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Variations on Taylor's Formula LECTURE 1 Variations on Taylor's Formula Numerical metho ds are ipso facto approximate metho ds. This b eing the case, it will b e imp ortant through- out this course to determine the accuracy of numerical results. We shall b egin by reviewing the analytic metho ds by whichwe approximate functions and howwe b ound the errors that arise from such approxima- tions. n Definition 1.1. We denote by C [a; b] the set of functions on the interval [a; b] R that have continuous n derivatives up to order n. We denote by C R the set of functions on the real line that have continuous 1 1 derivatives up to order n. We denote by C [a; b] and C R the sets of functions for which derivatives of al l orders exist on, respectively, [a; b] and R. 2 1 2 Example 1.2. If f x x sin1=x then f is in C R but not in C R. To see this, note 1 2 =0 lim f x = lim x sin x!0 x!0 x 1 2 0 since x ! 0 and sin is b ounded b etween 1 and 1. Thus f 2 C R. x df f x f 0 lim x!0 dx x x=0 1 x sin = lim x!0 x = 0 again b ecause it the limit of a function that is the pro duct of a function that vanishes at the limit p oint 1 and a function that remains b ounded. So f is also in C R. However, 2 0 0 d f f x f 0 lim 2 x!0 dx x x=0 1 2x sin1=x cos 0 x = lim x!0 x 1 cos x = lim x!0 x 2 do es not exist. Thus, f=2 C R. Remark 1.3. In general, wehave 1 3 2 1 0 C [a; b] C [a; b] C [a; b] C [a; b] C [a; b] n+1 Theorem 1.4. Taylor's Theorem with Integral Remainder. If f 2 C [a; b], then for any points x; x 2 [a; b], o n X 1 k k f x x x + R x f x= o o n k ! k =0 where Z x 1 n+1 n R x f tx t dt n n! x o 1 1. VARIATIONS ON TAYLOR'S FORMULA 2 Proof. This theorem is surprisingly easy to prove. We start by using integration by parts to evaluate the right hand side of the de nition of the error term R x: Setting n 1 n n+1 u = x t dv = f tdt n! 1 n1 n x t v = f t du = n1! and using the integration by parts formula Z Z x x x udv = uv j vdu x o x x o o we nd Z x x 1 1 n n n n1 x t f t f tx t dt + R x n n! n 1! x o x o 1 n n f x x x + R x = o o n1 n! Wethus have a recursive formula for R x. Using this recursive formula over and over again we can reduce n the right hand side to 1 1 n n n1 n1 R x = f x x x f x x x + n o o o o n! n 1! 1 1 0 n f x x x + R x o o 0 1! 0! Z n x X 1 n 0 0 n f tx t dt = f x x x + o o 0! x o n=1 Z n x X n 0 n = f tdt f x x x + o o x o n=1 n X n n = f x x x + f x f x o o o n=1 n X n n = f x x x + f x o o n=0 Solving the extreme sides for f x yields n X 1 k k f x= f x x x + R x o o n k ! k =0 Remark 1.5. Note that in this formulation of Taylor's Theorem, wehave an explicit formula for computing the error term R x. However, since the error term dep ends on the choice of x and x in a non-trivial way, n o we do not have at least at face value an understanding of how the error term changes as wevary x. Such an understanding would b e critical if we are to regard n X 1 k k f x f x x x n o o k ! k =0 as a p olynomial function approximate to the original function f x. Belowwe'll deduce b ounds on the error term R xasx ranges throughout the interval [a; b]. n Theorem 1.6. Mean Value Theorem for Integrals. Let u and v becontinuous real-valued functions on an interval [a; b], and suppose that v x 0 for al l x 2 [a; b]. Then there exists a point 2 [a; b] such that Z Z b b uxv xdx = u v xdx . a a 1. VARIATIONS ON TAYLOR'S FORMULA 3 Proof. Let and denote, resp ectively, the minimal and maximal values of uxon[a; b]. Then v x uxv x vx ; 8 x 2 [a; b] Integrating this relationship b etween a and b yields Z Z Z b b b v xdx uxv xdx v xdx a a a or R b uxv xdx a R b v xdx a Nowbyhyp othesis, = u x and = u x for some x ;x 2 [a; b] . By the Intermediate Value Theorem 1 2 1 2 for Continuous Functions, for anynumber between and there must exist a p oint 2 [a; b] such that u = . In particular, there must exist a p oint 2 [a; b] such that R b uxv xdx a u = R b v xdx a Hence there exists a p oint 2 [a; b] such that Z Z b b v xdx uxv xdx = u a a Remark 1.7. The usual Mean Value Theorem; i.e., the statement that if f xiscontinuous and di eren- tiable on [a; b] then there is a p oint 2 [a; b] such that f b f a 0 f = b a 0 is a sp ecial case of the formulation ab ove. To see this, note that if we take ux=f x and v x = 1, then the ab ove theorem implies Z Z b b 0 0 f xdx = f dx a a Carrying out the integrations on b oth sides yields f b f a 0 0 f b f a=f b a f = b a n n+1 Theorem 1.8. Taylor's Theorem with Lagrange Remainder. If f 2 C [a; b] and f x exists on a; b then for any point x and x in [a; b] o n X 1 k k f x= f x x x + E x o o n k ! k =0 where 1 n+1 n+1 E x= f x x n o n + 1! for some point between x and x : o Proof. According to Taylor's Theorem with Integral Remainder is Z n x X 1 1 k k n+1 n f x= f x x x + f tx t dt o o k ! n! x o k =0 By the Mean Value Theorem for Integrals Z Z x x 1 1 1 n+1 n+1 n+1 n+1 n n f f x x f tx t dt = x t dt = o n! n! n + 1! x x o o 1. PROBLEMS 4 for some between x and x . Hence, o n X 1 1 n+1 k k n+1 f x= f x x x + f x x o o o k ! n + 1! k =0 for some between x and x . o Remark 1.9. Note that in this formulation of Taylor's theorem the error term is still not precisely deter- mined; b ecause the p oint in the theorem statement is left undetermined. We know only that there is some p oint 2 [a; b] such that n X 1 1 n+1 k k n+1 f x x x = f x x f x o o o k ! n + 1! k =0 However, what we gain in this formulation are solid b ounds on the size of the error term over a range of x. n+1 To see this let M b e the maximum value of f x on the interval [a; b], then n X M 1 n+1 k k f x x x jx x j f x o o o k ! n + 1! k =0 Note that this b ound is indep endent of the choice of x and x in [a; b]. o n+1 Corollary 1.10. Alternative Form of Taylor's Theorem.Iff 2 C [a; b], then for any points x and x + h 2 [a; b], n X 1 k k f x + h= f xh + E h n k ! k =0 where 1 n+1 n+1 E h= f h n n + 1! in which the point lies between x and x + h. 1. Problems 1.1. Given that n d n1 ln jxj =1 n 1! n dx x=1 a Use the Taylor Theorem with Integral Remainder to nd the magnitude of the error term R 1:99 100 when one approximates ln[1:99] using the rst 101 terms of the Taylor expansion ab out 1 of ln jxj.
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