Question Video: Finding the Power Series Representation of a Sum of Rational Functions | Nagwa Question Video: Finding the Power Series Representation of a Sum of Rational Functions | Nagwa

# Question Video: Finding the Power Series Representation of a Sum of Rational Functions Mathematics • Higher Education

Consider the power series representations of π(π₯) = 1/(1 β π₯) and π(π₯) = 1/(1 + π₯). Use them, or otherwise, to calculate the first three nonzero terms, in ascending powers of π₯, for the power series of π(π₯) + π(π₯).

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### Video Transcript

Consider the power series representations of π of π₯ is equal to one divided by one minus π₯ and π of π₯ is equal to one divided by one plus π₯. Use them, or otherwise, to calculate the first three nonzero terms, in ascending powers of π₯, for the power series of π of π₯ plus π of π₯.

The question gives us the rational functions π of π₯ and π of π₯. And it wants us to find the power series representation of π of π₯ plus π of π₯. We could add these two rational functions together and then find a power series representation. However, we could also find the power series representation of each function separately and then add these together. To write these rational functions as a power series, we recall the following fact about geometric series. If the absolute value of the ratio of successive terms π is less than one, then the sum from π equals zero to β of π times π to the πth power is convergent and is equal to π divided by one minus π. We can see this gives us a method of turning a quotient into a power series, provided the absolute value of our ratio π is less than one.

We see that our function π of π₯, which is equal to one divided by one minus π₯, is already in this form. We set π equal to one and π equal to π₯. So by using our fact about infinite geometric series, we have that π of π₯ is equal to the sum from π equals zero to β of one times π₯ to the πth power if the absolute value of π₯ is less than one. And we can simplify one times π₯ to the πth power to just be π₯ to the πth power. We want to do the same with our function π of π₯ is equal to one divided by one plus π₯. We see the numerator is just equal to one, so weβll set π equal to one. And we want our denominator to be one minus π. So we have negative π is equal to π₯. Then, weβll just multiply both sides of this equation by negative one to see that π is equal to negative π₯.

Using our fact about infinite geometric series, we have that π of π₯ is equal to the sum from π equals zero to β of one times negative π₯ raised to the πth power if the absolute value of negative π₯ is less than one. And one times negative π₯ to the πth power is just equal to negative π₯ to the πth power. We can also see the same. The absolute value of negative π₯ is less than one is the same as saying the absolute value of π₯ is less than one. So weβve now found power series representations for both π of π₯ and π of π₯. We can find a power series representation for π of π₯ plus π of π₯ by adding our two power series together. However, we must be careful since this will only be necessarily true if both of our power series are convergent for that value of π₯.

If we have the absolute value of π₯ is less than one, both our power series for π of π₯ and π of π₯ converge. So we could just add these power series together. This gives us the sum from π equals zero to β of π₯ to the πth power plus negative π₯ to the πth power. We can then evaluate this power series term by term. Remember, we want to find the first three nonzero terms in this power series.

When π is zero, π₯ to the zeroth power is one and negative π₯ all raised to the zeroth power is one. When π is one, π₯ to the first power is equal to π₯. And negative π₯ all raised to the first power is equal to negative π₯. And of course these two terms cancel. When π is equal to two, π₯ to the πth power is equal to π₯ squared and negative π₯ all raised to the πth power is also equal to π₯ squared. In fact, we can start to see a pattern. When our value of π is odd, weβll get π₯ to the πth power minus π₯ to the πth power, so our terms will cancel. And when π is even, we just get π₯ to the πth power plus π₯ to the πth power.

In other words, anytime we have an odd power of π, the terms will cancel. And when π is even, we just always end up with two π₯ to the power of π. Therefore, weβve found the first three terms of this power series to be two plus two π₯ squared plus two π₯ to the fourth power. However, this was not the only way of approaching this problem. We couldβve just added the two rational functions together. To add two rational functions together, we need their denominators to be equal. So we multiply the first by one plus π₯ divided by one plus π₯ and the second by one minus π₯ divided by one minus π₯. We see that our denominator is a difference between squares, so itβs just one minus π₯ squared. And if we add our numerators together, we get a numerator of one plus π₯ plus one minus π₯. And of course the positive π₯ and the negative π₯ cancel.

So weβve shown that π of π₯ plus π of π₯ is equal to two divided by one minus π₯ squared. And we can see that this is a rational function. And itβs also very similar to the formula given in the summation of our infinite geometric series. We set our initial term π equal to two and our ratio of successive terms π equal to π₯ squared. So by using our fact of our infinite geometric series, we have that π of π₯ plus π of π₯ is equal to the sum from π equals zero to β of two times π₯ squared to the πth power if the absolute value of π₯ squared is less than one. And by using our laws of exponents, we can rewrite two times π₯ squared to the πth power as two times π₯ to the power of two π.

Expanding our series and writing out the first three terms, we get two times π₯ to the power of two times zero plus two times π₯ to the power of two times one plus two times π₯ to the power of two times two. And this sum goes on indefinitely. However, weβre only interested in the first three terms. And if we evaluate this expression, we see we also get two plus two π₯ squared plus two π₯ to the fourth power. Therefore, weβve shown two different ways of finding the power series representation of π of π₯ plus π of π₯, where π of π₯ is equal to one divided by one minus π₯ and π of π₯ is equal to one divided by one plus π₯. And by using both of these methods, we have shown that the first three nonzero terms in ascending powers of π₯ for the power series of π of π₯ plus π of π₯ are two plus two π₯ squared plus two π₯ to the fourth power.

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