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Macroeconomics Exam Review 215

# Macroeconomics Exam Review 215 - Solutions for Foundations...

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Note that 𝑔 ( 𝑥 1 ) = 𝑓 ( 𝑥 1 ) and 𝑔 ( 𝑥 0 ) = 𝑓 ( 𝑥 0 ) + 𝑓 [ 𝑥 0 ]( 𝑥 1 𝑥 0 ) + 1 2 𝑓 ′′ [ 𝑥 0 ]( 𝑥 1 𝑥 0 ) 2 + 1 3! 𝑓 (3) [ 𝑥 0 ]( 𝑥 1 𝑥 0 ) 3 + . . . + 1 𝑛 ! 𝑓 ( 𝑛 +1) [ 𝑥 0 ]( 𝑥 1 𝑥 0 ) 𝑛 + 𝑎 𝑛 +1 ( 𝑥 1 𝑥 0 ) 𝑛 +1 (4.51) is a polynomial approximation for 𝑓 near 𝑥 0 . If we require that 𝑎 𝑛 +1 be such that 𝑔 ( 𝑥 0 ) = 𝑓 ( 𝑥 1 ) = 𝑔 ( 𝑥 1 ), there exists (Theorem 4.1) some ¯ 𝑥 between 𝑥 0 and 𝑥 1 such that 𝑔 ( 𝑥 1 ) 𝑔 ( 𝑥 0 ) = 𝑔 𝑥 )( 𝑥 1 𝑥 0 ) = 0 which for 𝑥 1 = 𝑥 0 implies that 𝑔 𝑥 ) = 1 𝑛 ! 𝑓 𝑛 +1 𝑥 ] ( 𝑛 + 1) 𝑎 𝑛 +1 = 0 or 𝑎 𝑛 +1 = 1 ( 𝑛 + 1)! 𝑓 𝑛 +1 𝑥 ] Setting 𝑥 = 𝑥 1 𝑥 0 in (4.51) gives the required result. 4.58 By Taylor’s theorem (Exercise 4.57), for every 𝑥 𝑆 𝑥 0 , there exists ¯ 𝑥 between 0 and 𝑥 such that 𝑓 ( 𝑥 0 + 𝑥 ) = 𝑓 ( 𝑥 0 ) + 𝑓 [ 𝑥 0 ] 𝑥 + 1 2 𝑓 ′′ [ 𝑥 0 ] 𝑥 2 + 𝜖 ( 𝑥 ) where 𝜖 ( 𝑥 ) = 1 3! 𝑓 (3) 𝑥 ] 𝑥 3 and 𝜖 ( 𝑥 ) 𝑥 2 = 1 3! 𝑓 (3) 𝑥 ]( 𝑥 ) Since 𝑓 𝐶 3 , 𝑓 (3) 𝑥 ] is bounded on [0 , 𝑥
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