Video: Use Partial Fractions to Find a Power Series Representation

Use partial fractions to find the power series of the function 𝑓(π‘₯) = 3/((π‘₯ βˆ’ 2)(π‘₯ + 1)).

06:46

Video Transcript

Use partial fractions to find the power series of the function 𝑓 of π‘₯ is equal to three divided by π‘₯ minus two multiplied by π‘₯ plus one.

The question asks us to find the power series of the function 𝑓 of π‘₯ and asks us to do this by using partial factions. We recall that since the denominator in our function 𝑓 of π‘₯ has two unique roots, by using partial fractions, we can write three divided by π‘₯ minus two multiplied by π‘₯ plus one as some number 𝐴 divided by π‘₯ minus two. And then we add some number 𝐡 divided by π‘₯ plus one.

To help us find the values of 𝐴 and 𝐡, we’re going to multiply both sides of our equation by the denominator π‘₯ minus two multiplied by π‘₯ plus one. We can cancel the shared factors in both our numerator and our denominator on the left-hand side of our equation to just get three. And when we distribute over the parentheses, when we multiply our first term, we can cancel the shared factor of π‘₯ minus two to just get 𝐴 multiplied by π‘₯ plus one. And when we do the same of our second term, we can cancel the shared factor of π‘₯ plus one to get 𝐡 multiplied by π‘₯ minus two.

Since this is true for all values of π‘₯, we’re going to write that three is equivalent to 𝐴 multiplied by π‘₯ plus one plus 𝐡 multiplied by π‘₯ minus two. Since we know that this is true for any value of π‘₯, we can use substitution to help us eliminate our variables. Substituting π‘₯ is equal to negative one gives us that three is equal to 𝐴 multiplied by negative one plus one plus 𝐡 multiplied by negative one minus two. We have that negative one plus one is equal to zero and negative one minus two is equal to negative three. So we have that three is equal to 𝐴 multiplied by zero minus three 𝐡. And since 𝐴 multiplied by zero is equal to zero, we must have that three is equal to negative three 𝐡. So if we divide through by negative three, we have that negative one is equal to 𝐡.

Now we substitute π‘₯ is equal to two to get that three is equal to 𝐴 multiplied by two plus one plus 𝐡 multiplied by two minus two. We have that two plus one is three and two minus two is zero. So we get that three is equal to three 𝐴 plus 𝐡 multiplied by zero. Since 𝐡 multiplied by zero is just equal to zero, we must have that three is equal to three 𝐴. So we can divide both sides of this equation by three to get that one is equal to 𝐴.

So what we’ve shown is, by using partial fractions, we can rewrite three divided by π‘₯ minus two multiplied by π‘₯ plus one as one divided by π‘₯ minus two minus one divided by π‘₯ plus one. We’re now ready to turn these into a power series.

We recall that, for a geometric series with ratio π‘Ÿ, if the absolute value of π‘Ÿ is less than one, then the sum from 𝑛 equals zero to infinity of π‘Ž multiplied by π‘Ÿ to the 𝑛th power is equal to π‘Ž divided by one minus π‘Ÿ. What this means is if we were to set π‘Ž divided by one minus π‘Ÿ to be equal to each of our fractions, then we could use this property of geometric series to convert each of our fractions into a series.

Let’s start with one divided by π‘₯ minus two. We want a denominator of one minus π‘Ÿ. However, our denominator has a negative two. So we’re going to need to rearrange our fraction. If we were to take a factor of negative two outside of our denominator, then we would get a new denominator of negative two multiplied by negative π‘₯ over two plus one. And we can take this factor of negative a half out of our faction to get negative a half multiplied by one divided by one minus π‘₯ over two.

And now we have a fraction in the desired form. We have π‘Ž equal to one and π‘Ÿ equal to π‘₯ divided by two. So by substituting π‘Ž is equal to one and π‘Ÿ is equal to π‘₯ over two into our geometric series formula, we’ve shown that one divided by π‘₯ minus two is equal to negative a half multiplied by the sum from 𝑛 equals zero to infinity of one multiplied by π‘₯ over two raised to the 𝑛th power. And this is only true when the absolute value of our ratio π‘₯ divided by two is less than one.

Let’s now do the same for our second fraction negative one divided by π‘₯ plus one. There isn’t much manipulation needed for this fraction. If we set π‘Ž to equal negative one and π‘Ÿ to equal negative π‘₯, we can see it’s already in the correct form. So by substituting π‘Ž is equal to negative one and π‘Ÿ is equal to negative π‘₯ into our geometric series formula, we have that negative one divided by π‘₯ plus one is equal to the sum from 𝑛 equals zero to infinity of negative one multiplied by negative π‘₯ to the 𝑛th power. And we have that this is true when the absolute value of our ratio negative π‘₯ is less than one.

Earlier, we showed by using partial fractions that we could write our function 𝑓 of π‘₯ to be equal to one divided by π‘₯ minus two minus one divided by π‘₯ plus one. We can then substitute in our power series representations for one divided by π‘₯ minus two and negative one divided by π‘₯ plus one. We can bring the coefficient of negative a half inside our sum to change our first sum into the sum from 𝑛 equals zero to infinity of negative a half multiplied by π‘₯ over two to the 𝑛th power. We could also distribute our exponent over our parentheses to get π‘₯ to the 𝑛th power divided by two to the 𝑛th power.

Then we see our denominator is two multiplied by two to the 𝑛th power, which we can simplify to two to the power of 𝑛 plus one. So our first term is equal to the sum from 𝑛 equals zero to infinity of negative π‘₯ to the 𝑛th power divided by two to the power of 𝑛 plus one. In our second sum, we can distribute the exponent over our parentheses to get negative one to the 𝑛th power multiplied by π‘₯ to the 𝑛th power. Then we can notice that negative one multiplied by negative one to the 𝑛th power is equal to negative one to the power of 𝑛 plus one. This gives us that our second term is equal to the sum from 𝑛 equal zero to infinity of negative one to the power of 𝑛 plus one multiplied by π‘₯ to the 𝑛th power.

Now we notice that the limits of both of our sums start from 𝑛 equals zero to infinity. So we can combine both of these into one sum. This gives us the sum from 𝑛 equals zero to infinity of negative π‘₯ to the 𝑛th power divided by π‘₯ to the power of two 𝑛 plus one plus negative one to the power of 𝑛 plus one multiplied by π‘₯ to the 𝑛th power.

Next, we can notice that both terms inside our summand have a shared factor of π‘₯ to the 𝑛th power. So we can factor this out. The last thing we need to do is to notice that one divided by two to the power of 𝑛 plus one is equal to a half raised to the power of 𝑛 plus one. So we’ll write this term as a half raised to the power of 𝑛 plus one. And then we’ll switch the order of the terms inside our parentheses. This gives us that we can rewrite 𝑓 of π‘₯ is equal to three divided by π‘₯ minus two multiplied by π‘₯ plus one as the power series. The sum from 𝑛 equals zero to infinity of negative one to the power of 𝑛 plus one minus a half to the power 𝑛 plus one all multiplied by π‘₯ to the 𝑛th power.

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