204Lecture122007

204Lecture122007 - Economics 204 Lecture 12–Tuesday...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

This note was uploaded on 03/30/2009 for the course ECON 0204 taught by Professor Staff during the Summer '08 term at Berkeley.

Page1 / 10

204Lecture122007 - Economics 204 Lecture 12–Tuesday...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online