Video: Representing Rational Functions Using Power Series

In this video, we will learn how to use the formula of the infinite sum of geometric series to find a power series representation for some rational functions.

17:50

Video Transcript

Representing Rational Functions Using Power Series

In this lesson, we’re going to learn how to convert rational functions into power series by using what we already know about infinite geometric series. We’re going to discuss both the radius and interval of convergence of this series. And finally, we’ll go over some examples of turning a power series into a rational function.

We’ll start by recalling some facts we know about infinite geometric series. First for an infinite geometric series with first term π‘Ž and ratio of successive terms π‘Ÿ, the sum from 𝑛 equals zero to ∞ of π‘Ž multiplied by π‘Ÿ to the 𝑛th power is equal to π‘Ž divided by one minus π‘Ÿ, when the absolute value of our ratio π‘Ÿ is less than one. And we also recall that the same infinite geometric series, the sum from 𝑛 equals zero to ∞ of π‘Ž multiplied by π‘Ÿ to the 𝑛th power will diverge when the absolute value of our ratio is greater than one.

Now, imagine our ratio of successive terms instead of being represented by π‘Ÿ, it was actually some function of π‘₯. Let’s call this 𝑓 of π‘₯. So we have an expression which looks like this: π‘Ž divided by one minus some function of π‘₯. We can then use our facts about geometric series to write this as a power series, assuming that the absolute value of our ratio of successive terms is less than one. So we can rewrite π‘Ž divided by one minus 𝑓 of π‘₯ as the sum from 𝑛 equals zero to ∞ of π‘Ž multiplied by 𝑓 of π‘₯ to the 𝑛th power provided that the absolute value of the ratio of our successive terms, which in our case is 𝑓 of π‘₯ , is less than one.

And in particular, by using our second fact, we know that if the absolute value of our ratio of successive terms is greater than one, then this series will diverge. So when our ratio π‘Ÿ is some function of π‘₯, we know that our power series will diverge when the absolute value of 𝑓 of π‘₯ is greater than one. In particular, we notice that we don’t know what happens when the absolute value of 𝑓 of π‘₯ is equal to one. So if we’re asked to find an interval of convergence, we will need to check each of these cases separately.

Okay, so now, we’ve seen how to write the rational function π‘Ž divided by one minus 𝑓 of π‘₯ as a power series and we’ve also looked at the radius of convergence. However, this leaves us with the question, β€œWhat about a general rational function, some polynomial 𝑝 of π‘₯ divided by some polynomial π‘ž of π‘₯?” How would we write this as a power series? Since we’ve already seen how to write rational functions in the form π‘Ž divided by one minus 𝑓 of π‘₯ as a power series, we could try to write our rational function as the sum of fractions of this form.

There are numerous different ways of doing this. One such method is algebraic manipulation. One such way of using algebraic manipulation would be rewriting the fraction one divided by one plus π‘₯ as one divided by one minus negative π‘₯. And we see this is now in the form π‘Ž divided by one minus 𝑓 of π‘₯, where π‘Ž is one and 𝑓 of π‘₯ is equal to negative π‘₯. Another method we could use to rewrite our rational function is polynomial division. For example, using polynomial division, we have two π‘₯ squared plus three all divided by π‘₯ squared plus one is equal to two plus one divided by one plus π‘₯ squared. And we can write this into a power series by using the same trick we did before: one plus π‘₯ squared is equal to one minus negative π‘₯ squared.

The last method we’re going to discuss for rewriting our rational function is partial fractions. For example, we could rewrite π‘₯ plus three all divided by π‘₯ plus one multiplied by π‘₯ minus one as two divided by π‘₯ minus one minus one divided by π‘₯ plus one by using partial fractions. And then, we can use algebraic manipulation on the resulting rational functions to write them in the form π‘Ž divided by one minus 𝑓 of π‘₯. In our first rational function, if we multiply the numerator and the denominator by negative one, we get negative two divided by one minus π‘₯. And we can rewrite the denominator in our second fraction as one minus negative π‘₯.

