WEBVTT

00:00:01.246 --> 00:00:03.736 A:middle
>> Welcome to the Cypress
College math review

00:00:03.806 --> 00:00:05.476 A:middle
on power series solutions.

00:00:06.476 --> 00:00:09.186 A:middle
Verify that X equals
zero is an ordinary point

00:00:09.186 --> 00:00:10.386 A:middle
for the differential equation.

00:00:11.016 --> 00:00:13.226 A:middle
And express the general
solution in terms

00:00:13.226 --> 00:00:14.856 A:middle
of power series about
this point.

00:00:16.766 --> 00:00:19.886 A:middle
A solution Y equals
the sum N equals zero

00:00:19.886 --> 00:00:22.786 A:middle
to infinity A sub N X to the Nth

00:00:23.136 --> 00:00:26.856 A:middle
that contains two arbitrary
constants A sub zero

00:00:26.856 --> 00:00:27.746 A:middle
and A sub 1.

00:00:28.366 --> 00:00:31.816 A:middle
Converges inside a circle
centered at X equals zero.

00:00:32.336 --> 00:00:35.186 A:middle
And extends out to
the singular point

00:00:35.186 --> 00:00:37.406 A:middle
or points nearest X equals zero.

00:00:38.526 --> 00:00:40.386 A:middle
First let's find
the singular points.

00:00:41.506 --> 00:00:43.446 A:middle
So we take the differential
equation.

00:00:44.616 --> 00:00:47.566 A:middle
We divide by the coefficient
of the second derivative.

00:00:47.846 --> 00:00:50.566 A:middle
1 minus X squared and we see

00:00:50.566 --> 00:00:53.736 A:middle
that the singular points
are plus or minus 1.

00:00:54.436 --> 00:00:57.746 A:middle
Which tells us that X equals
zero is an ordinary point.

00:01:00.116 --> 00:01:02.966 A:middle
Since this series is
centered at X equals zero,

00:01:04.216 --> 00:01:06.586 A:middle
and the singular points
are plus or minus 1.

00:01:07.466 --> 00:01:10.876 A:middle
That tells us that this
series will converge

00:01:11.896 --> 00:01:14.476 A:middle
for absolute value
of X less than one.

00:01:15.776 --> 00:01:16.556 A:middle
Now let's go back

00:01:16.596 --> 00:01:18.886 A:middle
to the original version
of our equation.

00:01:19.306 --> 00:01:25.796 A:middle
Before we actually get started

00:01:25.796 --> 00:01:29.006 A:middle
on this problem let's
take a look

00:01:29.006 --> 00:01:31.746 A:middle
at power series solutions
in general.

00:01:37.876 --> 00:01:39.906 A:middle
They look like this
the general form.

00:01:42.246 --> 00:01:42.976 A:middle
So if we write them out.

00:01:54.626 --> 00:01:57.146 A:middle
And if we take the
derivative term by term.

00:02:02.146 --> 00:02:04.976 A:middle
This would be A sub 1
because we lose the constant.

00:02:19.546 --> 00:02:20.856 A:middle
And how about the
2nd derivative.

00:02:24.216 --> 00:02:35.406 A:middle
So 2 A sub 2 plus 3 times 2 A
sub 3 X plus 4 times 3 A sub 4 X

00:02:35.406 --> 00:02:37.236 A:middle
squared and so on and so forth.

00:02:38.486 --> 00:02:40.796 A:middle
So we're losing a term
each time aren't I?

00:02:42.286 --> 00:02:44.426 A:middle
So here we're starting at 1.

00:02:45.316 --> 00:02:46.946 A:middle
Here we're starting at 2.

00:02:49.556 --> 00:02:54.266 A:middle
Yes. We still have
A sub N each time.

00:02:57.946 --> 00:03:00.916 A:middle
And notice -- well
this is 1 times it.

00:03:01.126 --> 00:03:03.636 A:middle
This is 2 times it,
so we have N times it.

