204Lecture122009web

# 204Lecture122009web - Economics 204 Lecture 12Tuesday...

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Economics 204 Lecture 12–Tuesday, August 11, 2009 Inverse and Implicit Function Theorems, and Generic Methods: Section 4.3 (Conclusion), Regular and Critical Points and Values: Defnition 1 Suppose X R n is open. Suppose f : X R m is diferentiable at x X ,andl e t W = { e 1 ,...,e n } denote the standard basis oF R n .Th en df x L ( R n , R m ), and Rank x =d im I m ( ) s p a n { x ( e 1 ) ,...,df x ( e n ) } s p a n { Df ( x ) e 1 ,...,Df ( x ) e n } s p a n { column 1 oF Df ( x ) ,..., co lumnno F Df ( x ) } =R a n k Df ( x ) Thus, Rank ( x ) min { m, n } We say x is a regular point oF f iF Rank ( x )=m in { m, n } . x is a critical point oF f iF Rank ( x ) < min { m, n } . y is a critical value oF f iF there exists x X , f ( x )= y , x is a critical point oF f . y is a regular value oF f iF y is not a critical value oF f (notice this has the counterintuitive implication that iF y 6∈ f ( X ), then y is automatically a regular value oF f ). 1

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Remark: The defnition oF regular point and critical point in de la ±uente (as well as in Mas-Colell, Whinston, and Green) is di²erent: they use m rather than min { m, n } . I think the defnition I have given is more natural. IF m n , the two are equivalent. IF m>n , then since Rank ( df x ) min { m, n } ,then every x X will be a critical point in the de la ±uente and MWG defnitions, and every y f ( X ) will be a critical value. In the defnition I have given, a point is critical iF the rank is smaller than the largest it could possibly be. The two important theorems (Sard’s Theorem and the Transversality Theorem) concerning critical values are true with either defnition. Example: Consider the Function f :(0 , 2 π ) R defned by f ( x )=s in x Then f 0 ( x )=co s x ,so f 0 ( x )=0Fo r x = π/ 2and x =3 2. Df ( x )i sth e1 × 1 matrix ( f 0 ( x )), so Rank x =Rank Df ( x ) = 1 iF and only iF f 0 ( x ) 6 = 0. Thus, the critical points oF f are 2and3 2, so the set oF regular points oF f is (0 ,π/ 2) ( 2 , 3 2) (3 2 , 2 π ) The critical values oF f are f ( 2) = sin( 2) = 1 and f (3 2) = sin(3 2) = 1; the set oF regular values oF f is ( −∞ , 1) ( 1 , 1) (1 , ) Notice that 0 is not a critical value. Given α R , consider the perturbed Function f α ( x )= f ( x )+ α Notice that f 0 α ( x f 0 ( x ), so the critical points oF f α are the same as those oF f o r α close to zero, the solution to the equation f α ( x )=0 2

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near x = π moves smoothly with respect to changes in α ; the direction a solution moves is determined by the sign of f 0 α . Now, let α =1 . f 1 ( x )=0 sin x +1=0 sin x = 1 x = 3 π 2 Since 3 π/ 2 is a critical point of f 1 , 0 is a critical value of f 1 .
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204Lecture122009web - Economics 204 Lecture 12Tuesday...

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