extensionset7

extensionset7 - Extension Set 7 Lucas Manuelli February 25,...

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Extension Set 7 Lucas Manuelli February 25, 2007 Solution to Problem 1 In each iteration the number of circles is denoted by a n = n ( n +1) 2 We also know that one side of the triangle has length 1. Thus 1=2 r n ( n 1) + 2 3 r n 1= r n 2 ³ n 1+ 3 ´ r n = 1 2 ¡ n 3 ¢ So the area of a circle circle in any particular iteration is π 4 " 1 ¡ n 3 ¢ # 2 So A n = n ( n 2 π 4 " 1 ¡ n 3 ¢ # 2 And lim n →∞ A n = lim n →∞ n ( n 2 π 4 " 1 ¡ n 3 ¢ # 2 = π 8 Solution to Problem 2 Let a 1 be the right hand side of the topmost block. And a 1 =0 . Let left be the positive direction. So the center of gravity of block a 1 is a 1 + 1 2 = 1 2 . And this in turn tells us where block a 2 should start. And the block a 3 should start at the average of the center of gravities of a 1 and a 2 . So a 3 = a 1 + 1 2 + a 2 + 1 2 2 = 3 4 And in turn a n +1 = a 1 + a 2 + ··· + a n + n 2 n = a 1 + a 2 + + a n n + 1 2 1
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To see the pattern let us write out the f rst few terms in the sequence a 1 =0 a 2 = 1 2 = a 1 + 1 2 a 3 = 3 4 = a 2 + 1 4 a 4 = 11
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This note was uploaded on 05/03/2008 for the course ENG 111 taught by Professor Josephs during the Fall '07 term at MATC Madison.

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extensionset7 - Extension Set 7 Lucas Manuelli February 25,...

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