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

Consider the power series representations of 𝑓(π‘₯) = 1/(1 βˆ’ π‘₯) and 𝑔(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 𝑓(π‘₯) + 𝑔(π‘₯).

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Video Transcript

Consider the power series representations of 𝑓 of π‘₯ is equal to one divided by one minus π‘₯ and 𝑔 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 𝑓 of π‘₯ plus 𝑔 of π‘₯.

The question gives us the rational functions 𝑓 of π‘₯ and 𝑔 of π‘₯. And it wants us to find the power series representation of 𝑓 of π‘₯ plus 𝑔 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‘Ÿ is less than one, then the sum from 𝑛 equals zero to ∞ of π‘Ž times π‘Ÿ to the 𝑛th power is convergent and is equal to π‘Ž divided by one minus π‘Ÿ. We can see this gives us a method of turning a quotient into a power series, provided the absolute value of our ratio π‘Ÿ is less than one.

We see that our function 𝑓 of π‘₯, which is equal to one divided by one minus π‘₯, is already in this form. We set π‘Ž equal to one and π‘Ÿ equal to π‘₯. So by using our fact about infinite geometric series, we have that 𝑓 of π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of one times π‘₯ to the 𝑛th power if the absolute value of π‘₯ is less than one. And we can simplify one times π‘₯ to the 𝑛th power to just be π‘₯ to the 𝑛th power. We want to do the same with our function 𝑔 of π‘₯ is equal to one divided by one plus π‘₯. We see the numerator is just equal to one, so we’ll set π‘Ž equal to one. And we want our denominator to be one minus π‘Ÿ. So we have negative π‘Ÿ is equal to π‘₯. Then, we’ll just multiply both sides of this equation by negative one to see that π‘Ÿ is equal to negative π‘₯.

Using our fact about infinite geometric series, we have that 𝑔 of π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of one times negative π‘₯ raised to the 𝑛th power if the absolute value of negative π‘₯ is less than one. And one times negative π‘₯ to the 𝑛th power is just equal to negative π‘₯ to the 𝑛th 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 we’ve now found power series representations for both 𝑓 of π‘₯ and 𝑔 of π‘₯. We can find a power series representation for 𝑓 of π‘₯ plus 𝑔 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 𝑓 of π‘₯ and 𝑔 of π‘₯ converge. So we could just add these power series together. This gives us the sum from 𝑛 equals zero to ∞ of π‘₯ to the 𝑛th power plus negative π‘₯ to the 𝑛th 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 𝑛 is zero, π‘₯ to the zeroth power is one and negative π‘₯ all raised to the zeroth power is one. When 𝑛 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 𝑛 is equal to two, π‘₯ to the 𝑛th power is equal to π‘₯ squared and negative π‘₯ all raised to the 𝑛th power is also equal to π‘₯ squared. In fact, we can start to see a pattern. When our value of 𝑛 is odd, we’ll get π‘₯ to the 𝑛th power minus π‘₯ to the 𝑛th power, so our terms will cancel. And when 𝑛 is even, we just get π‘₯ to the 𝑛th power plus π‘₯ to the 𝑛th power.

In other words, anytime we have an odd power of 𝑛, the terms will cancel. And when 𝑛 is even, we just always end up with two π‘₯ to the power of 𝑛. Therefore, we’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 could’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 it’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 we’ve shown that 𝑓 of π‘₯ plus 𝑔 of π‘₯ is equal to two divided by one minus π‘₯ squared. And we can see that this is a rational function. And it’s also very similar to the formula given in the summation of our infinite geometric series. We set our initial term π‘Ž equal to two and our ratio of successive terms π‘Ÿ equal to π‘₯ squared. So by using our fact of our infinite geometric series, we have that 𝑓 of π‘₯ plus 𝑔 of π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of two times π‘₯ squared to the 𝑛th 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 𝑛th power as two times π‘₯ to the power of two 𝑛.

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, we’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, we’ve shown two different ways of finding the power series representation of 𝑓 of π‘₯ plus 𝑔 of π‘₯, where 𝑓 of π‘₯ is equal to one divided by one minus π‘₯ and 𝑔 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 𝑓 of π‘₯ plus 𝑔 of π‘₯ are two plus two π‘₯ squared plus two π‘₯ to the fourth power.

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