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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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00:04:04,200 --> 00:04:06,700
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
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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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00:08:27,560 --> 00:08:31,410
So here, if I start at
negative 1, I start here at 0.
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00:08:31,410 --> 00:08:33,920
The next term here
is 0, n equals 0,
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00:08:33,920 --> 00:08:36,620
but the next term
here is m equals 1.
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00:08:36,620 --> 00:08:39,729
The next term here
would be n equals 1,
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and the next term here
would be m equals 2.
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00:08:41,520 --> 00:08:43,510
So they're just off by
1, but they're still
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00:08:43,510 --> 00:08:45,530
catching every index.
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00:08:45,530 --> 00:08:47,480
Now, I don't have to
change what's up here.
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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.
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00:08:51,880 --> 00:08:53,840
So I don't have to
change what's up here.
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00:08:53,840 --> 00:08:57,420
But I do want to write the
formula, or the formula I have
202
00:08:57,420 --> 00:08:59,790
inside the sum in terms of m.
203
00:08:59,790 --> 00:09:01,180
But I've already got it.
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00:09:01,180 --> 00:09:04,340
Because I know m is
always equal to n plus 1,
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00:09:04,340 --> 00:09:08,916
so I can replace this
n plus 1 by an m.
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00:09:08,916 --> 00:09:10,290
And now we know
what that one is.
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00:09:10,290 --> 00:09:11,810
Right?
208
00:09:11,810 --> 00:09:14,670
m equals 0 to infinity
of x to the m.
209
00:09:14,670 --> 00:09:16,040
That's our geometric series.
210
00:09:16,040 --> 00:09:18,490
So it actually, even
though I wrote it
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00:09:18,490 --> 00:09:22,510
in kind of a funny way,
it was actually just still
212
00:09:22,510 --> 00:09:24,000
the geometric series.
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00:09:24,000 --> 00:09:26,430
I just moved the
indices a little bit
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00:09:26,430 --> 00:09:28,760
to make sure we could
play with those.
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00:09:28,760 --> 00:09:31,830
So the idea here, the whole
point of this exercise,
216
00:09:31,830 --> 00:09:33,740
just to recognize how
you can manipulate
217
00:09:33,740 --> 00:09:36,520
these series a little
bit, so that if they're
218
00:09:36,520 --> 00:09:38,770
in a form that looks kind
of like one of the functions
219
00:09:38,770 --> 00:09:42,407
you know, you can see if
it actually is, you know,
220
00:09:42,407 --> 00:09:44,490
a product of something
with the function you know,
221
00:09:44,490 --> 00:09:46,870
or the sum of two
functions you know,
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00:09:46,870 --> 00:09:49,390
or maybe one of the functions
you know is a power series,
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00:09:49,390 --> 00:09:52,210
and then you have to drop
off a couple of terms.
224
00:09:52,210 --> 00:09:54,930
So they each sort of
demonstrate a different idea
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00:09:54,930 --> 00:09:59,070
of how you can manipulate these
convergent power series, based
226
00:09:59,070 --> 00:10:00,820
on functions you already know.
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00:10:00,820 --> 00:10:02,370
So I'll stop there.