We want to represent the given function
f of x equals five divided by the quantity
one plus nine x squared as a power series,
then compute the first few coefficients of the power series
where c sub zero would be the coefficient
of the degree zero term or the constant term,
c sub one would be the coefficient of the degree one term
and so on.
We also want to find the radius of convergence.
Well notice how the given function somewhat resembles
the formula used to find the infinite sum
of a converging geometric series.
But it doesn't fit perfectly,
so we'll have to build the power series from what we know
about a more basic function.
For example, if we had the function f of x
equals one divided by the quantity,
let's say just one plus x,
we know we could write this as one divided by
the quantity one minus negative x.
So for this function,
notice how a would be equal to positive one
and the common ratio r would be equal to negative x.
So as a power series, we have the summation
from n equals zero to infinity of a, which is one,
times r raised to the power of n,
which, in this case, would be negative x
raised to the power of n.
We can change the form of this.
We can factor out negative one to the nth
and write this as the summation
from n equals zero to infinity of
just negative one to the nth times x to the nth.
Now from here let's look at the function f of x
equals one divided by the quantity one plus,
instead of x, let's say x squared,
which more closely resembles our function.
Well, we can take this given power series
and substitute x squared for x,
which would give us the summation
from n equals zero to infinity of negative one to the nth
times, instead of x to the nth,
we'd have x squared to the nth,
which we could also write as the summation
from n equals zero to infinity of negative one to the nth
times, let's write this as x raised to the power of two n.
So now, looking back at our function,
we could write our function as f of x
equals five times one
divided by the quantity one plus,
instead of nine x squared, we could write this as
three x to the second.
And now we can take this power series,
multiply it by five and substitute three x for x
to write a power series for our function f of x.
So we would have f of x,
again, equals five times one divided by
the quantity one plus three x squared,
which would be equal to five times the summation
from n equals zero to infinity of negative one to the nth
and then instead of x to the power of two n,
we'd have three x raised to the power of two n.
Now, if we bring the factor of five into our power series,
we can write this as the summation
from n equals zero to infinity of negative one to the nth
times five times three x raised to the power of two n.
So this would be the power series for the given function,
and now we'll generate the first several terms
to find the requested coefficients.
And let's do this on the next slide.
So notice when n equals zero,
we would have negative one to the zero, that's one,
times five times three x to the zero, which is also one,
so we'd have five.
When n is one, notice how the term would be negative,
so we'll have minus five times two x
to the power of two times one or squared.
When n is two, the term would be positive,
so we'll have plus
five times three x
to the power of two times two or four,
of course, this continues, but notice how
this would be the degree four term,
and the highest degree coefficient we're looking for
is the degree four term coefficient.
So let's go ahead and simplify these.
We have five.
This would be minus five times nine x squared,
that's 45x squared.
And here we'd have plus five times three x to the fourth,
that would be five times 81x to the fourth,
or 405x to the fourth.
So this tells us that c sub zero would be five.
Now c sub one would be the coefficient
of the degree one term.
There is no degree one term so c sub one is zero,
c sub two would be equal to negative 45,
c sub three would be the coefficient
of the degree three term.
There is no three degree terms, so c sub three is zero,
and c sub four is equal to positive 405.
Let's go ahead and record this on the previous slide.
So we have five, zero, negative 45, zero, and 405.
And now to find the radius of convergence
where we know the absolute value of r must be less than one.
In this case, we'd have the absolute value of three x
must be less than one.
So if we have the absolute value of three x
must be less than one.
Since the absolute value of three is three,
we can write this as three times the absolute value of x
less than one, divide both sides by three
and we have the absolute value of x less than 1/3,
and therefore the radius of convergence is 1/3.
Now before we go, let's look at the graph
of the original function f of x,
and the graph of the polynomial that would be formed
by using these first three terms of the power series.
And because this power series is centered at zero,
the corresponding polynomial would be called
a Maclaurin polynomial.
So here's the graph of the original function in blue
and here's the graph of the first three terms
of the power series graphed here in red.
Notice how the polynomial function
is a good representation of the given function
around x equals zero where the power series is centered.
And the interval of convergence for this power series
would be the open interval
from negative 1/3 to positive 1/3.
I hope you found this helpful.