solution4 - Solution of homework 4 1. (a) No, the sequence...

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Solution of homework 4 1. (a) No, the sequence { } 1, 1, 1, 1,. .., 1, 1,. .. n a =− + − + − + , will never approach a specific value. It does not blow up, but it oscillates. (b) Yes, the sequence n a is bounded above because 3 2 5, {1,2,3,. ..} n n −≤ ∀∈ . The sequence is increasing since 1 33 3 3 22 11 nn aa n n + >⇒ < −⇒ < ++ By the theorem, n a is increasing and bounded above, then it converges. Obviously, 3 lim lim 2 2 n a n →∞ →∞  =− =   . 2. () 2 * max , uv u v u v u v =≤ + 3. (a) 2 * * 3 2 3 2 3 * 2 3 2 * () () ˆˆ e x p2 e x 1 exp 2 exp 2 1 TT T T T km T T T T T ut d t utu td t tt uj k m d t k u j m d t T uu T ππ −− ∞∞ + =−∞ + = = = ∫∫ ∑∑ 3 2 3 2 3 * 2 3 2 exp 2 1 exp 2 T T T T T T T T t jk md t T t j k m d t π + + where 3 2 3 2 , exp 2 0, T T T T Tk m t t T + = −= Hence, 3 * 2 3 2 ˆ T T T kk k T t u + == (b)
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() 3 * 2 3 2 * 3 2 3 2 3 * 2 3 2 * () () 11 ˆˆ exp 2 exp 2 1 exp 2 exp 2 1 exp 2 T T T T T T T km T T T T T utv td t tt uj k vj m d t TT k v j m d t T t uv j k m ππ π + ∞∞ + =−∞ +  =   =− ∑∑ 3 2 3 2 3 * 2 3 2 * 1 exp 2 T T T T T T T T kk k dt t j k m + + = (c) 3 ** 2 3 2 ()() T T T T k utvt d t + = From triangle inequality, 12 1 2 x xx x
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solution4 - Solution of homework 4 1. (a) No, the sequence...

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