00:03:09.356 --> 00:03:12.436 A:middle
And then we have X to
the 1 smaller power

00:03:12.436 --> 00:03:13.496 A:middle
than it was before.

00:03:13.966 --> 00:03:15.596 A:middle
This is 2 now it's 1.

00:03:23.556 --> 00:03:25.146 A:middle
Power rule yes.

00:03:27.046 --> 00:03:28.276 A:middle
And we have N here.

00:03:29.246 --> 00:03:35.956 A:middle
And then we have N minus 1,
we have 3 times 2, 4 times 3,

00:03:36.526 --> 00:03:42.546 A:middle
and then we have N
minus 2 good cool.

00:03:53.506 --> 00:03:56.426 A:middle
Distributing we get
the 2nd derivative.

00:03:57.546 --> 00:03:59.876 A:middle
Minus X squared times
the 2nd derivative.

00:04:01.056 --> 00:04:03.586 A:middle
Minus 5X times the
1st derivative.

00:04:04.496 --> 00:04:07.056 A:middle
Minus 3Y is equal to zero.

00:04:07.976 --> 00:04:10.166 A:middle
Here are the series that
we just came up with.

00:04:11.776 --> 00:04:14.956 A:middle
And I'm going to use the
following shorthand notation

00:04:14.956 --> 00:04:15.916 A:middle
for this video.

00:04:16.976 --> 00:04:20.896 A:middle
Series starts at 2 so instead of
writing N equal 2 to infinity,

00:04:21.286 --> 00:04:22.706 A:middle
I'm simply going to write 2.

00:04:24.166 --> 00:04:31.276 A:middle
So N, N minus 1, A sub
N, X to the N minus 2,

00:04:32.766 --> 00:04:36.236 A:middle
and then we have minus X
squared times the next series.

00:04:38.126 --> 00:04:38.976 A:middle
Starting at 2.

00:04:47.206 --> 00:04:50.486 A:middle
And then we have minus
5X times the next series.

00:04:52.656 --> 00:04:53.976 A:middle
That series starts at 1.

00:05:01.506 --> 00:05:04.406 A:middle
And then we have minus 3
times the next series starting

00:05:04.406 --> 00:05:04.976 A:middle
at zero.

00:05:10.566 --> 00:05:11.696 A:middle
Equals zero.

00:05:13.566 --> 00:05:15.616 A:middle
Now we notice that in front of 2

00:05:15.616 --> 00:05:18.766 A:middle
of these series we
have variables.

00:05:18.766 --> 00:05:21.436 A:middle
So we have X squared in front
of this series and we have X

00:05:21.436 --> 00:05:22.566 A:middle
in front of this series.

00:05:23.056 --> 00:05:26.006 A:middle
We need to distribute
those inside the series.

00:05:26.006 --> 00:05:27.906 A:middle
So that's our next step.

00:05:35.956 --> 00:05:38.276 A:middle
So we distribute
the X squared inside

00:05:39.016 --> 00:05:42.806 A:middle
which will change the exponent.

00:05:44.126 --> 00:05:52.546 A:middle
Instead of N minus 2 it
will now be N. Here instead

00:05:52.546 --> 00:05:55.376 A:middle
of the exponent being
N minus 1 it will be N.

00:06:07.886 --> 00:06:08.526 A:middle
Next step.

00:06:11.226 --> 00:06:21.166 A:middle
We notice that all of the
exponents are N except

00:06:21.166 --> 00:06:23.166 A:middle
for the one on the left.

00:06:23.226 --> 00:06:26.806 A:middle
Over here we have an
exponent of N minus 2.

00:06:26.806 --> 00:06:28.546 A:middle
We need to make them all match.

00:06:29.216 --> 00:06:31.016 A:middle
So we're going to
make a substitution.

00:06:32.326 --> 00:06:38.586 A:middle
So let's let K equal N minus 2.

