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Consider the function 𝑓(π‘₯) = 𝑒^(2π‘₯). Find the Taylor series representation of 𝑓 about π‘₯ = 3.

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Video Transcript

Consider the function 𝑓 of π‘₯ equals 𝑒 to the power of two π‘₯. Find the Taylor series representation of 𝑓 about π‘₯ equals three. And in fact, there are two further parts of this question that we’ll consider in a moment.

We begin by recalling that the Taylor series of a function 𝑓 about π‘Ž is given by the sum of the 𝑛th derivative of 𝑓 evaluated at π‘Ž over π‘š factorial times π‘₯ minus π‘Ž to 𝑛th power values of π‘š between zero and infinity. And so what do we know about our function? Well, it’s given by 𝑓 of π‘₯ equals 𝑒 to the power of two π‘₯. We want to find the Taylor series about π‘₯ equals three. So we let π‘Ž be equal to three. Then we see that 𝑓 of π‘₯ is equal to the sum of the 𝑛th derivative of 𝑓 evaluated at three of 𝑛 factorial times π‘₯ minus three to the 𝑛th power for values of 𝑛 between zero and infinity. This is, of course, equal to 𝑓 of three plus 𝑓 prime of three over one factorial times π‘₯ minus three plus 𝑓 double prime of three over two factorial times π‘₯ minus three all squared and so on.

It’s quite clear that we’re going to need to work out 𝑓 of three, 𝑓 prime of three, 𝑓 double prime of three, and possibly look for a pattern. 𝑓 of three is quite straightforward. It’s 𝑒 to the power of two times three which is, of course, 𝑒 to the power of six. So let’s next look for the derivative of 𝑓 of π‘₯. The derivative of 𝑒 to the power of two π‘₯ is two 𝑒 to the power of two π‘₯. So 𝑓 prime of three must be two 𝑒 to the power of two times three, which is two 𝑒 to the power of six. We’ll differentiate once again. This time we get four 𝑒 to the power of two π‘₯. And 𝑓 double prime of three is for 𝑒 to the power of two times three, which is four 𝑒 to the sixth power.

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

The second part of this question asks us to find the interval of convergence of the Taylor series representation of 𝑓 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‘Ž. So here π‘₯ equals three. But that’s all we know right now. We’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‘Žπ‘›, and we defined 𝑙 to be equal to the limits as 𝑛 approaches infinity of π‘Žπ‘› plus one over π‘Žπ‘›. If 𝑙 is less than one, the series is absolutely convergent and hence convergent. This is the scenario we’re interested in. So our 𝑙 must be the limit as 𝑛 approaches infinity of the absolute value of two to the power of 𝑛 plus one times 𝑒 to the sixth power times π‘₯ minus three to the power of 𝑛 plus one over 𝑛 plus one factorial divided by two to the 𝑛th power times 𝑒 to the sixth power times π‘₯ minus three to the 𝑛th power over 𝑛 factorial.

Now of course, when we divide by a fraction, it’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 𝑒 to the power of six and two to the power of 𝑛. Similarly, we can cancel π‘₯ minus three to the power of 𝑛. And if we, finally, recall that 𝑛 plus one factorial is the same as 𝑛 plus one times 𝑛 factorial, we see that we can cancel by dividing through by 𝑛 factorial. Well, this all simplifies really nicely. We see that 𝑙 is equal to the limit as 𝑛 approaches infinity of the absolute value of two times π‘₯ minus three over 𝑛 plus one.

Now in fact, the expression two times π‘₯ minus three is independent of 𝑛. So we can factor this, remembering that we must write the absolute value of two times π‘₯ minus three. And so 𝑙 is equal to the absolute value of two times π‘₯ minus three times the limit as 𝑛 approaches infinity of the absolute value of one over 𝑛 plus one. Now, as 𝑛 gets larger, one over 𝑛 plus one gets smaller. So the limit approaches zero. This means 𝑙 is equal to zero. This is of course less than one. And it doesn’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 𝑓 about π‘₯ equals three?

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

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