### Video Transcript

Representing Rational Functions
Using Power Series

In this lesson, weβre going to
learn how to convert rational functions into power series by using what we already
know about infinite geometric series. Weβre going to discuss both the
radius and interval of convergence of this series. And finally, weβll go over some
examples of turning a power series into a rational function.

Weβll start by recalling some facts
we know about infinite geometric series. First for an infinite geometric
series with first term π and ratio of successive terms π, the sum from π equals
zero to β of π multiplied by π to the πth power is equal to π divided by one
minus π, when the absolute value of our ratio π is less than one. And we also recall that the same
infinite geometric series, the sum from π equals zero to β of π multiplied by π
to the πth power will diverge when the absolute value of our ratio is greater than
one.

Now, imagine our ratio of
successive terms instead of being represented by π, it was actually some function
of π₯. Letβs call this π of π₯. So we have an expression which
looks like this: π divided by one minus some function of π₯. We can then use our facts about
geometric series to write this as a power series, assuming that the absolute value
of our ratio of successive terms is less than one. So we can rewrite π divided by one
minus π of π₯ as the sum from π equals zero to β of π multiplied by π of π₯ to
the πth power provided that the absolute value of the ratio of our successive
terms, which in our case is π of π₯ , is less than one.

And in particular, by using our
second fact, we know that if the absolute value of our ratio of successive terms is
greater than one, then this series will diverge. So when our ratio π is some
function of π₯, we know that our power series will diverge when the absolute value
of π of π₯ is greater than one. In particular, we notice that we
donβt know what happens when the absolute value of π of π₯ is equal to one. So if weβre asked to find an
interval of convergence, we will need to check each of these cases separately.

Okay, so now, weβve seen how to
write the rational function π divided by one minus π of π₯ as a power series and
weβve also looked at the radius of convergence. However, this leaves us with the
question, βWhat about a general rational function, some polynomial π of π₯ divided
by some polynomial π of π₯?β How would we write this as a power
series? Since weβve already seen how to
write rational functions in the form π divided by one minus π of π₯ as a power
series, we could try to write our rational function as the sum of fractions of this
form.

There are numerous different ways
of doing this. One such method is algebraic
manipulation. One such way of using algebraic
manipulation would be rewriting the fraction one divided by one plus π₯ as one
divided by one minus negative π₯. And we see this is now in the form
π divided by one minus π of π₯, where π is one and π of π₯ is equal to negative
π₯. Another method we could use to
rewrite our rational function is polynomial division. For example, using polynomial
division, we have two π₯ squared plus three all divided by π₯ squared plus one is
equal to two plus one divided by one plus π₯ squared. And we can write this into a power
series by using the same trick we did before: one plus π₯ squared is equal to one
minus negative π₯ squared.

The last method weβre going to
discuss for rewriting our rational function is partial fractions. For example, we could rewrite π₯
plus three all divided by π₯ plus one multiplied by π₯ minus one as two divided by
π₯ minus one minus one divided by π₯ plus one by using partial fractions. And then, we can use algebraic
manipulation on the resulting rational functions to write them in the form π
divided by one minus π of π₯. In our first rational function, if
we multiply the numerator and the denominator by negative one, we get negative two
divided by one minus π₯. And we can rewrite the denominator
in our second fraction as one minus negative π₯.

Weβre now ready to try some
examples.

Consider the function π of π₯ is
equal to one divided by two plus π₯, find the power series for π of π₯. Identify its interval of
convergence.

The question gives us a rational
function π of π₯. It wants us to find the power
series for the function π of π₯ and then to find the interval of convergence for
this power series. We recall a fact about geometric
series. The sum from π equals zero to β of
π multiplied by π to πth power is equal to π divided by one minus π, when the
absolute value of our ratio π is less than one. And we also recall that this power
series will diverge when the absolute value of π is greater than one. If we could rewrite our function π
of π₯ in the form π divided by one minus π, then we could use this fact about
geometric series to rewrite our function as a power series.

We want the denominator of our
fraction to be in the form one minus the ratio π. However, in our fraction for π of
π₯, we have a two in the denominator. Since we want our denominator to be
one minus the ratio π, weβll take out our factor two in our denominator. We have that two multiplied by one
gives us two and then two multiplied by π₯ divided by two gives us π₯. We can then take the constant of a
half outside of our fraction. This gives us a half multiplied by
one divided by one plus π₯ over two. This is now almost in the form that
we need it. We have the constant of one in the
numerator. But instead of one minus π in the
denominator, we have one plus π₯ over two.

We can do this by using a little
bit of algebraic manipulation. Instead of adding π₯ over two, we
can subtract negative π₯ over two. Now, if we set the numerator of our
function to be equal to π and the denominator of our fraction to be one minus π,
so that setting π to be equal to negative π₯ over two. Then, weβve shown that our fraction
is in the form of the infinite sum of a geometric series with first term π equal to
one and ratio π equal to negative π₯ over two.