00:06:39.696 --> 00:06:42.776 A:middle
So here we're going
to have X to the K

00:06:43.816 --> 00:06:46.796 A:middle
so this would be A sub what?

00:06:47.426 --> 00:06:53.406 A:middle
Well if K is N minus 2 then
N is equal to K plus 2?

00:06:54.086 --> 00:07:04.456 A:middle
So that's going to be K plus 2 N
minus 1 Then would be K plus 1.

00:07:07.556 --> 00:07:10.996 A:middle
N is going to be K plus 2.

00:07:14.856 --> 00:07:18.706 A:middle
Our starting value
was N equaled 2.

00:07:19.796 --> 00:07:24.006 A:middle
Well if N starts out at 2 then
K is going to start out at zero.

00:07:27.046 --> 00:07:30.566 A:middle
Everything else on this line
is just being copied down.

00:07:31.606 --> 00:07:32.966 A:middle
That was the only thing
we did on that step.

00:08:02.556 --> 00:08:05.296 A:middle
We now wish to combine the
series into one series.

00:08:06.196 --> 00:08:08.736 A:middle
We notice that we have
different starting values.

00:08:09.526 --> 00:08:14.356 A:middle
So the latest starting value is
the one that we're going to use.

00:08:14.966 --> 00:08:18.876 A:middle
And any series that starts
earlier then that, we will pull

00:08:18.876 --> 00:08:20.426 A:middle
out those earlier terms.

00:08:20.926 --> 00:08:23.426 A:middle
So this first series
starts out at zero.

00:08:23.766 --> 00:08:27.346 A:middle
So we'll pull out the term with
N equals zero and N equal 1.

00:08:28.216 --> 00:08:30.406 A:middle
So when N is equal to
zero or K whatever.

00:08:31.286 --> 00:08:40.976 A:middle
So when we substitute in zero
we get 2 times 1 A sub 2 X

00:08:41.056 --> 00:08:41.656 A:middle
to the zero.

00:08:42.616 --> 00:08:50.006 A:middle
We substitute in 1 we get 3
times 2 X sub 3 X to the first.

00:08:51.526 --> 00:08:56.106 A:middle
The next series starts out at
2 so we have no extra terms.

00:08:57.116 --> 00:08:58.976 A:middle
The next series starts out at 1.

00:08:58.976 --> 00:09:02.096 A:middle
So we have to pull out
the term with N equal 1

00:09:02.446 --> 00:09:08.266 A:middle
so we have minus 5 times
1 A sub 1 X to the first.

00:09:08.906 --> 00:09:12.146 A:middle
The next series starts out at
zero so we pull out the term

00:09:12.526 --> 00:09:14.026 A:middle
where N is equal to zero.

00:09:14.166 --> 00:09:21.316 A:middle
So we have minus 3 A sub zero X
to the zero and when N is equal

00:09:21.316 --> 00:09:26.206 A:middle
to 1 we have minus 3 A
sub 1 X to the first.

00:09:27.596 --> 00:09:31.216 A:middle
Now we have our series
starting off at 2.

00:09:33.336 --> 00:09:47.646 A:middle
So then we have N plus 2 N plus
1 A sub N plus 2 minus N N minus

00:09:47.646 --> 00:10:01.706 A:middle
1 A sub N minus 5 N A sub
N minus 3 A sub N and all

00:10:01.706 --> 00:10:04.686 A:middle
of that is being multiplied
times X to the Nth power

00:10:05.096 --> 00:10:06.236 A:middle
and that's equal to zero.

00:10:07.146 --> 00:10:08.866 A:middle
Now we equate coefficients.

00:10:11.706 --> 00:10:20.796 A:middle
Alright the constant on the left
hand side is 2 A sub 2 minus 3 A

00:10:20.796 --> 00:10:24.796 A:middle
sub zero that's equal to
the constant on the right.

00:10:25.446 --> 00:10:31.606 A:middle
Which tells us that A sub 2
must be 3 halves A sub zero.

