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204Lecture112009web

# 204Lecture112009web - Economics 204 Lecture 11Monday...

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Economics 204 Lecture 11–Monday, August 10, 2009 Sections 4.1-4.3, Unified Treatment Definition 1 Let f : I R , where I R is an open interval. f is differentiable at x I if lim h 0 f ( x + h ) f ( x ) h = a for some a R . This is equivalent to lim h 0 f ( x + h ) ( f ( x ) + ah ) h = 0 ε> 0 δ> 0 0 < | h | < δ f ( x + h ) ( f ( x ) + ah ) h < ε ε> 0 δ> 0 0 < | h | < δ | f ( x + h ) ( f ( x ) + ah ) | | h | < ε lim h 0 | f ( x + h ) ( f ( x ) + ah ) | | h | = 0 Recall that the limit considers h near zero, but not h = 0 . Definition 2 If X R n is open, f : X R m is differentiable at x X if T x L ( R n , R m ) lim h 0 ,h R n | f ( x + h ) ( f ( x ) + T x ( h )) | | h | = 0 (Recall | · | denotes the Euclidean distance.) f is differentiable if it is differentiable at all x X . 1

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h is a small, nonzero element of R n ; h 0 from any direction, along a spiral, etc. One linear operator T x works no matter how h approaches zero. f ( x ) + T x ( h ) is the best linear approximation to f ( x + h ) for small h Notation: y = O ( | h | n ) as h 0 means K,δ> 0 | h | < δ ⇒ | y | ≤ K | h | n read y is big-Oh of | h | n y = o ( | h | n ) as h 0 means lim h 0 | y | | h | n = 0 read y is little-oh of | h | n Note that the statement y = O ( | h | n +1 ) as h 0 implies y = o ( | h | n ) as h 0. Note that f is differentiable at x ⇔ ∃ T x L ( R n , R m ) f ( x + h ) = f ( x ) + T x ( h ) + o ( h ) as h 0 Notation: df x is the linear transformation T x Df ( x ) is the matrix of df x with respect to the standard basis; called the Jacobian or Jacobian matrix of f at x E f ( h ) = f ( x + h ) ( f ( x ) + df x ( h )) (Error Term) f is differentiable at x E f ( h ) = o ( h ) as h 0 2
Let’s compute Df ( x ) = ( a ij ). Let { e 1 , . . . , e n } be the standard basis of R n . Look in direction e j ; | γe j | = | γ | . o ( γ ) = f ( x + γe j ) ( f ( x ) + T x ( γe j )) = f ( x + γe j ) f ( x ) + a 11 · · · a 1 j · · · a 1 n . . . . . . . . . . . . . . . a m 1 · · · a mj . . . a mn 0 . . . 0 γ 0 . . . 0 = f ( x + γe j ) f ( x ) + γa 1 j . . . γa mj For i = 1 , . . . , m , let f i denote the i th component of the function f : f i ( x + γe j ) f i ( x ) + γa ij = o ( γ ) so a ij = ∂f i ∂x j Theorem 3 (3.3) Suppose X R n is open and f : X R m is differentiable at x X . Then ∂f i ∂x j exists for 1 i m , 1 j n , and ( Df )( x ) = ∂f 1 ∂x 1 · · · ∂f 1 ∂x n .

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204Lecture112009web - Economics 204 Lecture 11Monday...

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