 Question Video: Identifying Today, Tomorrow, and Yesterday | Nagwa Question Video: Identifying Today, Tomorrow, and Yesterday | Nagwa

# Question Video: Identifying Today, Tomorrow, and Yesterday Mathematics

Consider the function 𝑓(𝑥) = 𝑒^(2𝑥). Find the Taylor series representation of 𝑓 about 𝑥 = 3.

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### Video Transcript

Consider the function 𝑓 of 𝑥 equals 𝑒 to the power of two 𝑥. Find the Taylor series representation of 𝑓 about 𝑥 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 𝑎 is given by the sum of the 𝑛th derivative of 𝑓 evaluated at 𝑎 over 𝑚 factorial times 𝑥 minus 𝑎 to 𝑛th power values of 𝑚 between zero and infinity. And so what do we know about our function? Well, it’s given by 𝑓 of 𝑥 equals 𝑒 to the power of two 𝑥. We want to find the Taylor series about 𝑥 equals three. So we let 𝑎 be equal to three. Then we see that 𝑓 of 𝑥 is equal to the sum of the 𝑛th derivative of 𝑓 evaluated at three of 𝑛 factorial times 𝑥 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 𝑥 minus three plus 𝑓 double prime of three over two factorial times 𝑥 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 𝑥. The derivative of 𝑒 to the power of two 𝑥 is two 𝑒 to the power of two 𝑥. 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 𝑥. 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 𝑥 is equal to two to the 𝑛th power times 𝑒 to the power of two 𝑥. 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 𝑥 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 𝑥 equals three.

We say that the intervals of all 𝑥s 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 𝑥 equals 𝑎. So here 𝑥 equals three. But that’s all we know right now. We’ll use the ratio test to determine the rest of the 𝑥s for which the power series will converge. This says suppose that we have some series, the sum of 𝑎𝑛, and we defined 𝑙 to be equal to the limits as 𝑛 approaches infinity of 𝑎𝑛 plus one over 𝑎𝑛. 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 𝑥 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 𝑥 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 𝑥 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 𝑥 minus three over 𝑛 plus one.

Now in fact, the expression two times 𝑥 minus three is independent of 𝑛. So we can factor this, remembering that we must write the absolute value of two times 𝑥 minus three. And so 𝑙 is equal to the absolute value of two times 𝑥 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 𝑥. And so we say our series is absolutely convergent for all values of 𝑥. 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 𝑥 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 𝑥 minus 𝑎 less than 𝑅 and will diverge for absolute values of 𝑥 minus 𝑎 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.