00:10:34.056 --> 00:10:41.396 A:middle
The coefficient of X on the left
hand side is 6 A sub 3 minus 8 A

00:10:41.396 --> 00:10:42.056 A:middle
sub 1.

00:10:42.396 --> 00:10:45.026 A:middle
The coefficient of X on the
right hand side is zero.

00:10:45.716 --> 00:10:50.816 A:middle
Which tell us that A sub 3
must be 4 thirds A sub 1.

00:10:51.976 --> 00:10:56.186 A:middle
The coefficient of
all other powers of X

00:10:56.186 --> 00:10:59.976 A:middle
on the left hand side is
what's inside the brackets.

00:11:19.496 --> 00:11:24.396 A:middle
That's equal to zero
and then do the algebra.

00:11:26.356 --> 00:11:32.126 A:middle
And eventually you'll get
that A sub N plus 2 is equal

00:11:32.126 --> 00:11:41.696 A:middle
to N plus 3 over N plus 2
times A sub N and this was

00:11:41.796 --> 00:11:45.956 A:middle
for N greater than
or equal to 2.

00:11:56.376 --> 00:12:05.006 A:middle
Initially we had that A sub 2
was 3 halves times A sub zero.

00:12:06.176 --> 00:12:11.906 A:middle
And that A sub 3 was 4
thirds times A sub 1.

00:12:13.176 --> 00:12:18.136 A:middle
That would be for N
equals zero and N equal 1.

00:12:19.146 --> 00:12:23.986 A:middle
Now our recurrence relation
gives us other values.

00:12:24.876 --> 00:12:28.986 A:middle
If N is equal to 2 we substitute
it into this recurrence relation

00:12:29.466 --> 00:12:39.746 A:middle
and we get A sub 4 is 5
fourths times A sub 2.

00:12:41.106 --> 00:12:49.466 A:middle
But we know that A sub 2
is 3 halves A sub zero.

00:12:50.866 --> 00:13:00.916 A:middle
If N is 3 A sub 5
is 6 fifths A sub 3.

00:13:05.426 --> 00:13:11.066 A:middle
But A sub 3 is 4 thirds A sub 1.

00:13:14.496 --> 00:13:25.146 A:middle
N equal 4 gives us the A sub
6 is 7 over 6 times A sub 4.

00:13:28.686 --> 00:13:34.926 A:middle
But A sub 4 was 5 times
3 over the product

00:13:34.926 --> 00:13:39.956 A:middle
of 4 times 2 times A sub zero.

00:13:43.516 --> 00:13:53.886 A:middle
N equal 5 if A sub
7 8 sevenths A sub 5

00:13:55.326 --> 00:13:56.966 A:middle
which gives us 8 sevenths.

00:14:00.756 --> 00:14:04.436 A:middle
And we have 6 times 4

00:14:05.516 --> 00:14:16.436 A:middle
over 5 times 3 times A sub
1 N equals 6 we have A sub 8

00:14:19.706 --> 00:14:31.036 A:middle
which is 9 over 8 times A sub
6 and we can replace A sub 6

00:14:32.066 --> 00:14:34.746 A:middle
with 7 times 5 times 3

00:14:35.876 --> 00:14:41.836 A:middle
over 6 times 4 times
2 times A sub zero.

00:14:42.016 --> 00:14:46.636 A:middle
And you can continue on as
long as you want to with this.

00:14:55.946 --> 00:14:57.976 A:middle
So in general our
solution looks like this.

00:15:06.046 --> 00:15:06.946 A:middle
Our series.

00:15:11.106 --> 00:15:12.896 A:middle
Substitute in the
values we know.

00:15:15.926 --> 00:15:16.396 A:middle
Here we go.

00:15:17.406 --> 00:15:23.516 A:middle
So replacing A sub 2
with 3 halves A sub zero.

00:15:25.486 --> 00:15:33.456 A:middle
Replacing A sub 3
with 4 thirds A sub 1.

00:15:37.496 --> 00:15:40.116 A:middle
Replace A sub 4.

