# Question Video: Identifying Today, Tomorrow, and Yesterday Mathematics • Higher Education

Consider the function π(π₯) = π^(2π₯). Find the Taylor series representation of π about π₯ = 3.

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### Video Transcript

Consider the function π of π₯ equals π to the power of two π₯. Find the Taylor series representation of π about π₯ equals three. And in fact, there are two further parts of this question that weβll consider in a moment.

We begin by recalling that the Taylor series of a function π about π is given by the sum of the πth derivative of π evaluated at π over π factorial times π₯ minus π to πth power values of π between zero and infinity. And so what do we know about our function? Well, itβs given by π of π₯ equals π to the power of two π₯. We want to find the Taylor series about π₯ equals three. So we let π be equal to three. Then we see that π of π₯ is equal to the sum of the πth derivative of π evaluated at three of π factorial times π₯ minus three to the πth power for values of π between zero and infinity. This is, of course, equal to π of three plus π prime of three over one factorial times π₯ minus three plus π double prime of three over two factorial times π₯ minus three all squared and so on.

Itβs quite clear that weβre going to need to work out π of three, π prime of three, π double prime of three, and possibly look for a pattern. π of three is quite straightforward. Itβs π to the power of two times three which is, of course, π to the power of six. So letβs next look for the derivative of π of π₯. The derivative of π to the power of two π₯ is two π to the power of two π₯. So π prime of three must be two π to the power of two times three, which is two π to the power of six. Weβll differentiate once again. This time we get four π to the power of two π₯. And π double prime of three is for π to the power of two times three, which is four π to the sixth power.

Now actually, we can see that the πth derivative of π of π₯ is equal to two to the πth power times π to the power of two π₯. And this means the πth derivative of π evaluated at three is two to the πth power times π to the sixth power. Our Taylor series expansion using sigma notation is, therefore, the sum of two to the πth power times π to the sixth power times π₯ minus three to the πth power all over π factorial for values of π between zero and infinity.

The second part of this question asks us to find the interval of convergence of the Taylor series representation of π about π₯ equals three.

We say that the intervals of all π₯s for which the power series converges is called the interval of convergence of the series. Now actually, we know that the power series will converge for π₯ equals π. So here π₯ equals three. But thatβs all we know right now. Weβll use the ratio test to determine the rest of the π₯s for which the power series will converge. This says suppose that we have some series, the sum of ππ, and we defined π to be equal to the limits as π approaches infinity of ππ plus one over ππ. If π is less than one, the series is absolutely convergent and hence convergent. This is the scenario weβre interested in. So our π must be the limit as π approaches infinity of the absolute value of two to the power of π plus one times π to the sixth power times π₯ minus three to the power of π plus one over π plus one factorial divided by two to the πth power times π to the sixth power times π₯ minus three to the πth power over π factorial.

Now of course, when we divide by a fraction, itβs the same as multiplying by the reciprocal of that fraction. So we can rewrite our limit as shown. Then we notice we can perform some simplification. We can cancel out π to the power of six and two to the power of π. Similarly, we can cancel π₯ minus three to the power of π. And if we, finally, recall that π plus one factorial is the same as π plus one times π factorial, we see that we can cancel by dividing through by π factorial. Well, this all simplifies really nicely. We see that π is equal to the limit as π approaches infinity of the absolute value of two times π₯ minus three over π plus one.

Now in fact, the expression two times π₯ minus three is independent of π. So we can factor this, remembering that we must write the absolute value of two times π₯ minus three. And so π is equal to the absolute value of two times π₯ minus three times the limit as π approaches infinity of the absolute value of one over π plus one. Now, as π gets larger, one over π plus one gets smaller. So the limit approaches zero. This means π is equal to zero. This is of course less than one. And it doesnβt change no matter the value of π₯. And so we say our series is absolutely convergent for all values of π₯. Its interval of convergence is negative infinity to infinity.

The third part of this question asks, what is the radius of convergence of the Taylor series representation of π about π₯ equals three?

Recall that there exist a number capital π, which we call the radius of convergence, so that the power series will converge for absolute values of π₯ minus π less than π and will diverge for absolute values of π₯ minus π greater than π. Well here, weβve seen that our power series is absolutely convergent. And thereβs a special case for the radius of convergence. In cases such as these, we say that the radius of convergence is infinity. Capital π is, therefore, equal to infinity.