### Video Transcript

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.