00:15:50.586 --> 00:15:51.976 A:middle
Replace A sub 5.

00:16:05.496 --> 00:16:06.496 A:middle
And continue on.

00:16:18.056 --> 00:16:20.546 A:middle
Substituting in the values
that we just determined.

00:16:30.476 --> 00:16:32.356 A:middle
Next let's factor
out our constants.

00:16:34.966 --> 00:16:39.016 A:middle
So we have 1 plus
3 halves X squared.

00:16:39.016 --> 00:16:45.686 A:middle
And then we have 5 times 3
over 4 times 2 X to the 4th.

00:16:46.916 --> 00:16:50.866 A:middle
For the next term we
have 7 times 5 times 3

00:16:51.776 --> 00:16:55.166 A:middle
over 6 times 4 times
2 X to the 6th.

00:16:56.006 --> 00:16:57.126 A:middle
And so on and so forth.

00:16:59.526 --> 00:17:03.076 A:middle
Then we factor out A sub
1 from the other terms.

00:17:04.486 --> 00:17:08.726 A:middle
We have X plus 4 thirds X cubed.

00:17:11.266 --> 00:17:16.866 A:middle
Then we have 6 times 4 over
5 times 3 X to the 5th.

00:17:16.866 --> 00:17:26.856 A:middle
Then we have 8 times 6 times
4 over 7 times 5 times 3 X

00:17:26.856 --> 00:17:28.836 A:middle
to the 7th and so
on and so forth.

00:17:32.596 --> 00:17:36.646 A:middle
They'll be times when
they'll be no obvious pattern.

00:17:37.306 --> 00:17:40.576 A:middle
And you'll be allowed
to just simplify

00:17:41.086 --> 00:17:43.806 A:middle
and write your answer like this.

00:17:47.276 --> 00:17:50.676 A:middle
For this problem we can
definitely determine

00:17:50.676 --> 00:17:51.196 A:middle
the pattern.

00:17:51.816 --> 00:17:52.766 A:middle
And we're going to do so.

00:18:02.386 --> 00:18:04.046 A:middle
We're going to go
ahead and start at 1.

00:18:06.496 --> 00:18:11.646 A:middle
Alright so this term
is N is equal to 1.

00:18:13.156 --> 00:18:14.336 A:middle
This is N equal 2.

00:18:15.066 --> 00:18:16.316 A:middle
This is N equal 3.

00:18:18.736 --> 00:18:22.116 A:middle
So let's write down the term
that has the most information

00:18:22.116 --> 00:18:24.746 A:middle
for us, which is the last term.

00:18:26.046 --> 00:18:35.656 A:middle
So 7 times 5 times 3 over 6
times 4 times 2 X to the 6th.

00:18:36.316 --> 00:18:40.296 A:middle
And that's when N is equal to 3.

00:18:41.976 --> 00:18:46.596 A:middle
So for the numerator we
have 7 when N is equal to 3.

00:18:47.456 --> 00:18:53.176 A:middle
So possibilities for
odd representations

00:18:54.686 --> 00:19:00.836 A:middle
of numbers are 2N plus 1, 2N
minus 1, 2N plus 3, 2N minus 5,

00:19:00.836 --> 00:19:03.886 A:middle
there's all kinds of different
representations of odd numbers.

00:19:04.466 --> 00:19:06.786 A:middle
Play around with it until
you find the one that works.

00:19:07.116 --> 00:19:09.046 A:middle
And yes it's 2N plus 1.

00:19:10.286 --> 00:19:17.446 A:middle
So we need a representation of
an odd integer where we get 7

00:19:17.446 --> 00:19:19.946 A:middle
as our result when
N is equal to 3.

00:19:21.046 --> 00:19:23.556 A:middle
Subtract 2 to go down
to the next odd integer.

00:19:24.846 --> 00:19:26.596 A:middle
And so on and so forth.

00:19:27.006 --> 00:19:31.056 A:middle
Down to 5 times 3
for our numerator.

