# Some Notes on Taylor Polynomials and Taylor Series

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1 Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited way. In these notes, we present these ideas in a condensed format. For students who wish to gain a deeper understanding of these concepts, we encourage a thorough reading of the chapter on Infinite Sequences and Series in the accompanying text by James Stewart. Taylor Polynomials We have considered a wide range of functions as we have explored calculus. The most basic functions have been the polynomial functions like p(x) = x 3 + 9x 2 3x + 2, which have particularly easy rules for computing their derivatives. As well, polynomials are evaluated using the simple operations of multiplication and addition, so it is relatively easy to compute their exact values given x. On the other hand, the transcendental functions (e.g., sin x or ln x) are more difficult to compute. (Though it seems trivial to evaluate ln(.75), say, by pressing a few buttons on your calculator, what really happens when you push the ln button is a bit more involved.) Question: Is it possible to approximate a given function by a polynomial? That is, can we find a polynomial of a given degree n that can be substituted in place of a more complex function without too much error? We have already considered this question in the specific case of linear approximation. There we took a specific function, f, and a specific point on the graph of that function, (a, f(a)), and approximated the function near x = a by its tangent line at x = a. Explicitly, we approximated the curve by the straight line y = f(x) y = L(x) = f(a) + f (a)(x a).

3 Now, T 2 (x) = 2c and so T 2 () = 2c. Also, f (x) = e x, so f () = e =. Hence, substituting these into (), we find that 2c =, which gives us c = 2. Hence, T 2 (x) = + x + 2 x2 is a second degree polynomial that agrees with f(x) = e x by having T 2 () = f(), T 2() = f (), and T 2 () = f (). We call T 2 (x) the second degree Taylor polynomial for e x about x =. Taylor polynomials generated by looking at data at x = are called also Maclaurin polynomials. There is nothing that says we need to stop the process of constructing a Taylor (or Maclaurin) polynomial after the quadratic term. For f(x) = e x, for example, we know that we can continue to take derivatives of f at x = as many times as we like (we say e x is infinitely differentiable in this case), and, indeed, its kth derivative is f (k) (x) = e x, and so f (k) () = e = for k =,, 2,.... So, if we construct an nth degree polynomial T n (x) = c + c x + c 2 x c n x n as an approximation to f(x) = e x by requiring that p (k) () = f (k) () for k =,, 2,..., n, then we find that e x T n (x) = + x + 2 x2 + 6 x x4 + + n! xn. You can derive Taylor s formula for the coefficients c k by using the fact that to show that T n (k) (x) = k! c k + terms of higher degree in x c k = f (k) (). (2) k! Note that the k! arises since (x k ) = kx k and so taking k successive derivatives of x k gives you k (k ) (k 2) 3 2 k!. Example: Let us construct the fifth degree Maclaurin polynomial for the function f(x) = sin x. That is, we wish to find the coefficients of First, we need the derivatives T 5 (x) = c + c x + c 2 x 2 + c 3 x 5 + c 4 x 4 + c 5 x 5. d k dx k sin x x= 3

4 for k =,, 2,..., 5: k = : sin() =, k = 3 : cos() =, k = : cos() =, k = 4 : sin() =, k = 2 : sin() =, k = 5 : cos() =. Using Taylor s formula (2) for the coefficients, we find that T 5 (x) = x x3 3! + x5 5!. Note that because sin() = and every even order derivative of sin x is ± sin x, we have only odd powers appearing with non-zero coefficients in T 5 (x). This is not surprising since sin x is an odd function; that is, sin( x) = sin x. In general, we wish to use information about a function f at points other than x = to construct an approximating polynomial of degree n. If we look at a function around the point given by x = a, Taylor polynomials look like where T n (x) = c + c (x a) + c 2 (x a) c n (x a) n, c k = f (k) (a). (3) k! This form of the polynomial may look a little strange at first since you are likely used to writing your polynomials in simple powers of x, but it is very useful to write this polynomial in this form. In particular, if we follow Taylor s program to construct the coefficients c k by making the derivatives T n (k) (a) = f (k) (a), then the calculation becomes trivial since T n (k) (x) = k! c k + higher order terms in powers of (x a), so that plugging in x = a makes all the high-order terms vanish. Example: Suppose we are asked to find the Taylor polynomial of degree 5 for sin x about x = π 2. This time, we lose the symmetry about the origin that gave us the expectation that we would only see odd terms. The derivatives at x = π 2 are k = : sin( π 2 ) =, k = 3 : cos( π 2 ) =, k = : cos( π 2 ) =, k = 4 : sin( π 2 ) =, k = 2 : sin( π 2 ) =, k = 5 : cos( π 2 ) =. In fact, we are only left with even-order terms and the required polynomial has no x 5 term: T 5 (x) = 2! (x π 2 )2 + 4! (x π 2 )4. Many of the basic functions you know have useful Maclaurin polynomial approximations. If you wish an approximation of degree n, then 4

