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Question Video: Determining the Interval of Convergence for a Rational Functions’ Power Series Mathematics • 12th Grade

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

06:14

Video Transcript

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

We’re given a rational function 𝑓 of π‘₯, and we’re asked two things. First, we need to find the power series representation of our rational function 𝑓 of π‘₯. We also need to identify its interval of convergence. And there are several different ways we could approach this. However, if we see our function 𝑓 of π‘₯, it’s given in a special form. Our function 𝑓 of π‘₯ is very similar to the following form: π‘Ž divided by one minus π‘Ÿ. And we know something very interesting about this involving infinite geometric series.

We know for a geometric sequence of initial value π‘Ž and ratio of successive terms π‘Ÿ, if we were to sum an infinite number of values of this sequence β€” the sum from 𝑛 equals zero to ∞ of π‘Ž times π‘Ÿ to the 𝑛th power β€” then this will converge when the absolute value of π‘Ÿ is less than one. And it will converge to be equal to π‘Ž divided by one minus π‘Ÿ. And we also know this must be divergent when the absolute value of π‘Ÿ is greater than one. And it’s also worth pointing out this does technically assume the value of π‘Ž is not equal to zero. And this is, in fact, useful in two ways. We can use this to find a power series representation of our function 𝑓 of π‘₯. However, we can also use this to find information about the convergence of this series.

So let’s start by using this to find the power series representation for 𝑓 of π‘₯. In our numerator, we have one, so we can just set the value of π‘Ž equal to one. And in our denominator, we have one minus nine π‘₯ squared, so we need to set the value of π‘Ÿ equal to nine π‘₯ squared. So if we set our value of π‘Ž equal to one and π‘Ÿ equal to nine π‘₯ squared, we have one over one minus nine π‘₯ squared is equal to the sum from 𝑛 equals zero to ∞ of one times nine π‘₯ squared all raised to the 𝑛th power, provided the absolute value of our ratio nine π‘₯ squared is less than one. And we know this will be divergent if the absolute value of nine π‘₯ squared is greater than one.

And of course, all of this is just equal to our function 𝑓 of π‘₯, provided that our series converges. And we can simplify this. First, multiplying by one inside of our summand doesn’t change anything. Next, we can distribute the exponent 𝑛 over our parentheses. This gives us the sum from 𝑛 equals zero to ∞ of nine to the 𝑛th power multiplied by π‘₯ squared all raised to the 𝑛th power. And we can simplify this even further. First, by using our laws of exponents, π‘₯ squared all raised to the 𝑛th power is just equal to π‘₯ to the power of two 𝑛. Next, we want to simplify nine to the 𝑛th power. Well, we know that nine is equal to three squared. And using the exact same reasoning we did before, three squared all raised to the 𝑛th power is equal to three to the power of two 𝑛.

And this gives us our power series representation for 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, provided this power series converges. But we’re not yet done. Remember, we also need to find the interval of convergence for this power series. We can do this by using what we found about infinite geometric series. First, we know our power series must be convergent when the absolute value of nine π‘₯ squared is less than one. This is because we wrote this as an infinite geometric series. And we know that all geometric series will be convergent when the absolute value of their ratio is less than one.

So let’s try and write this in interval notation. We’ll start by using what we know about absolute values. First, the absolute value of nine π‘₯ squared can be written as the absolute value of nine multiplied by the absolute value of π‘₯ squared. But actually, this can be simplified. First, nine is a constant. And in fact, it’s a positive constant. So the absolute value of nine is just equal to nine. And in fact, we can say something very similar about π‘₯ squared. π‘₯ squared is never negative. So its absolute value will just be equal to itself, π‘₯ squared.

So our power series must be convergent whenever nine π‘₯ squared is less than one. And we can just solve this inequality. First, we’ll divide through by nine. This gives us π‘₯ squared is less than one-ninth. So we need to find the values of π‘₯ where π‘₯ squared is less than one-ninth. There’s several different ways of doing this. For example, we could do this graphically or by using what we know about square roots. Using either method, since the square root of one-ninth is one-third, we get the values of π‘₯ which solve this inequality are when π‘₯ is greater than negative one-third and less than one-third. And we can write this in interval notation. It’s the open interval from negative one-third to one-third.

However, we’re not yet done. Remember, the interval of convergence must contain all values of π‘₯ where our power series is convergent. So we need to check all of the rest of the values of π‘₯. Well, first, we already know that it must be divergent when the absolute value of nine π‘₯ squared is greater than one. And we can solve this inequality just as we did before. First, the absolute value of nine π‘₯ squared is just equal to nine π‘₯ squared. Then, by dividing through by nine and either by using what we know about square roots or by using a graphical approach, we can show that this is true when π‘₯ is greater than one-third or when π‘₯ is less than negative one-third.

So now we’ve checked every single value of π‘₯ except when π‘₯ is equal to one-third and when π‘₯ is equal to negative one-third. So we’ll substitute both of these values directly into our power series. Let’s start with 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-third all raised to the power of two 𝑛. And since both of these are raised to the same exponent of two 𝑛, we can instead multiply three by one-third and then raise this to the power of two 𝑛. So this gives us the sum from 𝑛 equals zero to ∞ of three times one-third all raised to the power of two 𝑛. And of course, we can simplify three times one-third is just equal to one.

So this simplified to give us the sum from 𝑛 equals zero to ∞ of one to the power of two 𝑛. Of course, one to the power of two 𝑛 is just equal to one. And we can see that this sum is divergent. For example, its partial sum is approaching ∞. And we can do something very similar when π‘₯ is equal to negative one-third. We’ll do exactly the same thing we did with our exponents. This time we get the sum from 𝑛 equals zero to ∞ of three times negative one-third all raised to the power of two 𝑛. And again, we can simplify this. Three times negative one-third is equal to negative one. So this gives us the sum from 𝑛 equals zero to ∞ of negative one all raised to the power of two 𝑛. However, two 𝑛 is even for all values of 𝑛. So this is just equal to one. And once again, we can see that this series is divergent.

Therefore, we’ve shown the only values of π‘₯ where our series is convergent is when π‘₯ is in the open interval from negative one-third to one-third. So therefore, this must be the interval of convergence for our power series representation of 𝑓 of π‘₯. Therefore, we were able to show for 𝑓 of π‘₯ is equal to one divided by one minus nine π‘₯ squared, its power series representation will be the sum from 𝑛 equals zero to ∞ of three to the power of two 𝑛 times π‘₯ to the power of two 𝑛. And it will have an interval of convergence, which is the open interval from negative one-third to one-third.

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