TaylorRemainderProof

TaylorRemainderProof - ( x # ) n # 1 \$ % & ' ( ) # R N...

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Proof of the Taylor Remainder Theorem In this post we give a proof of the Taylor Remainder Theorem. It is a very simple proof and only assumes Rolle’s Theorem. Rolle’s Theorem. Let f ( x ) be differentiable on [ a , b ] and suppose that f ( a ) = f ( b ) . Then there is a point a < ! < b such that ! f ( " ) = 0 . Taylor Remainder Theorem. Suppose that f ( x ) is ( N + 1) times differentiable on the interval [ a , b ] with a < x 0 < b . Let a < x < b , then there is a point between x 0 and x such that the following holds. f ( x ) = f ( a ) + ! f ( a )( x " a ) + !! f ( a ) 2 ( x " a ) 2 + ! + f ( N ) ( a ) N ! ( x " a ) N + f ( N + 1) ( # ) ( N + 1)! ( x " a ) N + 1 = f ( n ) ( a ) n ! ( x " a ) n n = 0 N \$ + f ( N + 1) ( ) ( N + 1)! ( x " a ) N + 1 Proof. Let R be such that f ( x ) = f ( n ) ( a ) n ! ( x ! a ) n n = 0 N " + R ( N + 1)! ( x ! a ) N + 1 . Define F ( ) by the following formula. F ( ) = f ( n ) ( ) n ! ( x " ) n + R ( N + 1)! n = 0 N # ( x " ) N + 1 Then we have the following. ! F ( ) = ! f ( ) + f ( n + 1) ( ) n ! ( x # ) n # f ( n ) ( ) ( n # 1)!
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Unformatted text preview: ( x # ) n # 1 \$ % & ' ( ) # R N ! ( x # ) N n = 1 N * = f ( N + 1) ( ) N ! ( x # ) N # R N ! ( x # ) N = ( x # ) N N ! ( f ( N + 1) ( ) # R ) Clearly F ( x ) = F ( x ) = f ( x ) . So, by Rolle’s Theorem there is a point between x and x such that ! F ( ) = . At that point , we have R = f ( N + 1) ( ) and thus we have f ( x ) = f ( n ) ( a ) n ! ( x ! a ) n n = N " + f ( N + 1) ( ) ( N + 1)! ( x ! a ) N + 1 . This is what was to be proved....
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This note was uploaded on 07/14/2011 for the course MAC 3473 taught by Professor Block during the Spring '08 term at University of Florida.

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