Video Transcript
In this video, we’re going to learn
how we can find the power series that results from differentiating or integrating a
power series. We’ll also see how we can find a
power series representation for 𝑓 by integrating a power series for the
derivative. As well as looking at how we can
find the interval of convergence of a power series through using calculus.
We begin by recalling that the sum
of a power series is given by the function 𝑓 of 𝑥 equals the sum of 𝐶𝑛 times 𝑥
minus 𝑎 to the 𝑛th power for values of 𝑛 between zero and ∞. Whose domain is the interval of
convergence of the series. Ideally, we’d like to be able to
differentiate and integrate such functions. And whilst we’re not going to prove
it, the following theorem says that we can do so simply by differentiating or
integrating each individual term in the series. Just as we would for a
polynomial.
This theorem says that if our power
series has a radius of convergence 𝑅 is greater than zero. Then the function 𝑓 defined by 𝑓
of 𝑥 equals 𝐶 naught plus 𝐶 one times 𝑥 minus 𝑎 plus 𝐶 two times 𝑥 minus 𝑎
squared, and so on. Which can of course be written as
the sum of 𝐶𝑛 times 𝑥 minus 𝑎 to the 𝑛th power for values of 𝑛 between zero
and ∞ is differentiable and therefore continuous on the open interval 𝑎 minus 𝑅 to
𝑎 plus 𝑅.
We perform this differentiation
term by term such that 𝑓 prime of 𝑥 is equal to the sum of 𝑛 times 𝐶𝑛 times 𝑥
minus 𝑎 to the power of 𝑛 minus one for values of 𝑛 between one and ∞. Similarly, if we integrate our
function with respect to 𝑥, we find that it’s equal to the sum of 𝐶𝑛 times 𝑥
minus 𝑎 to the power of 𝑛 plus one divided by 𝑛 plus one for values of 𝑛 between
zero and ∞ plus of course the constant of integration 𝐶. Now we denote this capital 𝐶. And we must make sure that it’s
outside of the summation.
In each of these equations, the
radii of convergence are both 𝑅. It’s worth noting that whilst the
radius of convergence 𝑅 remains the same when a power series is differentiated or
integrated, the interval of convergence might not. In other words, the end points of
the interval may change from being included to not included, or vice versa. We can apply the idea that the
derivative or integral of a sum of functions is equal to the sum of the derivative
or integral of those respective functions to these infinite sums. And we can write the equations as
shown, providing we are working with power series. Let’s now have a look at a couple
of examples of where we might use this idea.
Consider the series 𝑓 of 𝑥 equals
𝑥 over one minus 𝑥 squared, which is equal to the sum of 𝑥 to the power of two 𝑛
plus one for values of 𝑛 between zero and ∞. Differentiate the given series
expansion of 𝑓 term by term to find the corresponding series expansion for the
derivative of 𝑓.
Let’s begin by using the power
series to write out the first few terms. The first term is when 𝑛 is equal
to zero. So it’s 𝑥 to the power of two
times zero plus one, which is one. We then let 𝑛 be equal to one. And we get two times one plus one,
which is three. Next, we let 𝑛 be equal to
two. And we find the exponent to be
equal to five. And we can continue in this
manner. We find the first few terms to be
𝑥 to the power of one plus 𝑥 cubed plus 𝑥 to the fifth power plus 𝑥 to the
seventh power, and so on.
We’re told to differentiate the
series expansion of 𝑓 term by term. And so we’re going to do just
that. The derivative of 𝑥 with respect
to 𝑥 is just one. And we recall that, for any
polynomial term, we differentiate by multiplying by the exponent and then reducing
that exponent by one. So the derivative of 𝑥 cubed is
three 𝑥 squared. The derivative of 𝑥 to the fifth
power is five 𝑥 to the fourth power. Differentiating 𝑥 to the seventh
power, and we get seven 𝑥 to the sixth power, and so on.