We’re now ready to try some examples.

Consider the function 𝑓 of π‘₯ is equal to one divided by two plus π‘₯, find the power series for 𝑓 of π‘₯. Identify its interval of convergence.

The question gives us a rational function 𝑓 of π‘₯. It wants us to find the power series for the function 𝑓 of π‘₯ and then to find the interval of convergence for this power series. We recall a fact about geometric series. The sum from 𝑛 equals zero to ∞ of π‘Ž multiplied by π‘Ÿ to 𝑛th power is equal to π‘Ž divided by one minus π‘Ÿ, when the absolute value of our ratio π‘Ÿ is less than one. And we also recall that this power series will diverge when the absolute value of π‘Ÿ is greater than one. If we could rewrite our function 𝑓 of π‘₯ in the form π‘Ž divided by one minus π‘Ÿ, then we could use this fact about geometric series to rewrite our function as a power series.

We want the denominator of our fraction to be in the form one minus the ratio π‘Ÿ. However, in our fraction for 𝑓 of π‘₯, we have a two in the denominator. Since we want our denominator to be one minus the ratio π‘Ÿ, we’ll take out our factor two in our denominator. We have that two multiplied by one gives us two and then two multiplied by π‘₯ divided by two gives us π‘₯. We can then take the constant of a half outside of our fraction. This gives us a half multiplied by one divided by one plus π‘₯ over two. This is now almost in the form that we need it. We have the constant of one in the numerator. But instead of one minus π‘Ÿ in the denominator, we have one plus π‘₯ over two.

We can do this by using a little bit of algebraic manipulation. Instead of adding π‘₯ over two, we can subtract negative π‘₯ over two. Now, if we set the numerator of our function to be equal to π‘Ž and the denominator of our fraction to be one minus π‘Ÿ, so that setting π‘Ÿ to be equal to negative π‘₯ over two. Then, we’ve shown that our fraction is in the form of the infinite sum of a geometric series with first term π‘Ž equal to one and ratio π‘Ÿ equal to negative π‘₯ over two.

So, using our fact about geometric series, we can write this as a half multiplied by the sum from 𝑛 equals zero to ∞ of one multiplied by negative π‘₯ over two raised to the 𝑛th power. And in particular, we’ll know that this is definitely true, when the absolute value of our ratio π‘Ÿ is less than one. And we know that it will diverge when the absolute value of our ratio π‘Ÿ is greater than one. In our case, we have that π‘Ÿ is equal to negative π‘₯ divided by two. So this power series must converge when the absolute value of negative π‘₯ divided by two is less than one.

We can simplify our power series slightly by noticing that negative π‘₯ over two is equal to a half multiplied by negative π‘₯. We can then distribute the exponent over the parenthesis. This gives us a half multiplied by the sum from 𝑛 equals zero to ∞ of a half to the 𝑛th power multiplied by negative π‘₯ to the 𝑛th power. Finally, we can bring the coefficient of a half inside of our summand. But then, a half multiplied by a half to the 𝑛th power is just equal to a half raised to the power of 𝑛 plus one. So the power series representation for our function 𝑓 of π‘₯ is the sum from 𝑛 equals zero to ∞ of a half raised to the power of 𝑛 plus one multiplied by negative π‘₯ to the 𝑛th power.

The second part of our question wants us to identify the interval of convergence of this power series. We recall that the interval of convergence is all of the values of π‘₯ where our series will converge. We’ve already shown that our series has to converge when the absolute value of our ratio π‘Ÿ is less than one. And it has to diverge when the absolute value of our ratio π‘Ÿ is greater than one. In our case, our ratio π‘Ÿ is equal to negative π‘₯ over two. So our power series must converge when the absolute value of negative π‘₯ divided by two is less than one. The absolute value of negative π‘₯ over two is just equal to the absolute value of π‘₯ over two.

