Video Transcript
Operations on Power Series
In this video, we’re going to
discuss the various different types of power series which can arise from adding,
subtracting, and multiplying to power series. We’ll discuss finding the radius of
convergence of a linear combination of power series. We’ll also discuss how to find a
power series representation of a function as a combination of power series. Finally, we’ll discuss what happens
when we multiply two power series together.
To answer these questions about
power series, let’s consider what we already know about regular series. If we have two series, the sum over
𝑛 of 𝑎 𝑛 and the sum over 𝑛 of 𝑏 𝑛, which converge and we have a constant 𝑐,
then we know the sum over 𝑛 of 𝑎 𝑛 plus the sum over 𝑛 of 𝑏 𝑛 is equal to the
sum over 𝑛 of 𝑎 𝑛 plus 𝑏 𝑛. And we also know that this will
converge. Similarly, since 𝑐 is a constant,
we know when we’re summing over 𝑛 𝑐 multiplied by 𝑎 𝑛, we can take the constant
of 𝑐 outside our sum. This gives us 𝑐 multiplied by the
sum over 𝑛 of 𝑎 𝑛. And this also converges.
Now, we want to ask the question,
what would’ve happened if instead of being given regular series, we were given power
series? We recall when we’re discussing
power series, they don’t converge or diverge. Instead, they have what we call a
radius of convergence. This is because 𝑥 is a
variable. So for each value of 𝑥, we’ll have
a different power series. And we want to know the values of
𝑥 for which these series converges. So let’s add radius of convergences
for both of our power series. Let’s call these 𝑅 one and 𝑅
two.
We now want to know what happens
when we try to add our power series together. We’ll do this by using our formula
for adding two series together. We’re trying to add two series
together. The summand in the first one is 𝑎
𝑛 multiplied by 𝑥 to the 𝑛th power. And the summand of the second one
is 𝑏 𝑛 multiplied by 𝑥 to the 𝑛th power. And our formula tells us that when
these series converges, we can just add the summands together. So for any value of 𝑥 which both
of our power series converge, we can use this formula. This gives us the sum over 𝑎 𝑛 of
𝑎 𝑛 multiplied by 𝑥 to the 𝑛th power plus 𝑏 𝑛 multiplied by 𝑥 to the 𝑛th
power. And we can factor out 𝑥 to the
𝑛th power. This gives us the sum over [𝑛] of
𝑎 𝑛 plus 𝑏 𝑛 multiplied by 𝑥 to the 𝑛th power.
We notice, in particular, that this
new formula will always work when the absolute value of 𝑥 is less than the smaller
of 𝑅 one and 𝑅 two. This is because if the absolute
value of 𝑥 is less than the smaller of 𝑅 one and 𝑅 two, then it’s 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 𝑅 one and 𝑅 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 𝑛 of 𝑥 to the 𝑛th
power divided by 𝑛 will converge when the absolute value of 𝑥 is less than
one. Similarly, the sum over 𝑛 of
negative 𝑥 to the 𝑛th power of 𝑛 will also converge when the absolute value of 𝑥
is less than one. However, adding these two power
series together gives us the sum of one over 𝑛 minus one over 𝑛 multiplied by 𝑥
to the 𝑛th power. Which simplifies to just give us
the sum over 𝑛 of zero, which converges for all values of 𝑥.
So we can use this formula to add
any two power series together. However, we’re going to have to
work hard to find the radius of convergence of our new power series. We’ll have to use techniques such
as inspection, the ratio test, or the integral test to find our new radius of
convergence. Let’s take a look at an
example.
Suppose the sum from 𝑛 equals zero
to ∞ of 𝑎 𝑛 multiplied by 𝑥 to the 𝑛th power is a power series whose interval of
convergence is the open interval from negative three to three. And the sum from 𝑛 equals zero to
∞ of 𝑏 𝑛 multiplied by 𝑥 to the 𝑛th 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 𝑛 equals zero to ∞ of 𝑎 𝑛 𝑥 to the 𝑛th power minus 𝑏
𝑛 𝑥 to the 𝑛th power. And find the [interval] of
convergence of the series the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 two the 𝑛th
power 𝑥 to the 𝑛th 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 𝑥 such that our series
converges. If both power series converges so
the absolute value of 𝑥 is less than the smaller of three or five. Therefore, since we’re 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 𝑛
equals zero to ∞ of 𝑎 𝑛 𝑥 to the 𝑛th power minus 𝑏 𝑛 𝑥 to the 𝑛th power will
converge when the absolute value of 𝑥 is less than the smaller of three and
five.