We need to find a way to write this
as a sum. So let’s have a look at what’s
happened. Each time, we’ve multiplied by the
exponent. And we saw that that exponent was
generated by using the expression two 𝑛 plus one. Then we reduced each power by one,
while the original exponent was two 𝑛 plus one. So our new exponent is two 𝑛 plus
one minus one. The values for our sum remain the
same. They’re from 𝑛 equals zero to
∞. And of course two 𝑛 plus one minus
one is simply two 𝑛. And so by differentiating term by
term, we found the corresponding series expansion for the derivative of 𝑓. It’s the sum of two 𝑛 plus one
times 𝑥 to the power of two 𝑛 for values of 𝑛 between zero and ∞.
Next, we’ll consider an example
where we might integrate.
For the given function 𝑓 of 𝑥
equals the natural log of one plus two 𝑥, find a power series representation for 𝑓
by integrating the power series for 𝑓 prime.
We don’t have a nice power series
expansion for the function 𝑓 of 𝑥 equals the natural log of one plus two 𝑥. But we should notice that the
derivative of 𝑓, 𝑓 prime of 𝑥, is two over one plus two 𝑥. So we’re going to start with an
equation that we’ve seen before. That is, one over one minus 𝑥 is
equal to one plus 𝑥 plus 𝑥 squared plus 𝑥 cubed, and so on. And we write this as the sum of 𝑥
to the 𝑛th power for values of 𝑛 between zero and ∞.
We write our derivative as two
times one over one plus two 𝑥. And then we’re going to use the
equation we’ve seen before, replacing 𝑥 with negative two 𝑥. And this is because we want it in
the form one minus something. And one minus negative two 𝑥 gives
us the one plus two 𝑥 we’re after. We can therefore use this equation
to write one over one plus two 𝑥 as one plus negative two 𝑥 plus negative two 𝑥
squared plus negative two 𝑥 cubed, and so on. Which can of course be written as
the sum of negative two 𝑥 to the 𝑛th power for values of 𝑛 between zero and
∞.
𝑓 prime of 𝑥 is two times one
over one plus two 𝑥. So that’s two times the sum of
negative two 𝑥 to the 𝑛th power for values of 𝑛 between zero and ∞. Two is independent of 𝑛. So we’re going to take it inside
the sum. And we’re going to rewrite negative
two 𝑥 as two times negative one times 𝑥.
We then distribute this exponent
across each term. And we see that we have the sum of
two times two to the 𝑛th power times negative one to the 𝑛th power times 𝑥 to the
𝑛th power. Two times two to the 𝑛th power can
be written as two to the power of 𝑛 plus one. And now we have our expression for
𝑓 prime of 𝑥.
We’re trying to find a power series
representation for 𝑓. So we recall that we can achieve
this by integrating our expression for 𝑓 prime of 𝑥. And we can do so by integrating
each individual term in the series. That’s called term-by-term
integration. Here that’s the integral of the sum
of two to the power of 𝑛 plus one times negative one to the 𝑛th power times 𝑥 to
the 𝑛th power between 𝑛 equals zero and ∞ with respect to 𝑥.
And then of course since we’re
dealing with a power series, we can write this as the sum of the integrals. So how are we going to integrate
two to the power of 𝑛 plus one times negative one to the 𝑛th power times 𝑥 to the
𝑛th power? Well, no matter the value of 𝑛,
two to the power of 𝑛 plus one times negative one to the 𝑛th power is a
constant. This means we can take it outside
of the integral and focus on integrating 𝑥 to the 𝑛th power itself.
Now when we integrate 𝑥 to the
𝑛th power, we know that 𝑛 is positive. So we simply add one to the
exponent and then divide by that new number. So we get 𝑥 to the power of 𝑛
plus one over 𝑛 plus one. And of course we had that constant
of integration 𝐶, which is outside the summation. So how do we find that constant of
integration 𝐶?
