### Video Transcript

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.