Video Transcript
Operations on Power Series
In this video, weβre going to
discuss the various different types of power series which can arise from adding,
subtracting, and multiplying to power series. Weβll discuss finding the radius of
convergence of a linear combination of power series. Weβll also discuss how to find a
power series representation of a function as a combination of power series. Finally, weβll discuss what happens
when we multiply two power series together.
To answer these questions about
power series, letβs consider what we already know about regular series. If we have two series, the sum over
π of π π and the sum over π of π π, which converge and we have a constant π,
then we know the sum over π of π π plus the sum over π of π π is equal to the
sum over π of π π plus π π. And we also know that this will
converge. Similarly, since π is a constant,
we know when weβre summing over π π multiplied by π π, we can take the constant
of π outside our sum. This gives us π multiplied by the
sum over π of π π. And this also converges.
Now, we want to ask the question,
what wouldβve happened if instead of being given regular series, we were given power
series? We recall when weβre discussing
power series, they donβt converge or diverge. Instead, they have what we call a
radius of convergence. This is because π₯ is a
variable. So for each value of π₯, weβll have
a different power series. And we want to know the values of
π₯ for which these series converges. So letβs add radius of convergences
for both of our power series. Letβs call these π
one and π
two.
We now want to know what happens
when we try to add our power series together. Weβll do this by using our formula
for adding two series together. Weβre trying to add two series
together. The summand in the first one is π
π multiplied by π₯ to the πth power. And the summand of the second one
is π π multiplied by π₯ to the πth power. And our formula tells us that when
these series converges, we can just add the summands together. So for any value of π₯ which both
of our power series converge, we can use this formula. This gives us the sum over π π of
π π multiplied by π₯ to the πth power plus π π multiplied by π₯ to the πth
power. And we can factor out π₯ to the
πth power. This gives us the sum over [π] of
π π plus π π multiplied by π₯ to the πth power.
We notice, in particular, that this
new formula will always work when the absolute value of π₯ is less than the smaller
of π
one and π
two. This is because if the absolute
value of π₯ is less than the smaller of π
one and π
two, then itβs smaller than
both of them. So both of our power series will
converge. We might be tempted at this point
to call the smaller of π
one and π
two the radius of convergence for the sum of
our two power series. However, this is not necessarily
the case.
To see why finding the radius of
convergence of this series is not so simple, consider the following example. The sum over π of π₯ to the πth
power divided by π will converge when the absolute value of π₯ is less than
one. Similarly, the sum over π of
negative π₯ to the πth power of π will also converge when the absolute value of π₯
is less than one. However, adding these two power
series together gives us the sum of one over π minus one over π multiplied by π₯
to the πth power. Which simplifies to just give us
the sum over π of zero, which converges for all values of π₯.
So we can use this formula to add
any two power series together. However, weβre going to have to
work hard to find the radius of convergence of our new power series. Weβll have to use techniques such
as inspection, the ratio test, or the integral test to find our new radius of
convergence. Letβs take a look at an
example.
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.
Letβs take a look at an example to
see how we can use this formula in practice.
Multiply the series one divided by
one plus π₯ is equal to the sum from π equals zero to β of negative one to the πth
power π₯ to the πth power by itself to construct a series for one divided by one
plus π₯ squared. Write the answer in sigma
notation.
We recall if we have two power
series, π of π₯ is equal to the sum over π of π π π₯ to the πth power and π of
π₯ is equal to the sum over π of π π π₯ to the πth power, which both converge
for this value of π₯. Then we can calculate the product
of these two power series, π of π₯ multiplied by π of π₯, as the sum over π of
all of our π₯ to the πth power terms with the coefficient the sum over π π π
multiplied by π π minus π. In our case, we want to multiply
one divided by one plus π₯ by itself to get the power series for one divided by one
plus π₯ squared. So we set both of our functions, π
of π₯ and π of π₯, to be equal to one divided by one plus π₯.
So we have both π of π₯ and π of
π₯ are equal to the sum of one over one plus π₯ which are equal to the sum from π
equals zero to β of negative one to the πth power π₯ to the πth power. So for the value of π₯ where our
power series for π of π₯ converges, we can use our formula to find an expression
for π of π₯ squared. The question tells us that one
divided by one plus π₯ is equal to the sum from π equals zero to β of negative one
to the πth power π₯ to the πth power. This gives us that π π is
negative one raised to the power of π. And π π minus π is negative one
raised to the power of π minus π.
Therefore, we can calculate π π
multiplied by π π minus π as negative one to the power of π multiplied by
negative one to the power of π minus π, which is just negative one to the πth
power. So we now have our coefficient of
π₯ to the power of π is the sum from π equals zero to π of negative one to the
πth power. And we can see thatβs our summand,
negative one to the πth power, is independent of π. So, in fact, our sum is negative
one to the πth power added to itself π plus one times. This is because π goes from zero
to π. And if weβre adding negative one to
the πth power to itself π plus one times, this is the same as saying π plus one
multiplied by negative one to the πth power.
Therefore, weβve shown for the
values of π₯ where a power series for one divided by one plus π₯ converges. We can multiply the power series of
one divided by one plus π₯ by itself to get that one divided by one plus π₯ squared
is equal to the sum from π equals zero to β of negative one to the πth power
multiplied by π plus one multiplied by π₯ to the πth power.
So to summarize what weβve shown in
this video, if we have two power series, the sum from π equals zero to β of π π
π₯ to the πth power and the sum from π equals zero to β of π π π₯ to the πth
power, which both converge for a particular value of π₯. Then we can add these two power
series together to give us the sum from π equals zero to β of π π plus π π all
multiplied by π₯ to the πth power. And, in particular, if our first
power series has a radius of convergence π
one and our second power series has a
radius of convergence π
two. Then we know we can add the power
series in this way for any value of π₯ such that the absolute value of π₯ is less
than the smaller of π
one and π
two.
And for a power series, the sum
from π equals zero to β of π π plus π π all multiplied by π₯ to the πth power,
if we call its radius of convergence π
, then the minimum of π
one and π
two gives
us a lower bound on π
. Finally, we were shown if we were
given two power series, the sum from π equals zero to β of π π π₯ to the πth
power and the sum from π equals zero to β of π π π₯ to the πth power, which both
converge for a value of π₯. Then the product of these two power
series is given by. We sum over all of our π₯ to the
πth terms. And their coefficient is the sum
from π equals zero to π of π π multiplied by π π minus π.