math-2011-note13

# Math-2011-note13 - DF x y x = However g prime = lim t → F 1 t x t y F x t ≥ lim t → 1 t F x tF y F x t = F y F x> a contrdiction 17.3

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Unformatted text preview: DF ( x * )( y- x * ) = . However, g prime ( ) = lim t → F (( 1- t ) x * + t y ))- F ( x * ) t ≥ lim t → ( 1- t ) F ( x * )+ tF ( y )- F ( x * ) t = F ( y )- F ( x * ) > , a contrdiction. 17.3 Calculus Criteria for Concavity ( § 21.1) Theorem 63 (Thm 21.3(p.511) and Thm 21.5(p.513)) . Let F : U → R 1 be a C 2 func- tion defined on a convex open set U ⊂ R n . Then the following conditions are equiva- lent. (i) F is a concave function on U , (ii) F ( y )- F ( x ) ≤ DF ( x )( y- x ) for all x , y ∈ U , (iii) D 2 F ( x ) is negative semidefinite for all x ∈ U . 18 Implicit Functions and Their Derivatives (ch.15) 18.1 Level Curves Definition 38. Given a function G : R 2 → R 1 , a level curve of G is the set of ( x , y ) satisfying G ( x , y ) = z , for some value z ∈ R 1 . Example: Q = 4 L 1 4 K 3 4 Figure 24: Graph and level curves of Q = 4 L 1 4 K 3 4 53 18.2 Implicit Function Theorem What is the slope of a level curve? • Explicit function: y = F ( x 1 , ··· , x n ) • Implicit function: G ( x 1 , ··· , x n , y ) = Example 15. x 2 + y 2 = 1 . Near ( , 1 ) one can find y for each x that satisfies x 2 + y 2 = 1 ; y = √ 1- x 2 . But near ( 1 , ) , one cannot represent y as a function of x . • Suppose that ( x , y ) solves G ( x , y ) = c and that there is a C 1 solution y ( x ) to the equation G ( x , y ) = c so that G ( x , y ( x )) = c holds for ( x , y ( x )) near ( x , y ) . • Differentiating G ( x , y ( x )) = c at x using the chain rule, we obtain ∂ G ∂ x ( x , y ( x )) dx dx + ∂ G ∂ y ( x , y ( x )) dy dx ( x ) = ∂ G ∂ x ( x , y ( x ))+ ∂ G ∂ y ( x , y ( x )) y prime ( x ) = • Solving for y prime ( x ) , we have y prime ( x ) =- ∂ G ∂ x ( x , y ) ∂ G ∂ y ( x , y ) . • The necessary condition for the existence of y ( x ) is that ∂ G ∂ y ( x , y ) negationslash = . Theorem 64 (15.1 Implicit Function Theorem, p.339) . Let G ( x , y ) be a C 1 function on a ball about ( x , y ) ∈ R 2 . Suppose that G ( x , y ) = c. If ∂ G ∂ y ( x , y ) negationslash = , then there exists a C 1 function y = y ( x ) defined on an interval I about x such that (a) G ( x , y ( x )) = c for all x ∈ I , (b) y ( x ) = y , (c) y prime ( x ) =- ∂ G ∂ x ( x , y ) / ∂ G ∂ y ( x , y ) . Example 16. G ( x , y ) = x 2- 3 xy + y 3 = 7 , ( x , y ) = ( 4 , 3 ) • ∂ G ∂ x ( x , y ) = 2 x- 3 y , ∂ G ∂ y ( x , y ) =- 3 x + 3 y 2 • ∂ G ∂ x ( 4 , 3 ) =- 1 , ∂ G ∂ y ( 4 , 3 ) = 15 , DG ( 4 , 3 ) = (- 1 , 15 ) • y prime ( 4 ) =- ∂ G ∂ x ( 4 , 3 ) / ∂ G ∂ y ( 4 , 3 ) = 1 15 54 Figure 25: Slope of a level curve • The change in G in the neighborhood of ( x , y ) is G ( x + Δ x , y + Δ y )- G ( x , y ) ≈ ∂ G ∂ x ( x , y ) Δ x + ∂ G ∂ y ( x , y ) Δ y ....
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## This note was uploaded on 03/26/2012 for the course ECON 205 taught by Professor Mr.lee during the Spring '11 term at Korea University.

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Math-2011-note13 - DF x y x = However g prime = lim t → F 1 t x t y F x t ≥ lim t → 1 t F x tF y F x t = F y F x> a contrdiction 17.3

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