Unformatted text preview: 1 to f q x around x (p). We obtain functions g defined around x (p) such that i1 n d f q x f = f(p) + S (x x (p))g q x and g ( x (p)) =( x (p)) . i i i i d x i=1 i In this case however we have n D(f) = D(f(p)) + S D((x x (p))) g ( x (p)) + (x (p)x (p)) D(g q x ) = i i i i i i i=11 n d f q x n # $ S D(x )( x (p)) = S D(x ) d (p) (f). i d x i 3 i 4 i=1 i i=1 To finish the proof, we only have to show that every derivation at the point p can be obtained as a speed vector of a curve passing through p. Define the curve g :[e , e ]L M by the formula1 g (t) := x ( x (p) + (t a ,...,t a )). 1 n n d Then obviously the speed vector g ’(0) is just S a(p). 44444 i d x i=1 i 5...
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 Spring '09
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 Derivative, i=1, ) dt =

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