Lecture36-rev

Lecture36-rev - MAT A26 Lecture 36 1 Lagrange Form of the...

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Unformatted text preview: MAT A26 Lecture 36 1 Lagrange Form of the Remainder We have already established that if f ( x ) is n + 1 times differentiable on an interval ( a- h, a + h ) then for x ( a- h, a + h ) we have f ( x ) = f ( a ) + f ( a )( x- a ) + + f ( n ) ( a ) ( x- a ) n n ! | {z } f n,a ( x ) + Z x a f ( n +1) ( t ) ( x- t ) n n ! dt | {z } E n,a ( x ) Theorem 1 (Lagrange form of the remainder) If f ( x ) is n + 1 times differ- entiable on an interval ( c, d ) and f ( n +1) ( x ) is continuous on ( c, d ) , then, for a and x in ( c, d ) we have E n,a ( x ) = f ( n +1) ( ) ( x- a ) n +1 ( n + 1)! for some between a and x so that f ( x ) = f ( a ) + f ( a )( x- a ) + + f ( n ) ( a ) ( x- a ) n n ! + f ( n +1) ( ) ( x- a ) n +1 ( n + 1)! Proof: Let us suppose that a < x . (The case x < a is similar.) Since f ( n +1) is continuous on [ a, x ] we let m denote the minimum and M the max- imum values there. (These exist by EVT) Thus m f ( n +1) ( t ) M for all t [ a, x ] and so Z x a m ( x- t ) n n ! dt Z x a f ( n +1) ( t ) ( x- t ) n n ! dt Z x a M ( x- t ) n n ! dt But Z x a ( x- t ) n n ! dt =- ( x- t ) n +1 ( n + 1)! x a = ( x- a ) n +1 ( n + 1)! so m ( x- a ) n +1 ( n + 1)! Z x a f ( n +1) ( t ) ( x- t ) n n !...
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This note was uploaded on 03/04/2012 for the course MATH 10250 taught by Professor Himonas during the Fall '08 term at Notre Dame.

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Lecture36-rev - MAT A26 Lecture 36 1 Lagrange Form of the...

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