# 6.8 Taylor and Maclaurin s Series

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1 6.8. TAYLOR AND MACLAURIN S SERIES Taylor and Maclaurin s Series Introduction The previous section showed us how to find the series representation of some functions by using the series representation of known functions. The methods we studied are limited since they require us to relate the function to which we want a series representation with one for which we already know a series representation. In this section, we develop a more direct approach. When dealing with functions and their power series representation, there are three fundamental questions one has to answer: 1. Does a given function have a power series representation? 2. If it does, how do we find it? 3. What is the domain in other words, for which values of x is the representation valid? Question 1 is more theoretical, we won t address it. We will concentrate on questions 2 and 3. Assuming f has a power series representation that is f (x = c 0 c 1 (x a c 2 (x a 2..., we want to find what the power series representation is, that is we need to find the coeffi cients c 0, c 1, c 2,... It turns out that it is not very diffi cult. The technique used is worth remembering. We first find c 0. Having found c 0 we next find c 1. Then, we find c 2 and so on. Finding c 0. Since f (x = c 0 c 1 (x a c 2 (x a 2 c 3 (x a 3 c 4 (x a 4... (6.6 it follows that f (a = c 0 c 1 (0 c 2 ( = c 0 Finding c 1. Differentiating equation 6.6 gives us Therefore, f (x = c 1 2c 2 (x a 3c 3 (x a 2 4c 4 (x a 3... f (a = c 1 Finding c 2. We proceed the same way. First, we compute f (x, then f (a. Therefore f (x = 2c 2 (2 (3 c 3 (x a (3 (4 c 4 (x a 2... f (a = 2c 2

2 358 CHAPTER 6. INFINITE SEQUENCES AND SERIES or c 2 = f (a 2 In general. Continuing this way, we can see that c n = f (n (a Definitions and Theorems Theorem If the function f has a power series representation, that is if f (x = c n (x a n = c 0 c 1 (x a c 2 (x a 2... for x a < R then its coeffi cients are given by: In other words c n = f (n (a f (x = f (a f (a (x a f (a (x a2 2 f (a (x a3... Remark From the formula in the theorem, we see that to find the Taylor series of a function f centered at a, we have to find allthe derivatives of f at a that is f (n (a for n = 1, 2, 3,... Definition The series in the previous theorem is called the Taylor series of the function f at a. ( f (n (a (x a n 2. The nth partial sum is called the nth order Taylor polynomial It is denoted T n. So, n f (i (a T n (x = (x a i i! i=0 3. In the special case a = 0, the series is called a Maclaurin s Series. f (n (0 So, a Maclaurin s series is of the form x n and the Maclaurin s series for a function f is given by f (x = f (n (0 x n = f (0 f (0 x f (0 x2 2 f (0 x3...

3 6.8. TAYLOR AND MACLAURIN S SERIES Examples We now look how to find the Taylor and Maclaurin s series of some functions. Example Find the Maclaurin s series for f (x = e x, find its domain. f (n (0 The series will be of the form x n, we simply need to find the coef- ficients f (n (0. This is easy. Since all the derivatives of e x are e x, it follows that f (n (x = e x, thus f (n (0 = e 0 = 1, hence e x = = f (n (0 x n To find where this series converges, we use the ration test and compute: x n1 lim a n1 n a n = lim (n 1! n x n x n1 = lim n (n 1! x n Thus the domain is all real numbers. x n x = lim n n 1 = 0 Example Find a Taylor series for f (x = e x centered at 2. f (n (2 The series will be of the form (x 2 n, we simply need to find the coeffi cients f (n (2. This is easy. Since all the derivatives of e x are e x, it follows that f (n (x = e x, thus f (n (2 = e 2, hence e x = = f (n (2 e 2 (x 2n (x 2 n Example Find the n th order Taylor polynomial for e x centered at 0 and centered at 2. Plot these polynomials for n = 1, 2, 3, 4, 5. What do you notice? We already computed the power series corresponding to these two situations. We found that e x = 1 x x2 x3 x4 4! x5 5! x6 6!...

