SAMPLE PROBLEMS Problem 1

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.
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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
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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 .
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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