00:19:31.826 --> 00:19:34.176 A:middle
Denominators a little
bit easier in my opinion.

00:19:34.966 --> 00:19:37.696 A:middle
We're going to pull out a 2
from each of these factors.

00:19:37.696 --> 00:19:39.086 A:middle
We have 2 times 3.

00:19:39.926 --> 00:19:41.496 A:middle
Here we have 2 times 2.

00:19:42.436 --> 00:19:44.176 A:middle
Here we have 2 times 1.

00:19:45.246 --> 00:19:52.776 A:middle
So I have 2 to the
third times 3 factorial,

00:19:53.016 --> 00:19:55.386 A:middle
ut that was when
N was equal to 3.

00:19:56.026 --> 00:20:01.006 A:middle
So my denominator is 2 to
the Nth times N factorial.

00:20:02.386 --> 00:20:04.456 A:middle
And what's my exponent for X?

00:20:05.646 --> 00:20:07.516 A:middle
Well it's 2N yes.

00:20:15.836 --> 00:20:17.136 A:middle
Alright. So that's
my first summation.

00:20:22.426 --> 00:20:25.026 A:middle
Second one.

00:20:26.746 --> 00:20:29.126 A:middle
So the term with the
most information for us.

00:20:29.226 --> 00:20:30.066 A:middle
So we have 8.

00:20:30.956 --> 00:20:39.026 A:middle
That's 8 times 6 times 4 over
7 times 5 times 3 X to the 7th.

00:20:39.536 --> 00:20:41.826 A:middle
And that's when N is
equal to 3 once again.

00:20:43.536 --> 00:20:46.496 A:middle
So if you look at
it the denominator

00:20:46.746 --> 00:20:49.796 A:middle
of this fraction is the
same as the numerator

00:20:49.796 --> 00:20:50.786 A:middle
for the last fraction.

00:20:50.786 --> 00:21:02.636 A:middle
So we don't have to
reinvent the wheel.

00:21:02.636 --> 00:21:10.276 A:middle
And the exponent it's
the same representation

00:21:10.976 --> 00:21:16.626 A:middle
because we have 7 here and
when N is equal to 3 we have 7

00:21:16.626 --> 00:21:18.046 A:middle
so it's 2N plus 1 again.

00:21:20.656 --> 00:21:21.466 A:middle
Now the numerator.

00:21:22.016 --> 00:21:24.916 A:middle
So we're going to pull out a
2 from each of those factors.

00:21:25.346 --> 00:21:34.926 A:middle
So we have 2 times 4, 2
times 3, and 2 times 2 yeah.

00:21:35.706 --> 00:21:39.226 A:middle
Let's clean that
up a little bit.

00:21:39.226 --> 00:21:40.786 A:middle
Looking just at that.

00:21:42.756 --> 00:21:48.196 A:middle
So I have 2 to the 3rd times
4 times 3 times 2 basically

00:21:48.196 --> 00:21:49.816 A:middle
times 1.

00:21:50.126 --> 00:21:53.536 A:middle
So I have 4 factorial
when N is equal to 3.

00:21:53.536 --> 00:21:57.256 A:middle
So I have 2 to the 3rd times 4
factorial when N is equal to 3.

00:21:57.786 --> 00:22:02.666 A:middle
So I have 2 to the Nth but
I don't have N factorial,

00:22:02.666 --> 00:22:06.656 A:middle
I have N plus 1 factorial.

00:22:08.606 --> 00:22:13.146 A:middle
And now I have my
solution written in terms

00:22:13.146 --> 00:22:16.976 A:middle
of summations with
sigma notation.

00:22:33.286 --> 00:22:38.096 A:middle
By the ratio test it can be
shown that these series converge

00:22:38.776 --> 00:22:41.386 A:middle
for the absolute value
of X less than 1.

00:22:42.116 --> 00:22:45.016 A:middle
Which is the result that
we originally expected.