4 360 CHAPTER 6. INFINITE SEQUENCES AND SERIES and e x = e 2 e 2 2 (x 22 2 (x 23 (x 2 e e... ( = e 2 1 (x 2 (x 22 (x 23 (x 24 4! (x 25 5!... Let T n be the n th order Taylor polynomial for e x centered at 0 and Q n the n th order Taylor polynomial for e x centered at 2. The polynomials and their graph are given below. The function e x is plotted in black and the polynomials with the indicated color. (blue T 1 (x = 1 x (red T 2 (x = 1 x x2 (green T 3 (x = 1 x x2 x3 (purple T 4 (x = 1 x x2 x3 x4 4! (yellow T 5 (x = 1 x x2 x3 x4 4! x5 5!

5 6.8. TAYLOR AND MACLAURIN S SERIES 361 and (blue Q 1 (x = e 2 (1 (x 2 (red Q 2 (x = e 2 ( 1 (x 2 (green Q 3 (x = e 2 ( (purple Q 4 (x = e 2 ( (yellow Q 5 (x = e 2 ( 1 (x 2 1 (x 2 1 (x 2 (x 22 (x 22 (x 22 (x 22 (x 23 (x 23 (x 23 (x 24 4! (x 24 4! (x 25 5! We can see the following: The Taylor polynomial approximates the functions well near the point at which the series is centered. As we move away from this point, the approximation deteriorates very quickly. The approximation is better the higher the degree of the Taylor polynomial. More specifically, the Taylor polynomial stay closer to the function over a larger interval, the higher its degree.

6 362 CHAPTER 6. INFINITE SEQUENCES AND SERIES Example Find a Maclaurin s series for f (x = sin x and find its domain f (n (0 The series will be of the form x n, we simply need to find the coeffi - cients f (n (0. That is we need to find all the derivatives of sin x and evaluate them at x = 0. We summarize our findings in the table below: n f (n (x f (n (0 0 sin x 0 1 cos x 1 2 sin x 0 3 cos x 1 4 sin x 0 So, we see that we will have coeffi cients only for odd values of n. In addition, the sign of the coeffi cients will alternate. Thus, since we have f (x = f (0 f (0 x f (0 x2 2 f (0 x3 f (4 (0 x4 4!... sin x = 0 x 0 x3 0 x5 5!... = x x3 x5 5! x7 7!... = ( 1 n x 2n1 (2n 1! To find where this series converges, we use the ration test and compute lim a n1 n a n. Since a n = x2n1 (2n 1!, a n1 = x2n3 (2n 3!. Therefore, x 2n3 lim a n1 n a n = lim (2n 3! n x 2n1 (2n 1! x 2n3 (2n 1! = lim n x 2n1 (2n 3! = x 2 (2n 1! lim n (2n 3! = x 2 1 lim n (2n 2 (2n 3 = 0 Thus the series representation is valid for all x.

7 6.8. TAYLOR AND MACLAURIN S SERIES 363 Here are some important functions and their corresponding Maclaurin s series 1 1 x = 1 x x2 x 3 x 4... ( 1, 1 e x = 1 x x2 x3... (, sin x = x x3 x5 5! x7... (, 7! cos x = 1 x2 x4 4! x6... (, 6! tan 1 x = x x3 3 x5 5 x7... [ 1, 1] 7 Remark The n th order Taylor polynomial (T n associated with the series expansion of a function f is the n th degree polynomial obtained by truncating the series expansion and keeping only the terms of degree less than or equal to n. In the case of sin x, since It follows that sin x = x x3 x5 5! x7 7!... T 1 = x T 2 = x T 3 = x x3 T 4 = x x3 T 5 = x x3 x5 5! and so on. T 6 = x x3 x5 5! Summary We now have four different techniques to find a series representation for a function. These techniques are: 1. Substitution 2. Differentiation 3. Integration