Next, solving the absolute value of π‘₯ over two is less than one is the same as solving negative one is less than π‘₯ over two is less than one. We can then multiply this inequality through by two. This gives us that negative two is less than π‘₯ is less than two. This is our radius of convergence. Our power series must converge for all values of π‘₯ between negative two and two. And we know it will diverge when π‘₯ is greater than two or when π‘₯ is less than negative two. However, we don’t know what will happen when π‘₯ is equal to two or when π‘₯ is equal to negative two.

To find out what happens at the bounds of our radius of convergence, let’s substitute in π‘₯ is equal to two and π‘₯ is equal to negative two into our power series. Substituting π‘₯ is equal to two gives us the sum from 𝑛 equals zero to ∞ of a half raised to the power of 𝑛 plus one multiplied by negative two to the power of 𝑛. There are several different ways of evaluating this series. For example, we could use the 𝑛th term divergence test. However, we’re going to look term by term at the expansion of our series.

Our first term when 𝑛 is equal to zero gives us a half raised to the power of zero plus one, which is a half. And then, we multiply this by negative two raised to the power of zero, which is one. We can do the same of our second term. When 𝑛 is equal to one, we get a half squared multiplied by negative two, which we can work out to be negative one-half. Our third term when 𝑛 is equal to two gives us a half cubed multiplied by negative two squared, which is plus a half. In fact, this series continues. We add a half. Then, we subtract a half. Then, we add a half and we subtract a half et cetera. We can then consider what the partial sums will look like.

Our first partial sum will be equal to a half. Our second partial sum, the sum of the first two terms of our series, will be a half minus a half which is zero. Our third partial sum, the sum of the first three terms of our series, will be a half. And our fourth partial sum, the sum of the first four terms of our series, will be zero. And we can see that this pattern will continue. In fact, this means that our partial sums won’t converge. They will fluctuate between a half and zero. Since the sequence of partial sums doesn’t converge, we can conclude that our power series does not converge when π‘₯ is equal to two. So π‘₯ is equal to two is not in our interval of convergence.

We can do the same when π‘₯ is equal to negative two. We substitute π‘₯ is equal to negative two into our power series and then calculate the series term by term. We see that every single term in our series is equal to a half. Therefore, our 𝑛th partial sum of this series will be a half plus a half plus a half 𝑛 times or 𝑛 divided by two. Therefore, since our 𝑛th partial sum is unbounded, we can conclude that the power series does not converge when π‘₯ is equal to negative two. Therefore, since we’ve shown that neither of the end points of our radius of convergence are in our interval of convergence, we can conclude that the interval of convergence for the power series of the function one divided by two plus π‘₯ is the open interval from negative two to two.

Let’s now look at an example, where our rational function 𝑓 of π‘₯ has a polynomial in the numerator.

Let us consider the function 𝑔 of π‘₯ is equal to π‘₯ divided by three minus π‘₯. Find a power series representation for the function 𝑔 of π‘₯. Find the radius of convergence.

The question gives us a rational function 𝑔 of π‘₯. It wants us to find the power series representation of the function 𝑔 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 𝑛 equal zero to ∞ of π‘Ž multiplied by π‘Ÿ to the 𝑛th power is equal to π‘Ž divided by one minus π‘Ÿ, when the absolute value of our ratio π‘Ÿ is less than one.