5 . e x + x + x2 2! + + xn n! ; 2. sin x x x3 3! + x5 5! + + ( )k x 2k+, where 2k + is the greatest odd (2k + )! integer less than or equal to n; 3. cos x x2 2! + x4 4! + + ( )k x 2k, where 2k is the greatest even integer less than or equal to (2k)! n; 4. ln( x) x x2 2 xn n ; 5. tan x x x3 3 + x ( )k x 2k+, where 2k + is the greatest 2k + odd integer less than or equal to n; 6. x + x + x2 + + x n. Exercises:. Find the third degree Maclaurin polynomial for f(x) = + x. 2. Find the Taylor polynomials of degree 3 for f(x) = 5x 2 3x + 2 (a) about x = and (b) about x = 2. What do you notice about them? If you expand each of these polynomials and collect powers of x, what do you notice? 3. If f(x) = ( + e x ) 2, show that f (k) () = k for any k. Write the Maclaurin polynomial of degree n for this function. 4. Find the Maclaurin polynomial of degree 5 for f(x) = tan x. 5. Find the Maclaurin polynomial of degree 3 for f(x) = e sin x. 2 Taylor s Formula with Remainder We constructed the Taylor polynomials hoping to approximate functions f by using information about the given function f at exactly one point x = a. How well does the Taylor polynomial of degree n approximate the function f? One way of looking at this question is to ask for each value x, what is the difference between f(x) and T n (x)? If we call this difference the remainder, R n (x), we can write f(x) = f(a) + f (a)(x a) + + f (n) (a) (x a) n + R n (x). (4) n! 5

6 The first thing we notice if we look at (4) is that by taking this as the definition of R n (x), Taylor s formula (the rest of the right-hand side of (4)) is automatically correct. (This might take you a little thought to appreciate.) Of course, we would like to be able to deal with this remainder, R n (x), quantitatively. It turns out that we can use the Mean Value Theorem to find an expression for this remainder. The proof of this formula is a bit of a diversion from where we wish to go, so we will state the result without proof. The Lagrange Remainder Formula: Suppose that f has derivatives of at least order n + on some interval [b, d]. Then if x and a are any two numbers in (b, d), the remainder R n (x) in Taylor s formula can be written as where c is some number between x and a. R n (x) = f (n+) (c) (n + )! (x a)n+, (5) (Remark: The n = case is the Mean Value Theorem itself.) First, note that c depends on both x and a. Now, if we could actually find this number c, we could know the remainder exactly for any given value of x. However, if you were to look at the proof of this formula, you would see that this number c comes into the formula because of the Mean Value Theorem. The Mean Value Theorem is very powerful, but all it tells us is that such a c exists, and not what its exact value is. Hence, we must figure out a way to use this Remainder Formula given our limited knowledge of c. One approach is to ask ourselves: What is the worst error we could make in approximating f(x) using a Taylor polynomial of degree n about x = a? To answer this question, we will focus our attention on R n (x), the absolute value of the remainder. If we look at (5), we notice that we know everything except f (n+) (c), and so, if our goal is to find a bound on the magnitude of the error, then we will need to find a bound on f (n+) (t) that works for all values of t in the interval containing x and a. That is, we seek a positive number M such that f (n+) (t) M. If we can find such an M, then we are able to bound the remainder, knowing x and a, as R n (x) M (n + )! x a n+. (6) Example: Suppose we wish to compute using a Taylor polynomial of degree n = (the linear approximation) for a = 9 and give an estimate on the 6