So, using our fact about geometric
series, we can write this as a half multiplied by the sum from π equals zero to β
of one multiplied by negative π₯ over two raised to the πth power. And in particular, weβll know that
this is definitely true, when the absolute value of our ratio π is less than
one. And we know that it will diverge
when the absolute value of our ratio π is greater than one. In our case, we have that π is
equal to negative π₯ divided by two. So this power series must converge
when the absolute value of negative π₯ divided by two is less than one.

We can simplify our power series
slightly by noticing that negative π₯ over two is equal to a half multiplied by
negative π₯. We can then distribute the exponent
over the parenthesis. This gives us a half multiplied by
the sum from π equals zero to β of a half to the πth power multiplied by negative
π₯ to the πth power. Finally, we can bring the
coefficient of a half inside of our summand. But then, a half multiplied by a
half to the πth power is just equal to a half raised to the power of π plus
one. So the power series representation
for our function π of π₯ is the sum from π equals zero to β of a half raised to
the power of π plus one multiplied by negative π₯ to the πth power.

The second part of our question
wants us to identify the interval of convergence of this power series. We recall that the interval of
convergence is all of the values of π₯ where our series will converge. Weβve already shown that our series
has to converge when the absolute value of our ratio π is less than one. And it has to diverge when the
absolute value of our ratio π is greater than one. In our case, our ratio π is equal
to negative π₯ over two. So our power series must converge
when the absolute value of negative π₯ divided by two is less than one. The absolute value of negative π₯
over two is just equal to the absolute value of π₯ over two.

Next, solving the absolute value of
π₯ over two is less than one is the same as solving negative one is less than π₯
over two is less than one. We can then multiply this
inequality through by two. This gives us that negative two is
less than π₯ is less than two. This is our radius of
convergence. Our power series must converge for
all values of π₯ between negative two and two. And we know it will diverge when π₯
is greater than two or when π₯ is less than negative two. However, we donβt know what will
happen when π₯ is equal to two or when π₯ is equal to negative two.

To find out what happens at the
bounds of our radius of convergence, letβs substitute in π₯ is equal to two and π₯
is equal to negative two into our power series. Substituting π₯ is equal to two
gives us the sum from π equals zero to β of a half raised to the power of π plus
one multiplied by negative two to the power of π. There are several different ways of
evaluating this series. For example, we could use the πth
term divergence test. However, weβre going to look term
by term at the expansion of our series.

Our first term when π is equal to
zero gives us a half raised to the power of zero plus one, which is a half. And then, we multiply this by
negative two raised to the power of zero, which is one. We can do the same of our second
term. When π is equal to one, we get a
half squared multiplied by negative two, which we can work out to be negative
one-half. Our third term when π is equal to
two gives us a half cubed multiplied by negative two squared, which is plus a
half. In fact, this series continues. We add a half. Then, we subtract a half. Then, we add a half and we subtract
a half et cetera. We can then consider what the
partial sums will look like.

Our first partial sum will be equal
to a half. Our second partial sum, the sum of
the first two terms of our series, will be a half minus a half which is zero. Our third partial sum, the sum of
the first three terms of our series, will be a half. And our fourth partial sum, the sum
of the first four terms of our series, will be zero. And we can see that this pattern
will continue. In fact, this means that our
partial sums wonβt converge. They will fluctuate between a half
and zero. Since the sequence of partial sums
doesnβt converge, we can conclude that our power series does not converge when π₯ is
equal to two. So π₯ is equal to two is not in our
interval of convergence.

We can do the same when π₯ is equal
to negative two. We substitute π₯ is equal to
negative two into our power series and then calculate the series term by term. We see that every single term in
our series is equal to a half. Therefore, our πth partial sum of
this series will be a half plus a half plus a half π times or π divided by
two. Therefore, since our πth partial
sum is unbounded, we can conclude that the power series does not converge when π₯ is
equal to negative two. Therefore, since weβve shown that
neither of the end points of our radius of convergence are in our interval of
convergence, we can conclude that the interval of convergence for the power series
of the function one divided by two plus π₯ is the open interval from negative two to
two.

Letβs now look at an example, where
our rational function π of π₯ has a polynomial in the numerator.

Let us consider the function π of
π₯ is equal to π₯ divided by three minus π₯. Find a power series representation
for the function π of π₯. Find the radius of convergence.

The question gives us a rational
function π of π₯. It wants us to find the power
series representation of the function π of π₯. And it wants us to then find the
radius of convergence for this power series. To do this, we recall a fact about
the sum of an infinite geometric series. The sum from π equal zero to β of
π multiplied by π to the πth power is equal to π divided by one minus π, when
the absolute value of our ratio π is less than one.

