SAMPLE PROBLEMS Problem 1 Scratch work. At first glance this may seem a bit complicated so lets first unpack the definitions. The sequence |an |1/n might remind one of the root test for the convergence of series. The problem, however, is that the sequence |an |1/n might not converge (which is an assumption needed in the version of the root test that you have been taught). The problem is asserting that it is still possible to figure out what the radius of convergence for the power series is. This is done by defining another sequence Tk which converges to the “largest” limit of any subsequence of |ak |1/n that does converge. P We want to show that the radius of convergence for the power series an xn is 1/T . So we want to show that when |x| > 1/T the series diverges and that when |x| < 1/T the series converges. Lets start with the former. First note that T1 ≥ T2 ≥ T3 ≥ . . . ≥ T because the Tk are defined by taking the least upper bounds of smaller and smaller sets. Since the Tk are defined to be the “largest” of the |an |1/n where n ≥ k and Tk ≥ T , this means no matter how far along the sequence {ai } we go we can always find terms ak such that |ak |1/k > T . But then |x| > 1/T we would have |ak xk | > T k |xk | > 1 and this will not allow the series to converge. In the other direction we now assume that |x| < 1/T . Since the Tk are defined to be the “largest” of the |an |1/n where n ≥ k and Tk ≥ T we must have that all terms of the sequence |ak |1/k must eventually be less than T + ε no matter how small ε is. Since |x| < 1/T we can say |x| = 1/T − r. Then for sufficiently large k k we would then have |ak xk | < (T + ε)k (1/T − r)k = (1 + ε(1/T − r) − T r) so if k k we choose an appropriate ε we would get |ak x | < (something less than 1) and we would be done by comparing with a geometric series. Solution. |x| > 1/T ⇒ the series diverges. P We will denote by Sn the nth partial sum of the series ak xk . Define a subsequence ki of {ak } as follows. Let ak0 be any element of the sequence {ak } so that |ak |1/k ≥ T . We know such an element exists because Ti =l.u.b{|an |1/n |n ≥ i} ≥ T for any i. Similarly given aki define aki+1 to be any element of the sequence {ak } past aki so that |aki+1 |1 /ki+1 > T . Now note that |Ski − Ski −1 | = |aki xki | > |T |ki |1/T |ki = 1 P So the sequence given by ak xk is not Cauchy and thus does not converge. Date: October 30, 2013. 1 2 SAMPLE PROBLEMS |x| < 1/T ⇒ the series converges absolutely Let |x| = 1/T − r where r > 0. Then choose ε so that ε|x| < T r. Now since Tk → T we can pick N so that for all n > N we have Tn < T + ε. Since Tn =l.u.b{|ak |1/k |k ≥ n} we must have that for all n > N, |an |1/n < T + ε. So it follows that ∞ X n=1 |an xn | < ∞ X (T + ε)n (1/T − r)n = n=1 ∞ X (1 + ε(1/T − r) − T r)n n=1 = ∞ X (1 + (T r − ε|x|))n n=1 Recall that we chose ε so that ε|x| < T r and it follows that T r − ε|x| > 0 also Pso ∞ since 0 ≥ |x| = (1/T − r) we get T r < 1. So n=1 (1 + (T r − ε|x|))n is a converging geometric series and we are done. Problem 2 Scratch work. To show that f ◦ g is continuous we need to figure out how close x and y need to be to bound the distance between f (g(x) and f (g(y)). Since f is continuous we know that for any given ε > 0 we can find a δ > 0 so that when |x − y| < δ we have |f (x) − f (y)| < ε and the same is true for g. So all we have to do is put these two facts together. Solution. Given ε > 0 we want to show that there is a δ > 0 so that when |x − y| < δ we have |f (g(x)) − f (g(y))| < ε. Since f is continuous we can pick δ1 so that whenever |x − y| < δ1 we have |f (x) − f (y)| < ε. Now choose δ so that when |x − y| < δ we have |g(x) − g(y)| < δ1 then it follows by assumption that |f (g(x)) − f (g(y))| < ε as desired. Problem 3 Scratch work. This problem is stating that if the terms in a sequence eventually get closer to one another then the terms of any subsequence of that sequence must also get closer to one another. All we have to do is carefully formalize this intuition. To do this we need to understand some notation. Suppose we are given a sequence {xj } = {x1 , x2 , x3 , . . .} then a subsequence of this sequence is commonly denoted by {xji }. By this we mean there is a sequence ji of natural numbers so that ji → ∞ and we are only looking at the terms of {xj } that are indexed by elements of the sequence ji . So for instance if {ji } = {j1 = 1, j2 = 1, j3 = 2, j4 = 3, j5 = 5, . . .} then by {xji } we mean the sequence {x1 , x1 , x2 , x3 , x5 , . . .