and I get that, I get that this series,

00:04:58.350 --> 00:05:00.517
this g of x is equal to

00:05:01.631 --> 00:05:05.127
the natural log of the absolute
value of one minus two x,

00:05:05.127 --> 00:05:06.961
is equal to the natural
log of the absolute value

00:05:06.961 --> 00:05:08.895
of one minus two x.

00:05:08.895 --> 00:05:11.232
Now, if we keep this restriction

00:05:11.232 --> 00:05:12.707
that we're in this open interval,

00:05:12.707 --> 00:05:14.330
then this is always going to be positive.

00:05:14.330 --> 00:05:17.183
And we wouldn't have to
write the absolute value,

00:05:17.183 --> 00:05:21.154
but we could be safe by
writing the absolute value.

00:05:21.154 --> 00:05:22.021
But there you go.

00:05:22.021 --> 00:05:24.157
Using the fact above,
we found the function

00:05:24.157 --> 00:05:25.977
that corresponds to this following series.

00:05:25.977 --> 00:05:28.238
And I guess you could
say that the trick of it

00:05:28.238 --> 00:05:30.446
was recognizing that

00:05:30.446 --> 00:05:34.113
the series up here is a derivative
of the series down here.

00:05:34.113 --> 00:05:36.775
And so this is going to be the derivative

00:05:36.775 --> 00:05:38.775
of the following series,

00:05:40.879 --> 00:05:42.838
and so we just took the
antiderivative of both sides,

00:05:42.838 --> 00:05:45.857
or the indefinite integral of both sides.


Function as a geometric series

WEBVTT
Kind: captions
Language: en

00:00:00.040 --> 00:00:02.951
- [Instructor] We're asked
to find a power series for f,

00:00:02.951 --> 00:00:05.656
and they've given us f of
x is equal to six over one

00:00:05.656 --> 00:00:08.630
plus x to the third power.

00:00:08.630 --> 00:00:11.988
Now, since they're letting
us pick which power series,

00:00:11.988 --> 00:00:14.639
you might say, well, let me
just find the Maclaurin series

00:00:14.639 --> 00:00:17.020
because the Maclaurin series
tends to be the simplest

00:00:17.020 --> 00:00:19.993
to find 'cause we're centered at zero.

00:00:19.993 --> 00:00:22.598
And so you might immediately
go out and say all right,

00:00:22.598 --> 00:00:24.340
well, let me evaluate
this function at zero,

00:00:24.340 --> 00:00:25.775
evaluate its derivative at zero,

00:00:25.775 --> 00:00:28.523
its second derivative at
zero, so on and so forth.

00:00:28.523 --> 00:00:30.476
And then I can use the formula

00:00:30.476 --> 00:00:32.901
for the Maclaurin series
to just expand it out.

00:00:32.901 --> 00:00:35.794
But very quickly, you
will run into roadblocks

00:00:35.794 --> 00:00:37.656
because the first,

00:00:37.656 --> 00:00:39.004
evaluating this at f,

00:00:39.004 --> 00:00:40.778
at x equals zero is
pretty straightforward.

00:00:40.778 --> 00:00:42.915
Evaluating the first derivative
is pretty straightforward.

00:00:42.915 --> 00:00:44.118
But then once you start taking the second

00:00:44.118 --> 00:00:47.344
and third derivatives, it
gets very hairy, very fast.

00:00:47.344 --> 00:00:49.878
You could do a simplification,
where you could say,

00:00:49.878 --> 00:00:52.065
well, let me find the Maclaurin series

00:00:52.065 --> 00:00:55.648
for f of u is equal to
six over one plus u,

00:00:56.804 --> 00:00:59.775
where u is equal to x to the third.

00:00:59.775 --> 00:01:02.625
So you find this Maclaurin
expansion, in terms of u,

00:01:02.625 --> 00:01:04.742
and then you substitute
for x to the third.

00:01:04.742 --> 00:01:06.624
And actually, that makes
it a good bit simpler,

00:01:06.624 --> 00:01:08.699
so that is another way to approach it.

00:01:08.699 --> 00:01:11.101
But the simplest way to
approach it is to say,

00:01:11.101 --> 00:01:13.979
hey, you know what, this
form right over here,

00:01:13.979 --> 00:01:16.796
this rational expression,
it looks similar.

00:01:16.796 --> 00:01:20.282
It looks like the sum
of a geometric series.

00:01:20.282 --> 00:01:21.985
Let's just remind ourselves what the sum

00:01:21.985 --> 00:01:23.908
of a geometric series looks like.

00:01:23.908 --> 00:01:25.241
If I have a plus

00:01:26.356 --> 00:01:30.837
a times r, so a is my first
term, r is my common ratio,

00:01:30.837 --> 00:01:32.744
plus, I'm gonna multiply it times r again,

00:01:32.744 --> 00:01:34.661
plus a times r squared,

00:01:36.748 --> 00:01:39.412
plus a times r to the third power,

00:01:39.412 --> 00:01:42.699
and I keep going on,
and on, and on forever.

00:01:42.699 --> 00:01:46.602
We know that this is going to
be equal to a, our first term,

00:01:46.602 --> 00:01:48.883
over one minus our common ratio,

00:01:48.883 --> 00:01:53.604
and this just comes from a,
the sum of a geometric series.

00:01:53.604 --> 00:01:55.861
And notice that what we
have here, our f of x,

00:01:55.861 --> 00:01:57.584
our definition of f of x,

00:01:57.584 --> 00:02:01.049
and the sum of a geometric
series look very, very similar.

00:02:01.049 --> 00:02:04.761
If we say that this,
right over here, is a,

00:02:04.761 --> 00:02:06.873
so a is equal to six.

00:02:06.873 --> 00:02:11.043
And if negative r is
equal to x to the third,

00:02:11.043 --> 00:02:13.181
or we could say, let me rewrite this.

00:02:13.181 --> 00:02:15.785
I could write this
denominator as one minus

00:02:15.785 --> 00:02:18.403
negative x to the third.

00:02:18.403 --> 00:02:21.579
And so now, you could say, okay, well, r

00:02:21.579 --> 00:02:24.785
could be equal to negative x to the third.

00:02:24.785 --> 00:02:26.828
And just like that, we can expand it out.

00:02:26.828 --> 00:02:29.078
Well, if a is equal to six,

00:02:30.015 --> 00:02:32.644
and r is equal to negative
x to the third, well,

00:02:32.644 --> 00:02:35.360
then we could just write this
out as a geometric series,

00:02:35.360 --> 00:02:36.775
which is very straightforward.

00:02:36.775 --> 00:02:38.093
So let's do that.

00:02:38.093 --> 00:02:41.553
And I will do this in, I'll do
this in this nice pink color.

