Question Video: Finding the Power Series for a Rational Function and Identifying Its Interval of Convergence Mathematics • Higher Education

Consider the function 𝑓(π‘₯) = π‘₯/(1 βˆ’ 9π‘₯Β²). Find the power series for 𝑓(π‘₯). Identify its interval of convergence.

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Video Transcript

Consider the function 𝑓 of π‘₯ is equal to π‘₯ divided by one minus nine π‘₯ squared. Find the power series for 𝑓 of π‘₯. Identify its interval of convergence.

The question gives us a rational function 𝑓 of π‘₯. And it wants us to find the power series for our function 𝑓 of π‘₯. Since we want to find the power series of a rational function, we can recall the following fact about geometric series. The sum from 𝑛 equals zero to ∞ of π‘Ž times π‘Ÿ to the 𝑛th power is equal to π‘Ž divided by one minus π‘Ÿ if the absolute value of π‘Ÿ is less than one. And this series will be divergent if the absolute value of π‘Ÿ is greater than one. We see this gives us a way of writing a quotient as a power series.

So if we can rewrite our function 𝑓 of π‘₯ to be in the form π‘Ž divided by one minus π‘Ÿ, then we can use this to rewrite it as a power series. Since our numerator only has a single term of π‘₯, we’ll take it outside of our fraction. So our function 𝑓 of π‘₯ is equal to π‘₯ multiplied by one divided by one minus nine π‘₯ squared. And we now see that this fraction is of the form π‘Ž divided by one minus π‘Ÿ, where π‘Ž is equal to one and π‘Ÿ is equal to nine π‘₯ squared.

This means we can now apply our fact about infinite geometric series. Giving us that 𝑓 of π‘₯ is equal to π‘₯ times the sum from 𝑛 equals zero to ∞ of one times nine π‘₯ squared to the 𝑛th power if the absolute value of nine π‘₯ squared is less than one. And it’s also worth reiterating that this series will be divergent if the absolute value of nine π‘₯ squared is greater than one since our value of π‘Ž is nonzero.

We can simplify our power series by noticing that nine π‘₯ squared is equal to three squared times π‘₯ squared. We can then distribute the exponent over our parentheses to get π‘₯ multiplied by the sum from 𝑛 equals zero to ∞ of three to the power of two 𝑛 times π‘₯ to the power of two 𝑛. We can then bring the factor of π‘₯ inside of our sum. And then we see that π‘₯ to the power of two 𝑛 multiplied by π‘₯ is just equal to π‘₯ to the power of two 𝑛 plus one. So we’ve shown our function 𝑓 of π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of three to the power of two 𝑛 times π‘₯ to the power of two 𝑛 plus one.

The question then wants us to identify the interval of convergence for our power series. And we recall the interval of convergence is an interval of all of the values of π‘₯ which will make our power series convergent. When we used our fact about geometric series to find a power series for our function 𝑓 of π‘₯, we knew that the power series would be convergent when the absolute value of the ratio was less than one. And it must be divergent when the absolute value of π‘Ÿ was greater than one. In other words, our power series converges when the absolute value of nine π‘₯ squared is less than one and diverges when the absolute value of nine π‘₯ squared is greater than one.

There’s a few different ways of finding the value of π‘₯ where the absolute value of nine π‘₯ squared is less than one and those where it’s greater than one. We’re going to do this by sketching 𝑦 is equal to the absolute value of nine π‘₯ squared and 𝑦 is equal to one. We know π‘₯ squared is never negative. Multiplying it by nine is a vertical stretch by nine, so it’s still never negative. So the absolute value of nine π‘₯ squared doesn’t change anything because nine π‘₯ squared is always positive. This means we can find our intersects by just solving the equation nine π‘₯ squared equals one. We divide both sides of the equation by nine to get π‘₯ squared is equal to one-ninth. Then we can take the square root to see that π‘₯ is equal to positive or negative one-third.

Now, by adding the intersects into our sketch, we can see the curve 𝑦 is equal to the absolute value of nine π‘₯ squared is below the line 𝑦 is equal to one when π‘₯ is between negative one-third and one-third. And we can see that the curve is above the line when π‘₯ is bigger than one-third or when π‘₯ is less than negative one-third. So because our power series converges when the absolute value of nine π‘₯ squared is less than one, it converges when π‘₯ is bigger than negative one-third and π‘₯ is less than one-third. And we know that our series diverges when the absolute value of nine π‘₯ squared is greater than one. This is when π‘₯ is less than negative one-third or when π‘₯ is greater than one-third.

Remember, we’re trying to find the interval of convergence, which is all of the values of π‘₯ where our power series converges. We’ve checked every single value of π‘₯ except when π‘₯ is equal to negative one-third and π‘₯ is equal to one-third. To see what happens to our power series at these two values of π‘₯, we’re going to substitute them into our power series. Let’s start with π‘₯ is equal to negative one-third.

Substituting π‘₯ is equal to negative one-third into our power series gives us the sum from 𝑛 equals zero to ∞ of three to the power of two 𝑛 multiplied by negative one-third to the power of two 𝑛 plus one. We split negative one-third into negative one multiplied by one-third. Then, by using our laws of exponents, we have that our power series is equal to the sum from 𝑛 equals zero to ∞ of three to the power of two 𝑛 multiplied by negative one to the power of two 𝑛 plus one multiplied by one divided by three to the power of two 𝑛 plus one.

We see that three to the power of two 𝑛 divided by three to the power of two 𝑛 plus one is just one-third. Giving us that our power series is equal to the sum from 𝑛 equals zero to ∞ of negative one to the power of two 𝑛 plus one multiplied by one-third. And this is divergent. There’s a few different ways of seeing this. For example, if we write our series out term by term, we see that every single term in our series is negative one-third. And so the sequence of partial sums is approaching negative ∞. So our power series is not convergent when π‘₯ is equal to negative one-third.

We can do the same for π‘₯ is equal to one-third. We get the sum from 𝑛 equals zero to ∞ of three to the power of two 𝑛 times one over three to the power of two 𝑛 plus one, which simplifies to the sum from 𝑛 equals zero to ∞ of one-third. And writing this out term by term, we just see it’s one-third plus one-third plus one-third to the sequence of partial sums approaches ∞. And our power series is divergent when π‘₯ is equal to one-third. So we’ve shown the only values of π‘₯ where our power series is convergent is when π‘₯ is greater than negative one-third and when π‘₯ is less than one-third. And this is the same as saying the only values of π‘₯ where our power series is convergent is when π‘₯ is in the open interval from negative one-third to one-third.

Therefore, we’ve shown the function 𝑓 of π‘₯ is equal to π‘₯ divided by one minus nine π‘₯ squared has the power series representation the sum from 𝑛 equals zero to ∞ of three to the power of two 𝑛 times π‘₯ to the power of two 𝑛 plus one. With the interval of convergence the open interval from negative one-third to one-third.

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