7 size of the error R (). First, we note that Taylor s formula for f(x) = x at a = 9 is given by and so f(x) = f(9) + f (9)(x 9) + R (x), x = (x 9) + R (x). Thus, 3 6. We now estimate R (). We first find M so that f (t) M for all t in [9, ]. Now, f (t) = 4t 3/2 = 4t. 3/2 So, we want to make this function as big as possible on the interval [9, ]. As t gets larger, /4t 3/2 gets smaller, so it is largest at the left-hand endpoint, at t = 9. Hence, any value of M such that M 4 9 = 3/2 8 will work. We might as well choose M = /8 (though if you don t have a calculator, choosing M = / would make the computations easier if you wished to use decimal notation) and substitute this into (6) with a = 9 and x = to get R () /8 9 2 = 2! 26. Hence, we know that = 3 6 ± 26. In fact, we can make a slightly stronger statement by noticing that the value of the second derivative, f (t), is always negative for t in the interval [9, ] and so we know that this tangent line always lies above the curve y = x, and hence we are overestimating the value of by using this linear approximation. Thus, Example: We approximate sin(.5) by using a Maclaurin polynomial of degree 3. Recall that sin x = x x3 3! + R 3(x), so sin(.5).5.53 = 3! 2 48 = To estimate the error in this approximation, we look for M > such that d 4 dt 4 sin(t) = sin(t) M 7

8 for t in [,.5]. The easiest choice for M is since we know that sin(t) never gets larger than. However, we can do a bit better since we know that the tangent to sin(t) at t = is y = t, which lies above the graph of sin(t) on [,.5]. Thus, if we choose M.5 we will get an appropriate bound. In this case, R 3 (.5).5 4!.5 4 = = 768. Exercises:. Find the second degree Taylor polynomial about a = for f(x) = /x and use it to compute /.5 to as many decimal places as is justified by this approximation. 2. What degree Maclaurin polynomial do you need to approximate cos(.25) to 5 decimal places of accuracy? 3. Show that the approximation e = + + 2! + + 7! gives the value of e to within an error of ** If f(x) = + x, show that R 2 (x), the remainder term associated to the second degree Maclaurin polynomial for f(x), satisfies for x >. 3 Taylor Series x 3 6( + x) R 2(x) x3 5/2 6 We can use the Lagrange Remainder Formula to see how many functions can actually be represented completely by something called a Taylor Series. Stewart gives a fairly complete discussion of series, but we will make use of a simplified approach that is somewhat formal in nature since it suits our limited purposes. We begin by considering something we will call a power series in (x a): c + c (x a) + c 2 (x a) c n (x a) n +, (7) where we choose the coefficients c k to be real numbers. Of course, we are particularly interested in the case where we choose these coefficients to be those given by the Taylor formula, but there are also more general power series. At first glance, the formula (7) looks very much like the polynomial formulae we have considered in the previous sections. However, the final indicate that 8

