# Lesson Video: Taylor Series Mathematics

In this video, we will learn how to find Taylor series of a function and find the radius of convergence of the series.

16:41

### Video Transcript

In this video, we’ll learn how to find a Taylor representation of an arbitrary function by using the Taylor series expansion. We’ll consider how we can find the radius of convergence of these series as well as intervals of convergence, primarily using the ratio test for convergence. We begin by supposing that 𝑓 is a function that can be represented by a power series 𝑓 of 𝑥 equals 𝑐 zero plus 𝑐 one 𝑥 minus 𝑎 plus 𝑐 two 𝑥 minus 𝑎 all squared and so on, where the absolute value of 𝑥 minus 𝑎 is less than some value capital 𝑅. We can begin to work out what 𝑐𝑛, the coefficients, must be in terms of 𝑓. For example, if we let 𝑥 be equal to 𝑎, we see that 𝑥 minus 𝑎 becomes 𝑎 minus 𝑎, which is zero. And then we find that 𝑓 of 𝑎 equals 𝑐 zero.

Okay, well, that’s great. But there’s nothing else we can do with this function as it stands to help us represent the other coefficients in terms of 𝑓. So instead, we differentiate our function 𝑓 with respect to 𝑥, remembering that we can do so term by term. And 𝑐 zero is a constant so its derivative is zero. Then the derivative of 𝑐 one times 𝑥 minus 𝑎 is just 𝑐 one. And we could achieve this result by using the product rule or distributing our parentheses and differentiating as normal. Similarly, we find the derivative of 𝑐 two times 𝑥 minus 𝑎 all squared to be two times 𝑐 two times 𝑥 minus 𝑎. The derivative of 𝑐 three times 𝑥 minus 𝑎 all cubed is three 𝑐 three times 𝑥 minus 𝑎 all squared and so on.

Once again, we let 𝑥 be equal to 𝑎. And this time we obtain 𝑓 prime of 𝑎 to be equal to 𝑐 one. Now, it follows that we can repeat this process. We differentiate our function once again with respect to 𝑥. 𝑓 double prime of 𝑥 is two 𝑐 two plus two times 𝑐 three times 𝑥 minus 𝑎 plus another term, three times four 𝑐 four times 𝑥 minus 𝑎 all squared and so on. And this time when we let 𝑥 be equal to 𝑎 we find that 𝑓 double prime of 𝑎 is equal to two 𝑐 two. We’ll repeat this one more time to help us establish a pattern. The third derivative is two times three 𝑐 three plus two times three times four 𝑐 four times 𝑥 minus 𝑎 plus three times four times five 𝑐 five times 𝑥 minus 𝑎 all squared and so on. And letting 𝑥 be equal to 𝑎 now, we see that 𝑓 triple prime of 𝑎, the third derivative of our function with respect to 𝑥, evaluated at 𝑥 is equal to 𝑎 is two times three times 𝑐 three, which we could write alternatively as three factorial 𝑐 three.

Now, if we were to continue to different and let 𝑥 be equal to 𝑎, we would find that the 𝑛th derivative of our function evaluated at 𝑎 is two times three times four times five, all the way up to 𝑛 lots of 𝑐𝑛. In other words, it’s 𝑛 factorial times 𝑐𝑛. We’re going to solve the equation the 𝑛th derivative of 𝑓 evaluated at 𝑎 equals 𝑛 factorial 𝑐𝑛 for 𝑐𝑛. When we do, we find that 𝑐𝑛 is equal to the 𝑛th derivative of 𝑓 evaluated at 𝑎 over 𝑛 factorial. It’s useful to remember that this formula does remain valid even if 𝑛 is equal to zero, since zero factorial is just one and the zeroth derivative of 𝑓 is simply 𝑓. So we obtain our first theorem. If 𝑓 has a power series representation at 𝑎 ⁠— that is, if 𝑓 of 𝑥 is equal to the sum of 𝑐𝑛 times 𝑥 minus 𝑎 to the 𝑛th power for values of 𝑛 between zero and infinity ⁠— where the modules of 𝑥 minus 𝑎 is less than sum capital 𝑅. Then its coefficients are given by the formula 𝑐𝑛 equals the 𝑛th derivative of 𝑓 evaluated at 𝑎 over 𝑛 factorial.