00:02:41.553 --> 00:02:44.545
So the first term would be six,

00:02:44.545 --> 00:02:47.603
plus six times our common ratio,

00:02:47.603 --> 00:02:50.875
six times negative x to the third.

00:02:50.875 --> 00:02:52.744
And so actually, let me just write that as

00:02:52.744 --> 00:02:56.142
negative six x to the third,
and then we're gonna multiply

00:02:56.142 --> 00:02:58.526
by negative x to the third again.

00:02:58.526 --> 00:03:00.666
So that's going to be, if I multiply this

00:03:00.666 --> 00:03:02.359
times negative x to the third,

00:03:02.359 --> 00:03:04.542
that's gonna be positive six

00:03:04.542 --> 00:03:07.042
times x to the sixth power.

00:03:08.263 --> 00:03:10.127
And then I'm gonna multiply
it by times negative x

00:03:10.127 --> 00:03:14.995
to the third again, so it's
gonna be minus six times x

00:03:14.995 --> 00:03:16.578
to the ninth power,

00:03:17.767 --> 00:03:19.886
and I'm gonna go on, and on, and on.

00:03:19.886 --> 00:03:21.660
So it's gonna be, and
then I could keep going,

00:03:21.660 --> 00:03:24.207
I multiply it times
negative x to the third.

00:03:24.207 --> 00:03:28.210
I will get six x to the twelfth power.

00:03:28.210 --> 00:03:32.099
Now, we can go on, and on,
and on, and on forever.

00:03:32.099 --> 00:03:33.826
And so the key here was,

00:03:33.826 --> 00:03:37.914
and this is the Maclaurin
series expansion for our f of x,

00:03:37.914 --> 00:03:39.809
but the key is to not have to go through

00:03:39.809 --> 00:03:41.618
all of this business and just to recognize

00:03:41.618 --> 00:03:43.923
that hey, the way this
function was defined

00:03:43.923 --> 00:03:47.350
is it looks a lot like the
sum of a geometric series.

00:03:47.350 --> 00:03:49.732
And it can be considered the
sum of a geometric series,

00:03:49.732 --> 00:03:54.324
and we can use that to find
the power series expansion

00:03:54.324 --> 00:03:55.306
for our function.

00:03:55.306 --> 00:03:58.639
This is a very, very, very useful trick.


Support Khan Academy and double your impact

WEBVTT
Kind: captions
Language: en

00:00:00.480 --> 00:00:03.012
- Hi, I'm Sal Khan, founder
of the Khan Academy,

00:00:03.012 --> 00:00:06.005
and as you might know, we are
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00:00:06.005 --> 00:00:08.914
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00:00:08.914 --> 00:00:10.825
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00:00:10.825 --> 00:00:12.484
And the way that we're able to do that

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Trick involving Maclaurin expansion of cosx

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:02.095
- [Instructor] What are the
first three non-zero terms

00:00:02.095 --> 00:00:03.746
of the Maclaurin series for the function

00:00:03.746 --> 00:00:06.993
f of x is equal to x cosine of x?

00:00:06.993 --> 00:00:09.347
So, one thing that you're
immediately gonna find,

00:00:09.347 --> 00:00:10.180
let's just remind ourselves

00:00:10.180 --> 00:00:12.210
what a Maclaurin series looks like.

00:00:12.210 --> 00:00:15.872
So, our f of x can be
approximated by the polynomial,

00:00:15.872 --> 00:00:18.218
we're gonna evaluate f at zero, that's why

00:00:18.218 --> 00:00:20.342
it's a Maclaurin series, it's
gonna be centered at zero,

00:00:20.342 --> 00:00:22.342
it's a Taylor series centered at zero.

00:00:22.342 --> 00:00:23.342
Plus f prime

00:00:24.552 --> 00:00:25.802
of zero times x

00:00:27.184 --> 00:00:31.149
plus the second derivative
evaluated at zero

00:00:31.149 --> 00:00:33.684
times x squared over two

00:00:33.684 --> 00:00:36.745
plus the third derivative
evaluated at zero

00:00:36.745 --> 00:00:39.470
times x to the third over three factorial,

00:00:39.470 --> 00:00:41.527
we could view this as
two factorial as well,

00:00:41.527 --> 00:00:43.729
and we keep going on
and on and on forever.

00:00:43.729 --> 00:00:45.139
And, what you're
immediately going to find,

00:00:45.139 --> 00:00:47.715
is when you start taking
the derivative of f

00:00:47.715 --> 00:00:48.837
and then the second and the third

00:00:48.837 --> 00:00:50.401
and the fourth and the fifth derivative,

00:00:50.401 --> 00:00:51.708
you're actually gonna have to take

00:00:51.708 --> 00:00:52.837
a bunch of derivatives here.

00:00:52.837 --> 00:00:54.690
Every time you're going to
have to apply the product rule,

00:00:54.690 --> 00:00:57.372
this is going to get very
hairy, very, very fast.

00:00:57.372 --> 00:00:59.938
So, there's actually a
trick to this problem.

00:00:59.938 --> 00:01:02.238
What we wanna do, we
could find the Maclaurin,

00:01:02.238 --> 00:01:04.872
we can find the first three non-zero terms

00:01:04.872 --> 00:01:07.649
of the Maclaurin series for cosine of x,

00:01:07.649 --> 00:01:11.661
and then just multiply that by x, because,

00:01:11.661 --> 00:01:12.754
x cosine of x,

00:01:12.754 --> 00:01:15.767
you could view as the same
thing as x times the--

00:01:15.767 --> 00:01:17.547
or another way of saying it is:

00:01:17.547 --> 00:01:21.329
the first three terms of the
Maclaurin expansion for this

00:01:21.329 --> 00:01:24.447
is just going to be x
times the first three terms

00:01:24.447 --> 00:01:27.390
of the Maclaurin
expansion for cosine of x.

00:01:27.390 --> 00:01:29.086
So, let's do that.

00:01:29.086 --> 00:01:30.828
So, to do that, let's just
think about the various

00:01:30.828 --> 00:01:33.356
derivatives of cosine of x.

00:01:33.356 --> 00:01:34.848
So, if f of x,

00:01:34.848 --> 00:01:37.112
I'm gonna use a different
letter here for the function.

00:01:37.112 --> 00:01:38.862
If g of x is equal

00:01:39.858 --> 00:01:41.564
to cosine of x,

00:01:41.564 --> 00:01:42.731
then g of zero

00:01:44.399 --> 00:01:47.315
is going to be equal to one.

00:01:47.315 --> 00:01:49.283
And then the first derivative

00:01:49.283 --> 00:01:53.016
would be equal to negative sine of x,

00:01:53.016 --> 00:01:56.529
so the first derivative
evaluated at zero is zero

00:01:56.529 --> 00:01:58.193
sine of anything, or
negative sine of anyth--

00:01:58.193 --> 00:02:00.789
of zero I should say, is gonna be zero.