9 we want to think about what happens if we keep adding more and more terms (i.e. we let n go to infinity). The question is whether or not we can make sense of this potentially troublesome situation. To make things as easy as possible, we will focus completely on the case where we generate the coefficients of the series (7) using Taylor s formula. Suppose that you have a function f which is infinitely differentiable, which means you can take derivatives of all possible orders. (Examples of such functions are f(x) = e x and f(x) = sin x.) Then consider the Taylor polynomial for f about the point x = a with the remainder term: f(x) = f(a) + f (a)(x a) + f (a) (x a) f (n) (a) + R n (x). 2! n! In a naive way, we can think of the Taylor series that corresponds to this as the object that results when we let n. That is, we keep adding more and more terms to generate polynomials of higher and higher degrees. (We can do this since we have assumed we can take derivatives to as high an order as we need.) Now, in order for this process to produce a finite value for a given value of x, it must be that R n (x) as n. Example: Consider the degree n Maclaurin polynomial, with remainder, for f(x) = e x : e x = + x + x2 2! + x3 3! + + xn n! + R n(x), Now, we would like to show that lim n R n (x) = for any real x. The remainder formula says R n (x) = e c x n+ (n + )!, for some c between and x. Since e c e x, we can say R n (x) e x x n+ (n + )!. Does this last expression go to zero as n goes to infinity? First, since x a fixed real number, e x is constant. Moreover, x n+ is the product of n + x s, whereas (n+)! is the product of, 2, 3,..., n+. Since x is fixed, in the limit (n + )! grows faster than x n+. (This may be surprising if you look just at the first few values of n for, say, x = ). So, x n+ /(n + )! does indeed go to as n is made arbitrarily large. Now, because of the way we have constructed these Taylor series, we are guaranteed to have R n (a) = and so the Taylor series always represents the function at x = a. It is more interesting to think about the question: For what values of x does the Taylor series represent the value of f(x)? (Mathematicians usually say the series converges to f(x) for such values of x.) 9

10 In general, there is a symmetric interval of values around the centre x = a of the Taylor series for which the series converges and hence where it is a valid representation of the function. We call this interval the interval of convergence for the Taylor series centred at x = a. There are various techniques used to find out where series converge. These are discussed in Stewart, but we will not delve into these technicalities in this course. We will simply make use of the fact that Taylor series can be used in place of the functions that gave rise to them wherever they converge. The basic functions we considered in section have the following intervals of convergence:. e x = + x + x2 2! + + xn + for x in (, ); n! 2. sin x = x x3 3! + x5 5! + + ( )k x 2k+ + for x in (, ); (2k + )! 3. cos x = x2 2! + x4 4! + + ( )k x 2k + for x in (, ); (2k)! 4. ln( x) = x x2 2 xn + for x in [, ); n 5. tan x = x x3 3 + x ( )k x 2k+ + for x in [, ]; 2k + 6. x = + x + x2 + + x n + for x in (, ). Note that e x, sin x, and cos x can all be defined by their Maclaurin series everywhere, a fact which is sometimes useful. It is also useful to know that when a function is equal to its power series (7) in some interval centred at x = a, then this is the only such power series formula for f(x) on this interval. In particular, the coefficients are uniquely determined. This can be useful to know since it sometimes makes it possible to find the Taylor series coefficients using some computational trick rather than computing them directly. It is also possible to get the derivative for a function f(x) from its power series whenever the series converges. In this case, you simply differentiate the power series term-by-term to compute f (x): If then f(x) = c + c (x a) + c 2 (x a) 2 + c 3 (x a) 3 + f (x) = c + 2c 2 (x a) + 3c 3 (x a) 2 +. We can use this to generate new series from ones we know. (There are actually a few subtleties to what happens with convergence in this, but they won t affect our limited use of series. You can consult Stewart s more detailed discussion if you wish to learn more about these.)

11 Besides differentiating power series, it is possible to multiply or divide them, to substitute functions into them and, as we shall see in the next section, to integrate them. Effectively, once we understand how power series represent the functions that generate them, the series can be manipulated in place of using the original functions. Example: If we wish to find the Maclaurin series for xe x, then we use the fact that e x = + x + x 2 /2 + + x n /n! + to find that ) xe x = x ( + x + x xn n! + = x + x 2 + x xn+. n! Example: We can find the Maclaurin series for e sin x by using the series for e x and the series for sin x: e sin(x) = + sin x + sin2 x + sin3 x ) = + (x x ) 2 (x x = + x + 2 x2 8 x4 5 x5 +. Taylor series can also be useful for computing limits. Example: We wish to evaluate lim x e x cos x. sin x We substitute the Maclaurin series for each of e x, cos x, and sin x to get ( + x + x2 x2 2 + ) ( 2 lim + x ) x (x x3 6 + ) = lim (x + x2 + x ) x (x x3 6 + ) Exercises: ( + x + x 6 = lim + ) x ( x2 6 + ) =. 2