Well, let’s go back to 𝑓 of
𝑥. We’re told that it’s equal to the
natural log of one plus two 𝑥. And we know that if we let 𝑥 be
equal to zero, we get quite a nice value for 𝑓 of zero. It’s the natural log of one plus
two times zero, which is the natural log of one, which is of course zero. By replacing 𝑓 of 𝑥 with zero and
𝑥 with zero, we see that zero is equal to the sum of two to the power of 𝑛 plus
one times negative one to the 𝑛th power plus zero to the power of 𝑛 plus one over
zero plus one plus 𝐶.
Now zero to the power of 𝑛 plus
one over zero plus one is always zero. And so we have the sum of zeros,
which is zero. And we find the constant of
integration itself is zero. And so by integrating the power
series for 𝑓 prime, we found a power series representation for 𝑓. It’s the sum of two to the power of
𝑛 plus one times negative one to the 𝑛th power times 𝑥 to the power of 𝑛 plus
one over 𝑛 plus one.
Let’s have a look at a further
example.
Consider the series 𝑓 of 𝑥 equals
one over one plus 𝑥, which equals the sum of negative one to the 𝑛th power times
𝑥 to the 𝑛th power, for values of 𝑛 between zero and ∞. Differentiate the given series
expansion of 𝑓 term by term to find the corresponding series expansion for the
derivative of 𝑓. Then use the result of the first
part to evaluate the sum of the series the sum of negative one to the 𝑛 plus one
times 𝑛 plus one over three to the 𝑛th power for values of 𝑛 between zero and
∞.
We’ve been given the series
expansion of 𝑓. So let’s work out the first few
terms. The first term is when 𝑛 is equal
to zero. It’s negative one to the power of
zero times 𝑥 to the power of zero, which is equal to one. The second term is when 𝑛 is equal
to one. So it’s negative one to the power
of one times 𝑥 to the power of one, which is negative 𝑥. We then add negative one squared
times 𝑥 squared, which is 𝑥 squared. And we continue in this manner. And we see that 𝑓 of 𝑥 is equal
to one minus 𝑥 plus 𝑥 squared minus 𝑥 cubed plus 𝑥 to the fourth power minus 𝑥
to the fifth power, and so on.
The question tells us to
differentiate the given series expansion of 𝑓 that will give us 𝑓 prime. And of course we can simply do this
term by term. The derivative of one is zero, and
the derivative of negative 𝑥 is negative one. The derivative of 𝑥 squared is two
𝑥. And then our next term is negative
three 𝑥 squared.
We continue in this way,
multiplying each term by its exponent and then reducing that exponent by one. Without writing the zero, we see
that we can write 𝑓 prime of 𝑥 as negative 𝑥 to the power of zero — remember,
that’s just negative one — plus two 𝑥 to the power of one minus three 𝑥 squared
plus four 𝑥 cubed, and so on.
Let’s think about how we can write
this as a sum. We know that we start with a
negative term, and then the sign alternates. To achieve this, we need negative
one to the power of 𝑛 plus one. This will work for values of 𝑛
between zero and ∞. We then multiply each term by 𝑛
plus one. So the first term is when 𝑛 is
equal to zero. And we’re multiplying that by
one. The second term is when 𝑛 is equal
to one. And we’re multiplying that by two,
and so on. And this is all times 𝑥 to the
𝑛th power.
And so we found the series
expansion for the derivative of 𝑓. It’s the sum of negative one to the
power of 𝑛 plus one times 𝑛 plus one times 𝑥 to the 𝑛th power for values of 𝑛
between zero and ∞.
Part two says to use the result of
the first part to evaluate the sum of the series the sum of negative one to the
power of 𝑛 plus one times 𝑛 plus one over three to the 𝑛th power for values of 𝑛
between zero and ∞. Let’s compare this sum to the one
we generated in the first part of this question. If we rewrite this as the sum of
negative one to the power of 𝑛 plus one times 𝑛 plus one times a third to the 𝑛th
power, then we see it’s of the same form. Just 𝑥 equals a third.