00:02:00.789 --> 00:02:03.338
And the the second derivative,

00:02:03.338 --> 00:02:05.095
well, derivative of sine is cosine

00:02:05.095 --> 00:02:05.948
so it's negative sine,

00:02:05.948 --> 00:02:08.853
so it's negative cosine of x.

00:02:08.853 --> 00:02:12.770
And so, the second
derivative evaluated at zero

00:02:14.141 --> 00:02:16.821
is going to be equal to negative one,

00:02:16.821 --> 00:02:18.749
you might start seeing a pattern here.

00:02:18.749 --> 00:02:20.832
The third derivative of x

00:02:23.265 --> 00:02:25.015
is equal to sine of x

00:02:26.745 --> 00:02:29.926
and so if I evaluate the
third derivative at zero

00:02:29.926 --> 00:02:31.386
I'm gonna get zero again.

00:02:31.386 --> 00:02:32.842
And then, when I take
the derivative of this,

00:02:32.842 --> 00:02:34.563
when I take the fourth derivative,

00:02:34.563 --> 00:02:36.682
I get back to cosine of
x, the fourth derivative

00:02:36.682 --> 00:02:38.430
is the same thing as the function.

00:02:38.430 --> 00:02:39.994
So, the fourth derivative
evaluated at zero,

00:02:39.994 --> 00:02:42.077
this is also going to be
the fourth derivative--

00:02:42.077 --> 00:02:44.414
I'll write it like
that--evaluated at zero.

00:02:44.414 --> 00:02:47.490
This is also going to
be the fifth derivative

00:02:47.490 --> 00:02:49.607
evaluated at zero, this cycles.

00:02:49.607 --> 00:02:51.931
This is going to be the sixth derivative

00:02:51.931 --> 00:02:53.567
evaluated at zero.

00:02:53.567 --> 00:02:55.960
This is going to be the seventh derivative

00:02:55.960 --> 00:02:57.787
evaluated at zero.

00:02:57.787 --> 00:03:01.551
And so, what are the first
three non-zero terms?

00:03:01.551 --> 00:03:03.882
Well, let's see, it's gonna be this one.

00:03:03.882 --> 00:03:05.828
It's going to be this one.

00:03:05.828 --> 00:03:08.495
And, it is going to be this one.

00:03:09.350 --> 00:03:12.267
And so, we're gonna have this term,

00:03:13.975 --> 00:03:15.993
or let me just do it for cosine of x.

00:03:15.993 --> 00:03:18.456
So, let me write this
down, so I'm gonna write--

00:03:18.456 --> 00:03:21.241
and you might already know
the first three non-zero

00:03:21.241 --> 00:03:23.286
terms of the Maclaurin
series for cosine of x,

00:03:23.286 --> 00:03:25.828
but I'm finding it here for us--

00:03:25.828 --> 00:03:28.511
so cosine of x is approximately equal to

00:03:28.511 --> 00:03:32.406
the first term here is
just cosine of zero,

00:03:32.406 --> 00:03:34.060
so that's just one.

00:03:34.060 --> 00:03:35.464
And then, the next one is going to be

00:03:35.464 --> 00:03:37.279
our second derivative,

00:03:37.279 --> 00:03:38.696
so it's minus one

00:03:40.206 --> 00:03:41.706
over two factorial

00:03:43.464 --> 00:03:46.774
times x squared--this
is the one that involves

00:03:46.774 --> 00:03:49.746
the second derivative--x squared.

00:03:49.746 --> 00:03:52.791
And then, we're going to
involve the fourth derivative,

00:03:52.791 --> 00:03:54.431
and the coefficient's one so it's gonna be

00:03:54.431 --> 00:03:55.264
plus one

00:03:56.293 --> 00:03:57.876
over four factorial

00:03:59.138 --> 00:04:00.708
times x to the fourth.

00:04:00.708 --> 00:04:02.764
If it involves the fourth derivative,

00:04:02.764 --> 00:04:04.927
it's gonna be divided by four factorial,

00:04:04.927 --> 00:04:07.048
it's going to be times x to the fourth.

00:04:07.048 --> 00:04:07.881
And there you have it.

00:04:07.881 --> 00:04:10.517
These are the first three non-zero terms

00:04:10.517 --> 00:04:12.092
of the Maclaurin--oh sorry!

00:04:12.092 --> 00:04:15.825
These are the first three
non-zero terms of the Maclaurin

00:04:15.825 --> 00:04:18.215
series for cosine of x,

00:04:18.215 --> 00:04:21.597
and so then we can say x cosine of x.

00:04:21.597 --> 00:04:23.847
So, x times cosine of x.

00:04:25.396 --> 00:04:28.838
I'm just gonna multiple
each of these things by x.

00:04:28.838 --> 00:04:29.671
So, one way to think about it,

00:04:29.671 --> 00:04:31.460
I'm multiplying both
sides of this times x.

00:04:31.460 --> 00:04:36.076
So, I'm getting x minus
one over two factorial,

00:04:36.076 --> 00:04:39.220
times x to the third power,

00:04:39.220 --> 00:04:42.063
plus one over four factorial,

00:04:42.063 --> 00:04:44.968
times x to the fifth power.

00:04:44.968 --> 00:04:45.801
And we're done.

00:04:45.801 --> 00:04:48.010
And even though this
might have seemed a little

00:04:48.010 --> 00:04:50.110
hairy and a bit of a long process,

00:04:50.110 --> 00:04:52.546
some people have it somewhat
committed to memory,

00:04:52.546 --> 00:04:55.876
what the Maclaurin series
expansion of cosine of x

00:04:55.876 --> 00:04:58.937
and sine of x are, and
sometimes e to the x, as well.

00:04:58.937 --> 00:05:01.325
And then, this could
be a very fast process

00:05:01.325 --> 00:05:03.139
where you just multiply that by x.

00:05:03.139 --> 00:05:05.376
But, as you can see, even
finding that Maclaurin series

00:05:05.376 --> 00:05:08.824
expansion for cosine of x isn't too bad.

00:05:08.824 --> 00:05:10.106
If you'd done it the other
way, if you'd just tried

00:05:10.106 --> 00:05:11.847
to take the first, second,
third, fourth, fifth,

00:05:11.847 --> 00:05:14.827
or all the way to the
fourth derivative of this

00:05:14.827 --> 00:05:16.205
and evaluated zero it would've gotten

00:05:16.205 --> 00:05:18.122
very, very, very hairy.


Worked example: coefficient in Taylor polynomial

WEBVTT
Kind: captions
Language: en

00:00:00.028 --> 00:00:01.311
- [Teacher] We're given
an f of x and they say,

00:00:01.311 --> 00:00:02.787
what is the coefficient for the term

00:00:02.787 --> 00:00:07.286
containing x plus two to the fourth power

00:00:07.286 --> 00:00:10.105
in the Taylor polynomial,

00:00:10.105 --> 00:00:13.605
centered at x equals negative two of f?