}. Solution. Let {xj } be a Cauchy sequence and let {xji } be some subsequence of this sequence. We want to show that for any ε > 0 we can find an N so that ji1 , ji2 > N implies |xji1 −xji2 | < ε. Since the original sequence is Cauchy we already know that we can find an N 0 so that whenever n, m > N 0 we have |xn −xm | < ε. Since ji → ∞ we pick N so that i1 , i2 > N ⇒ ji1 , ji2 > N 0 and so it follows that |xji1 − xji2 | < ε and so we are done. SAMPLE PROBLEMS 3 Problem 4 Scratch work. We are given intervals Ij that satisfy I1 ⊃ I2 ⊃ I3 ⊃ . . . and also limj→∞ lj = 0 where lj is the length of Ij . So it follows that xk ∈ Ij whenever k ≥ j and so the distance between xk1 and xk2 can be bounded by lj whenever k1 , k2 > j and we can use this to prove that {xj } is a Cauchy sequence. The second part of the problem is very close to problem 2 on homework 2 and we will solve it similarly. Solution. We want to show that for any ε > 0 we can pick N so that n, m > N implies |xn − xm | < ε. Since the sequence {lj } converges to 0 we can pick N so that for any k ≥ N we have lk < ε. Then for any n, m > N we have that xn and xm are in the interval IN and so |xn − xm | < lN < ε. So we have that the sequence {xn } is Cauchy as desired. For the second part of the problem we suppose that there is another sequence {yj } that also satisfies yj ∈ Ij . We want to show that {xj } and {yj } converge to the same limit. Let L be the limit of the sequence {xj }. We want to show that L is also the limit of the sequence {yj }. Given ε > 0 pick N1 so that n > N1 implies |xn − L| < ε/2. Similarly pick N2 so that n > N2 implies ln < ε/2. Set N = max{N1 , N2 } Then we get that for n > N |yn − L| < |yn − xn | + |xn − L| < ln + ε/2 < ε/2 + ε/2 = ε So we are done. Problem 5 Thoughts. This problem was essentially solved in class. However I wanted to show that rational powers can be defined using the intermediate value theorem, which may be something you have not thought about previously Solution. Showing that xp/q does not depend on the expression for p/q We want to show that (l.u.b{y : y rq < x})rp = (l.u.b{y : y q < x})p for any integer r. In this case f (y) = y rq is a monotone function defined on (0, ∞) . If rq < 0 we know that y rq is unbounded as y → 0 so pick some y0 so that y0rq > x. If rq > 0 we can still pick such a y0 since in this case f (y) is unbounded as y increases. Similarly if rq < 0 we know that f (y) gets arbitrarily close to 0 as y → ∞ so we can pick y1 so that f (y1 ) < x. We can also pick such a y1 if rq > 0 since then f (0) = 0 < x. In other words we can find an interval [y0 , y1 ] (or perhaps [y1 , y0 ]) where f (y0 ) > x > f (y1 ). Thus by the intermediate value theorem (which we can use since f is the product of continuous functions and thus continuous), there must be some z0 that satisfies f (z0 ) = x. Since no two values of f (z) are the same (since it is strictly monotone) when z > 0 we also get that z0 is the unique real number that satisfies f (z0 ) = x. Since z0rq = x we get that z0 = (l.u.b{y : y rq < x}). Also note that z0r satisfies (z0r )q ) = x and be the same argument as before it is the only number to do so. It immediately follows that z0r = (l.u.b{y : y q < x}). From here it is immediate that (l.u.b{y : y rq < x})r p = (l.u.b{y : y q < x})p . 4 SAMPLE PROBLEMS Showing that xα exists Now we want to show that l.u.b{xp/q |p/q < α} exists. To do this all we have to do is show that the set {xp/q |p/q < α} is nonempty and bounded above. The set is clearly nonempty since there are always rationals less between α and 0, take for instance tn (α). Now to get an upper bound on the set let y be bigger than both x and 2 and let n be a natural number bigger than α. We claim y n bounds the set. If x < 1 then xq < 1 for any natural number q and so it follows that x1/q < 1 and so xp/q < 1 when p/q > 0. Since y > 2 we have that in this case y n ≥ 2n > 1 > xp/q . Now for the case x > 1. Note that for p/q < α we have p/q < n ⇒ p < qn. So if x > 1 we get xp < xqn ⇒ xp/q < xn < y n . So y n is indeed an upper bound for the set and so the least upper bound must exist. Showing that xα is a continuous function of α Now we want to show that f (α) = xα is a continuous function depending when α ∈ [0, ∞) and x > 0. To do this we will first show that lim x1/n = 1 n→∞ . Note that there are 2 cases either x > 1 or 0 < x < 1. If x > 1 then x1/n > 1 so we write x1/n = 1 + bn . Then x = (1 + bn )n > 1 + nbn ⇒ bn ≤ (x − 1)/n. Picking n large enough we can make (x − 1)/n < ε for any ε > 0. Thus |x1/n − 1| = bn < ε. Similarly if 0 < x < 1 then x1/n < 1. So we write x1/n = 1/(1 + cn ) ⇒ x = 1/(1 + cn )n < 1/(1 + ncn ) ⇒ cn < 1/n(1/x − 1) so we can pick n large enough so that 1/n(1/x − 1) < ε for any ε. Then 1 − x1/n = 1 − 1/(1 + ε) = ε/(1 + ε) < ε. Now finally we can finish the problem. For a fixed α ≥ 0 and given any ε > 0 we want to be able to find a δ so that |α − β| < δ ⇒ |xα − xβ | < ε. Now we can show that that xα−β converges to 1 as β approaches α because if α − β > 0 we can squeeze it between x1/n and x1/(n+1) where 1/n > α − β > 1/(n + 1). Similarly if α − β < 0 we can squeeze it between x−1/n and x−1/(n+1) . Finally we just choose δ > 0 so that |α − β| < δ implies |xα−β − 1| < ε/|xβ |. Then |xα − xβ | = |xβ ||xα−β − 1| < ε