So we’ve shown that the power
series converges when the absolute value of 𝑥 is less than three. But what about when it’s equal to
three? What about when it’s greater than
three? Let’s consider the case when the
absolute value of 𝑥 is equal to three. And let’s assume that our series
converges. Then, since the absolute value of
𝑥 is equal to three, the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 𝑥 to the 𝑛th 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, we’ve assumed that our first
power series will converge when the absolute value of 𝑥 is equal to three. And if the absolute value of 𝑥 is
equal to three, this is less than five. So our second power series must
also converge.
This means we’re trying to add
together two power series which both converge. So we can just add the coefficients
of 𝑥 to the 𝑛th power together. We see that negative 𝑏 𝑛 and 𝑏
𝑛 cancel. So this simplifies to give us the
sum from 𝑛 equals zero to ∞ of 𝑎 𝑛 multiplied by 𝑥 to the 𝑛th power. However, the sum from 𝑛 equals
zero to ∞ of 𝑎 𝑛 𝑥 to the 𝑛th power will only converge when the absolute value
of 𝑥 is less than three. So this series must diverge. The only way this could’ve happened
is if the sum over 𝑛 of 𝑎 𝑛 𝑥 to the 𝑛th power minus 𝑏 𝑛 𝑥 to the 𝑏 𝑛 𝑥
to the 𝑛th power was divergent when the absolute value of 𝑥 was equal to
three.
Therefore, since neither three nor
negative three are in our interval of convergence, we’ve shown that the sum from 𝑛
equals zero to ∞ of 𝑎 𝑛 𝑥 to the 𝑛th power minus 𝑏 𝑛 𝑥 to the 𝑛th power has
the interval of convergence the open interval from negative three to three. So let’s clear some space and work
on the second case of finding our interval of convergence.
We’ll let the function 𝑓 of 𝑥 be
equal to the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 𝑥 to the 𝑛th power. And this will be true whenever the
absolute value of 𝑥 is less than five. We want to make this power series
look like the one given to us in the question. We’ll start by multiplying and
dividing our summands by two to the 𝑛th power. Next, we can use our laws of
exponents to notice that a half to the 𝑛th power multiplied by 𝑥 to the 𝑛th power
is equal to 𝑥 over two all raised to the 𝑛th power. We want our summands to have 𝑥 to
the 𝑛th power instead of 𝑥 over two to the 𝑛th power. So we’ll let 𝑢 be equal to 𝑥
divided by two. This gives us that the sum from 𝑛
equals zero to ∞ of 𝑏 𝑛 two to the 𝑛th power multiplied by 𝑢 to the 𝑛th power
will converge when the absolute value of two 𝑢 is less than five.
If the absolute value of two 𝑢 is
less than five, this is the same as saying the absolute value of 𝑢 is less than
five over two. So we’ve shown that the radius of
convergence of the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 multiplied by two to the
𝑛th power multiplied by 𝑢 to the 𝑛th power is five over two. Since it doesn’t matter if we call
our variable 𝑢 or 𝑥, we’ve shown that our second power series, the sum from 𝑛
equals zero to ∞ of 𝑏 𝑛 two to the 𝑛th power 𝑥 to the 𝑛th power, will have the
interval of convergence the open interval from negative five over two to five over
two.
Let’s now take a look at what would
happen if we tried to multiply two power series together.
So let’s say we wanted to multiply
two power series together, how would we do this? Well, let’s 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 𝑎 nought multiplied
by 𝑏 nought.
Now, to find all our terms of 𝑥,
we multiply the constant terms by the coefficients with 𝑥. This gives us 𝑎 nought multiplied
by 𝑏 one plus 𝑎 one multiplied by 𝑏 nought. We can then do the same for the
coefficient of 𝑥 squared. To get our terms of 𝑥 squared in
the product, we’ll be multiplying the constant terms by the terms of 𝑥 squared. And we’ll also be multiplying the
terms of a single 𝑥 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 𝑥 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 𝑥. Let’s now take a closer look at our
second term. We could write this as the sum from
𝑗 equals zero to one of 𝑎 𝑗 multiplied by 𝑏 one minus 𝑗 all multiplied by 𝑥 to
the first power. And again, this is because we
wanted the sum of our indexes to be equal to the power of 𝑥. We could do exactly the same for
the coefficient of 𝑥 squared. In fact, we could make a similar
sum as the coefficient of any power of 𝑥. We can finally write this as a sum
over our different powers of 𝑥. Where each coefficient of our power
of 𝑥 will be equal to one of the coefficients we found earlier.
As we discussed earlier, to find
the coefficient of 𝑥 to the 𝑛th power, we need to add together the products of all
terms whose indexes add to 𝑛. And this gives us a motivation of
how to multiply two power series together. If we’re given two power series
which converge for a certain value of 𝑥, then the product of these two power series
is equal to. We sum over 𝑛 to find the
coefficient of 𝑥 to the 𝑛th power. And this coefficient is the sum
over 𝑗 of 𝑎 𝑗 multiplied by 𝑏 𝑛 minus 𝑗.