00:00:15.510 --> 00:00:18.315
So like always, see if you
can take a stab at this video

00:00:18.315 --> 00:00:21.032
on your own before we
work through it together.

00:00:21.032 --> 00:00:22.665
Alright, now let's do this.

00:00:22.665 --> 00:00:26.764
So in general, our Taylor
polynomial, p of x,

00:00:26.764 --> 00:00:28.317
it's going to have the form and remember,

00:00:28.317 --> 00:00:30.894
we're centering at x equals negative two

00:00:30.894 --> 00:00:34.522
so this means we're going
to evaluate our function

00:00:34.522 --> 00:00:36.774
at where we're centering it.

00:00:36.774 --> 00:00:39.516
We are going to divide
it by zero factorial

00:00:39.516 --> 00:00:40.419
which is just one.

00:00:40.419 --> 00:00:42.849
I'm just gonna write 'em all
out just so you see the pattern

00:00:42.849 --> 00:00:45.261
and we could even say
that's gonna be times

00:00:45.261 --> 00:00:47.636
x minus where we're centering it

00:00:47.636 --> 00:00:48.942
but if we're subtracting a negative two,

00:00:48.942 --> 00:00:51.765
it's gonna be x plus two and I
could write to the zero power

00:00:51.765 --> 00:00:53.159
but once again, that's
just going to be one

00:00:53.159 --> 00:00:55.462
so a lot of times, you won't
see someone write this and this

00:00:55.462 --> 00:00:56.625
but I'm writing it just to show

00:00:56.625 --> 00:00:58.235
that there's a consistent pattern.

00:00:58.235 --> 00:01:01.226
So then you're gonna have
plus the first derivative

00:01:01.226 --> 00:01:03.351
evaluated at negative two

00:01:03.351 --> 00:01:06.139
divided by one factorial
which is still just one

00:01:06.139 --> 00:01:10.071
times x plus two to the first power

00:01:10.071 --> 00:01:14.649
plus the second derivative
evaluated at negative two

00:01:14.649 --> 00:01:19.154
over two factorial times
x plus two squared.

00:01:19.154 --> 00:01:20.952
I think you see where this is going

00:01:20.952 --> 00:01:22.876
and really all we care about is the one

00:01:22.876 --> 00:01:24.431
that has a fourth degree term and,

00:01:24.431 --> 00:01:26.274
well, actually, let me just
write the third degree term too

00:01:26.274 --> 00:01:28.157
just so we get fluent doing this.

00:01:28.157 --> 00:01:32.539
So the third derivative
evaluated at negative two

00:01:32.539 --> 00:01:36.706
over three factorial times x
plus two to the third power

00:01:38.010 --> 00:01:40.993
and now, this is the part
that we really care about,

00:01:40.993 --> 00:01:43.896
plus the fourth derivative.

00:01:43.896 --> 00:01:45.247
I could have just written a four there

00:01:45.247 --> 00:01:47.692
but I think you get what I'm saying.

00:01:47.692 --> 00:01:50.690
And then evaluate at x equals negative two

00:01:50.690 --> 00:01:54.909
divided by four factorial times x plus two

00:01:54.909 --> 00:01:56.874
to the fourth power.

00:01:56.874 --> 00:01:58.594
So what's the coefficient here?

00:01:58.594 --> 00:02:01.788
Well, the coefficient is this business.

00:02:01.788 --> 00:02:04.423
So we'd have to take the fourth derivative

00:02:04.423 --> 00:02:06.512
of our original function.

00:02:06.512 --> 00:02:07.738
We gotta take the fourth derivative

00:02:07.738 --> 00:02:09.784
of that original function,
evaluate at negative two

00:02:09.784 --> 00:02:12.677
and divide it by four
factorial so let's do that.

00:02:12.677 --> 00:02:14.824
So our function,

00:02:14.824 --> 00:02:18.422
so our first derivative, f
prime of x is just going to be,

00:02:18.422 --> 00:02:19.978
just gonna use the power rule a lot,

00:02:19.978 --> 00:02:23.895
six x to the fifth minus three x squared.

00:02:25.215 --> 00:02:28.410
Second derivative is going to be equal to

00:02:28.410 --> 00:02:31.521
five times six is 30 x to the fourth.

00:02:31.521 --> 00:02:35.654
Two times three, minus
six x to the first power.

00:02:35.654 --> 00:02:37.820
Third derivative.

00:02:37.820 --> 00:02:40.724
Third derivative of x
is going to be equal to

00:02:40.724 --> 00:02:44.391
four times 30 is 120 x to the third power

00:02:46.386 --> 00:02:50.339
minus six and then the fourth derivative

00:02:50.339 --> 00:02:52.109
which is what we really care about

00:02:52.109 --> 00:02:54.646
is going to be three times 120

00:02:54.646 --> 00:02:57.229
is 360 x to the second power

00:02:58.943 --> 00:03:01.116
and the derivative of a
constant is just zero.

00:03:01.116 --> 00:03:04.674
So if we were to evaluate
this at x equals negative two,

00:03:04.674 --> 00:03:07.007
so f, the fourth derivative,

00:03:08.035 --> 00:03:10.202
evaluated when x equals negative two

00:03:10.202 --> 00:03:14.334
is going to be 360 times
negative two squared is four.

00:03:14.334 --> 00:03:16.431
I'm just gonna keep
that as 360 times four.

00:03:16.431 --> 00:03:18.324
We can obviously evaluate that

00:03:18.324 --> 00:03:20.194
but we're gonna have to
divide it by four factorial

00:03:20.194 --> 00:03:24.085
so the whole coefficient is
going to be 360 times four

00:03:24.085 --> 00:03:25.297
which is the numerator here

00:03:25.297 --> 00:03:26.730
divided by four factorial,

00:03:26.730 --> 00:03:30.290
divided by four times
three times two times one.

00:03:30.290 --> 00:03:31.725
Well, four divided by four,

00:03:31.725 --> 00:03:33.877
those are just gonna be one.

00:03:33.877 --> 00:03:37.859
360 divided by three, maybe
I'll think of it this way,

00:03:37.859 --> 00:03:40.859
360 divided by six is going to be 60

00:03:42.396 --> 00:03:43.420
and so that's all we have.

00:03:43.420 --> 00:03:45.994
We have 60 and then in the
denominator we just have a one

00:03:45.994 --> 00:03:49.067
so this is going to simplify to 60.

00:03:49.067 --> 00:03:52.150
That's the coefficient for this term.


Worked example: coefficient in Maclaurin polynomial

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:03.891
- [Instructor] Nth derivative
of g at x equals zero is given

00:00:03.891 --> 00:00:08.075
by, so the nth derivative
of g evaluated x equals zero

00:00:08.075 --> 00:00:10.194
is equal to square root of n
plus seven over n to the third

00:00:10.194 --> 00:00:12.699
for n is greater than or equal to one.

