Video Transcript
Using partial fractions, calculate the power series of the function 𝑓 of 𝑥 is equal to 𝑥 divided by 𝑥 squared minus one.
The question gives us the rational function 𝑓 of 𝑥. It wants us to find the power series of this rational function by using partial fractions. To use partial fractions, we first need to fully factor the denominator of our rational function 𝑓 of 𝑥. We see that the denominator 𝑥 squared minus one is a difference between squares. So we can factor this to give us 𝑥 minus one multiplied by 𝑥 plus one.
Since we now have two unique factors in our denominator, by using partial fractions, we can rewrite 𝑓 of 𝑥 as 𝐴 divided by 𝑥 minus one plus 𝐵 divided by 𝑥 plus one for some constants 𝐴 and 𝐵. To find the values of 𝐴 and 𝐵, we’re going to multiply both sides of our equation by 𝑥 minus one times 𝑥 plus one.
Multiplying through by 𝑥 minus one times 𝑥 plus one gives us the following equation. Canceling the shared factors in our numerator and our denominator on the left-hand side leaves us with 𝑥. And canceling the shared factors in our numerator and our denominator on the right-hand side leaves us with 𝐴 times 𝑥 plus one plus 𝐵 times 𝑥 minus one. So we have 𝑥 is equal to 𝐴 times 𝑥 plus one plus 𝐵 times 𝑥 minus one.
However, this equation should be true for all values of 𝑥. We represent this by using an equivalent sign. Since this is true for all values of 𝑥, we can eliminate one of our variables 𝐴 or 𝐵 by substituting specific values of 𝑥. If we choose 𝑥 is equal to negative one, we remove the variable 𝐴 from our equation. If we substitute 𝑥 is equal to one, we remove the variable 𝐵 from our equation.
Substituting 𝑥 is equal to negative one, we get that negative one is equal to 𝐴 times negative one plus one plus 𝐵 times negative one minus one. Evaluating this expression, we get that negative one is equal to negative two times 𝐵, which we can rearrange to see that 𝐵 is equal to one-half. Substituting 𝑥 is equal to one, we get that one is equal to 𝐴 times one plus one plus 𝐵 times one minus one. This time, the one minus one cancels to give us zero. So we eliminate the variable 𝐵. And one plus one simplifies to give us two. So we have that one is equal to two times 𝐴, which we can rearrange to see that 𝐴 is also equal to one-half.
Since 𝐴 is equal to one-half and 𝐵 is also equal to one-half, by using partial fractions, we’ve shown we can rewrite our function 𝑓 of 𝑥 as one divided by two times 𝑥 minus one plus one divided by two times 𝑥 plus one. This means instead of finding the power series for our function 𝑓 of 𝑥, we can instead find the power series for one divided by two times 𝑥 minus one and one divided by two times 𝑥 plus one. We can then just add these power series together. And as long as both of these series converges, we will have a power series for 𝑓 of 𝑥.
To help us find the power series of these two rational functions, we recall the following fact about geometric series. If the absolute value of 𝑟 is less than one, then 𝑎 divided by one minus 𝑟 is equal to the sum from 𝑛 equals zero to ∞ of 𝑎 times 𝑟 to the 𝑛th power. In other words, provided the absolute value of 𝑟 is less than one, we can turn a quotient into a series.
Let’s start by trying to write our first fraction one divided by two times 𝑥 minus one in the form 𝑎 divided by one minus 𝑟. We can move the two in our denominator up into our numerator. In other words, this fraction is equal to one-half divided by 𝑥 minus one. We want the denominator of this fraction to be in the form one minus 𝑟. However, we have a denominator of 𝑥 minus one. We can fix this by multiplying both our numerator and our denominator by negative one. We can then multiply out the parentheses in our denominator to get negative one-half divided by one minus 𝑥.
And we can now see that this is in the form of 𝑎 divided by one minus 𝑟, where we have 𝑎 is equal to negative one-half and 𝑟 is equal to 𝑥. Using our formula for the infinite sum of a geometric series, we have if the absolute value of 𝑥 is less than one, then we can rewrite this rational function as the sum from 𝑛 equals zero to ∞ of negative one-half multiplied by 𝑥 to the 𝑛th power.
We now want to do the same to find a power series representation of one divided by two times 𝑥 plus one. We do the same thing we did before. We bring the two in our denominator up into our numerator. And instead of writing the denominator as 𝑥 plus one, we’ll write it as one plus 𝑥.
Remember, we want our denominator to be in the form one minus 𝑟. And we can do this by writing 𝑥 as negative one multiplied by negative 𝑥. And we now see this is in the form 𝑎 divided by one minus 𝑟, where 𝑎 is equal to one-half and 𝑟 is equal to negative 𝑥. And again, by using our formula for the infinite sum of a geometric series. We have if the absolute value of negative 𝑥 is less than one, then we can rewrite this rational function as the sum from 𝑛 equals zero to ∞ of one-half multiplied by negative 𝑥 to the 𝑛th power.
So we’ve now found power series representations for both of our functions, which adds to give us 𝑓 of 𝑥. So let’s clear some space and write what we know. We know, by partial factions, 𝑓 of 𝑥 is equal to one divided by two times 𝑥 minus one plus one divided by two times 𝑥 plus one. And we’ve found power series representations of both of these two rational functions.
We want to add these two power series together term by term. But we can only guarantee this will work if both of our power series are convergent. We can see that our first power series is convergent if the absolute value of 𝑥 is less than one. And our second power series is convergent if the absolute value of negative 𝑥 is less than one. This is actually the same as saying the absolute value of 𝑥 is less than one. So both of our power series are convergent if the absolute value of 𝑥 is less than one.
So if the absolute value of 𝑥 is less than one, both of these series converge. This means we can just add the series together term by term. Giving us the sum from 𝑛 equals zero to ∞ of negative one-half times 𝑥 to the 𝑛th power plus one-half times negative 𝑥 to the 𝑛th power.
We could leave our answer like this. However, we can simplify our answer a little bit. We see that both terms in our summand share a factor of one-half. We can factor this out. In fact, we can take it outside of our sum altogether. This gives us one-half times the sum from 𝑛 equals zero to ∞ of negative one times 𝑥 to the 𝑛th power plus negative 𝑥 to the 𝑛th power.
We can simplify this series further by noticing that negative 𝑥 raised to the 𝑛th power is equal to negative one to the 𝑛th power times 𝑥 to the 𝑛th power. This gives us the following expression. And we can see that both terms in our summand now share a factor of 𝑥 to the 𝑛th power. By taking out the shared factor of 𝑥 to the 𝑛th power and rearranging the terms in our summand, we get one-half times the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power minus one multiplied by 𝑥 to the 𝑛th power.
And so by using partial fractions, we’ve shown that the power series of 𝑥 divided by 𝑥 squared minus one is one-half times the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power minus one times 𝑥 to the 𝑛th power.
