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Macroeconomics Exam Review 205

Macroeconomics Exam Review 205 - Solutions for Foundations...

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(4.43) implies that 𝑓 𝑗 ( x 0 + x ) 𝑓 𝑗 ( x 0 ) 𝐷𝑓 𝑗 [ x 0 ] x x 0 as x ∥ → 0 for every 𝑗 . 4.19 If 𝐷𝑓 [ x 0 ] has full rank, then it is one-to-one (Exercise 3.25) and onto (Exercise 3.16). Therefore 𝐷𝑓 [ x 0 ] is nonsingular. The Jacobian 𝐽 𝑓 ( x 0 ) represents 𝐷𝑓 [ x 0 ], which is therefore nonsingular if and only if det 𝐽 𝑓 ( x 0 ) = 0. 4.20 When 𝑓 is a functional, rank 𝑋 𝑟𝑎𝑛𝑘𝑌 = 1. If 𝐷𝑓 [ x 0 ] has full rank (1), then 𝐷𝑓 [ x 0 ] maps 𝑋 onto (Exercise 3.16), which requires that 𝑓 ( x 0 ) = 0 . 4.21 4.23 If 𝑓 : 𝑋 × 𝑌 𝑍 is bilinear 𝑓 ( x 0 + x , y 0 + y ) = 𝑓 ( x 0 , y 0 ) + 𝑓 ( x 0 , y ) + 𝑓 ( x , y 0 ) + 𝑓 ( x , y ) Defining 𝐷𝑓 [ x 0 , y 0 ]( x , y ) = 𝑓 ( x 0 , y ) + 𝑓 ( x , y 0 ) 𝑓 ( x 0 + x , y 0 + y ) = 𝑓 ( x 0 , y 0 ) + 𝐷𝑓 [ x 0 , y 0 ]( x , y ) + 𝑓 ( x , y ) Since 𝑓 is continuous, there exists 𝑀 such that 𝑓 ( x , y ) 𝑀 x ∥ ∥ y for every x 𝑋 and y 𝑌 and therefore NOTE This is not quite right. See Spivak p. 23. Avez (Tilburg) has 𝑓 (
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