00:00:12.699 --> 00:00:15.890
What is the coefficient for
the term containing x squared

00:00:15.890 --> 00:00:18.307
in the Maclaurin series of g?

00:00:19.150 --> 00:00:23.450
Let's just think about the
Maclaurin series for g.

00:00:23.450 --> 00:00:26.729
If I were to have my function g of x.

00:00:26.729 --> 00:00:29.583
The Maclaurin series, I could
say approximately equal to

00:00:29.583 --> 00:00:31.938
especially if I'm not gonna
list out all of the terms,

00:00:31.938 --> 00:00:35.630
is going to be equal to,
well it's going to be equal

00:00:35.630 --> 00:00:39.047
to g of zero plus g prime of zero times x

00:00:43.789 --> 00:00:46.956
plus g prime prime of zero divided by,

00:00:48.854 --> 00:00:51.489
I could say two factorial
but that's just two,

00:00:51.489 --> 00:00:54.416
times x squared, and that's
about as far as we go.

00:00:54.416 --> 00:00:56.259
Because we just have to think
about what is the coefficient

00:00:56.259 --> 00:00:57.972
for the term containing x squared.

00:00:57.972 --> 00:00:59.680
If they said what's the coefficient

00:00:59.680 --> 00:01:01.865
for the term containing x to
the third, I would keep going.

00:01:01.865 --> 00:01:02.934
I'd go g prime.

00:01:02.934 --> 00:01:05.512
I would take the third
derivative evaluated at zero

00:01:05.512 --> 00:01:07.151
over three factorial.

00:01:07.151 --> 00:01:08.504
I could do this as a factorial too,

00:01:08.504 --> 00:01:10.027
but that just evaluates to two.

00:01:10.027 --> 00:01:12.042
I could do this as one factorial.

00:01:12.042 --> 00:01:14.849
I could do this as zero
factorial just so you see

00:01:14.849 --> 00:01:17.101
it's a consistent idea here.

00:01:17.101 --> 00:01:20.831
I could, of course, keep on
going, but we just care about,

00:01:20.831 --> 00:01:22.491
they're just asking us
what is the coefficient

00:01:22.491 --> 00:01:25.436
for the term containing x squared?

00:01:25.436 --> 00:01:27.831
They just want us to figure out this.

00:01:27.831 --> 00:01:29.845
What is this thing right over here?

00:01:29.845 --> 00:01:31.217
To know that, we need to figure out

00:01:31.217 --> 00:01:35.408
what is the second derivative
of g evaluated x equals zero?

00:01:35.408 --> 00:01:36.539
Well they tell us that over here.

00:01:36.539 --> 00:01:38.474
It's a little bit unconventional

00:01:38.474 --> 00:01:41.814
where they give us a
formula, a general formula

00:01:41.814 --> 00:01:44.586
for any derivative
evaluated x equals zero,

00:01:44.586 --> 00:01:45.674
but that's what they're telling us here.

00:01:45.674 --> 00:01:48.596
So in this case, the n isn't zero.

00:01:48.596 --> 00:01:50.717
The n is the derivative we're
taking, and that's going

00:01:50.717 --> 00:01:54.166
to be our second derivative, so this is,

00:01:54.166 --> 00:01:56.749
so if I wanted to figure out g.

00:01:57.645 --> 00:02:01.896
If I am figuring out
the second derivative,

00:02:01.896 --> 00:02:04.462
and I could write it
like that evaluated zero,

00:02:04.462 --> 00:02:06.304
or I could write it like this

00:02:06.304 --> 00:02:08.582
just so the notation is consistent.

00:02:08.582 --> 00:02:10.549
I could write it like that.

00:02:10.549 --> 00:02:14.383
The second derivative evaluated
x equals zero is going

00:02:14.383 --> 00:02:16.984
to be equal to, well our n is two,

00:02:16.984 --> 00:02:18.868
so this is going to be the square root

00:02:18.868 --> 00:02:22.701
of two plus seven over
two to the third power.

00:02:24.217 --> 00:02:27.668
Two plus seven is nine, take
the principle root of that.

00:02:27.668 --> 00:02:30.505
It's gonna give us positive
three over two to the third

00:02:30.505 --> 00:02:34.477
which is eight, so this
part right over here is 3/8,

00:02:34.477 --> 00:02:38.339
so the whole coefficient
is going to be 3/8,

00:02:38.339 --> 00:02:42.901
that's this numerator,
divided by, divided by two.

00:02:42.901 --> 00:02:47.420
Which of course is equal
to 3/16, and we're done.

00:02:47.420 --> 00:02:50.052
They didn't want us to
figure out a couple of terms

00:02:50.052 --> 00:02:52.810
of this which we could call
the Maclaurin polynomial

00:02:52.810 --> 00:02:54.427
and nth degree Maclaurin polynomial.

00:02:54.427 --> 00:02:55.973
They didn't want us to
find the entire, you know,

00:02:55.973 --> 00:02:57.769
keep going with this series.

00:02:57.769 --> 00:02:59.771
They just wanted to find
one coefficient right here.

00:02:59.771 --> 00:03:02.276
The coefficient on the second degree term

00:03:02.276 --> 00:03:04.873
which we just figured out is 3/16.


Worked example: Maclaurin polynomial

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:01.974
- [Instructor] We're
told that f of x is equal

00:00:01.974 --> 00:00:04.527
to one over the square root of x plus one,

00:00:04.527 --> 00:00:06.399
and what we want to figure out is what is

00:00:06.399 --> 00:00:10.451
the second degree
Maclaurin polynomial of f?

00:00:10.451 --> 00:00:12.421
And like always, pause this video

00:00:12.421 --> 00:00:14.629
and see if you could have a go at it.

00:00:14.629 --> 00:00:17.516
so let's remind ourselves what
a Maclaurin polynomial is,

00:00:17.516 --> 00:00:20.370
a Maclaurin polynomial is
just a Taylor polynomial

00:00:20.370 --> 00:00:24.876
centered at zero, so the
form of this second degree

00:00:24.876 --> 00:00:27.684
Maclaurin polynomial,
and we just have to find

00:00:27.684 --> 00:00:30.584
this Maclaurin expansion
until our second degree term,

00:00:30.584 --> 00:00:32.615
it's going to look like this.

00:00:32.615 --> 00:00:35.652
So p of x, I'm using p for polynomial,

00:00:35.652 --> 00:00:38.152
it's going to be our f of zero

00:00:39.338 --> 00:00:42.911
plus, and we could do
that as f of zero times x

00:00:42.911 --> 00:00:45.478
to the zero power, well,
that's just f of zero.

