Question Video: Finding the Radius of Convergence for the Taylor Series Expansion of a Rational Function | Nagwa Question Video: Finding the Radius of Convergence for the Taylor Series Expansion of a Rational Function | Nagwa

# Question Video: Finding the Radius of Convergence for the Taylor Series Expansion of a Rational Function Mathematics • Higher Education

Find the radius of convergence for the Taylor series of π(π₯) = 1/(2π₯ + 1) about π₯ = 1.

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### Video Transcript

Find the radius of convergence for the Taylor series of π of π₯ is equal to one divided by two π₯ plus one about π₯ is equal to one.

Weβre given a rational function π of π₯, and weβre asked to find the radius of convergence of the Taylor series for this rational function π of π₯ about π₯ is equal to one. We have several different methods we could use to answer this question. For example, we could directly find the Taylor series for π of π₯ about π₯ is equal to one by using either the chain rule or the general power rule. However, to use this method, we would need to find an expression for the πth derivative of π evaluated at one. And then we would need to use something like the ratio test to evaluate the radius of convergence.

So this method will require a lot of work. However, there is an easier method involving what we know about infinite geometric series. We need to recall the following facts about the sum of an infinite geometric series. If our initial value is π and ratio of successive terms is π, then the sum from π equals zero to β of π times π to the πth power will be equal to π divided by one minus π, provided the absolute value of π is less than one. And if the absolute value of π is greater than or equal to one, then this series will be divergent. And itβs also worth pointing out, technically, this assumes the value of π is nonzero.

We want to use this to find the Taylor series of our function π of π₯ centered around π₯ is equal to one. The first thing we need to remember is our Taylor series being centered around π₯ is equal to one means that our power series needs to be in terms of π₯ minus one. So what we want to do is rewrite our function π of π₯ in the form π divided by one minus π. However, π needs to include π₯ minus one because then our power series will have powers of π₯ minus one. So letβs discuss how weβre going to do this.

Well, we want to write π₯ minus one in our dominator. If we were to just write this directly, we would get two times π₯ minus one. Of course if we distribute this, we see we get two π₯ minus two. However, our denominator is supposed to be two π₯ plus one. So of course, we need to add an extra value of three. So we have one over two π₯ plus one is equal to one over two times π₯ minus one plus three.

Remember though, weβre trying to write this in the form π divided by one minus π. To make this easier to see, letβs start by rearranging the two terms in our denominator. This gives us the following expression. Now, we see in our denominator we need a constant one. So weβre going to need to take a factor of three outside of our denominator. Taking out a factor of three from our denominator, our new denominator will be three times one plus two-thirds multiplied by π₯ minus one.

And now weβre almost in the required form. However, instead of dividing by three in our denominator, weβll instead move one-third into our numerator. So by doing this, weβve rewritten π of π₯ to be equal to one-third divided by one plus two-thirds times π₯ minus one. And now this is exactly in the required form. Our value of π will be equal to one-third, and our value of π will be equal to two-thirds times π₯ minus one.

All we need to do then is substitute our value of π and π into our formula for the sum of an infinite geometric series. This gives us that π of π₯ is equal to the sum from π equals zero to β of one-third times two-thirds multiplied by π₯ minus one all raised to the πth power, which of course will be convergent provided the absolute value of our ratio two-thirds times π₯ minus one is less than one and will be divergent if the absolute value of two-thirds times π₯ minus one is greater than or equal to one.

So letβs discuss exactly what weβve just done. We found a power series representation for our function π of π₯. And we can see itβs centered at π₯ is equal to one. But we know all power series representations centered about a value for a function must be equal to their Taylor series. So this must be the Taylor series of our function π of π₯ centered at π₯ is equal to one. So we just need to find the radius of convergence of this series. And in fact, we already have information about the radius of convergence of this series.

We can see that our power series converges if the absolute value of two-thirds times π₯ minus one is less than one and will diverge if the absolute value of two-thirds times π₯ minus one is greater than or equal to one. So all we need to do is rearrange these inequalities to help us find the radius of convergence. Weβll start with our first inequality. We know two-thirds is a positive constant. So we can take it just outside of our absolute value sign. But two-thirds is just a positive constant. So we can just divide through by two over three or alternatively multiply through by three over two.

So we were able to rewrite this inequality as the absolute value of π₯ minus one must be less than three over two. And we can do exactly the same thing to find the values of π₯ where this is divergent. We see our power series is divergent when the absolute value of π₯ minus one is greater than or equal to three over two. And this is exactly what it means to say the radius of convergence of our power series is three over two.

Therefore, by using what we know about geometric series, we were able to show the Taylor series of π of π₯ is equal to one over two π₯ plus one about π₯ is equal to one has a radius of convergence three over two.

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