We’re going to substitute this back into our original series. We find that if 𝑓 has a power series expansion at 𝑎, then the Taylor series of the function 𝑓 at 𝑎 or about 𝑎 is the sum of the 𝑛th derivative of 𝑓 evaluated at 𝑎 over 𝑛 factorial times 𝑥 minus 𝑎 to the 𝑛th power for values of 𝑛 between zero and infinity. And this can alternatively be written as 𝑓 of 𝑎 plus 𝑓 prime of 𝑎 over one factorial times 𝑥 minus 𝑎 plus 𝑓 double prime of 𝑎 over two factorial times 𝑥 minus 𝑎 all squared and so on. It’s useful to know that if 𝑎 is equal to zero, we do have a special case that has its own name. It’s called the Maclaurin series. And we won’t be investigating its properties in this video. Let’s now have a look at a couple of examples of how to apply the formula for the Taylor series.

What are the first four terms of the Taylor series of the function 𝑓 of 𝑥 equals the square root of 𝑥 about 𝑥 equals four?

We recall that the Taylor series of a function 𝑓 about 𝑎 is given by the sum of the 𝑛th derivative of 𝑓 evaluated at 𝑎 of 𝑛 factorial times 𝑥 minus 𝑎 to the 𝑛th power for values of 𝑛 between zero and infinity. So what do we know about our function? It’s given as 𝑓 of 𝑥 equals the square root of 𝑥, which we might alternatively say is 𝑥 to the power of one-half. We want to find the Taylor series about 𝑥 equals four, so we’re going to let 𝑎 be equal to four. Using the second form of the Taylor series, we find that 𝑓 of 𝑥 is equal to 𝑓 of four plus 𝑓 prime of four over one factorial times 𝑥 minus four plus 𝑓 double prime of four over two factorial times 𝑥 minus four squared plus 𝑓 triple prime of four over three factorial times 𝑥 minus four cubed.

We can work out 𝑓 of four fairly easily. We’ll substitute four into our original function. But to be able to work out 𝑓 prime of four, 𝑓 double prime of four, and 𝑓 triple prime of four, we’re going to need to differentiate our function 𝑓 of 𝑥 with respect to 𝑥 three times. We said that 𝑓 of 𝑥 was equal to 𝑥 to the power of one-half and we recall that to differentiate a polynomial term, we multiply the entire term by the exponent and then reduce that exponent by one. This means 𝑓 prime of 𝑥, the first derivative of 𝑓 is a half 𝑥 to the power of negative one-half. 𝑓 double prime of 𝑥 is negative a half times a half 𝑥 to the power of negative three over two, which simplifies to negative one-quarter 𝑥 to the power of negative three over two. Finally, 𝑓 triple prime of 𝑥 is negative three over two times negative a quarter 𝑥 to the power of negative five over two, which is three-eighths 𝑥 to the power of negative five over two.

We’re now going to substitute 𝑥 equals four into our functions for 𝑓 of 𝑥, 𝑓 prime of 𝑥, 𝑓 double prime of 𝑥, and 𝑓 triple prime of 𝑥. 𝑓 of four is the square root of four; it’s simply two. 𝑓 prime of four is a half times four to the power of negative one-half which is a quarter. 𝑓 double prime of four is negative one-quarter times four to the power of negative three over two, which is negative one over 32. And 𝑓 triple prime of four is three-eighths times four to the power of negative five over two, which is three over 256. Our final job is simply to substitute these into our expansion. When we do, we find the first four terms of the Taylor series of the function 𝑓 of 𝑥 equals the square root of 𝑥 about 𝑥 equals four to be two plus a quarter times 𝑥 minus four minus one over 64 times 𝑥 minus four squared plus one over 512 times 𝑥 minus four cubed.

We’ll have a look at another example of this type.

Consider the function 𝑓 of 𝑥 equals cos of 𝑥. Find the Taylor series expansion of 𝑓 of 𝑥 equals cos of 𝑥 at 𝑥 equals 𝜋 and write the Taylor series expansion of 𝑓 of 𝑥 in sigma notation.

And we recall that the Taylor series expansion of a function about 𝑥 equals 𝑎 is given by 𝑓 of 𝑥 equals 𝑓 of 𝑎 plus 𝑓 prime of 𝑎 over one factorial times 𝑥 minus 𝑎 plus 𝑓 double prime of 𝑎 over two factorial times 𝑥 minus 𝑎 all squared and so on. In this question, our function is cos of 𝑥. And we’re looking to find the Taylor series expansion at 𝑥 equals 𝜋. So we’re going to let 𝑎 be equal to 𝜋. And so we see that 𝑓 of 𝑥 here is equal to 𝑓 of 𝜋 plus 𝑓 prime of 𝜋 over one factorial times 𝑥 minus 𝜋 and so on. Now, we can quite easily evaluate 𝑓 of 𝜋. We would substitute 𝜋 into the function cos of 𝑥. But what about 𝑓 prime of 𝜋, 𝑓 double prime of 𝜋, and so on?

