Intro to Analysis in-class_Part_37

Intro to Analysis in-class_Part_37 - 5.4 Higher derivatives...

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5.4 Higher derivatives and Taylor’s Thm 79 Proof. Suppose p is a polynomial in ( x - a ): p ( x ) = c 0 + c 1 ( x - a ) + ··· + c n ( x - a ) n . After k differentiations, the terms c 0 ,...,c k - 1 vanish: p ( k ) ( x ) = k ! c k + (terms with ( x - a ) as a factor) . So p ( k ) ( a ) = k ! c k . If f ( x ) and p ( x ) have n th -order agreement, f ( k ) ( a ) = k ! c k = c k = f ( k ) ( a ) k ! , k = 1 ,...,n. Theorem 5.4.5 (Taylor’s Thm) . Suppose f C n +1 ( I ) for some open interval I contain- ing a and x . Then for some c between a and x , f ( x ) = f ( a ) + f 0 ( a )( x - a ) + 1 2 f 00 ( a )( x - a ) 2 + ··· + 1 n ! f ( n ) ( a )( x - a ) n + R n ( x ) , R n ( x ) = f ( n +1) ( c ) ( n + 1)! ( x - a ) n +1 . NOTE: c depends on x , so RHS of f is not a polynomial ; f ( n +1) ( c ) is not a constant. Proof. We show the theorem holds at x = b . Let P be defined by P ( x ) = T n ( a,x ) + C ( x - a ) n +1 , where C is chosen so that f ( b ) = P ( b ), i.e., C = f ( b ) - T n ( a,b
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This note was uploaded on 11/26/2011 for the course MAT 4944 taught by Professor Wong during the Fall '11 term at FSU.

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Intro to Analysis in-class_Part_37 - 5.4 Higher derivatives...

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