00:00:45.478 --> 00:00:48.228
F of zero plus f prime of zero x,

00:00:50.382 --> 00:00:52.632
plus f prime prime of zero,

00:00:55.419 --> 00:00:57.160
divided by, we could think of it

00:00:57.160 --> 00:00:59.179
as two factorial, but
it's really just two,

00:00:59.179 --> 00:01:01.178
we could think of this as
dividing by one factorial,

00:01:01.178 --> 00:01:03.305
which is just one, this
dividing by zero factorial,

00:01:03.305 --> 00:01:06.214
well that's just one again,
so we have f prime prime

00:01:06.214 --> 00:01:08.092
of zero, the second derivative evaluated

00:01:08.092 --> 00:01:10.945
zero divided by two, x squared.

00:01:10.945 --> 00:01:13.154
Now if we wanted a higher
degree we could keep on going,

00:01:13.154 --> 00:01:15.144
but remember, they're just
asking us for the second degree.

00:01:15.144 --> 00:01:16.745
So this is the form that we're gonna need,

00:01:16.745 --> 00:01:18.558
we're gonna have these three terms.

00:01:18.558 --> 00:01:21.972
So let's see if we can evaluate

00:01:21.972 --> 00:01:25.572
the function and its derivatives at zero.

00:01:25.572 --> 00:01:29.239
So f of zero, f of zero
is equal to one over

00:01:30.181 --> 00:01:33.185
the square root of zero plus one,

00:01:33.185 --> 00:01:35.587
well that's one over
the square root of one,

00:01:35.587 --> 00:01:37.732
the principle root of one,
which is positive one,

00:01:37.732 --> 00:01:39.744
so that's just going to be equal to one.

00:01:39.744 --> 00:01:42.015
So that right over there is equal to one.

00:01:42.015 --> 00:01:44.682
Now let's evaluate f prime of x,

00:01:45.675 --> 00:01:48.677
and then I'll evaluate f prime of zero.

00:01:48.677 --> 00:01:50.774
F prime of x is equal to,

00:01:50.774 --> 00:01:52.796
well, one over the square
root of x plus one,

00:01:52.796 --> 00:01:55.948
this is the same thing as x plus one,

00:01:55.948 --> 00:01:57.894
let me write it this way,

00:01:57.894 --> 00:02:00.781
this is the same thing as,

00:02:00.781 --> 00:02:03.244
actually let me write it down this way.

00:02:03.244 --> 00:02:06.961
Another way of writing f of
x is this is the same thing

00:02:06.961 --> 00:02:10.461
as x plus one to the negative one half.

00:02:11.344 --> 00:02:15.267
And so if I'm thinking
the first derivative of f,

00:02:15.267 --> 00:02:17.452
well, I could use the chain rule here,

00:02:17.452 --> 00:02:20.035
the derivative of x plus
one with respect to x,

00:02:20.035 --> 00:02:21.936
well, that's just going to be one,

00:02:21.936 --> 00:02:23.789
and then I'll take the
derivative of this whole thing

00:02:23.789 --> 00:02:25.518
with respect to x plus
one and I'll just use

00:02:25.518 --> 00:02:28.630
the power rule there, this is
gonna be negative one half,

00:02:28.630 --> 00:02:31.517
times x plus one, to the,

00:02:31.517 --> 00:02:34.845
and I decrement the exponent,
negative three halves.

00:02:34.845 --> 00:02:38.513
And so the first derivative
evaluated of zero

00:02:38.513 --> 00:02:41.692
is just negative one half, times,

00:02:41.692 --> 00:02:44.144
if this is zero, zero
plus one is just one,

00:02:44.144 --> 00:02:46.579
one to the negative three halves,

00:02:46.579 --> 00:02:49.577
one to the negative three half power,

00:02:49.577 --> 00:02:51.375
well, that's just going to be one.

00:02:51.375 --> 00:02:53.186
So this whole thing, f prime of zero

00:02:53.186 --> 00:02:55.100
is just negative one half.

00:02:55.100 --> 00:02:59.916
So that is this right over
here is negative one half.

00:02:59.916 --> 00:03:03.266
And now let's figure out
the second derivative.

00:03:03.266 --> 00:03:07.099
All right, let me do
this in this green color.

00:03:08.199 --> 00:03:12.561
So the second derivative
with respect to x,

00:03:12.561 --> 00:03:14.129
well I did the same thing again,

00:03:14.129 --> 00:03:15.736
the derivative of x plus
one with respect to x,

00:03:15.736 --> 00:03:17.284
that's just one, so I just have to take

00:03:17.284 --> 00:03:18.661
the derivative of the whole thing

00:03:18.661 --> 00:03:20.137
with respect to x plus one.

00:03:20.137 --> 00:03:22.246
So I take my exponent, bring it out front,

00:03:22.246 --> 00:03:24.594
negative three halves times one half,

00:03:24.594 --> 00:03:26.909
times negative one half, is going to be

00:03:26.909 --> 00:03:30.595
positive three fourths times x plus one,

00:03:30.595 --> 00:03:33.877
and then I decrement the
exponent here by one,

00:03:33.877 --> 00:03:37.671
or by two halves, so that's
gonna be negative five halves,

00:03:37.671 --> 00:03:41.314
and so the second derivative
evaluated at zero,

00:03:41.314 --> 00:03:43.936
well if this is equal to
zero you're gonna have

00:03:43.936 --> 00:03:46.301
one to the negative five
halves which is just one,

00:03:46.301 --> 00:03:49.174
times three fourths, is
gonna be three fourths.

00:03:49.174 --> 00:03:53.033
So this part right over
here is three fourths.

00:03:53.033 --> 00:03:56.370
And so you're gonna have
three fourths divided by two,

00:03:56.370 --> 00:03:59.068
three fourths divided
by two is three eighths.

00:03:59.068 --> 00:04:02.758
So our Taylor, I should say,
our Maclaurin polynomial,

00:04:02.758 --> 00:04:05.163
our second degree Maclaurin polynomial,

00:04:05.163 --> 00:04:07.978
p of x is going to be equal to,

00:04:07.978 --> 00:04:09.846
and I'll do it in the same colors,

00:04:09.846 --> 00:04:11.887
going to be equal to one,

00:04:11.887 --> 00:04:15.832
plus, maybe I'll just
write it as minus one half,

00:04:15.832 --> 00:04:17.249
minus one half x,

00:04:21.250 --> 00:04:23.917
plus three eighths x squared,

00:04:25.290 --> 00:04:27.957
plus three eighths x squared,

00:04:28.899 --> 00:04:31.000
and we are done, there you have it.

00:04:31.000 --> 00:04:34.287
We have our second degree
Maclaurin polynomial of f,

00:04:34.287 --> 00:04:36.583
which could be used to
provide an approximation

00:04:36.583 --> 00:04:40.750
for our function, especially
if our x is near zero.