8 364 CHAPTER 6. INFINITE SEQUENCES AND SERIES 4. Taylor/Maclaurin s series. It may appear that the last technique is much more powerful, as it gives us a direct way to derive the series representation. In contrast, the first three techniques require we start from a known series representation. However, the first three techniques should not be ignored. In many cases, they make the work easier. We illustrate this with a few examples. Example Find a Maclaurin s series for f (x = e x2. Of course, this can be done directly. But consider the problem of finding all the derivative of e x2. One can also start from e x and substitute x 2 for x. Since it follows that e x = x n = 1 x x2 2 x3 x4 4! e x2 = = ( x 2 n ( 1 n x 2n This was pretty painless. To convince yourself that this is the way to do it, try the direct approach instead! Example Find a Maclaurin s series for cos x. Again, one can try the direct approach as in example It is not too diffi cult. One can also realize that (sin x = cos x. Thus, one can start with the series for sin x (which we already derived and find the series for cos x by differentiating it. We get cos x = (sin x ( = ( 1 n x 2n1 (2n 1! = (( 1 n x 2n1 (2n 1! = ( 1 n x 2n (2n! = 1 x2 2 x4 4! x6 6!

9 6.8. TAYLOR AND MACLAURIN S SERIES Things to Remember Given a function, be able to find its Taylor or Maclaurin s series. Be able to find the radius and interval of convergence of Taylor or Maclaurin s series Problems 1. If f (n (0 = (n 1!, find the Maclaurin s series for f and its radius of convergence. 2. Find the Maclaurin s series and its radius of convergence for each function below. (a f (x = cos x (b f (x = sin (2x (c f (x = e 5x 3. Find the Taylor series at the given point for each function below. (a f (x = 1 x x 2 at a = 2. (b f (x = e x at a = 3. (c f (x = cos x at a = π. (d f (x = ln x at a = Using a combination of the techniques studied (substitution, differentiation, integration, Taylor and Maclaurin s series, find a Maclaurin s series representation for the functions below. (a f (x = cos πx (b f (x = x tan 1 x (c f (x = x 2 e x (d f (x = sin 2 x (hint: use the identity sin 2 x = Answers 1 cos 2x 2 1. If f (n (0 = (n 1!, find the Maclaurin s series for f and its radius of convergence. f (x = (n 1 x n and R = 1 2. Find the Maclaurin s series using the definition of a Maclaurin s series and its radius of convergence for each function below.

10 366 CHAPTER 6. INFINITE SEQUENCES AND SERIES (a f (x = cos x (b f (x = sin (2x f (x = f (x = ( 1 n x 2n (2n! and R = ( 1 n 2 2n1 x 2n1 (2n 1! and R = (c f (x = e 5x 5 n x n f (x = and R = 3. Find the Taylor series at the given point for each function below. (a f (x = 1 x x 2 at a = 2. f (x = 7 5 (x 2 (x 2 2 and R = (b f (x = e x at a = 3. f (x = (c f (x = cos x at a = π. 3 (x 3n e and R = f (x = ( 1 n1 (x π 2n (2n! and R = (d f (x = ln x at a = 2. f (x = ln 2 ( 1 n1 2 n (x 2 n and R = n n=1 4. Using a combination of the techniques studied (substitution, differentiation, integration, Taylor and Maclaurin s series, find a Maclaurin s series representation for the functions below. (a f (x = cos πx cos πx = ( 1 n π 2n (2n! x2n and R =

11 6.8. TAYLOR AND MACLAURIN S SERIES 367 (b f (x = x tan 1 x (c f (x = x 2 e x x tan 1 x = x 2 e x = ( 1 n x 2n2 2n 1 and R = 1 ( 1 n x n2 and R = (d f (x = sin 2 x (hint: use the identity sin 2 x = sin 2 x = n=1 1 cos 2x 2 ( 1 n1 2 2n 1 (2n! x2n and R =

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