So if we can write our rational
function π of π₯ in the form π divided by one minus π, then we can use this fact
about geometric series to write it as a power series with a radius of convergence
the absolute value of π is less than one. Weβll start by noticing that we can
factor a coefficient of π₯ outside of our rational function. This gives us π₯ multiplied by one
over three minus π₯. We want our denominator to be in
the form one minus the ratio π. However, we have three minus π₯ as
our denominator. To have a denominator in the form
one minus π, weβre going to factor out a coefficient of three in our
denominator. This gives us three multiplied by
one minus π₯ divided by three. We can then simplify this by taking
the coefficient of three in our denominator and putting it outside of our rational
function.

If we then set π equal to one and
π equal to π₯ divided by three, we can use our infinite geometric series to rewrite
our expression. Using our infinite geometric series
formula, we have that π of π₯ is equal to π₯ divided by three multiplied by the sum
from π equals zero to β of one multiplied by π₯ over three to the πth power, when
the absolute value of our ratio π₯ divided by three is less than one. We can simplify this further.

First, we can remove the
multiplication by one. Next, we can take our coefficient
of π₯ divided by three inside of our sum to give us π₯ over three multiplied by π₯
over three to the πth power. However, we can simplify this
further to just be π₯ over three all raised to the power of π plus one. This gives us that our function π
of π₯ is equal to the sum from π equals zero to β of π₯ over three to the power of
π plus one, when the absolute value of π₯ over three is less than one.

Next, the question wants us to find
the radius of convergence of this power series. We recall that we called π the
radius of convergence of a power series if the power series converges when the
absolute value of π₯ is less than π and diverges when the absolute value of π₯ is
greater than π. Unless the power series converges
for all values of π₯, in which case we call the radius of convergence β. We recall that for a geometric
series, the sum from π equals zero to β of π multiplied by π to the πth power
will diverge whenever the absolute value of our ratio π is greater than one. In our case, our ratio π is π₯
divided by three. So our power series converges when
the absolute value of π₯ over three is less than one and diverges when the absolute
value of π₯ over three is greater than one.

We can then rearrange both of these
expressions. The absolute value of π₯ over three
being less than one is the same as saying the absolute value of π₯ is less than
three. And if the absolute value of π₯
over three is greater than three, this is the same as saying the absolute value of
π₯ is greater than three. So our power series converges when
the absolute value of π₯ is less than three and diverges when the absolute value of
π₯ is greater than three. Therefore, our radius of
convergence π is equal to three. And our power series for π of π₯
will converge whenever the absolute value of π₯ is less than three.

Letβs now look at an example of
converting a power series into a rational function.

Convert the sum from π equals zero
to β of two multiplied by π₯ to the πth power into a rational function.

The question gives us a power
series. And it wants us to convert this
into a rational function. And we recall that a rational
function π of π₯ is the quotient of polynomials π of π₯ and π of π₯. To convert our power series into a
rational function, we recall the following fact about a geometric series: the sum
from π equals zero to β of π multiplied by π to the πth power is equal to π
divided by one minus π, when the absolute value of π is less than one.

We see that our power series is
already written as a geometric series, where π is equal to two and where π is
equal to π₯. Substituting π is equal to two and
π is equal to π₯ into a geometric series formula gives us the sum from π equals
zero to β of two multiplied by π₯ to the πth power is equal to two divided by one
minus π₯. And this is valid when the absolute
value of our ratio π₯ is less than one.

Weβll do one more example of
converting a power series into a rational function.

Convert the sum from π equals zero
to β of two multiplied by negative π₯ over three to the power of three π plus one
into a rational function.

We recall from infinite geometric
series the sum from π equals zero to β of π multiplied by π to the πth power is
equal to π divided by one minus π, when the absolute value of π is less than
one. Since the question wants us to
convert this into a rational function, we notice if we can manipulate our series to
be in the form of a geometric series, then we can use our fact about geometric
series to write it as a rational function.

Weβll start by simplifying the
exponent inside our summand. We can then take this coefficient
of negative π₯ over three outside of our sum. Instead of raising negative π₯ over
three to the power of three π, we can first cube it and then raise it to the πth
power. Evaluating our exponent of three
gives us negative π₯ cubed over 27. Then, we can see that this is a
geometric series with π equal to two and π equal to negative π₯ cubed over 27. Using our infinite geometric series
formula, we get negative π₯ over three multiplied by two divided by one minus
negative π₯ cubed over 27 which is equal to negative two π₯ divided by three plus π₯
cubed over nine. And this is valid when the absolute
value of our ratio is less than one.

So to summarize what weβve done, we
can use our infinite geometric series formula to write rational functions as power
series and to write power series as rational functions. Sometimes, weβll need to manipulate
our rational function or weβll need to manipulate our power series. And weβve seen multiple ways of
doing this. We can also find the radius of
convergence of our power series by using that our series converges when the absolute
value of our ratio π is less than one. Finally, we can find the interval
of convergence by checking the end points of our radius of convergence.