Commonly Used Taylor Series series 1 1 x when is valid/true = = 1 + x + x2 + x3 + x4 + . . . 1 X note this is the geometric series. just think of x as r xn n=0 x 2 ( 1, 1) so: 1 1 1 e = 1 + 1 + 2! + 3! + 4! + ... 1 X P1 (17x)n 17n xn (17x) e = n=0 n! = n! n=0 ex = 1 + x + 1 X x x x + + + ... 2! 3! 4! 2 3 4 = xn n! n=0 x2R w cos x = 1 1 X x2 x4 + 2! 4! ( 1)n x2n (2n)! x6 x8 + 6! 8! ... note y = cos x is an even function (i.e., cos( x) = + cos(x)) and the taylor seris of y = cos x has only even powers. = n=0 x2R sin x = x 1 X x3 x5 + 3! 5! ( 1)(n x7 x9 + 7! 9! ... note y = sin x is an odd function (i.e., sin( x) = sin(x)) and the taylor seris of y = sin x has only odd powers. = 1) n=1 1 x2n 1 or X x2n+1 = ( 1)n (2n 1)! (2n + 1)! n=0 x2R question: is y = ln(1 + x) even, odd, or neither? ln (1 + x) = = x 1 X x2 x3 + 2 3 ( 1)(n 1) x4 x5 + 4 5 ... n=1 1 xn or X xn = ( 1)n+1 n n n=1 x 2 ( 1, 1] question: is y = arctan(x) even, odd, or neither? tan 1 x = = x 1 X x3 x5 + 3 5 ( 1)(n 1) x7 x9 + 7 9 ... n=1 1 x2n 1 or X x2n+1 = ( 1)n 2n 1 2n + 1 n=0 x 2 [ 1, 1] 1 Math 142 Taylor/Maclaurin Polynomials and Series x0 2 I . 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., 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 . De?nition 1. The N th -order Taylor polynomial for y = f (x) at x0 is: f 00 (x0 ) (x 2! which can also be written as (recall that 0! = 1) pN (x) = f (x0 ) + f 0 (x0 )(x x0 ) + pN (x) = x0 )2 + · · · + f (N ) (x0 ) (x N! x0 )N , (open form) f (0) (x0 ) f (1) (x0 ) f (2) (x0 ) f (N ) (x0 ) + (x x0 )+ (x x0 )2 +· · ·+ (x x0 )N - a ?nite sum, i.e. the sum stops . 0! 1! 2! N! Formula (open form) is in open form. It can also be written in closed form, by using sigma notation, as pN (x) = N X f (n) (x0 ) (x n! n=0 x0 )n . (closed form) So y = pN (x) is a polynomial of degree at most N and it has the form pN (x) = N X cn (x x0 )n where the constants cn = n=0 f (n) (x0 ) n! are specially chosen so that derivatives match up at x0 , i.e. the constants cn ’s are chosen so that: pN (x0 ) = f (x0 ) pN (x0 ) = f (1) (x0 ) pN (x0 ) = f (2) (x0 ) . . . pN (x0 ) = f (N ) (x0 ) . The constant cn is the nth Taylor coe cient of y = f (x) about x0 . The N th -order Maclaurin polynomial for y = f (x) is just the N th -order Taylor polynomial for y = f (x) at x0 = 0 and so it is pN (x) = De?nition 2. 1 N X f (n) (0) n x . n! n=0 (N ) (2) (1) The Taylor series for y = f (x) at x0 is the power series: x0 ) + f 00 (x0 ) (x 2! x0 )2 + · · · + f (n) (x0 ) (x n! x0 )n + . . . (open form) P1 (x) = f (x0 ) + f 0 (x0 )(x which can also be written as f (0) (x0 ) f (1) (x0 ) f (2) (x0 ) f (n) (x0 ) P1 (x) = + (x x0 )+ (x x0 )2 +· · ·+ (x x0 )n +. . . 0! 1! 2! n! The Taylor series can also be written in closed form, by using sigma notation, as 1 X f (n) (x0 ) P1 (x) = (x x0 )n . n ! n=0 - the sum keeps on going and going. (closed form) The Maclaurin series for y = f (x) is just the Taylor series for y = f (x) at x0 = 0. 1Here we are assuming that the derivatives y = f (n) (x) exist for each x in the interval I and for each n 2 N ? {1, 2, 3, 4, 5, . . . } . 2 Big Questions 3. For what values of x does the power (a.k.a. Taylor) series 1 X f (n) (x0 ) P1 (x) = (x x0 )n n ! n=0 f (x) = P1 (x) ? (1) 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 (2) Question (2) is going to take some thought. De?nition 4. The N th -order Remainder term for y = f (x) at x0 is: RN (x) = f (x) th def PN (x) where y = PN (x) is the N -order Taylor polynomial for y = f (x) at x0 . So that is f (x) = PN (x) + RN (x) within an error of RN (x) . - a ?nite sum, the sum stops at N . (3) f (x) ? PN (x) We often think of all this as: N X f (n) (x0 ) f (x) ? (x x0 )n n ! n=0 We would LIKE TO HAVE THAT 1 X f (n) (x0 ) ?? f (x) = (x n! n=0 In other notation: x0 )n - the sum keeps on going and going . ?? f (x) ? PN (x) and the question is where y = P1 (x) is the Taylor series of y = f (x) at x0 . ?? f (x) = P1 (x) 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 X f (n) (x0 ) f (x) = (x x0 )n - the sum keeps on going and going . n ! n=0 if and only if N !1 lim RN (x) = 0 . Recall 6. limN !1 RN (x) = 0 if and only if limN !1 |RN (x)| = 0 . So 7. If then N !1 lim |RN (x)| = 0 1 X f (n) (x0 ) (x n! n=0 (4) x0 )n . f (x) = 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 co?eehouse 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 Version 1: for a ?xed point x 2 I and a ?xed N 2 N. There exists c between x and x0 so that RN (x) def 3 f (N +1) (c) (x x0 )(N +1) . (5) (N + 1)! 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. = f (x) PN (x) theorem = Note that formula (5) implies that |RN (x)| = f (N +1) (c) |x (N + 1)! x0 | (N +1) . (6) Version 2: for the whole interval I and a ?xed N 2 N.3 Assume we can ?nd M so that the maximum of f (N +1) (x) on the interval I i.e., Then for each x 2 I . Remark: This follows from formula (6). Version 3: for the whole interval I and all N 2 N. 4 Now assume that we can ?nd a sequence {MN }1 N =1 so that max f (N +1) (c) ? MN for each N 2 N and also so that c2I ? M , max f (N +1) (c) ? M . c2I |RN (x)| ? M |x (N + 1)! x0 | N +1 (7) MN N +1 |x x0 | = 0 N !1 (N + 1)! for each x 2 I . Then, by formula (7) and the Squeeze Theorem, lim for each x 2 I . Thus, by So 7, for each x 2 I . N !1 lim |RN (x)| = 0 1 X f (n) (x0 ) (x n! n=0 f (x) = x0 )n 4Here we assume that y = f (N ) (x), exists for each x 2 I and each N 2 N. 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