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Unformatted text preview: Economics 204 Lecture 12–Tuesday, August 14, 2007 Revised 8/14/07, revisions indicated by ** Section 4.4 (Cont.): Taylor’s Theorem in R n Definition 1 X ⊆ R n , X open, f : X → R m . f is continu ously differentiable on X if • f is differentiable on X and • df x is a continuous function of x from X to L ( R n , R m ), with operator norm k df x k f is C k if all partial derivatives of order ≤ k exist and are contin uous in X . Theorem 2 (4.3) Suppose X ⊆ R n , X open, f : X → R m . Then f is continuously differentiable on X if and only if f is C 1 . Notational Problem in Taylor’s Theorem: If f : R n → R m , the quadratic terms are OK for m = 1; for m > 1, handle each component separately. For cubic and higher order terms, notation is a mess. Linear Terms: Theorem 3 Suppose X ⊆ R n , X is open, x ∈ X . If f : X → R m is differentiable, then f ( x + h ) = f ( x ) + Df ( x ) h + o ( h ) as h → The previous theorem is essentially a restatement of the definition of differentiability. 1 Theorem 4 (Corollary of 4.4) Suppose X ⊆ R n , X is open, x ∈ X . If f : X → R m is C 2 , then f ( x + h ) = f ( x ) + Df ( x ) h + O  h  2 as h → Quadratic Terms: Treat each component of the function separately, so consider f : X → R , X ⊆ R n an open set. Let D 2 f ( x ) = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ∂ 2 f ∂x 2 1 ∂ 2 f ∂x 2 ∂x 1 ··· ∂ 2 f ∂x n ∂x 1 ∂ 2 f ∂x 1 ∂x 2 ∂ 2 f ∂x 2 2 ··· ∂ 2 f ∂x n ∂x 2 . . . . . . . . . . . . ∂ 2 f ∂x 1 ∂x n ··· ··· ∂ 2 f ∂x 2 n ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ f ∈ C 2 ⇒ ∂ 2 f ∂x i ∂x j = ∂ 2 f ∂x j ∂x i ⇒ D 2 f ( x ) is symmetric ⇒ D 2 f ( x ) has an orthonormal basis of eigenvectors and thus can be diagonalized Theorem 5 (Stronger Version of 4.4) Let X ⊆ R n be open, f : X → R , f ∈ C 2 ( X ) , x ∈ X . Then f ( x + h ) = f ( x ) + Df ( x ) h + 1 2 h T ( D 2 f ( x )) h + o  h  2 as h → If f ∈ C 3 , f ( x + h ) = f ( x ) + Df ( x ) h + 1 2 h T ( D 2 f ( x )) h + O  h  3 as h → 2 Remark: De la Fuente assumes X is convex which he has not yet defined. X is said to be convex if, for every x, y ∈ X and α ∈ [0 , 1], αx + (1 − α ) y ∈ X . We don’t need this. Since....
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 Summer '08
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 Economics, Critical Point, Continuous function, Inverse function, Lebesgue measure, det Df

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