So if we can write our rational function 𝑔 of π‘₯ in the form π‘Ž divided by one minus π‘Ÿ, 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‘Ÿ is less than one. We’ll 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‘Ÿ. However, we have three minus π‘₯ as our denominator. To have a denominator in the form one minus π‘Ÿ, we’re 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‘Ž equal to one and π‘Ÿ 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 𝑔 of π‘₯ is equal to π‘₯ divided by three multiplied by the sum from 𝑛 equals zero to ∞ of one multiplied by π‘₯ over three to the 𝑛th 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 𝑛th power. However, we can simplify this further to just be π‘₯ over three all raised to the power of 𝑛 plus one. This gives us that our function 𝑔 of π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of π‘₯ over three to the power of 𝑛 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‘Ÿ the radius of convergence of a power series if the power series converges when the absolute value of π‘₯ is less than π‘Ÿ and diverges when the absolute value of π‘₯ is greater than π‘Ÿ. Unless the power series converges for all values of π‘₯, in which case we call the radius of convergence ∞. We recall that for a geometric series, the sum from 𝑛 equals zero to ∞ of π‘Ž multiplied by π‘Ÿ to the 𝑛th power will diverge whenever the absolute value of our ratio π‘Ÿ is greater than one. In our case, our ratio π‘Ÿ 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‘Ÿ is equal to three. And our power series for 𝑔 of π‘₯ will converge whenever the absolute value of π‘₯ is less than three.

Let’s now look at an example of converting a power series into a rational function.

Convert the sum from 𝑛 equals zero to ∞ of two multiplied by π‘₯ to the 𝑛th power into a rational function.

The question gives us a power series. And it wants us to convert this into a rational function. And we recall that a rational function 𝑓 of π‘₯ is the quotient of polynomials 𝑝 of π‘₯ and π‘ž of π‘₯. To convert our power series into a rational function, we recall the following fact about a geometric series: the sum from 𝑛 equals zero to ∞ of π‘Ž multiplied by π‘Ÿ to the 𝑛th power is equal to π‘Ž divided by one minus π‘Ÿ, when the absolute value of π‘Ÿ is less than one.

We see that our power series is already written as a geometric series, where π‘Ž is equal to two and where π‘Ÿ is equal to π‘₯. Substituting π‘Ž is equal to two and π‘Ÿ is equal to π‘₯ into a geometric series formula gives us the sum from 𝑛 equals zero to ∞ of two multiplied by π‘₯ to the 𝑛th power is equal to two divided by one minus π‘₯. And this is valid when the absolute value of our ratio π‘₯ is less than one.

We’ll do one more example of converting a power series into a rational function.

Convert the sum from 𝑛 equals zero to ∞ of two multiplied by negative π‘₯ over three to the power of three 𝑛 plus one into a rational function.

We recall from infinite geometric series the sum from 𝑛 equals zero to ∞ of π‘Ž multiplied by π‘Ÿ to the 𝑛th power is equal to π‘Ž divided by one minus π‘Ÿ, when the absolute value of π‘Ÿ is less than one. Since the question wants us to convert this into a rational function, we notice if we can manipulate our series to be in the form of a geometric series, then we can use our fact about geometric series to write it as a rational function.

We’ll start by simplifying the exponent inside our summand. We can then take this coefficient of negative π‘₯ over three outside of our sum. Instead of raising negative π‘₯ over three to the power of three 𝑛, we can first cube it and then raise it to the 𝑛th power. Evaluating our exponent of three gives us negative π‘₯ cubed over 27. Then, we can see that this is a geometric series with π‘Ž equal to two and π‘Ÿ equal to negative π‘₯ cubed over 27. Using our infinite geometric series formula, we get negative π‘₯ over three multiplied by two divided by one minus negative π‘₯ cubed over 27 which is equal to negative two π‘₯ divided by three plus π‘₯ cubed over nine. And this is valid when the absolute value of our ratio is less than one.

So to summarize what we’ve done, we can use our infinite geometric series formula to write rational functions as power series and to write power series as rational functions. Sometimes, we’ll need to manipulate our rational function or we’ll need to manipulate our power series. And we’ve seen multiple ways of doing this. We can also find the radius of convergence of our power series by using that our series converges when the absolute value of our ratio π‘Ÿ is less than one. Finally, we can find the interval of convergence by checking the end points of our radius of convergence.

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