Welcome back to recitation.
In this video, I'd like us
to do some basic manipulation
of some power series that
you saw in the lecture.
So I have four problems here,
and what I'd like us to do,
is figure out what function
each of these series represents.
Now, you can assume that
the x-values that we
are going to insert
into this function
are only x-values that
let the sums converge.
So you don't have to
worry about anything
to do with convergence
of these sums.
Just assume the sums
converge, that we've
picked good x-values.
OK?
And what I'd like you
to have in the end,
for (a), (b), (c), and
(d), is something like,
this sum is equal to
a specific function.
It should not include
a sum anymore.
So why don't you work on those
for a bit, pause the video,
and when you're done,
restart the video,
and I'll come back, and show
you how I work with them.
All right.
Welcome back.
Well, hopefully these were
a little bit fun for you.
I always liked them,
the first time I
saw them, to see how one
could manipulate these series.
So again, what
we're trying to do
is figure out what function
these series represent.
We're assuming convergence.
And we're going to try
and manipulate them
to look like things
we already know.
So I'm going to start.
I'm just going to go straight
through (a), (b), (c), and (d).
And I'm going to rewrite the
problem each time, because I
don't want to keep coming back.
So (a), we had the sum n equals
0 to infinity, x to the n
plus 2 over n factorial.
Well, this looks very
close to something we know.
It looks a lot like the one,
the function e to the x.
The difference is that e to the
x just has a power x to the n.
But the good news is that
I am really close to that.
What I really have in the
numerator is x to the n times
x squared.
Because this x to
the n plus 2, I
can write as x to the
n times x squared.
So every term has an x squared.
So as was mentioned
in the lecture,
you can treat this
really like polynomials.
In some way, you
can factor this out.
So I can rewrite this as-- well,
I'll rewrite it in two steps.
0 to infinity of x squared,
x to the n over n factorial.
So x squared times x to the
n gives me x to the n plus 2.
Right?
But I can actually now, because
this belongs to every sum,
I can pull that all
the way out in front.
Right?
And if I pull that all
the way out in front,
if I move this out
to the front, I
have an x squared
times this sum.
Well, what is the sum?
The sum is e to the x.
It's n equals 0 to infinity of
x to the n over n factorial.
That's just e to the x.
So this function is
x squared e to the x.
That's what (a) is.
If you were worried
about it, you
could write e to
the x as a series,
and then you could multiply
through by x squared,
and see if that's what you get.
But you'll see, that is
indeed how this problem works.
OK.
Let's look at (b).
OK, (b) is equal to the sum--
that's a weird summation
sign, sorry about that-- (b)
is equal to the sum n equals 2
to infinity of x to the n.
That's what we
wanted to look at.
Well, this looks very close
to the geometric series,
but it's missing some terms.
Right?
The geometric series starts
at n equals 0 to infinity.
But the point I
want to make here
is that I can rewrite this
as the geometric series,
and then I can take
away what I've added in.
OK?
So these do not agree right now.
This equals sign
is not true yet.
But what do I notice?
I notice that this one has two
more terms at the beginning
than this one has.
What are those terms?
Those are when n equals
0 and when n equals 1.
Right?
There's no n equals
0 or n equals 1 here.
The formulas are
exactly the same.
Right?
x to the n, x to the n.
They both go to infinity, but
one is starting at n equals 2,
and one's starting
at n equals 0.
Again, I want to remind you.
Why did I bother to
write this thing at all?
Because this is a
function we know.
But this equals sign
is not true right now.
Right?
I have to get rid of
the extra stuff I added.
What did I add?
I added-- let me
get this as a sum.
What did I add on?
I added on x to the
zero and x to the first.
So now I guess I should have
written subtract here, right?
I should subtract
x to the 0, and I
should subtract x to the first.
I've already said
it once, but I'm
going to, just to make
sure everybody follows,
say it one more time.
I now-- I've taken the
summation I started with,
which went from n
equals 2 to infinity.
I added two more
terms to the sum.
