Commonly Used Taylor Series series when is valid/true 1 note this is the geometric series. = 1 + x + x2 + x3 + x4 + ::: 1 − x just think of x as r 1 X = xn x 2 (−1; 1) n=0 so: 1 1 1 2 3 4 e = 1 + 1 + 2! + 3! + 4! + ::: x x x x 1 e = 1 + x + + + + ::: n 2! 3! 4! (17x) P1 (17x) X 17nxn e = n=0 n! = n! n=0 1 X xn = x 2 n! R n=0 note y = cos x is an even function x2 x4 x6 x8 (i.e., cos(−x) = + cos(x)) and the cos x = 1 − + − + − ::: 2! 4! 6! 8! taylor seris of y = cos x has only even powers. 1 X x2n = (−1)n x 2 (2n)! R n=0 note y = sin x is an odd function x3 x5 x7 x9 (i.e., sin(−x) = − sin(x)) and the sin x = x − + − + − ::: 3! 5! 7! 9! taylor seris of y = sin x has only odd powers. 1 2n−1 1 2n+1 X x or X x = (−1)(n−1) = (−1)n x 2 (2n − 1)! (2n + 1)! R n=1 n=0 x2 x3 x4 x5 ln (1 + x) = x − + − + − ::: question: is y = ln(1 + x) even, 2 3 4 5 odd, or neither? 1 n 1 n X x or X x = (−1)(n−1) = (−1)n+1 x 2 (−1; 1] n n n=1 n=1 x3 x5 x7 x9 tan−1 x = x − + − + − ::: question: is y = arctan(x) even, 3 5 7 9 odd, or neither? 1 2n−1 1 2n+1 X x or X x = (−1)(n−1) = (−1)n x 2 [−1; 1] 2n − 1 2n + 1 n=1 n=0 1 Math 142 Taylor/Maclaurin Polynomials and Series Prof. Girardi Fix an interval I in the real line (e.g., I might be (−17; 19)) and let x0 be a point in I, i.e., x0 2 I: Next consider a function, whose domain is I, f : I ! R and whose derivatives f (n) : I ! R exist on the interval I for n = 1; 2; 3;:::;N. th Definition 1. The N -order Taylor polynomial for y = f(x) at x0 is: f 00(x ) f (N)(x ) p (x) = f(x ) + f 0(x )(x − x ) + 0 (x − x )2 + ··· + 0 (x − x )N ; (open form) N 0 0 0 2! 0 N! 0 which can also be written as (recall that 0! = 1) f (0)(x ) f (1)(x ) f (2)(x ) f (N)(x ) p (x) = 0 + 0 (x−x )+ 0 (x−x )2+···+ 0 (x−x )N - a finite sum, i.e. the sum stops : N 0! 1! 0 2! 0 N! 0 Formula (open form) is in open form. It can also be written in closed form, by using sigma notation, as N (n) X f (x0) p (x) = (x − x )n : (closed form) N n! 0 n=0 So y = pN (x) is a polynomial of degree at most N and it has the form N (n) X f (x0) p (x) = c (x − x )n where the constants c = N n 0 n n! n=0 are specially chosen so that derivatives match up at x0, i.e. the constants cn's are chosen so that: pN (x0) = f(x0) (1) (1) pN (x0) = f (x0) (2) (2) pN (x0) = f (x0) . (N) (N) pN (x0) = f (x0) : th th The constant cn is the n Taylor coefficient of y = f(x) about x0. The N -order Maclaurin polynomial for y = f(x) th is just the N -order Taylor polynomial for y = f(x) at x0 = 0 and so it is N X f (n)(0) p (x) = xn : N n! n=0 1 Definition 2. The Taylor series for y = f(x) at x0 is the power series: f 00(x ) f (n)(x ) P (x) = f(x ) + f 0(x )(x − x ) + 0 (x − x )2 + ··· + 0 (x − x )n + ::: (open form) 1 0 0 0 2! 0 n! 0 which can also be written as f (0)(x ) f (1)(x ) f (2)(x ) f (n)(x ) P (x) = 0 + 0 (x−x )+ 0 (x−x )2+···+ 0 (x−x )n+::: - the sum keeps on going and going: 1 0! 1! 0 2! 0 n! 0 The Taylor series can also be written in closed form, by using sigma notation, as 1 (n) X f (x0) P (x) = (x − x )n : (closed form) 1 n! 0 n=0 The Maclaurin series for y = f(x) is just the Taylor series for y = f(x) at x0 = 0. 1 (n) Here we are assuming that the derivatives y = f (x) exist for each x in the interval I and for each n 2 N ≡ f1; 2; 3; 4; 5;::: g . 