We're going to represent the given function
f of x equals one divided by the quantity
one minus five x squared as a power series.
Compute the first few coefficients
of the power series.
And also find the radius of convergence.
Notice how the given function here
does not resemble the formula used
to find the infinite sum
of a converging geometric series.
But the antiderivative of f of x will.
So now we'll integrate f of x
with respect to x.
So we'll begin by integrating f of x
with respect to x.
So we want to find the integral of
one divided by the quantity
one minus five x squared
with respect to x.
Let's write this as the integral of
the quantity one minus five x
to the power of negative two.
We'll have to perform u substitution here,
where u would be equal to one minus five x,
and therefore differential u is equal to
negative five dx.
So if we solved this for dx,
we'd have negative one-fifth u
equals dx.
So again, think of this as
u to the power of negative two,
integrate with respect to u,
we'll have an extra factor of negative one fifth.
So the antiderivative would be
negative one fifth
times u to the power of negative one
which would be the quantity
one minus five x to the power of negative one
divided by negative one plus c.
Simplifying, we would have
notice positive one fifth
times
we have one divided by the quantity
one minus five x
plus c.
And notice how one divided by the quantity
one minus five x
fits the form of the formula used here
to find the sum of a converging geometric series.
So because this is the antiderivative of f of x,
the derivative of this antiderivative would be f of x,
which means f of x,
the given function,
or in this case the integrand,
is equal to the derivative
with respect to x of one fifth times
one divided by the quantity one minus five x.
Of course, we can leave off the constant here,
which means f of x is equal to
the derivative with respect to x
of one fifth
times the power series for this function
which would be the summation from
n equals zero to infinity
of a times r, raised to the power of n,
where a is one
and r would be five x.
We'll simplify this power series
and then find the derivative with respect to x
on the next slide.
We can write five x to the power of n
as five to the nth times x to the nth.
So we can write this as the derivative
with respect to x of the summation
from n equals zero to infinity of
we'd have five to the nth
times x to the nth
divided by five.
This is five to the first,
so we'll simplify here
by subtracting the exponents.
So we have f of x equals the derivative
with respect to x
of the summation from n equals zero to infinity of
five to the power of n minus one
times x to the power of n.
Now we'll find the derivative of the power series
with respect to x,
which would be f of x equals the summation from
n equals zero to infinity of
five to the power of n minus one,
then we'd multiply by n
and then subtract one from the exponent,
so we'd have x to the power of n minus one.
So this would be the power series for our function.
And now we'll generate the first several terms
to find the desired coefficients.
Notice when n is zero,
the first term would be zero
because of this n here.
When n is one, we'd have five to the zero
times one
times x to the zero,
which is one,
plus when n is two,
we'd have five to the first
times two
times x to the first
plus when n is three,
we'd have five squared
times three
times x squared
plus when n is four,
we'd have five to the third,
times four
times x to the third,
and looking back at our question
just for a moment,
we defined one more term,
the degree four term,
to find c sub four,
the coefficient of a degree four term.
So when n is five, we'd have
five to the fourth
times five
times x to the fourth.
Of course, this continues.
Let's go and simplify each term.
Here we have one plus
this would be 10 x
plus this would be 25 times three times x squared,
75 x squared,
plus five to the third times four times x cubed,
that would be 125 times four times x cubed,
that would be 500 x cubed.
Then finally we have five to the fourth
times five
times x to the fourth.
That'd be plus 3,125 x to the fourth.
So c sub zero is one,
c sub one is 10,
c sub two is 75,
c sub three is 500,
and c sub four is 3,125.
Let's go ahead and record that.
Again, one,
10,
75,
500,
3,125.
And now to find the radius of convergence.
Notice in this form here we can recognize that
r, the common ratio, is equal to five x,
and therefore,
in order for the series to converge,
the absolute value of five x must be less than one,
which is equal to five times
the absolute value of x
less than one.
Dividing both sides by five,
we have the absolute value of x as
less than one fifth,
and therefore the radius of convergence
is one fifth.
This also tells us the interval of convergence
would be the open interval
from negative one fifth
to positive one fifth.
Again, we're only asked to find the
radius of convergence,
which again is one fifth.
But before we go,
let's take a look at the graph of divisional function
and the graph of the polynomial function
that could be formed using the first five terms
of our power series.
And because this power series is centered at zero,
we would call this a Maclaurin polynomial.
We have divisional function graphed here in blue
and the polynomial function graphed here in red.
Notice around x equals zero,
the red polynomial function
is a good representation
of the original function,
again graphed in blue.
We can also see
the power series is centered
at x equals zero.
I hope you found this helpful.