12 . By squaring the Maclaurin series for cos x, show that x cos x sin x 2. Evaluate lim x x 2. tan x 3. Evaluate lim x cos 2 x = x x4. + x2 + cos x 2 x Use term-by-term differentiation of the Maclaurin series for sin(x) to show that its derivative is cos x. 5. Differentiate the series for x to find a series for ( x) **Find the first 3 terms of the Maclaurin series for tan x by using the series for sin x and cos x. 4 Taylor Series and Integration If we have a power series representation of a function, we may integrate the series term-by-term to generate a new series, which converges in the same interval as the original series (though interesting things may happen at the endpoints of this interval). This can be useful for two things: () generating the series of a function by using the series of its derivative, and (2) approximating the value of a definite integral. Example: We start with a situation where we know what the result should be. We know that sin x is the antiderivative of cos x, and we also know the Maclaurin series for each of these functions. So, we compute, using the Fundamental Theorem of Calculus, x x ) cos t dt = ( t2 2! + t4 4! + dt ] x = [t t3 3 2! + t5 5 4! + = x x3 3! + x5 5! + = sin x. Here, x is allowed to be any real number since these series converge for all values of x. Example: We know that ln( + x) = x dt. (8) + t 2

13 Now, + x = ( x) Substituting this into (8) gives = + ( x) + ( x) 2 + ( x) 3 + = x + x 2 x 3 + x 4. ln( + x) = = x ( t + t 2 t 3 + t 4 + ) dt [ ] x t t2 2 + t3 3 t4 4 + t5 5 + = x x2 2 + x3 3 + ( )n x n +. n Example: Consider the function f(x) = e x2. Suppose we wish to find an antiderivative of f. It turns out that there is no elementary way to write down this antiderivative in terms of basic functions like polynomials, exponentials, logarithms, or trignometric functions. However, we can represent f(x) by substituting x 2 into the Maclaurin series for e x : Thus, x e x2 = + ( x 2 ) + ( x2 ) 2 2! + ( x2 ) 3 3! + + ( x2 ) n n! = x 2 + x4 2! x6 3! + + ( )n x 2n +. n! e t2 dt = x + ( t 2 + t4 2! t6 3! + + ( )n t 2n ) + dt n! = x x3 3 + x5 5 2! x7 7 3! + + ( )n x 2n+ (2n + ) n! + The series we have considered in this section are all examples of alternating series, which have successive terms with alternating plus and minus signs. It is particularly easy to estimate the error involved if we approximate the functions represented by these series by using the polynomials we get when we truncate them after a finite number of terms. In his chapter on Infinite Sequences and Series, Stewart presents the Alternating Series Estimation Theorem. We use it here for power series in the situation where we plug in a specific value for x to generate a series of real numbers. While we have not dealt with how to 3

14 make sense of general series, it is possible to define what it means for them to converge to a finite number. It will suffice for our purposes for you to simply accept that such a series can represent a finite number. Alternating Series Estimation Theorem: Let b b 2 + b 3 b ( ) n b n + ( ) n b n+ + be an alternating series that satisfies the conditions (a) b n+ b n and (b) lim n b n =. Then this series converges to a finite number, S. Moreover, if we write S = b b 2 + b 3 b ( ) n b n + R n, then we have that R n b n+. Example: We can use the Alternating Series Estimation Theorem to decide how many terms of the series we need to approximate ln(.5) to within 4. First, we know that ln(.5) =.5 + x dx =.5 (.5)2 2 + (.5)3 3 + ( )n (.5) n n +, and so we want to find n so that R n (.5) < 4. Well, we know that R n (.5) ( ) n (.5) n+ n + from the Alternating Series Estimation Theorem. Thus, we want n so that ( ) n (.5) n+ n + < 4. The easiest method for solving this inequality is to guess-and-check. This gives us n =. Exercises:. Find an approximation for 2. Evaluate.5 dx good to 2 decimal places. + x4 cos(x 2 ) dx to 6 decimal places using series. 4

15 3. We know that tan () = π 4. Thus, π = 4 + x 2 dx. Is the series method of computing this integral a good way to evaluate π to million decimal places? 5

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