21-356 Lecture 8

m ulti index 0m where lim xx0 1 f x0 x

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Unformatted text preview: ous together with # $m their partial derivatives up to order m. We set C # (U ) := C (U ). m=0 Theorem 30 (Taylor’s Formula) Let U $ RN be an open set, let f " C m (U ), m " N, and let x0 " U . Then for every x " U , f (x) = ! " m ulti-index, 0$|"|$m where lim x%x0 1 ""f " (x0 ) (x # x0 ) + Rm (x) , !! " x" Rm (x) m = 0. %x # x0 % Example 31 Let’s calculate the limit y (1 + x) # 1 % . (x,y )%(0,0) x2 + y 2 lim By substituting we get 0 . Consider the function 0 y y f (x, y ) = (1 + x) # 1 = elog(1+x) # 1 = ey log(1+x) # 1, & ' which is defined in the set U := (x, y ) " R2 : 1 + x > 0 . The function f is of class C # . Let’s use Taylor’s formula of order m = 1 at (0, 0), f (x, y ) = f (0, 0) + (% ) "f "f (0, 0) (x # 0) + (0, 0) (y # 0) + o x2 + y 2 . "x "y 19 We have "f (x, y ) = "x "f (x, y ) = "y ) " ( y log(1+x) 1 e # 1 = ey log(1+x) y , "x 1+x ( ) " ey log(1+x) # 1 = ey log(1+x) y log (1 + x) , "y and so f (x, y ) = 0 + 0 (x # 0) + 0 (y # 0) + o which means that (% ) x2 + y 2 , (%...
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