lecture3

10 0 x y 0 claim the directional derivative du f 0

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Unformatted text preview: 5), consider the function f : R2 → R defined by �2 xy (x, y ) = (0, 0) � 2 4, f (x, y ) = x +y (2.10) 0, x = y = 0. Claim. The directional derivative Du f (0) exists for all u. Proof. Let u = (h, k ). Then lim t→0 f (tu) − f (0) f (tu) = lim t→0 t �t 3 2 � t hk 1 = lim 2 2 4k4 t→0 t h +t t � 0, h=0 = 2 k /h, h �= 0. (2.11) So the limit exists for every u. However, the function is a non­zero constant on a parabola passing through the 4 origin: f (t2 , t) = 2tt4 = 1 , except at the origin where f (0, 0) = 0. The function f 2 is discontinuous at the origin despite the existence of all directional derivatives. • Guess 3. This guess will turn out to be correct. Remember than in one­dimensional calculus we defined f (a + t) − f (a) , t→0 t f � (a) = lim (2.12) for a function f : I → R and a point a ∈ I . Now consider the function λ : R → R defined by λ(t) = f � (a)t. (2.13) 2 Then, f (a + t) − f (a) − λ(t) f (a + t) − f (a) − f � (a) = lim t→0 t→0 t t = 0. lim (2.14) So, λ(t) ≈ f (a + t) − f (a) when t is small. Now we generalize to n dimensions. Definition 2.3. Given an open subset U of Rn...
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This note was uploaded on 10/02/2010 for the course MAT unknown taught by Professor Unknown during the Fall '04 term at MIT.

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