WEBVTT

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Welcome back to recitation.

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In this video, I'd like us
to do some basic manipulation

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of some power series that
you saw in the lecture.

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So I have four problems here,
and what I'd like us to do,

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is figure out what function
each of these series represents.

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Now, you can assume that
the x-values that we

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are going to insert
into this function

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are only x-values that
let the sums converge.

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So you don't have to
worry about anything

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to do with convergence
of these sums.

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Just assume the sums
converge, that we've

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picked good x-values.

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OK?

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And what I'd like you
to have in the end,

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for (a), (b), (c), and
(d), is something like,

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this sum is equal to
a specific function.

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It should not include
a sum anymore.

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So why don't you work on those
for a bit, pause the video,

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and when you're done,
restart the video,

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and I'll come back, and show
you how I work with them.

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All right.

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Welcome back.

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Well, hopefully these were
a little bit fun for you.

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I always liked them,
the first time I

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saw them, to see how one
could manipulate these series.

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So again, what
we're trying to do

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is figure out what function
these series represent.

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We're assuming convergence.

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And we're going to try
and manipulate them

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to look like things
we already know.

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So I'm going to start.

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I'm just going to go straight
through (a), (b), (c), and (d).

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And I'm going to rewrite the
problem each time, because I

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don't want to keep coming back.

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So (a), we had the sum n equals
0 to infinity, x to the n

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plus 2 over n factorial.

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Well, this looks very
close to something we know.

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It looks a lot like the one,
the function e to the x.

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The difference is that e to the
x just has a power x to the n.

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But the good news is that
I am really close to that.

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What I really have in the
numerator is x to the n times

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x squared.

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Because this x to
the n plus 2, I

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can write as x to the
n times x squared.

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So every term has an x squared.

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So as was mentioned
in the lecture,

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you can treat this
really like polynomials.

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In some way, you
can factor this out.

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So I can rewrite this as-- well,
I'll rewrite it in two steps.

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0 to infinity of x squared,
x to the n over n factorial.

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So x squared times x to the
n gives me x to the n plus 2.

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Right?

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But I can actually now, because
this belongs to every sum,

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I can pull that all
the way out in front.

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Right?

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And if I pull that all
the way out in front,

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if I move this out
to the front, I

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have an x squared
times this sum.

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Well, what is the sum?

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The sum is e to the x.

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It's n equals 0 to infinity of
x to the n over n factorial.

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That's just e to the x.

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So this function is
x squared e to the x.

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That's what (a) is.

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If you were worried
about it, you

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could write e to
the x as a series,

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and then you could multiply
through by x squared,

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and see if that's what you get.

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But you'll see, that is
indeed how this problem works.

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OK.

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Let's look at (b).

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OK, (b) is equal to the sum--
that's a weird summation

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sign, sorry about that-- (b)
is equal to the sum n equals 2

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to infinity of x to the n.

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That's what we
wanted to look at.

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Well, this looks very close
to the geometric series,

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but it's missing some terms.

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Right?

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The geometric series starts
at n equals 0 to infinity.

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But the point I
want to make here

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is that I can rewrite this
as the geometric series,

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and then I can take
away what I've added in.

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OK?

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So these do not agree right now.

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This equals sign
is not true yet.

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But what do I notice?

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I notice that this one has two
more terms at the beginning

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than this one has.

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What are those terms?

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Those are when n equals
0 and when n equals 1.

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Right?

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There's no n equals
0 or n equals 1 here.

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The formulas are
exactly the same.

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Right?

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x to the n, x to the n.

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They both go to infinity, but
one is starting at n equals 2,

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and one's starting
at n equals 0.

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Again, I want to remind you.

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Why did I bother to
write this thing at all?

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Because this is a
function we know.

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But this equals sign
is not true right now.

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Right?

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I have to get rid of
the extra stuff I added.

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What did I add?

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I added-- let me
get this as a sum.

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What did I add on?

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I added on x to the
zero and x to the first.

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So now I guess I should have
written subtract here, right?

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I should subtract
x to the 0, and I

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should subtract x to the first.

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I've already said
it once, but I'm

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going to, just to make
sure everybody follows,

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say it one more time.

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I now-- I've taken the
summation I started with,

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which went from n
equals 2 to infinity.

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I added two more
terms to the sum.

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I made it go from 0 to infinity.

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So to keep equality,
I subtracted off

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the two things I added in.

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I subtracted off x to
the 0 and x to the first.

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So now, what was the point
of this part of the exercise?

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Well, the point is that
I know what this sum is.

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That's the geometric series.

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That's 1 over 1 minus x.

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And this x to the 0, it's minus
1; x to the first, minus x.

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So this I broke
up into something

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I knew as a power series,
and then other pieces

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that I had added in to make
it look like something I knew.

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I had to subtract those off.

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So that's the idea.

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That one, if you wanted to
go on and simplify further,

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you could do that.

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But I'm willing to
leave it just as is.

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'Cause the idea was really
this part up here, and then

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translating it down to this.

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All right.

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Let me write (c) again.

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The sum n equals
0 to infinity, x

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to the n over n factorial
plus x to the n.

