Video: Finding the Power Series for a Function and Identifying Its Interval of Convergence

Let us consider 𝑓(π‘₯) = 1/(1 + π‘₯Β²). Find a power series representation for 𝑓(π‘₯). Find its interval of convergence.

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Video Transcript

Let us consider the function 𝑓 of π‘₯ is equal to one divided by one plus π‘₯ squared. Find a power series representation for 𝑓 of π‘₯. Find its interval of convergence.

The question wants us to find a power series representation for a rational function, one divided by one plus π‘₯ squared, and then wants us to find the interval of convergence for this power series. To help us find a power series of a rational function, we recall the following facts about geometric series. If the absolute value of the ratio π‘Ÿ is less than one, then the infinite sum of the geometric series the sum from 𝑛 equals zero to ∞ of π‘Žπ‘Ÿ to the 𝑛th power is just equal to π‘Ž divided by one minus π‘Ÿ. And we know that this sum will diverge if the absolute value of π‘Ÿ is greater than one. This is a useful result because it gives us a method of turning a quotient into a series.

So, if we can rewrite our function one divided by one plus π‘₯ squared in the form π‘Ž divided by one minus π‘Ÿ, then we can rewrite it as a power series on the condition the absolute value of the ratio of successive terms, π‘Ÿ, is less than one. In fact, one divided by one plus π‘₯ squared is almost already in this form. We just set π‘Ž equal to one, and then we see that negative π‘Ÿ should be equal to π‘₯ squared. And if negative π‘Ÿ is equal to π‘₯ squared, this is the same as saying that π‘Ÿ is equal to negative π‘₯ squared. So, what we’ve shown is we can rewrite our function 𝑓 of π‘₯ in the form π‘Ž divided by one minus π‘Ÿ, where π‘Ž is equal to one and π‘Ÿ is equal to negative π‘₯ squared.

We can now apply our formula for the sum of an infinite geometric series. We set our initial value π‘Ž equal to one and our ratio of successive terms π‘Ÿ equal to negative π‘₯ squared. This gives us if the absolute value of negative π‘₯ squared is less than one, then the sum from 𝑛 equals zero to ∞ of one times negative π‘₯ squared raised to the 𝑛th power is equal to one divided by one minus negative π‘₯ squared. And, of course, one divided by one minus negative π‘₯ squared is one divided by one plus π‘₯ squared, which is our function 𝑓 of π‘₯. So, we now have a power series representation of our function 𝑓 of π‘₯.

We can actually simplify this power series. We know that negative π‘₯ squared is equal to negative one multiplied by π‘₯ squared. Since we’re raising this to the 𝑛th power inside of our sum, we can distribute this over our parentheses, giving us negative one to the 𝑛th power multiplied by π‘₯ to the power of two 𝑛. This means we can rewrite our sum as the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power times π‘₯ to the power of two 𝑛. And this power series is equal to 𝑓 of π‘₯ if the absolute value of negative π‘₯ squared is less than one. So, we found our power series representation for our function 𝑓 of π‘₯.

The question then wants us to find the interval of convergence for this power series. We recall that an interval of convergence for a power series is an interval which contains all of the values of π‘₯ such that the power series is convergent. So, we need to find the values of π‘₯ which make our power series convergent and those which make it divergent. When we used our fact about geometric series, we knew that our series would be convergent if the absolute value of π‘Ÿ was less than one, and divergent if the absolute value of π‘Ÿ was greater than one. In our power series, we used π‘Ÿ is equal to negative π‘₯ squared. So, our power series must be convergent if the absolute value of negative π‘₯ squared is less than one. And it must be divergent if the absolute value of negative π‘₯ squared is greater than one. So, our power series is convergent if the absolute value of negative π‘₯ squared is less than one, and divergent if the absolute value of negative π‘₯ squared is greater than one.

Let’s start by simplifying the case for the convergent values of π‘₯. We know that π‘₯ squared is never negative. So, if we multiply this by negative one, we’ll get a value which is never positive. So, taking the absolute value of this would just leave us with π‘₯ squared. To find the values of π‘₯ where π‘₯ squared is less than one, we’ll start by sketching a graph of 𝑦 is equal to π‘₯ squared and a graph of 𝑦 is equal to one. The values of the intersects of these two curves will be when π‘₯ squared is equal to one. And if π‘₯ squared is equal to one, we can just take the square root to see that π‘₯ is equal to positive or negative one. From the sketch, we can see that the curve 𝑦 is equal to π‘₯ squared is below the line 𝑦 is equal to one when π‘₯ is between negative one and one. So, the values of π‘₯ where our power series is convergent is when π‘₯ is greater than negative one and π‘₯ is less than one.

We can use a very similar argument for the values of π‘₯ where our power series will be divergent. The absolute value of negative π‘₯ squared is still just π‘₯ squared, so we want the values of π‘₯ where π‘₯ squared is greater than one. And from our sketch, we can see that the curve π‘₯ squared is above the line 𝑦 is equal to one when π‘₯ is greater than one or when π‘₯ is less than negative one. So, we know that our power series must be divergent if π‘₯ is less than negative one or if π‘₯ is greater than one. However, remember the interval of convergence contains all of the values of π‘₯ such that the power series is convergent. We’ve checked all of the values of π‘₯ between negative one and one. We’ve checked all of the values of π‘₯ less than negative one and all of the values greater than one. But we’ve not checked π‘₯ is equal to negative one or π‘₯ is equal to one.

So, to check whether π‘₯ is equal to one or π‘₯ is equal to negative one make our power series convergent, we’ll substitute these into our power series. We start with π‘₯ is equal to one. This gives us the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power multiplied by one to the power of two 𝑛. We know that one to the power of two 𝑛 is always equal to one. So, we have the question, is the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power convergent? There’s a few different ways of seeing that this is a divergent sum. For example, we could use the 𝑛th-term divergence test. We could also write it out term by term. We get one minus one plus one minus one, and this goes on to ∞.

Let’s then calculate the sequence of partial sums. Our first partial sum is one. We then add negative one to find our second partial sum, and we get zero. To get our third partial sum, we add one and we get one. And we see that our sequence of partial sums is just alternating between one and zero. And the sequence is not getting closer to any particular value. So, our series is divergent. We get a very similar story if we set π‘₯ equal to negative one. Our series becomes the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power multiplied by negative one to the power of two 𝑛. And if we work this out term by term, we see we get the exact same sum. So, we can conclude this series is also divergent. Therefore, the only values of π‘₯ where our power series is convergent is when π‘₯ is greater than negative one and when π‘₯ is less than one. And this is the same as saying the absolute value of π‘₯ is less than one.

Therefore, we’ve shown the function 𝑓 of π‘₯ is equal to one divided by one plus π‘₯ squared has the power series representation the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power times π‘₯ to the power of two 𝑛, which has an interval of convergence the absolute value of π‘₯ is less than one.

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