I made it go from 0 to infinity.
So to keep equality,
I subtracted off
the two things I added in.
I subtracted off x to
the 0 and x to the first.
So now, what was the point
of this part of the exercise?
Well, the point is that
I know what this sum is.
That's the geometric series.
That's 1 over 1 minus x.
And this x to the 0, it's minus
1; x to the first, minus x.
So this I broke
up into something
I knew as a power series,
and then other pieces
that I had added in to make
it look like something I knew.
I had to subtract those off.
So that's the idea.
That one, if you wanted to
go on and simplify further,
you could do that.
But I'm willing to
leave it just as is.
'Cause the idea was really
this part up here, and then
translating it down to this.
All right.
Let me write (c) again.
The sum n equals
0 to infinity, x
to the n over n factorial
plus x to the n.
OK.
So the sum n equals
0 to infinity,
x to the n over n
factorial plus x to the n.
And the point to
recognize here is, again,
as Professor Jerison
mentioned, you're
really treating these
almost like polynomials.
So you're taking a series, and
you're adding these terms up.
But really, what you
can think about this,
is the two separate
series added up.
And so this is the
sum from n equals 0
to infinity of x to the n over
n factorial, which we know,
plus the sum from n
equals 0 to infinity
of x to the n, which we know.
The first one is
just e to the x,
and the second one,
we've been dealing
a lot with these already today,
is just 1 over 1 minus x.
So the point I want
to make, is if you
have this convergent
series, you can split it up
into pieces over the sum.
And these two are both
convergent, we know,
separately, and so we
can write what they are.
x to the n over n factorial from
0 to infinity is e to the x,
x to the n from n equals 0 to
infinity is 1 over 1 minus x.
So now we just have one more.
Let me write that one. (d)
was summation n equals minus 1
to infinity x to the n plus 1.
All right.
This was not meant
to scare you, but it
was meant to test your
understanding of sigma
notation.
So the problem is, we're
starting at n equals minus 1,
but x is starting-- the
exponent on x is at n plus 1.
So I want to write it in
some form that I know.
Well, let's try and get
the subscript to be 0.
so what I'm going to do,
is I'm going to change
the name of the subscript.
I'm going to let it be
m equals 0 to infinity.
OK?
And I want m to count up by
1 just the way n counts up.
So notice what I did.
When m equals 0, n
equals negative 1.
That's what we've set
up as the first term.
That means n plus 1 is
equal to 0 when m equals 0.
And since I'm going
up by one every time
in my summation, my
iteration, here that
means that for every m I have,
I just have to take n plus 1
to figure out what m is.
So I should have said
it the other way.
For every n I have, I just
have to add 1 to get m.
OK?
So here, if I start at
negative 1, I start here at 0.
The next term here
is 0, n equals 0,
but the next term
here is m equals 1.
The next term here
would be n equals 1,
and the next term here
would be m equals 2.
So they're just off by
1, but they're still
catching every index.
Now, I don't have to
change what's up here.
Because infinity--
if I add 1 to it,
I'm still going off to infinity.
So I don't have to
change what's up here.
But I do want to write the
formula, or the formula I have
inside the sum in terms of m.
But I've already got it.
Because I know m is
always equal to n plus 1,
so I can replace this
n plus 1 by an m.
And now we know
what that one is.
Right?
m equals 0 to infinity
of x to the m.
That's our geometric series.
So it actually, even
though I wrote it
in kind of a funny way,
it was actually just still
the geometric series.
I just moved the
indices a little bit
to make sure we could
play with those.
So the idea here, the whole
point of this exercise,
just to recognize how
you can manipulate
these series a little
bit, so that if they're
in a form that looks kind
of like one of the functions
you know, you can see if
it actually is, you know,
a product of something
with the function you know,
or the sum of two
functions you know,
or maybe one of the functions
you know is a power series,
and then you have to drop
off a couple of terms.
So they each sort of
demonstrate a different idea
of how you can manipulate these
convergent power series, based
on functions you already know.
So I'll stop there.