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OK.

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So the sum n equals
0 to infinity,

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x to the n over n
factorial plus x to the n.

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And the point to
recognize here is, again,

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as Professor Jerison
mentioned, you're

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really treating these
almost like polynomials.

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So you're taking a series, and
you're adding these terms up.

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But really, what you
can think about this,

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is the two separate
series added up.

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And so this is the
sum from n equals 0

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to infinity of x to the n over
n factorial, which we know,

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plus the sum from n
equals 0 to infinity

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of x to the n, which we know.

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The first one is
just e to the x,

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and the second one,
we've been dealing

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a lot with these already today,
is just 1 over 1 minus x.

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So the point I want
to make, is if you

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have this convergent
series, you can split it up

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into pieces over the sum.

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And these two are both
convergent, we know,

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separately, and so we
can write what they are.

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x to the n over n factorial from
0 to infinity is e to the x,

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x to the n from n equals 0 to
infinity is 1 over 1 minus x.

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So now we just have one more.

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Let me write that one. (d)
was summation n equals minus 1

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to infinity x to the n plus 1.

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All right.

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This was not meant
to scare you, but it

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was meant to test your
understanding of sigma

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notation.

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So the problem is, we're
starting at n equals minus 1,

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but x is starting-- the
exponent on x is at n plus 1.

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So I want to write it in
some form that I know.

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Well, let's try and get
the subscript to be 0.

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so what I'm going to do,
is I'm going to change

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the name of the subscript.

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I'm going to let it be
m equals 0 to infinity.

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OK?

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And I want m to count up by
1 just the way n counts up.

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So notice what I did.

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When m equals 0, n
equals negative 1.

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That's what we've set
up as the first term.

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That means n plus 1 is
equal to 0 when m equals 0.

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And since I'm going
up by one every time

00:08:12.280 --> 00:08:15.630
in my summation, my
iteration, here that

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means that for every m I have,
I just have to take n plus 1

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to figure out what m is.

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So I should have said
it the other way.

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For every n I have, I just
have to add 1 to get m.

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OK?

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So here, if I start at
negative 1, I start here at 0.

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The next term here
is 0, n equals 0,

00:08:33.920 --> 00:08:36.620
but the next term
here is m equals 1.

00:08:36.620 --> 00:08:39.729
The next term here
would be n equals 1,

00:08:39.729 --> 00:08:41.520
and the next term here
would be m equals 2.

00:08:41.520 --> 00:08:43.510
So they're just off by
1, but they're still

00:08:43.510 --> 00:08:45.530
catching every index.

00:08:45.530 --> 00:08:47.480
Now, I don't have to
change what's up here.

00:08:47.480 --> 00:08:50.100
Because infinity--
if I add 1 to it,

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I'm still going off to infinity.

00:08:51.880 --> 00:08:53.840
So I don't have to
change what's up here.

00:08:53.840 --> 00:08:57.420
But I do want to write the
formula, or the formula I have

00:08:57.420 --> 00:08:59.790
inside the sum in terms of m.

00:08:59.790 --> 00:09:01.180
But I've already got it.

00:09:01.180 --> 00:09:04.340
Because I know m is
always equal to n plus 1,

00:09:04.340 --> 00:09:08.916
so I can replace this
n plus 1 by an m.

00:09:08.916 --> 00:09:10.290
And now we know
what that one is.

00:09:10.290 --> 00:09:11.810
Right?

00:09:11.810 --> 00:09:14.670
m equals 0 to infinity
of x to the m.

00:09:14.670 --> 00:09:16.040
That's our geometric series.

00:09:16.040 --> 00:09:18.490
So it actually, even
though I wrote it

00:09:18.490 --> 00:09:22.510
in kind of a funny way,
it was actually just still

00:09:22.510 --> 00:09:24.000
the geometric series.

00:09:24.000 --> 00:09:26.430
I just moved the
indices a little bit

00:09:26.430 --> 00:09:28.760
to make sure we could
play with those.

00:09:28.760 --> 00:09:31.830
So the idea here, the whole
point of this exercise,

00:09:31.830 --> 00:09:33.740
just to recognize how
you can manipulate

00:09:33.740 --> 00:09:36.520
these series a little
bit, so that if they're

00:09:36.520 --> 00:09:38.770
in a form that looks kind
of like one of the functions

00:09:38.770 --> 00:09:42.407
you know, you can see if
it actually is, you know,

00:09:42.407 --> 00:09:44.490
a product of something
with the function you know,

00:09:44.490 --> 00:09:46.870
or the sum of two
functions you know,

00:09:46.870 --> 00:09:49.390
or maybe one of the functions
you know is a power series,

00:09:49.390 --> 00:09:52.210
and then you have to drop
off a couple of terms.

00:09:52.210 --> 00:09:54.930
So they each sort of
demonstrate a different idea

00:09:54.930 --> 00:09:59.070
of how you can manipulate these
convergent power series, based

00:09:59.070 --> 00:10:00.820
on functions you already know.

00:10:00.820 --> 00:10:02.370
So I'll stop there.