Worked example: interval of convergence

WEBVTT
Kind: captions
Language: en

00:00:00.254 --> 00:00:01.839
- [Instructor] So we have
an infinite series here,

00:00:01.839 --> 00:00:03.590
and the goal of this video is to try

00:00:03.590 --> 00:00:06.520
to figure out the interval of convergence

00:00:06.520 --> 00:00:08.004
for this series.

00:00:08.004 --> 00:00:10.260
And that's another way of
saying for what x values,

00:00:10.260 --> 00:00:12.263
what range of x values,

00:00:12.263 --> 00:00:14.994
is this series going to converge?

00:00:14.994 --> 00:00:16.779
And like always pause this video

00:00:16.779 --> 00:00:19.230
and see if you can figure it out.

00:00:19.230 --> 00:00:20.545
When you look at this series,

00:00:20.545 --> 00:00:22.509
it doesn't fit cleanly into something

00:00:22.509 --> 00:00:25.521
like a geometric series
or an alternating series.

00:00:25.521 --> 00:00:27.014
When I see something like this,

00:00:27.014 --> 00:00:28.742
I think about the ratio test,

00:00:28.742 --> 00:00:31.227
because it tends to be pretty general.

00:00:31.227 --> 00:00:33.252
To apply the ratio test,

00:00:33.252 --> 00:00:36.275
we want to think about the limit,

00:00:36.275 --> 00:00:39.108
the limit as n approaches infinity

00:00:40.481 --> 00:00:42.564
of the n plus oneth term,

00:00:44.253 --> 00:00:46.244
divided by the nth term

00:00:46.244 --> 00:00:48.761
and the absolute value of that.

00:00:48.761 --> 00:00:51.344
If this thing is less than one,

00:00:52.670 --> 00:00:56.260
so when this thing is less than one,

00:00:56.260 --> 00:00:58.760
then we are going to converge.

00:01:00.214 --> 00:01:04.254
And the x values that make
this thing greater than one,

00:01:04.254 --> 00:01:06.711
we are going to diverge.

00:01:06.711 --> 00:01:10.042
And the x values that
make this equal to one,

00:01:10.042 --> 00:01:12.022
well then we're going to be inconclusive

00:01:12.022 --> 00:01:14.030
and so we're going to have
to use other techniques

00:01:14.030 --> 00:01:18.739
to think about whether we're
going to converge or diverge.

00:01:18.739 --> 00:01:21.491
So let's just think about
this, let's just evaluate this,

00:01:21.491 --> 00:01:22.527
so let's do that.

00:01:22.527 --> 00:01:23.360
Limit

00:01:24.243 --> 00:01:26.243
as n approaches infinity

00:01:27.493 --> 00:01:31.320
of the absolute value of a sub-n plus one

00:01:31.320 --> 00:01:34.020
well that's going to
be x to the n plus one.

00:01:34.020 --> 00:01:36.547
Let me color code this, just
so we know what we're doing.

00:01:36.547 --> 00:01:40.505
So this thing right over
here is going to be x

00:01:40.505 --> 00:01:41.922
to the n plus one

00:01:42.778 --> 00:01:44.028
over n plus one

00:01:45.386 --> 00:01:47.803
times five to the n plus one.

00:01:48.788 --> 00:01:51.283
We're going to divide that by

00:01:51.283 --> 00:01:52.516
the nth term.

00:01:52.516 --> 00:01:54.058
We're going to divide
that by the nth term.

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And that's just going to be x to the n

00:01:57.067 --> 00:01:59.317
over n times five to the n.

00:02:01.920 --> 00:02:04.526
We're going to take the absolute
value of this whole thing.

00:02:04.526 --> 00:02:05.800
Now let me simplify this.

00:02:05.800 --> 00:02:07.252
I'll simplify it down here.

00:02:07.252 --> 00:02:09.174
So this is the same thing as

00:02:09.174 --> 00:02:11.174
x to the n plus one over

00:02:12.463 --> 00:02:15.796
n plus one times five to the n plus one,

00:02:16.779 --> 00:02:19.482
times the reciprocal of this.

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So it's going to be n

00:02:21.273 --> 00:02:23.507
times five to the n,

00:02:23.507 --> 00:02:25.265
over x to the n.

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And we could simplify this.

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This is going to be equal to,

00:02:29.035 --> 00:02:31.011
let's see, you divide
numerator and denominator

00:02:31.011 --> 00:02:32.317
by x to the n,

00:02:32.317 --> 00:02:34.469
you're left with just an x.

00:02:34.469 --> 00:02:36.304
Then divide numerator and denominator

00:02:36.304 --> 00:02:37.933
by five to the n.

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That is going to be,
that's a one, this is a one

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and then this is just going to be,

00:02:42.045 --> 00:02:44.028
five to the n plus one
divided by five to the n,

00:02:44.028 --> 00:02:46.017
that's just going to be a five.

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And so what do we have?

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We have x times n.

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x times n,

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over, distribute the five,

00:02:54.778 --> 00:02:55.611
five n,

00:02:56.929 --> 00:02:58.262
five n plus one.

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Oh, let me be careful there.

00:03:00.771 --> 00:03:01.791
Let me distribute the five.

00:03:01.791 --> 00:03:03.208
Five n plus five,

00:03:04.240 --> 00:03:05.948
five times n, five times one.

00:03:05.948 --> 00:03:07.448
Five n plus one.

00:03:07.448 --> 00:03:09.506
Five n plus five (chuckles).

00:03:09.506 --> 00:03:12.333
All right, so let me just rewrite that.

00:03:12.333 --> 00:03:14.745
This is going to be equal to

00:03:14.745 --> 00:03:15.578
the limit

00:03:16.884 --> 00:03:19.363
as n approaches infinity

00:03:19.363 --> 00:03:22.863
of the absolute value of this thing.

00:03:22.863 --> 00:03:24.509
And actually to help us with this limit,

00:03:24.509 --> 00:03:26.021
let me rewrite it a little bit.

00:03:26.021 --> 00:03:28.487
Let me divide the numerator
and the denominator both by n.

00:03:28.487 --> 00:03:30.012
I'm not changing the value,
I'm doing the same thing

00:03:30.012 --> 00:03:31.543
to the numerator and the denominator.

00:03:31.543 --> 00:03:33.007
I'm dividing it by the same value.

00:03:33.007 --> 00:03:35.508
So if I divide the numerator
and the denominator by n,

00:03:35.508 --> 00:03:38.758
this is going to be the same thing as x

00:03:39.771 --> 00:03:40.604
over

00:03:41.446 --> 00:03:42.932
five

00:03:42.932 --> 00:03:44.515
plus five to the n.

00:03:45.663 --> 00:03:47.826
So when you divided the numerator
and the denominator by n,

00:03:47.826 --> 00:03:50.742
it becomes very clear what
happens as n approaches infinity,

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As n approaches infinity,
x doesn't change,

00:03:52.998 --> 00:03:54.051
five doesn't change,

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but five over n goes to zero.

