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mathCompStat4-07-draft

# mathCompStat4-07-draft - ARE 211 FALL 2007 LECTURE#27 TUE...

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Preliminary draft only: please check for Fnal version ARE 211, FALL 2007 LECTURE #27: TUE, DEC 4, 2007 PRINT DATE: AUGUST 21, 2007 (COMPSTAT4) AND LECTURE #28: DECEMBER 05, 2007 Contents 7. Implicit Function Theorem and the Envelope Theorem (cont) 1 7.5. Inverse function theorem 1 7. Implicit Function Theorem and the Envelope Theorem (cont) 7.5. Inverse function theorem The inverse function theorem is a special case of the implicit function theorem. It applies to the case where n = m , i.e., same number of α ’s as x ’s and f ( α ; x ) = α - η ( x ), i.e., for each i , f i = α i - η i ( x ) or α i = η i ( x ). Theorem: Given η : R m R m and ¯ α R m , If the determinant of α ) is not zero, then there exists a neighborhood of ¯ α and a di±erentiable function η - 1 : R m R m such that on this neighborhood η - 1 ( x ) = x and ( η - 1 ) 1 ( x ) ∂x 1 ( η - 1 ) 1 ( x ) ∂x 2 ··· ( η - 1 ) 1 ( x ) ∂x m ( η - 1 ) 2 ( x ) ∂x 1 ( η - 1 ) 2 ( x ) ∂x 2 ··· ( η - 1 ) 2 ( x ) ∂x m . . . . . . . . . . . . ( η - 1 ) m ( x ) ∂x 1 ( η - 1 ) m ( x ) ∂x 2 ··· ( η - 1 ) m ( x ) ∂x m = p α ) P - 1 1

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2LECTURE #27: TUE, DEC 4, 2007 PRINT DATE: AUGUST 21, 2007 (COMPSTAT4) AND LECTURE #28: DECEMBER 05, We’ll do it intuitively frst, in one dimension. Suppose you have a Function η : R 1 R 1 that is invertible, ( x = η ( α )); i.e., every point in the range is associated with a unique point in the domain. In this case, you can certainly write α as a Function oF x ; that is, defne the Function η - 1 : R 1 R 1 such that η - 1 ( x ) picks out the α value that η took to x ; mathematically, η - 1 is defned by the condition that η - 1 ( η ( α )) = α The closed Form oF the inverse may be very hard to compute, e.g., suppose x = exp α × r sin( α ) / cos( α ). Could in principle defne α as a Function oF x , but it could get messy. Easier to use the inverse Function theorem, which says that ∂η - 1 ( · ) /∂x = ( ∂η ( · ) /∂α ) - 1 .
ARE 211, FALL 2007 3 Example: x = η ( α ) = 1 , so that ∂η ( · ) /∂α = - 1 2 ; frst we’ll take the derivative oF the inverse by hand: we have α = η - 1 ( x ) = 1 /x , ∂η - 1 ( · ) /∂x = - 1 /x 2 ; now use the inverse Function theorem: ∂η ( · ) /∂α = - 1 2 ; applying the inverse Function theorem ∂η - 1 ( · ) /∂x = ( ∂η ( · ) /∂α ) - 1 = ( - 1 2 ) - 1 = - α 2 ; substitute x For α to obtain ∂η - 1 ( · ) /∂x = -

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mathCompStat4-07-draft - ARE 211 FALL 2007 LECTURE#27 TUE...

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