Let’s take a look at an example to
see how we can use this formula in practice.
Multiply the series one divided by
one plus 𝑥 is equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th
power 𝑥 to the 𝑛th power by itself to construct a series for one divided by one
plus 𝑥 squared. Write the answer in sigma
notation.
We recall if we have two power
series, 𝑓 of 𝑥 is equal to the sum over 𝑛 of 𝑎 𝑛 𝑥 to the 𝑛th power and 𝑔 of
𝑥 is equal to the sum over 𝑛 of 𝑏 𝑛 𝑥 to the 𝑛th power, which both converge
for this value of 𝑥. Then we can calculate the product
of these two power series, 𝑓 of 𝑥 multiplied by 𝑔 of 𝑥, as the sum over 𝑛 of
all of our 𝑥 to the 𝑛th power terms with the coefficient the sum over 𝑗 𝑎 𝑗
multiplied by 𝑏 𝑛 minus 𝑗. In our case, we want to multiply
one divided by one plus 𝑥 by itself to get the power series for one divided by one
plus 𝑥 squared. So we set both of our functions, 𝑓
of 𝑥 and 𝑔 of 𝑥, to be equal to one divided by one plus 𝑥.
So we have both 𝑓 of 𝑥 and 𝑔 of
𝑥 are equal to the sum of one over one plus 𝑥 which are equal to the sum from 𝑛
equals zero to ∞ of negative one to the 𝑛th power 𝑥 to the 𝑛th power. So for the value of 𝑥 where our
power series for 𝑓 of 𝑥 converges, we can use our formula to find an expression
for 𝑓 of 𝑥 squared. The question tells us that one
divided by one plus 𝑥 is equal to the sum from 𝑛 equals zero to ∞ of negative one
to the 𝑛th power 𝑥 to the 𝑛th power. This gives us that 𝑎 𝑗 is
negative one raised to the power of 𝑗. And 𝑏 𝑛 minus 𝑗 is negative one
raised to the power of 𝑛 minus 𝑗.
Therefore, we can calculate 𝑎 𝑗
multiplied by 𝑏 𝑛 minus 𝑗 as negative one to the power of 𝑗 multiplied by
negative one to the power of 𝑛 minus 𝑗, which is just negative one to the 𝑛th
power. So we now have our coefficient of
𝑥 to the power of 𝑛 is the sum from 𝑗 equals zero to 𝑛 of negative one to the
𝑛th power. And we can see that’s our summand,
negative one to the 𝑛th power, is independent of 𝑗. So, in fact, our sum is negative
one to the 𝑛th power added to itself 𝑛 plus one times. This is because 𝑗 goes from zero
to 𝑛. And if we’re adding negative one to
the 𝑛th power to itself 𝑛 plus one times, this is the same as saying 𝑛 plus one
multiplied by negative one to the 𝑛th power.
Therefore, we’ve shown for the
values of 𝑥 where a power series for one divided by one plus 𝑥 converges. We can multiply the power series of
one divided by one plus 𝑥 by itself to get that one divided by one plus 𝑥 squared
is equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power
multiplied by 𝑛 plus one multiplied by 𝑥 to the 𝑛th power.
So to summarize what we’ve shown in
this video, if we have two power series, the sum from 𝑛 equals zero to ∞ of 𝑎 𝑛
𝑥 to the 𝑛th power and the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 𝑥 to the 𝑛th
power, which both converge for a particular value of 𝑥. Then we can add these two power
series together to give us the sum from 𝑛 equals zero to ∞ of 𝑎 𝑛 plus 𝑏 𝑛 all
multiplied by 𝑥 to the 𝑛th power. And, in particular, if our first
power series has a radius of convergence 𝑅 one and our second power series has a
radius of convergence 𝑅 two. Then we know we can add the power
series in this way for any value of 𝑥 such that the absolute value of 𝑥 is less
than the smaller of 𝑅 one and 𝑅 two.
And for a power series, the sum
from 𝑛 equals zero to ∞ of 𝑎 𝑛 plus 𝑏 𝑛 all multiplied by 𝑥 to the 𝑛th power,
if we call its radius of convergence 𝑅, then the minimum of 𝑅 one and 𝑅 two gives
us a lower bound on 𝑅. Finally, we were shown if we were
given two power series, the sum from 𝑛 equals zero to ∞ of 𝑎 𝑛 𝑥 to the 𝑛th
power and the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 𝑥 to the 𝑛th power, which both
converge for a value of 𝑥. Then the product of these two power
series is given by. We sum over all of our 𝑥 to the
𝑛th terms. And their coefficient is the sum
from 𝑗 equals zero to 𝑛 of 𝑎 𝑗 multiplied by 𝑏 𝑛 minus 𝑗.