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And so this limit is going
to be equal to x over five.

00:04:02.155 --> 00:04:04.707
So that's a pretty neat, clean thing.

00:04:04.707 --> 00:04:06.533
Now we can use this to think about,

00:04:06.533 --> 00:04:07.502
and actually let me write this.

00:04:07.502 --> 00:04:11.538
This is going to be the
absolute value of x over five.

00:04:11.538 --> 00:04:13.493
Now we can think about
under which conditions

00:04:13.493 --> 00:04:15.311
is the absolute value of x over five

00:04:15.311 --> 00:04:16.554
going to be less than one

00:04:16.554 --> 00:04:18.286
and we're definitely gonna converge?

00:04:18.286 --> 00:04:19.560
Under what conditions are we going to be

00:04:19.560 --> 00:04:21.603
greater than one and definitely diverge?

00:04:21.603 --> 00:04:24.666
And then under what
conditions is it inconclusive?

00:04:24.666 --> 00:04:27.388
So let's just see when
we know we can converge.

00:04:27.388 --> 00:04:30.728
So the absolute value of x over five

00:04:30.728 --> 00:04:32.662
is less than one.

00:04:32.662 --> 00:04:34.745
This is our convergence situation.

00:04:34.745 --> 00:04:36.489
Well, that's the same thing as saying

00:04:36.489 --> 00:04:38.322
that x negative one

00:04:39.465 --> 00:04:41.798
is less than x over five,

00:04:43.459 --> 00:04:45.376
which is less than one.

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And you multiply all the sides by five.

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This is the same thing as negative five

00:04:51.906 --> 00:04:53.696
is less than x,

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which is less than five.

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So if we know that this is true,

00:04:57.022 --> 00:04:58.446
this is definitely going to be part

00:04:58.446 --> 00:05:00.652
of the interval of convergence.

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We know that if x meets these constraints,

00:05:02.984 --> 00:05:05.504
then our series is going to converge.

00:05:05.504 --> 00:05:06.554
But we're not done yet.

00:05:06.554 --> 00:05:09.951
We have to think about
the inconclusive case.

00:05:09.951 --> 00:05:11.190
So let's think about the scenario

00:05:11.190 --> 00:05:14.805
where the absolute value of x over five,

00:05:14.805 --> 00:05:18.537
absolute value of x over
five is equal to one.

00:05:18.537 --> 00:05:20.778
Or another way of thinking about this,

00:05:20.778 --> 00:05:22.750
that means x over five is equal to one

00:05:22.750 --> 00:05:26.643
or x over five is equal to negative one.

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And this means that x is equal to five

00:05:29.752 --> 00:05:32.585
or x is equal to negative five.

00:05:34.047 --> 00:05:36.040
So these are the two inconclusive cases

00:05:36.040 --> 00:05:37.514
using the ratio test.

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So let's test them out individually

00:05:39.048 --> 00:05:40.531
by looking back at the series

00:05:40.531 --> 00:05:42.300
and just substituting x equals five

00:05:42.300 --> 00:05:44.048
or x equals negative five.

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So in the first scenario,

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let me find a new color
here, let me use red.

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So the first scenario of x equals five,

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let's go to our series.

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Then the series is going to be the sum

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from n equals one to infinity

00:06:00.475 --> 00:06:02.392
of five to the n over n

00:06:04.733 --> 00:06:06.400
times five to the n.

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Well, this is just going
to be equal to the sum

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n equals one to infinity

00:06:12.799 --> 00:06:13.966
of one over n.

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This is a harmonic series.

00:06:16.477 --> 00:06:20.989
This is the p-series
where p is equal to one.

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And we know our p-series
of p is equal to one.

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That's going to diverge.

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And we know the harmonic series
we've done in other videos,

00:06:28.001 --> 00:06:30.042
this definitely diverges.

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So this diverges.

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You could do that by
p-series convergence test.

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If the p for a p-series is one,

00:06:36.798 --> 00:06:38.492
well you're gonna diverge.

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So now let's think about,

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so five is definitely not part

00:06:41.053 --> 00:06:44.159
of our interval of convergence.

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Now let's think about
x equals negative five.

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When x equals negative five,

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let me get a another color going here.

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When x is equal to negative five,

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then this thing is going to be equal to

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the sum from n equals one

00:07:00.058 --> 00:07:03.225
to infinity of negative five to the n.

00:07:04.489 --> 00:07:06.232
Actually let me just write that as,

00:07:06.232 --> 00:07:09.290
I'll write it out negative five to the n

00:07:09.290 --> 00:07:11.540
over n times five to the n.

00:07:12.414 --> 00:07:15.017
This is the same thing as the sum

00:07:15.017 --> 00:07:18.023
from n equals one to infinity.

00:07:18.023 --> 00:07:20.813
We could write this as
negative one to the n

00:07:20.813 --> 00:07:22.313
times five to the,

00:07:23.781 --> 00:07:24.614
times

00:07:26.006 --> 00:07:27.995
five to the n,

00:07:27.995 --> 00:07:30.245
over n times five to the n.

00:07:31.260 --> 00:07:32.570
And now this thing,

00:07:32.570 --> 00:07:35.269
this is an alternating harmonic series.

00:07:35.269 --> 00:07:37.044
And so you could actually use the,

00:07:37.044 --> 00:07:38.810
you might already know
that that converges,

00:07:38.810 --> 00:07:42.025
or you could use the
alternating series test.

00:07:42.025 --> 00:07:43.319
The alternating series test,

00:07:43.319 --> 00:07:46.050
it might be a little bit
clearer if I write it like this.

00:07:46.050 --> 00:07:48.954
That this is an alternating series.

00:07:48.954 --> 00:07:51.704
So in an alternating series test,

00:07:53.237 --> 00:07:57.541
if we see that this thing
is monotonically decreasing

00:07:57.541 --> 00:08:00.492
and the limit as n
approaches infinity is zero,

00:08:00.492 --> 00:08:01.506
this thing converges.

00:08:01.506 --> 00:08:04.256
The alternating harmonic
series actually converges.

00:08:04.256 --> 00:08:05.756
So this converges.

00:08:06.853 --> 00:08:08.562
So given that this converges,

00:08:08.562 --> 00:08:10.317
you could view this as this boundary here.

00:08:10.317 --> 00:08:13.215
We would include that in
our interval of convergence.

00:08:13.215 --> 00:08:14.641
So x doesn't just have to be

00:08:14.641 --> 00:08:16.168
strictly greater than negative five,

00:08:16.168 --> 00:08:18.521
it could be greater than
or equal to negative five,

00:08:18.521 --> 00:08:20.515
but it has to be less than five.

00:08:20.515 --> 00:08:23.932
This is our true interval of convergence.