2 Big Questions 3. For what values of x does the power (a.k.a. Taylor) series 1 (n) X f (x0) P (x) = (x − x )n (1) 1 n! 0 n=0 converge (usually the Root or Ratio test helps us out with this question). If the power/Taylor series in formula (1) does indeed converge at a point x, does the series converge to what we would want it to converge to, i.e., does f(x) = P1(x) ? (2) Question (2) is going to take some thought. th Definition 4. The N -order Remainder term for y = f(x) at x0 is: def RN (x) = f(x) − PN (x) th where y = PN (x) is the N -order Taylor polynomial for y = f(x) at x0. So f(x) = PN (x) + RN (x) (3) that is f(x) ≈ PN (x) within an error of RN (x) : We often think of all this as: N (n) X f (x0) f(x) ≈ (x − x )n - a finite sum, the sum stops at N: n! 0 n=0 We would LIKE TO HAVE THAT 1 (n) ?? X f (x0) f(x) = (x − x )n - the sum keeps on going and going : n! 0 n=0 In other notation: ?? f(x) ≈ PN (x) and the question is f(x) = P1(x) where y = P1(x) is the Taylor series of y = f(x) at x0. ?? Well, let's think about what needs to be for f(x) = P1(x), i.e., for f to equal to its Taylor series. Notice 5. Taking the limN!1 of both sides in equation (3), we see that 1 (n) X f (x0) f(x) = (x − x )n - the sum keeps on going and going : n! 0 n=0 if and only if lim RN (x) = 0 : N!1 Recall 6. limN!1 RN (x) = 0 if and only if limN!1 jRN (x)j = 0 . So 7. If lim jRN (x)j = 0 (4) N!1 then 1 (n) X f (x0) f(x) = (x − x )n : n! 0 n=0 So we basically want to show that (4) holds true. How to do this? Well, this is where Mr. Taylor comes to the rescue! 2 2According to Mr. Taylor, his Remainder Theorem (see next page) was motivated by coffeehouse conversations about works of Newton on planetary motion and works of Halley (of Halley's comet) on roots of polynomials. 3 Taylor's Remainder Theorem 3 Version 1: for a fixed point x 2 I and a fixed N 2 N. There exists c between x and x0 so that f (N+1)(c) R (x) def= f(x) − P (x) theorem= (x − x )(N+1) : (5) N N (N + 1)! 0 So either x ≤ c ≤ x0 or x0 ≤ c ≤ x. So we do not know exactly what c is but atleast we know that c is between x and x0 and so c 2 I. Remark: This is a Big Theorem by Taylor. See the book for the proof. The proof uses the Mean Value Theorem. Note that formula (5) implies that f (N+1)(c) jR (x)j = jx − x j(N+1) : (6) N (N + 1)! 0 Version 2: for the whole interval I and a fixed N 2 N.3 Assume we can find M so that (N+1) the maximum of f (x) on the interval I ≤ M; i.e., max f (N+1)(c) ≤ M: c2I Then M jR (x)j ≤ jx − x jN+1 (7) N (N + 1)! 0 for each x 2 I. Remark: This follows from formula (6). 4 Version 3: for the whole interval I and all N 2 N. 1 Now assume that we can find a sequence fMN gN=1 so that (N+1) max f (c) ≤ MN c2I for each N 2 N and also so that MN N+1 lim jx − x0j = 0 N!1 (N + 1)! for each x 2 I. Then, by formula (7) and the Squeeze Theorem, lim jRN (x)j = 0 N!1 for each x 2 I. Thus, by So 7, 1 (n) X f (x0) f(x) = (x − x )n n! 0 n=0 for each x 2 I. 3Here we assume that the (N + 1)-derivative of y = f(x), i.e. y = f (N+1)(x), exists for each x 2 I. 4 (N) Here we assume that y = f (x), exists for each x 2 I and each N 2 N. 4.