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# 115sols4 - HW 4 Solutions Math 115 Winter 2009 Prof Yitzhak...

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HW 4 Solutions Math 115, Winter 2009, Prof. Yitzhak Katznelson 9.6 Let x 1 = 1 and x n +1 = 3 x 2 n for n 1. a) If a = lim x n , then we have that a = lim x n +1 = lim 3 x 2 n = 3 lim( x n · x n ) = 3(lim x n ) 2 = 3 a 2 , where the later steps use various limit laws. Solving this equation yields two solutions; either a = 0 or a = 1 / 3. b) Yes, the limit exists because x n diverges to . To prove this, we can first show by induction that x n 3 n 1 ; the basis for induction is x 1 = 1, which is true. And if x n 3 n 1 , then x n +1 = 3 x 2 n > 3 x n 3 · 3 n 1 = 3 n , so that is the induction step. Now pick an M > 0; we want to find N such that n > N x n > M . Let N = log 3 ( M ) + 1. Then if n > N , x n 3 n 1 > 3 N 1 = 3 log 3 ( M ) = M . squaresolid c) There is no contradiction, because when we assume that lim x n = a and do algebra to figure out what a is, we’re implicitly assuming that a is not ±∞ . As you may remember from calculus, the rules of algebra don’t always work with ±∞ . 9.11 a) Suppose that lim s n = + and inf { t n : n N } > −∞ ; we want to show that lim( s n + t n ) = + . To do this, let t = inf { t n : n N } ; then t > −∞ . Now pick M > 0. Since lim s n = , there exists N such that n > N implies that s n > max { M t, 1 } . Then s n > M t , so s n + t > M . But t = inf { t n } , and so t n t ; thus s n + t n s n + t > M whenever n > N . This is what we wanted. squaresolid (Aside: The reason we use max { M t, 1 } instead of just M t is that technically the definition 9.8 only guarantees the existence of such an N when M t > 0, and M t might not be positive. That requirement is absolutely unnecessary, though, since the only thing we really care about is large M , and any N that works for, say, 1 will also work for any negative number. So we can replace “for M > 0” in 9.8 with “for any M ”, and similarly with M < 0 in the second part of the definition. I will not worry about whether M > 0 again.) b) Show that if lim s n = + and lim t n > −∞ , then lim( s n + t n ) = + . To do this, notice that t n is a convergent sequence, so it is bounded (Theorem 9.1). Then there is an M such that | t n | ≤ M for all n , so t n ≥ − M for all n , so inf { t n } ≥ − M > −∞ . Then part a) immediately gives us what we want. squaresolid c) If lim s n = + and t n is bounded, then just as in b), inf { t n } > −∞ . Thus by part a), we are done immediately, and lim( s n + t n ) = + . squaresolid 1

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2 9.12. Assume all s n negationslash = 0 and that the limit L = lim | s n +1 s n | exists. a) Show that if L < 1, then lim s n = 0. First select a so that L < a < 1; notice that, say, the average of L and 1 would work. Now, lim | s n +1 s n | = L < a , so let ǫ = a L ; then there exists an N where n > N ⇒ || s n +1 s n | − L | < ǫ . In particular, if we let N = N + 1, then n N implies that | s n +1 s n | < L + ǫ = a , i.e. that | s n +1 | < a | s n | . So certainly | s N +1 | < a | s N | . Now we see that | s N +2 | < a | s N +1 | < a 2 | s N | . This pattern continues, and we see that | s N + k | < a k | s N | for any k > 0. (Technically this requires induction, but this is obvious enough that you could just state it and say it’s proved by induction). Changing variables and letting n = N + k , we see that for n > N , we have | s n | < a n
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115sols4 - HW 4 Solutions Math 115 Winter 2009 Prof Yitzhak...

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