# Video: Finding the Maximum and Minimum Values of a Function Involving Trigonometric Functions

Consider the power series β_(π = 0)^(β) π (2π₯)^(π). Determine the radius of convergence of the power series. Determine the interval of convergence of the power series.

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### Video Transcript

Consider the power series the sum from π equals zero to β of π times two π₯ to the πth power. Determine the radius of convergence of the power series and determine the interval of convergence of the power series.

To test for convergence of this power series, we could use the ratio or the root test. Letβs look at the root test. The part of the test weβre interested in says suppose we have a series the sum of π π. If the limit as π approaches β of the πth root of the absolute value of π π, which can of course be alternatively written as the absolute value of π π to the power of one over π, is less than one. Then the series is absolutely convergent and hence convergent.

So we define π π for our series to be π times two π₯ to the πth power. Weβre looking to establish where the limit as π approaches β of the absolute value of π times two π₯ to the πth power to the power of one over π is less than one. We know that finding an πth root doesnβt change the sign. So we can rewrite this and say that the limit as π approaches β of the absolute value of π to the power of one over π times two π₯ to the πth power to the power of one over π must be less than one. Two π₯ to the πth power to the power of one over π is simply two π₯. And so our limit is as shown.

We know that two π₯ is independent of π. So we can take the absolute value of two π₯ outside of our limit. The problem we have now is that if we try to evaluate the limit as π approaches β of the absolute value of π to the power of one over π. We get the absolute value of β to the power of one over β, which is β to the power of zero. And we know thatβs indeterminate form. But we might recall that we can write π to the power of one over π as π to the power of the natural log of π over π. And we know that as π approaches β, the natural logarithm of π over π approaches zero. And so the limit as π approaches β of the absolute value of π to the power of one over π is π to the power of zero, which is simply one.

Of course we donβt need the absolute value signs for one. And so we see weβre interested in when the absolute value of two π₯ is less than one. Since two is a purely positive number, we can divide through. And we see that the absolute value of π₯ is less than one-half. These are the values of π₯ for which the series converges. And we can therefore say that the radius convergence of our power series is one-half.

Another way of representing the absolute value of π₯ being less than one-half is to say that π₯ must be greater than negative one-half and less than one-half. And so weβve determined an interval for convergence. But we do need to check if the power series converges or diverges at the end points of our interval. In other words, when π₯ is equal to negative one-half or π₯ is equal to one-half.

Weβre going to plug these into our original power series and see if those series converge or diverge using alternative tests. Letβs clear some space. Weβll begin by letting π₯ equal negative one-half. The series is the sum from π equals zero to β of π times two minus negative one-half to the πth power. And this becomes π times negative one to the power of π.

To establish whether this series converges or diverges, we could use the alternating series test with π π equals π. Since the limit as π approaches β of π π is not equal to zero, we can say that this series ultimately diverges. So letβs try when π₯ is equal to one-half. We have the sum from π equals zero to β of π times two times a half to the πth power. Thatβs π times one to the πth power. And of course one to the πth power will always be one. So we have the sum from π equals zero to β of π.

This time, we can actually apply the πth term divergence test. The limit as π approaches β of π is β. Thatβs not equal to zero. And so, once again, when π₯ is equal to a half, our series diverges. And so the interval of convergence of our power series is the open interval negative one-half to one-half.

In this video, we saw that a series of the form the sum of π π times π₯ minus π to the πth power for values of π between zero and β is called a power series about π. We also saw that we can use the ratio test and root test to find the radius of convergence and interval of convergence of our power series. But that itβs also important that we should test the end points of the interval for convergence or divergence.