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18.100practicefinalsolutions

# 18.100practicefinalsolutions - 18.100B/C Fall 2008...

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18.100B/C: Fall 2008 Solutions to Practice Final Exam 1. Suppose for sake of contradiction that x > 0. Then 1 2 · x > 0 because the product of two positive quantities is positive. Thus x 2 + 0 < x 2 + x 2 (because y < z implies x + y < x + z for all x ), i.e., x 2 < x . Also, since x 2 > 0 we have by the givens that x x 2 . But the results of the previous two sentences are a contradiction with trichotomy. Thus x ± > 0. Since x 0, this implies x = 0, as desired. 2.(a) Let s k = k n =1 a n . Then s k = a 1 2 + a k 2 + k - 1 X n =1 a n +1 + a n 2 > k - 1 X n =1 a n a n +1 , where the last inequality follows because a n + a n +1 2 - a n a n +1 = 1 2 ( a n - a n +1 ) 2 0 . But then n =1 a n is an upper bound on the partial sums of n =1 a n a n +1 . Thus these partial sums are an increasing sequence bounded above and so they converge, as desired. (b) Deﬁne a 0 = a 1 and let b n = a n a n - 1 for n N . Then n =1 b n converges and b n a n because a n is decreasing, so n =1 converges by the comparison test. (Challenge: can you ﬁnd a counter-example to the converse when ( a n ) is positive but not necessarily decreasing?) 3.(a) Both f ( x ) = 4 x (1 - x ) and f ( x ) = 1 - | 2 x - 1 | work nicely. (b) No function: continuous functions take connected sets to connected sets. (c) Deﬁne f ( x ) = ( 0 , x 1 , 1 , x 2 . This function is continuous on [0 , 1] [2 , 3] and f ([0 , 1] [2 , 3]) = { 0 , 1 } . (d) No function: suppose such a function f exists. There exists x 1 for which f ( x 1 ) = 1 and x 2 for which f ( x 2 ) = 2, so by the Intermediate Value Theorem there is x between x 1 and x 2 for which f ( x ) = 2, a contradiction. (Or, use connectedness again.) (e) No function: continuous functions take compact sets to compact sets. 4. By uniform convergence of ( f n ), we can choose some N N such taht n N implies | f n ( x ) - f ( x ) | < ε 3 for all x E . By uniform continuity of f N , we can choose some δ such that d ( x, y ) < δ implies | f N ( x ) - f N ( y ) | < ε 3 . Then for any x, y E such that d ( x, y ) < δ we have | f ( x ) - f ( y ) | = | f

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18.100practicefinalsolutions - 18.100B/C Fall 2008...

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