So what we’re going to do is we’re
going to differentiate our original function one over one plus 𝑥 and evaluate that
when 𝑥 is equal to one-third. Let’s write 𝑓 of 𝑥 as one plus 𝑥
to the power of negative one. And then we use the chain rule. When we do, we find its derivative
to be equal to negative one over one plus 𝑥 squared. The sum of our series will be the
value of the derivative when 𝑥 is equal to one-third. So that’s negative one over one
plus one-third squared. That’s negative one over 16 over
nine, which is simply negative nine sixteenths.
In our final example, we’re going
to look at finding the interval of convergence of the derivative of a power
series.
Consider the series 𝑓 of 𝑥 equals
one over one minus 𝑥 squared, which equals the sum of 𝑥 to the power of two 𝑛 for
values of 𝑛 between zero and ∞. Find the interval of convergence
for the derivative of the given series.
We’ve been given a power series
expansion for our function. And we’re looking to find the
interval of convergence for its derivative. So we begin by recalling that the
derivative of the sum of 𝑥 to the power of two 𝑛 for values of 𝑛 between zero and
∞ is equal to the sum of the derivative of 𝑥 to the power of two 𝑛 for values of
𝑛 between zero and ∞.
Now of course when we’re
differentiating a polynomial term, we multiply the entire term by its exponent and
then reduce the exponent by one. So in this case, the derivative 𝑓
prime of 𝑥 is the sum of two 𝑛 times 𝑥 to the power of two 𝑛 minus one for
values of 𝑛 between zero and ∞. So how do we test for
convergence?
Well, we recall the ratio test. We’re specifically looking for
convergence. So we can use the first part of
this test. And it says that if the limit as 𝑛
approaches ∞ of the absolute value of 𝑎 𝑛 plus one over 𝑎 𝑛 is less than one,
then the series the sum of 𝑎 𝑛 converges. In our case, 𝑎 𝑛 is two 𝑛 times
𝑥 to the power of two 𝑛 minus one. So 𝑎 𝑛 plus one is two times 𝑛
plus one times 𝑥 to the power of two 𝑛 plus one minus one. We divide through by two and
distribute our parentheses. And we see that we’re looking for
the limit as 𝑛 approaches ∞ of 𝑛 plus one times 𝑥 to the power of two 𝑛 plus one
over 𝑛 times 𝑥 to the power of two 𝑛 minus one.
Then we recall that when we divide
𝑥 to the power of two 𝑛 plus one by 𝑥 to the power of two 𝑛 minus one, we
subtract their exponents. And this simplifies to 𝑥
squared. 𝑥 is independent of 𝑛. So we can rewrite this as the
absolute value of 𝑥 squared times the limit as 𝑛 approaches ∞ of 𝑛 plus one all
over 𝑛. We can divide each part of 𝑛 plus
one by 𝑛. And then we see as 𝑛 approaches ∞,
one plus 𝑛 approaches zero. And this means that the limit as 𝑛
approaches ∞ of one plus one over 𝑛 is simply one. And so we have the absolute value
of 𝑥 squared.
Remember, we’re interested in where
this converges. So we need to know where this is
less than one. So we recall that, for the absolute
value of 𝑥 squared to be less than one, the absolute value of 𝑥 itself must be
less than one. Which means that 𝑥 must be greater
than negative one or less than one. And thus we found the interval of
convergence for the derivative of our series. It’s the open interval negative one
to one.
In this video, we’ve seen that we
can differentiate or integrate power series by individually differentiating or
integrating each term, just as we would for a polynomial. We call this term-by-term
differentiation and integration. We saw that we can write this as
the derivative of the sum of 𝐶𝑛 times 𝑥 minus 𝑎 to the 𝑛th power for values of
𝑛 between zero and ∞. Equals the sum of the derivative of
𝐶𝑛 times 𝑥 minus 𝑎 to the 𝑛th power between those same values. And we have a similar formula for
integration.
