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9 Pages

### TAYLORnNEWTON

Course: MATH 135, Fall 2008
School: Allan Hancock College
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Word Count: 456

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Notes Math135 - Error Estimates C. Meaney April 2006 Taylors Formula Suppose that f is repeatedly dierentiable on an interval J that contains x and a. Dene F (t) = f (x ) f (t) f (t)(x t). Notice that F (x ) = 0 and F (t) = (x t)f (t). Now let G(t) = F (t) This has G(x ) = 0 = G(a) and G (t) = F (t) + 2(x t) F (a). (x a)2 (x t)2 F (a). (x a)2 The Mean Value Theorem says that there is a point c between...

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Notes Math135 - Error Estimates C. Meaney April 2006 Taylors Formula Suppose that f is repeatedly dierentiable on an interval J that contains x and a. Dene F (t) = f (x ) f (t) f (t)(x t). Notice that F (x ) = 0 and F (t) = (x t)f (t). Now let G(t) = F (t) This has G(x ) = 0 = G(a) and G (t) = F (t) + 2(x t) F (a). (x a)2 (x t)2 F (a). (x a)2 The Mean Value Theorem says that there is a point c between x and a where G (c) = 0. At this point F (c) = (x c)f (c) = This gives (x a)2 f (c), 2 and going back to the denition of F , shows that F (a) = f (x ) = f (a) + f (a)(x a) + (x a)2 f (c). 2 2(x c) F (a). (x a)2 Estimating the error in linear approximations Suppose that J is an interval containing x and a. If we use a linear approximation to f (x ) based around a, then the error is E (x ) = f (x ) f (a) + f (a)(x a) and we see from Taylors formula that |E (x )| where M = max f (c) cJ M |x a|2 , 2 Example If we want to approximate around 9/4, let 2 using a linear approximation based x, f (x ) = a = 9/4, and J = [2, 9/4]. The derivatives are f (x ) = x 1/2 /2 and f (x ) = x 3/2 /4. In particular, M = max The linear approximation is 2 3 1 2 9 + ( )(2 ) = 1.41666 . . . 2 2 3 4 1 1 c 3/2 = < . 4 8 2c9/4 8 2 and the error is less than M 9 2 2 4 2 < 1 16 2 = . 0.003906 . . Estimating Newtons Method The case where r is a solution of the equation f (x ) = 0. 0 = f (r ) = f (x ) + f (x )(r x ) + for some c between r and x . If f (r ) = 0 and f (x ) = 0, Taylors Formula tells us that f (x ) (r x )2 f (c) +r x = , f (x ) 2f (x ) and so r x f (x ) f (x ) = (r x )2 f (c) . 2f (x ) (r x )2 f (c) 2 Replace x with xn , and use the Newton recurrence xn+1 = xn f (xn ) f (xn ) Theorem If J is an interval containing r , xn , and xn+1 with f (r ) = 0 and xn+1 = xn f (xn ) , f (xn ) f (x ) K , f (x ) L, then |xn+1 r | for x J, for x J, K |xn r |2 . 2L Example Let f (x ) = x 2 2 and J = [1, 2]. Since f (1) = 1 < 0 f (2) = 2 > 0, there is a point and 1 < 2 < 2 where f ( 2) = 0. The derivatives satisfy: f (x ) ...

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