Video: Finding the Radius of Convergence of the Maclaurin Series of a Rational Function

Find the radius of convergence for the Maclaurin series for 𝑓(π‘₯) = 1/(10π‘₯ + 1).

04:58

Video Transcript

Find the radius of convergence for the Maclaurin series for 𝑓 of π‘₯ is equal to one divided by 10π‘₯ plus one.

The question wants us to find the radius of convergence of the Maclaurin series for the function 𝑓 of π‘₯. To start, we need to recall what a radius of convergence is and what the Maclaurin series of a function is. Let’s start with the Maclaurin series.

The Maclaurin series of a function 𝑓 of π‘₯ is the sum from 𝑛 equals zero to ∞ of the 𝑛th derivative of 𝑓 of π‘₯ with respect to π‘₯ evaluated at zero divided by 𝑛 factorial times π‘₯ to 𝑛th power. And this, of course, will only be convergent for certain values of π‘₯. And in fact, we’ve shown certain conditions must be true for the values of π‘₯, which can converge for certain Maclaurin series. Either our Maclaurin series will converge for all real values of π‘₯, there will exist some value 𝑅 such that π‘₯ converges whenever π‘₯ is in the open interval from negative 𝑅 to 𝑅 or a Maclaurin series will only converge when π‘₯ is equal to zero.

And this is how we define our radius of convergence. In the second instance, we call 𝑅 the radius of convergence. If instead our Maclaurin series converges for all real values of π‘₯, then we call our radius of convergence ∞. And finally, if our Maclaurin series only converges for π‘₯ is equal to zero, then we say our radius of convergence is equal to zero. Now, to find the radius of convergence of the Maclaurin series of the function given to us in the question, we need to find the Maclaurin series of this function.

We might be tempted to try and do this directly using our formula. We would then need to find the 𝑛th derivative of 𝑓 of π‘₯ with respect to π‘₯ evaluated at zero. And we could do this; we could use the quotient rule, the chain rule, or the general power rule. However, there’s a lot of steps to this, and it’s very easy to make a mistake. And even after all of this, we would then need to apply the ratio test to find our radius of convergence. Instead, let’s recall we can find a power series representation of functions in this form by using our rules for the infinite sum of geometric series.

We know for constants π‘Ž and π‘Ÿ, where π‘Ž is not equal to zero, 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. And we know this series will be convergent when the absolute value of π‘Ÿ is less than one and it will diverge otherwise. We want to represent our function 𝑓 of π‘₯ as the right-hand side of this expression. Then, we will get a power series representation for our function 𝑓 of π‘₯. The easiest way to do this is to see we already have our constant π‘Ž in our numerator; we just need to be subtracting π‘Ÿ.

And we can write our function 𝑓 of π‘₯ in this form by just writing it as one divided by one minus negative 10π‘₯. Now, we just set our value of π‘Ž equal to one and our value of π‘Ÿ equal to negative 10π‘₯. This then gives us that 𝑓 of π‘₯ is equal to one divided by one minus 10π‘₯ which is equal to the sum from 𝑛 equals zero to ∞ of one times negative 10π‘₯ raised to the 𝑛th power. And we also know this series will be convergent whenever the absolute value of negative 10π‘₯ is less than one and it will be divergent otherwise.

So, now, we have a power series representation of our function 𝑓 of π‘₯. In fact, this power series must be equal to the Maclaurin series of our function 𝑓 of π‘₯. This is because any power series of a function centered at a point must be equal to its Taylor series centered at this point. So because this power series is centered at zero, this power series for 𝑓 of π‘₯ must be equal to its Taylor series centered at zero, which we know is equivalent to its Maclaurin series. But remember, the question is not asking us to find the Maclaurin series of this function; it’s asking us to find the radius of convergence of this Maclaurin series. So, we need to work out which of our three categories we’re in.

In fact, we already know this from our infinite sum of a geometric series. We know the only values of π‘₯ this power series converge will be when the absolute value of negative 10π‘₯ is less than one. We know it must be divergent otherwise. So, to find our radius of convergence π‘Ÿ, we need to solve the inequality the absolute value of negative 10π‘₯ is less than one. The first thing we’ll do to solve this inequality is use our laws of absolute values. The absolute value of negative 10π‘₯ is the same as the absolute value of negative 10 multiplied by the absolute value of π‘₯. And we know the absolute value of negative 10 is just equal to 10. In fact, we can then simplify this further. We’ll divide both sides of our inequality through by 10.

So, we have the absolute value of π‘₯ must be less than one-tenth. Once again, by using our rules of absolute values, we know this is equivalent to saying that π‘₯ is greater than negative one-tenth and π‘₯ is less than one-tenth. In other words, π‘₯ is in the open interval from negative one-tenth to one-tenth. So, we’ve shown we’re in our second condition. And that means the radius of convergence of the Maclaurin series for 𝑓 of π‘₯ is equal to one-tenth.

Therefore, we were able to show the radius of convergence for the Maclaurin series of the function 𝑓 of π‘₯ is equal to one divided by 10π‘₯ plus one is equal to one-tenth.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.