- [Voiceover] What I would like us to do
in this video is find the power series representation
of or find the power series approximitation (chuckles)
the power series approximation of arctangent
of two x centered at zero
and let's just say we want the first four nonzero terms
of the power series approximation of arctangent
of two x centered at zero
so it's essentially the Maclaurin Series of arctangent
of two x, the first four nonzero terms of it.
If you feel good about it,
I encourage you to pause the video
and try to work through it yourself.
You might have tried to work through it
and you've probably took the first
derivative of it.
You probably saw that hey, the derivative
with respect to x of arctangent
of two x is equal to and this is a refresher
if you didn't realize it the first time.
It's going to be the derivative of arctangent of x
is one over one plus x squared
so this is going to be
the derivative of this which is two
over one plus this whole thing squared.
One plus, I would say one plus four x squared
and then as you tried to find more of the terms
of the Maclaurin Series,
you would have taken more derivatives of this
and it would have gotten very hairy,
very fast if especially if you're looking
for the first four nonzero terms.
You probably realize hey,
there must be some type of an insight
that I hadn't fully appreciated yet
when I just tried to just power through
or no pun intended, power through
finding the power series,
the first four nonzero terms of the power series
centered zero of arc tangent of two x.
You are right, there is a key insight here.
The key insight here is well,
instead of doing it directly,
let's see if we could find the power series
representation, the first four terms of this thing
right over here and then we can take the antiderivative
of that to find the power series
of arctangent of two x making sure
that we get the constant right
that it satisfies the fact that we are being centered
at zero, so I know what you're thinking now.
Well that seems like it's getting us
to the exact same issue.
If I want to find the power series representation of this,
the first four terms of it,
I'm still having to take the derivative of this
multiple times it seems just as hard
but this is the key insight I guess you could say.
The key insight is let's just say f of x
which of course is the derivative
of arctangent of two x
is two over one
plus four x squared.
Now, if we had another function
that cleans it up a little bit
so that we don't get all these hairiness
when we take the derivatives.
Let's say we had another function g of x,
I'm using the color that I have not used.
Let's say, that I have g of x
is equal to one over one plus x.
This is an interesting thing because it's really easy
and this is the same thing as one plus x
to the negative one power.
g of x is interesting because it's really easy
to take its derivatives.
For example, g prime of x
is going to be equal to chain rule
derivative one plus x is just one
so it's going to be equal to negative one plus x
to the negative two power
if I want to take the second derivative of that.
g prime prime of x that's going to be negative two
times negative one it's two times one plus x
to the negative three power.
If I want to take the third derivative of that,
that's going to be, let's see.
Negative three times two its negative six
times one plus x to the negative four power.
I know you're saying, "Sal,
"aren't we worried about this?
"Why are you doing this?"
Well just bare with me for a second.
Just said quickly, I was able to find
the first three derivatives of g of x
and it's very easy then to find the first four terms
of its power series representation
especially it's Maclaurin,
that's Maclaurin series
if the power series centered at zero.
We just have to evaluate each of these at zero.
g of zero is equal to one,
g prime of zero
is equal to negative one.
g prime, prime of zero,
so one plus zero and then the negative three,
that's just one times two is equal to two
and then the third derivative evaluated at zero
is equal to negative six.
I could write the g of x
is approximately equal,
so I'm just going to do the first four terms here
is going to be g of zero which is one
minus g prime of zero times x.
That's minus one times x, so that's negative x
plus g prime, prime of zero
two over two factorial times x squared.
Well this is just one times x squared
so let me just write that.
That's just going to be plus x squared
and then we have plus g prime, prime, prime of zero
which is negative six over three factorial
times x to the third.
Well three factorials are six
so negative six divided by six is just negative one.
That's going to be negative x to the third,
negative x to the third and I know what you're thinking.
"All right Sal, so you just started with a hard problem
"and you gave yourself a much easier problem
"to find the power series representation.
"How is this useful?"
Well this is the key insight that I've been promising
throughout this video so far.
The key insight, the long promised key insight
is that and I'm finding a suitable color for a key insight
is that we can write f of x.
Notice f of x is just two times,
f of x is two times
g of four x squared.
Notice you replace your x's with a four x squared.
You're going to have one over one plus four x squared
and then you multiply that whole thing times two,
you get this thing right over here.
If f of x is equal to that
then f of x is power series representation
its just going to be taking this power series
or at least the first four terms of it
and replacing the x's with four x squares
and then multiplying the whole thing times two,
so let's do that.
We can write
that f of x,
so we could write the f of x
is going to be approximately equal to
two times this thing,
a value when x is equal to four x squared.
It is one minus instead of an x,
I'm going to write a four x squared
plus x squared but instead of an x,
I have a four x squared squared.
This is plus four x squared squared.
Well that's going to be 16 x to the fourth,
so let me just write that,
that's going to be plus 16 x to the fourth
and finally minus x to the third
but now an x is four x squared
so it's minus four x squared to the third power.
Well that's going to be 64 x to the sixth.
Let me write that, minus 64
x to the sixth power
and then we could say that this is going to be
so f of x is approximately equal to,
distribute the two, two minus eight x squared
plus 32 x to the fourth
minus 128 x to the sixth.
Just like that with a little bit of a substitution,
I was able to reasonably a non hairy fashion
find the first four nonzero terms
of the power series of two over one plus four x squared
which is the derivative,
which is going to be the derivative
of the power series of arctangent
of two x.
Let's write this down, so I'm going to write,
I'm going to write down,
so arctangent of two x,
arctangent of two x
which is equal to an antiderivative
of f of x,
dx which is going to be equal to an antiderivative
of this whole, all of this business.
It's the antiderivative of two minus eight,
minus eight x squared plus 32
x to the fourth minus 128
x to the sixth.
Actually let me make this approximate
because now of course we're doing approximation
with the power series.
Dx and what is this going to be equal to?
We get approximately equal to
well I'm going to have a constant there.
Let me write the constant first
because when we write our power series or Maclaurin Series,
the first term is the constant term.
It's our function evaluated to add zero.
We're going to have a constant,
if I take the antiderivative of two,
that's going to be plus two x.
The antiderivative of this, let's see, x to the third
divide eight by three
so it's negative 8/3 x to the third power
then plus 32 x to the fifth
over five
minus 128 x to the seventh
over seven.
We're in the home stretch.
We've got at least four nonzero terms,
if this is nonzero
there's going to be five nonzero terms
but now let's just make sure
that our constant is appropriate for arctangent
of two x.
This is essentially evaluate to what arctangent
of when this function, when x is equal to zero.
What is arctangent of zero?
Remember this is centered at zero
so we better get it right there.
That's just the most basic thing
if we're doing the Maclaurin Series representation.
We're centering at zero
so our approximation evaluated zero
better be the same thing as a function of evaluated zero.
Well arctangent of two times zero
is just going to be zero
and so this thing when you evaluate it at zero,
this gets to C, so this gets to C
must be equal to zero.
C must be equal to zero if we want this thing to be zero
when X is equal to zero.
Just like that we're done.
We've been able to figure out that arctangent
of two x
is approximately equal to
two x minus 8/3 x to the third power
plus 32 over five x to the fifth
minus 128 over seven x to the seven.
If we wanted more terms,
we could have gotten more terms
by just doing what we just did
but doing it for more terms.
Hopefully you enjoyed that fairly hairy problem
but as you saw it's not as hairy as we thought
it was going to be.