Practice Test 2b Solutios

Practice Test 2b Solutios - Solution to Practice Test IIB...

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Solution to Practice Test IIB I: a) Has a solution since f ( x ) is continuous, f (2 ) = 2 /π > 1 / 2 and f (3 / (2 π )) = - 3 / (2 π ) < 1 / 2. b) False c) False, in fact we always have (1 + a ) 1 /n 1 + a/n . d) True e) True II: a) 1 / 2, b) 2, c) 0, d) 1 /e 2 , e) 0 III: First start with some hypothetical considerations. Solving the equation c = 1 - (1 - c ) 2 the limit will be presumably either 0 or 1. a) The sequence a n 1 and hence is bounded above. The sequence is also bounded below by zero. Clearly a 1 = 1 / 2 > 0. Assume that a n 0 for some n then a n +1 = 1 - (1 - a n ) 2 0 since 1 a n 0. b) Is the sequence monotone decreasing? a n - a n +1 = a n - 1 + (1 - a n ) 2 = ( a n - 1)(2 - a n ) 0 so a n is not decreasing but increasing. c) Therefor the sequence converges and has a limit which we call c . Clearly, since the function 1 - (1 - x ) 2 is continuous we have that c = 1 - (1 - c ) 2 which has the solution c = 1 since c > 1 / 2. 0 1
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Practice Test 2b Solutios - Solution to Practice Test IIB...

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