# aGadX1-07 - MAT 473 17 7 Taylor's Theorem The Inverse...

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MAT 473 17 7. Taylor’s Theorem; The Inverse Function Theorem. Theorem 7.1 (Taylor’s Theorem) . Let U R n be open and convex, and let f : U R be C N . For each a U and each h such that a + h U , there exists c on the line segment from a to a + h such that f ( a + h ) = f ( a ) + N 1 k =1 1 k ! n i 1 ,...,i k =1 D i 1 ··· i k f ( a ) h i 1 · · · h i k + 1 N ! n i 1 ,...,i N =1 D i 1 ··· i N f ( c ) h i 1 · · · h i N . The polynomial p N 1 ( x ) = f ( a ) + N 1 k =1 1 k ! n i 1 ,...,i k =1 D i 1 ··· i k f ( a )( x i 1 a i 1 ) · · · ( x i k a i k ) is called the N 1 st -order Taylor polynomial for f at a . The proof is more complicated orthographically than conceptually. In fact, you can even forget the statement of the theorem if you can apply Taylor’s Theorem in one variable to the function t f ( a + t h ) and then carefully use the Chain Rule in the resulting formula. Proof. Define g : R R n by g ( t ) = a + t h , so that g j ( t ) = h j for each 1 j n . Now put V = g 1 ( U ) (an open set in R which contains [0 , 1]), and consider f g : V R . By the Chain Rule, ( f g ) ( t ) = f ( g ( t )) g ( t ) = n j =1 D j f ( g ( t )) g j ( t ) = n j =1 D j f ( g ( t )) h j , which is a continuous function of t because f is C N (hence C 1 ). We claim that ( f g ) ( k ) exists and is continuous on V for all 1 k N , and that ( f g ) ( k ) ( t ) = n i 1 ,...,i k =1 D i 1 ··· i k f ( g ( t )) h i 1 · · · h i k Date : January 26, 2012 S. Kaliszewski, Department of Mathematics and Statistics, Arizona State University.

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