21-356 Lecture 8

x 1 1 x y 1 2f 2 0 0 y 0 o x2 y 2

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Unformatted text preview: ) y o x2 + y 2 (1 + x) # 1 % % lim = lim = 0. (x,y )%(0,0) (x,y )%(0,0) x2 + y 2 x2 + y 2 Note that if we had to calculate the limit y (1 + x) # 1 , x2 + y 2 (x,y )%(0,0) lim then we would need Taylor’s formula of order m = 2 at (0, 0), "f "f (0, 0) (x # 0) + (0, 0) (y # 0) "x "y 1 "2f 1 "2f 2 + (0, 0) (x # 0) + (0, 0) (x # 0) (y # 0) + 2 (2, 0)! " x (1, 1)! " x" y * + 1 "2f 2 + (0, 0) (y # 0) + o x2 + y 2 . 2 (0, 2)! " y f (x, y ) = f (0, 0) + 20 Wednesday, January 26, 2011 Proof of Taylor’s formula. Step 1: Since x0 " U and U is open, there exists B (x0 , r) $ U . We prove Taylor’s formula with Lagrange’s reminder, that is, f (x) = ! 0$|"|<m ! 1 ""f 1 ""f " " (x0 ) (x # x0 ) + (x0 + c (x # x0 )) (x # x0 ) , " !! " x !! " x" |"|=m (6) for all x " B (x0 , r) and where c " (0, 1). Fix x " B (x0 , r), let h := x # x0 and consider the function g (t) := f (x0 + th) defined for t " [0, 1]. By Lemma 21 and Remark 22, we have that N ! "f dg (t) = (x0 + th) hi = (h · &) f (x0 + th) dt " xi i=1 with for all t " [0, 1]. By repeated applications of Lemma 21 and Remark 22, we get that d ( n) g n (t) = (h · &) f (x0 + th) dtn n for all n = 1, . . . , m, where (h · &) means that we apply the operator h · & = h1 " " + · · · + hN " x1 " xN n times to f . By the multinomial theorem, and the fact that for functions in C m partial derivatives commute, " #n " " n (h · &) = h1 + · · · + hN " x1 " xN ! n! " " " = h , !! " x" " multi-index, |"|=n and so d ( n) g (t) = dtn ! " multi-index, |"|=n n! " " " f h (x0 + th) . !! " x" Using Taylor’s formula for g , we get g (1) = g (0) + m!1 ! n=1 1 d ( n) g 1 d(m) g n m (0) (1 # 0) + (c) (1 # 0) n n! dt m! dtm for some c " (0, 1). Substituting, we obtain f (x0 + h) = ! 0$|"|=m!1 ! 1 ""f 1 ""f (x0 ) h" + (x0 + ch) h" . " !! " x !! " x" |"|=m 21 (7)...
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