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Transform!

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[MUSIC]

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We have some series that we already know
about.

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Well, for example I know that the sum n
goes from 0 to infinity of x to the n.

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That's a power series for a function we
already know.

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Its just the power series for one over one
minus x.

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And we've got some operations that we can
apply to power series.

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Well, we can differentiate a power series,
we can integrate

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a power series term by term, I can
multiply together

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a power series and I can divide power
series.

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Now we can put together the series that we
know and the

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operations that we can do in order to
build new power series.

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Well, here's our goal.

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I'd like to find a power series for the
function 1/(1-x)^2.

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We can do this in two different ways.
Let's notice something about this.

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This e is the derivative of 1/(1-x).
Why is that?

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Well, let's just work it out right, this
is the derivative I'm

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claiming of 1 minus x to the minus 1st
power, well differentiate this

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using the power rule that's negative 1
times 1 minus x to

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the negative 2nd power times the
derivative of the inside by the chain

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rule which is just multiplying this by
another copy of minus 1.

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So this is minus 1, times 1 minus x to the
negative second power, times minus 1.

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Well, what happens, right, these minus 1s
cancel.

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And what I'm left with, is just 1 minus x
to the minus second power.

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Well, I could write that as 1 over 1 minus
x squared.

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That's exactly what I've got here.

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So the derivative of 1 over 1 minus

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x is this function that I'm interested in?

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So it's differentiate the power series
term by term.

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So this is the derivative of the sum n
goes from 0 to infinity of x to the n.

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And, like I said, differentiating term by
term.

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This is the sum n goes from 1 to infinity.

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Of the derivative of nth term the
derivative of x to the n,

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note that I changed from n equals 0, n
equals 1 because when derivative

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the n equals 0 terms that's the derivative
of a constant which is just 0.

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So when I start here with n equals 1,
alright

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so this is the sum and it goes from 1 to

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infinity what's the derivative of x to the
n by the

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power rule that's n times x to the n minus
1.

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And now if I don't like having the n minus
1 there I

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could re-index this series, right.
What happens when I plug in n equals 1?

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Well, that's 1 times x to the 0.

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Right.

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And then what happens when I plug in n
equals 2?

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Well, that's 2 times x to the 1st.
What happens when I plug in n equals 3?

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That's 3 times x squared, and so on.
So I could rewrite

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this series as just the sum n goes from 0
to infinity

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of (n+1)*x^n.
And there we go, I've written down a power

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series for the function 1/(1-x)^2.
Of course, if I

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don't like thinking like this, if I don't
like differentiating, I can also approach

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this problem by just trying to multiply
together two copies of the original power

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series.
Well, what I mean, is that the sum of x^n

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n goes from zero to infinity is 1/(1-x).

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That means that if I square this side and
square this side I

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should get a formula then, a power series
for 1 over 1 minus x quantity squared.

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But how do I write that as a power series?

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Well, lets at least get started, lets at

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least figure out what the first few terms
are.

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So I'm just going to write down the sum of

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x to the n or you know the first few terms
of it.

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So that's 1 plus x plus x squared plus x
cubed plus...

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and I'm squaring that so multiplying it by
itself so I'll write down

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the same thing again, one plus x plus x
squared, plus x cubed plus...

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Now I want to figure out what happens when
I multiply these things together.

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So what do I get?
Well, 1 times 1, thats 1.

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How many x

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terms do I get here?

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Well, I've got a 1 times and x, or I've
got an x times a 1.

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Now there's no other way to multiply these
things and get and x.

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So theres plus 2 x.

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How many x squareds do I get out?

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Well, I can multiply this 1 times this x
squared,

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multiply this x times this x or I can
multiply this

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x squared times this 1 but I got 3x
squared

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and there's no other way to get an x
squared out.

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how can

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I get an x cubed?
Well, I can multiply 1 times x cubed.

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Or x times x squared or this x squared
times this x or this x cubed times 1, so

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that gives me 4 ways of getting an x cubed
term and alright plus...

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Now, maybe you believe that pattern
continues.

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Well, it certainly looks like this is
giving me

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the sum n goes from 0 to infinity of

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n plus 1 times x to the nth power.
I mean, n equals 0 is 1

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times x to the 0, n equals 1 is 2 times x
to the first, n

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equals 2 is 3 times x squared, n equals 3
is 4 times x cubed and so on.

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To make that more rigorous, we probably
have to talk about induction.

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Or we could just bring up a theorem on
multiplying power series.

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Well, here's out theorem for multiplying
power series.

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The product of this power series and this

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power series is given by this power
series.

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And it's a little bit complicated to see
how the coefficient's affected, right?

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Here the coefficient for a suborn, here
the coefficients for

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b suborn and when I multiple these
together I get

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this concolved series the sum i goes from
0 to

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n of a suborn times b suborn n minus i.

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Now we can apply that theorem,

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well in this case our formula for 1 over 1

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minus x is just for all the coefficients
are 1.

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So if I multiply 1 over 1 minus x by 1
over

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1 minus x, then I get 1 over 1 minus x
squared.

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So that means all the a sub i's and b sub
n minus i's are

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all just 1, so here is a power series for
1 over 1 minus x squared.

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Lets simply, well the sum i goes from 0 to
n of just

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1 or that's 1 plus 1 plus 1 but its n

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plus 1 once.
So I can really write this power series

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which is the sum n goes to 0 from infinity
of n plus 1 times x to the n and

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that's exactly what we got before by

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using derivatives.

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