Rezolvate 3

# Rezolvate 3 - Problema 1 Fie diagrama Venn din figur˘ a...

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Unformatted text preview: Problema 1 Fie diagrama Venn din figur˘ a. Ar˘ atat¸i c˘ a funct¸iile de la u 1 la u 9 , definite mai jos, sunt corect pozit¸ionate ˆ ın diagram˘ a. Toate funct¸iile sunt nule pentru t < 0. u 1 ( t ) = 1 / √ t, dac˘ a t ≤ 1 , dac˘ a t > 1 u 2 ( t ) = 1 /t 1 / 4 , dac˘ a t ≤ 1 , dac˘ a t > 1 u 3 ( t ) = 1 u 4 ( t ) = 1 / (1 + t ) u 5 ( t ) = u 2 ( t ) + u 4 ( t ) u 6 ( t ) = 0 u 7 ( t ) = u 2 ( t ) + 1 Pentru u 8 , consider˘ am v k ( t ) = k, dac˘ a k < t < k + k- 3 , ˆ ın rest ¸ si definim u 8 ( t ) = ∞ X 1 v k ( t ) ˆ In fine, fie u 9 = 1 ˆ ın intervalele [2 2 k , 2 2 k +1 ] , k = 0 , 1 , 2 , ... ¸ si zero ˆ ın rest. Solut¸ie Prin calcul direct, aplicˆ and definit¸iile normelor sau seminormelor cerute a fi calculate, obt¸inem succesiv: k u 1 k 1 = 2 k u 1 k 2 = ∞ k u 1 k ∞ = ∞ pow ( u 1 ) = ∞ k u 2 k 1 = 4 3 k u 2 k 2 = 2 k u 2 k ∞ = ∞ pow ( u 2 ) = 0 1 k u 3 k 1 = ∞ k u 3 k 2 = ∞ k u 3 k ∞ = 1 pow ( u 3 ) = 1 √ 2 k u 4 k 1 = ∞ k u 4 k 2 = 1 k u 4 k ∞ = 1 pow ( u 4 ) = 0 k u 5 k 1 = ∞ k u 5 k 2 = 3 k u 5 k ∞ = ∞ pow ( u 5 ) = 0 k u 6 k 1 = 0 k u...
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## This note was uploaded on 11/05/2011 for the course SYSTEM THE 14 taught by Professor Oaracristian during the Spring '11 term at UPB Colombia.

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Rezolvate 3 - Problema 1 Fie diagrama Venn din figur˘ a...

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