### Video Transcript

Suppose the sum from π equals zero
to β of π π multiplied by π₯ to the πth power is a power series whose interval of
convergence is the open interval from negative three to three. And the sum from π equals zero to
β of π π multiplied by π₯ to the πth power is a power series whose interval of
convergence is the open interval from negative five to five. Find the interval of convergence of
the series the sum from π equals zero to β of π π π₯ to the πth power minus π
π π₯ to the πth power. And find the [interval] of
convergence of the series the sum from π equals zero to β of π π two the πth
power π₯ to the πth power.

The question wants us to find the
interval of convergence for two different power series. We recall that an interval of
convergence is an interval which contains all values of π₯ such that our series
converges. If both power series converges so
the absolute value of π₯ is less than the smaller of three or five. Therefore, since weβre adding
together two convergent power series, we can combine the summands just as we would
in a regular series. This gives us the sum from π
equals zero to β of π π π₯ to the πth power minus π π π₯ to the πth power will
converge when the absolute value of π₯ is less than the smaller of three and
five.

So weβve shown that the power
series converges when the absolute value of π₯ is less than three. But what about when itβs equal to
three? What about when itβs greater than
three? Letβs consider the case when the
absolute value of π₯ is equal to three. And letβs assume that our series
converges. Then, since the absolute value of
π₯ is equal to three, the sum from π equals zero to β of π π π₯ to the πth power
must also converge because its radius of convergence is five. Now, what would happen if we tried
to add these two series together? Well, weβve assumed that our first
power series will converge when the absolute value of π₯ is equal to three. And if the absolute value of π₯ is
equal to three, this is less than five. So our second power series must
also converge.

This means weβre trying to add
together two power series which both converge. So we can just add the coefficients
of π₯ to the πth power together. We see that negative π π and π
π cancel. So this simplifies to give us the
sum from π equals zero to β of π π multiplied by π₯ to the πth power. However, the sum from π equals
zero to β of π π π₯ to the πth power will only converge when the absolute value
of π₯ is less than three. So this series must diverge. The only way this couldβve happened
is if the sum over π of π π π₯ to the πth power minus π π π₯ to the π π π₯
to the πth power was divergent when the absolute value of π₯ was equal to
three.

Therefore, since neither three nor
negative three are in our interval of convergence, weβve shown that the sum from π
equals zero to β of π π π₯ to the πth power minus π π π₯ to the πth power has
the interval of convergence the open interval from negative three to three. So letβs clear some space and work
on the second case of finding our interval of convergence.

Weβll let the function π of π₯ be
equal to the sum from π equals zero to β of π π π₯ to the πth power. And this will be true whenever the
absolute value of π₯ is less than five. We want to make this power series
look like the one given to us in the question. Weβll start by multiplying and
dividing our summands by two to the πth power. Next, we can use our laws of
exponents to notice that a half to the πth power multiplied by π₯ to the πth power
is equal to π₯ over two all raised to the πth power. We want our summands to have π₯ to
the πth power instead of π₯ over two to the πth power. So weβll let π’ be equal to π₯
divided by two. This gives us that the sum from π
equals zero to β of π π two to the πth power multiplied by π’ to the πth power
will converge when the absolute value of two π’ is less than five.

If the absolute value of two π’ is
less than five, this is the same as saying the absolute value of π’ is less than
five over two. So weβve shown that the radius of
convergence of the sum from π equals zero to β of π π multiplied by two to the
πth power multiplied by π’ to the πth power is five over two. Since it doesnβt matter if we call
our variable π’ or π₯, weβve shown that our second power series, the sum from π
equals zero to β of π π two to the πth power π₯ to the πth power, will have the
interval of convergence the open interval from negative five over two to five over
two.

Letβs now take a look at what would
happen if we tried to multiply two power series together.

So letβs say we wanted to multiply
two power series together, how would we do this? Well, letβs start by writing each
out by term by term. Writing these out term by term
gives us the following expression. Well, we remember that both of
these expressions go on infinitely. If we look at the two expressions
we have, these just seem like two polynomials. And we know how to multiply two
polynomials together. To find the constant term, we
multiply both of our constant terms together. This gives us π nought multiplied
by π nought.

Now, to find all our terms of π₯,
we multiply the constant terms by the coefficients with π₯. This gives us π nought multiplied
by π one plus π one multiplied by π nought. We can then do the same for the
coefficient of π₯ squared. To get our terms of π₯ squared in
the product, weβll be multiplying the constant terms by the terms of π₯ squared. And weβll also be multiplying the
terms of a single π₯ in them together. And we could continue doing this
process indefinitely.

We can now notice when we were
choosing our coefficient to multiply to get the term π₯ squared in our product, we
were choosing them so that when we sum their indexes, we get two. And this makes sense because the
index of our coefficient is exactly the same as the power of π₯. Letβs now take a closer look at our
second term. We could write this as the sum from
π equals zero to one of π π multiplied by π one minus π all multiplied by π₯ to
the first power. And again, this is because we
wanted the sum of our indexes to be equal to the power of π₯. We could do exactly the same for
the coefficient of π₯ squared. In fact, we could make a similar
sum as the coefficient of any power of π₯. We can finally write this as a sum
over our different powers of π₯. Where each coefficient of our power
of π₯ will be equal to one of the coefficients we found earlier.

As we discussed earlier, to find
the coefficient of π₯ to the πth power, we need to add together the products of all
terms whose indexes add to π. And this gives us a motivation of
how to multiply two power series together. If weβre given two power series
which converge for a certain value of π₯, then the product of these two power series
is equal to. We sum over π to find the
coefficient of π₯ to the πth power. And this coefficient is the sum
over π of π π multiplied by π π minus π.