Adv Alegbra HW Solutions 8

# Adv Alegbra HW - g j x 1.16 Prove for every n ∈ N that 1 x n ≥ 1 nx whenever x ∈ R and 1 x> 0 Solution We prove the inequality by induction

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8 A 1 A 3 A 2 1 2 1 4 1 8 1 B 0 B 1 B 2 B 3 1 1 1 1 Figure 1.4 Oresme s Dissections 1.15 Let g 1 ( x ),. .., g n ( x ) be differentiable functions, and let f ( x ) be their prod- uct: f ( x ) = g 1 ( x ) ··· g n ( x ) . Prove, for all integers n 2, that the deriva- tive f 0 ( x ) = n X i = 1 g 1 ( x ) ··· g i 1 ( x ) g 0 i ( x ) g i + 1 ( x ) ··· g n ( x ). Solution. Base step .If n = 2, this is the usual product rule for derivatives. Inductive step .De f ne h ( x ) = g 1 ( x ) ··· g n ( x ) = f ( x )/ g n + 1 ( x ) . Rewrite what has to be shown: f 0 ( x ) = n + 1 X j = 1 g 0 j ( x ) f ( x ) g j ( x ) . Now f 0 ( x ) = ( h ( x ) g n + 1 ( x )) 0 = h 0 ( x ) g n + 1 ( x ) + h ( x ) g 0 n + 1 ( x ) = n X = 1 i = 1 · g 0 i ( x ) h ( x ) g i ( x ) ¸ g n + 1 ( x ) + · f ( x ) g n + 1 ( x ) ¸ g 0 n = n + 1 X j = 1 g 0 j ( x ) f ( x
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Unformatted text preview: ) g j ( x ) . 1.16 Prove, for every n ∈ N , that ( 1 + x ) n ≥ 1 + nx whenever x ∈ R and 1 + x > 0. Solution. We prove the inequality by induction on n ≥ 1. The base step n = 1 says 1 + x ≥ 1 + x , which is obviously true. For the inductive step,...
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## This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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