Well, we’re going to differentiate our function with respect to 𝑥. We might recall the first derivative of cos of 𝑥 to be negative sin of 𝑥. Then to get the second derivative, we differentiate negative sin 𝑥 with respect to 𝑥 and we get negative cos 𝑥. Differentiating once more, we find that 𝑓 triple prime of 𝑥 is sin 𝑥. And in fact, we’ll repeat this process one more time because when we differentiate sin 𝑥, we get cos 𝑥 again. And you might notice, we have a cycle. The fifth derivative of 𝑓 will be negative sin 𝑥, the sixth derivative will be negative cos 𝑥, and so on. And let’s use all of these to evaluate 𝑓 of 𝜋, 𝑓 prime of 𝜋, 𝑓 double prime of 𝜋, and so on and look for a pattern. 𝑓 of 𝜋 is cos 𝜋, which is negative one. 𝑓 prime of 𝜋 is negative sin 𝜋, which is zero. 𝑓 double prime of 𝜋 is negative cos 𝜋, which is one. And 𝑓 triple prime of 𝜋 is sin 𝜋, which is once again zero.

It, of course, follows that the fourth derivative of 𝜋 will be cos of 𝜋 again, which is negative one. Let’s substitute all of these back into our Taylor series expansion. Now of course, every other term is going to be equal to zero as we’ve seen. And so we find the Taylor series expansion to be negative one plus a half times 𝑥 minus 𝜋 all squared minus one over 24 times 𝑥 minus 𝜋 to the fourth power plus one over 720 times 𝑥 minus 𝜋 to the sixth power and so on. The second part of this question asks us to write the Taylor series expansion of 𝑓 of 𝑥 in sigma notation. And so let’s see if there’s a way we can find a pattern. We saw that 𝑓 prime of 𝜋 was equal to zero, while 𝑓 triple prime of 𝜋 is equal to zero. And extending the pattern, we would see that the fifth derivative, the seventh derivative, and so on would also be equal to zero. And so our derivative alternate between negative one, one, negative one, and so on.

Similarly, our denominators are factorials of ascending even numbers each time this even number is the same as the exponent. So let’s define that even number as two 𝑚. Then the denominator is two 𝑚 factorial and the exponent of 𝑥 minus 𝜋 is two 𝑚. To achieve alternating powers of negative one, we write negative one times 𝑚 plus one. That means when 𝑚 is zero, negative one to the power of one is negative one, when 𝑚 is one, negative one to the power of two gives us one, and so on. And so we find that 𝑓 of 𝑥 equals the sum of negative one to the power of 𝑚 plus one times 𝑥 minus 𝜋 to the power of two 𝑚 over two 𝑚 factorial for values of 𝑚 between zero and infinity.

We’re also able to find the radius of convergence of a Taylor series. We say that there exists the number capital 𝑅 such 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. And this number capital 𝑅 is called the radius of convergence of the series. And note that the series may or may not converge, if the absolute value of 𝑥 minus 𝑎 is equal to 𝑅. We then say that the interval of convergence is an interval containing all the values of 𝑥, for which the series converges. Now, this may or may not include the end points of the interval and even be the whole set of real numbers. And we say that if we know the radius of convergence of a power series is 𝑅, then for values of 𝑥 greater than 𝑎 minus 𝑅 and less than 𝑎 plus 𝑅, the power series converges. For values of 𝑥 less than 𝑎 minus 𝑅 and greater than 𝑎 plus 𝑅, the power series diverges. It also follows that the power series will always converge for 𝑥 is equal to 𝑎. Let’s have a look at an application of this theory.

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.

In this video, we’ve seen that if 𝑓 has a power series expansion at 𝑎, then the Taylor series of the function 𝑓 at 𝑎 or about 𝑎 is the sum of the 𝑛th derivative of 𝑓 evaluated at 𝑎 over 𝑛 factorial times 𝑥 minus 𝑎 to the 𝑛th power for values of 𝑛 between zero and infinity. We saw that capital 𝑅 represents the radius of convergence such 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. And finally, we learned that we can use the ratio test or other suitable tests to find the intervals of convergence. The power series converges for values of 𝑥 greater than 𝑎 minus 𝑅 and less than 𝑎 plus 𝑅. And it follows that the power series will always converge when 𝑥 is equal to 𝑎.