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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.
 Spring '08
 BLOCK
 Calculus, Remainder Theorem, Remainder

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