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\centerline{Mathematics 242, section 1 -- Principles of Analysis}
\centerline{Solutions For Final Examination -- May 15, 2013}
\bigskip
\noindent
I. (20) Let $A = \{x^2 - 2 : -1 < x < 2\}$ and $B = \{x : |x - 1| < 4\}$.
Find ${\rm lub} (A \cup B)$ and ${\rm glb}(A \cap B)$.
\bigskip
\noindent
{\it Solution:}  By the definitions, we see $A = [-2,2)$ and $B = (-3,5)$.
Hence, since $A \subset B$,  $A \cup B = B = (-3,5)$ and $A\cap B = A = [-2,2)$.
Hence ${\rm lub}(A \cup B) = 5$ and ${\rm glb} (A\cap B) = -2$.  
\bigskip
\noindent
II.
\bigskip
\item{A)} (10) State the $\varepsilon$, $n_0$ definition of convergence for 
a sequence of real numbers.
\bigskip
\item{} {\it Solution:}  The sequence $x_n$ converges to $c$ if 
for all $\varepsilon > 0$ there exist $n_0 \in {\bf N}$ such that
$|x_n - c| < \varepsilon$ for all $n \ge n_0$.  
\bigskip
\item{B)} (10) Identify $\lim_{n\to\infty} {5n^2+1\over n^2 + n + 4}$.
\bigskip
\item{} {\it Solution:}  Using the limit theorems, we have 
$$\eqalign{
\lim_{n\to\infty} {5n^2+1\over n^2 + n + 4} &=
\lim_{n\to\infty} {(5n^2+1)\cdot {1\over n^2} \over (n^2 + n + 4)\cdot {1\over n^2}}\cr
&= \lim_{n\to\infty} {5+{1\over n^2} \over 1 + {1\over n} + {4\over n^2}}\cr
&= {5+ \lim_{n\to\infty}{1\over n^2} \over 1 + \lim_{n\to\infty}{1\over n} + \lim_{n\to\infty}{4\over n^2}}\cr
&= {5 + 0\over 1 + 0 + 0}\cr
&= 5.\cr}$$
\item{C)} (10) Show that your result in part B is correct using the 
definition.
\bigskip
\item{} {\it Solution:}  Given $\varepsilon > 0$, let $n_0 > {\rm max}(20,{6\over
\varepsilon})$.  (No matter how small $\varepsilon$ is, this is possible
because ${\bf N}$ is not bounded above.)  Then for all $n \ge n_0$, we have
$$\left|{5n^2+1\over n^2 + n + 4} - 5\right| = \left| {-5n - 19 \over n^2 + n + 4}\right| < {5n + 19 \over n^2}$$
Since $n \ge 20$, we have $5n + 19 < 6n$, so this is
$$ < {6n\over n^2} = {6\over n} \le {6\over n_0} < \varepsilon.$$
Hence by the definition, ${5n^2+1\over n^2 + n + 4}$ converges to $5$.  
\bigskip
\noindent
III.  
\bigskip
\item{A)} (15) Show that if $x_n$ is a monotone increasing sequence
of real numbers that is bounded above, then $x_n$ converges
to some real number.
\bigskip
\item{} {\it Solution:}  The set $X = \{x_n\}$ is bounded above
in ${\bf R}$.  Hence the LUB axiom implies that it has a 
least upper bound $c$.  Given any $\varepsilon > 0$, $c - \varepsilon$
is not an unpper bound for $X$, therefore there exists some $n_0$
such that $c -\varepsilon < x_{n_0} \le c$.  But the sequence is monotone increasing
so $c - \varepsilon < x_{n_0} \le x_n \le c$ for all $n \ge n_0$.  That implies
$|x_n - c| < \varepsilon$ for all $n \ge n_0$, and hence $x_n \to c$
by the definition.  
\bigskip
\item{B)}  (10) True/False and prove/give a reason: The sequence
$$x_n = \cases{-1 + {1\over n^2} & if $n$ is a prime integer $> 1$\cr
                1 - {1\over n^2} & if $n$ is not prime\cr}
$$
has convergent subsequences.
\bigskip
\item{} {\it Solution:}  This is TRUE. From the definition, 
$x_n < 1$ for all $n$ and $x_n > -1$ for all $n$.  Since
$\{x_n\}$ is a bounded sequence, the Bolzano-Weierstrass theorem
implies it must have convergent subsequences.  (In fact, we can also
see directly from the above that the subsequences for $n_k = k$th prime number and
$n_k = 2k$, $k \ge 1$ are convergent.)
\bigskip
\item{C)}  True/False and prove/give a reason:  
The infinite ``continued radical''
$$\sqrt{3 + \sqrt{3 + \sqrt{3 + \sqrt{3 + \sqrt{ 3 + \cdots }}}}}$$
represents a finite real number.  (Hint: If so, that number would be 
the limit of a sequence defined by $x_1 = \sqrt{3}$ and $x_n = \sqrt{3 + x_{n-1}}$
for all $n\ge 2$.)
\bigskip
\item{} {\it Solution:}  This is TRUE.  First we notice that 
in the sequence described in the Hint, 
$$\eqalign{
x_2 &= \sqrt{3 + \sqrt{3}}\cr
x_3 &= \sqrt{3 + \sqrt{3 + \sqrt{3}}},\cr}$$
etc.  So as $n\to\infty$, if the sequence $x_n$ converges, it converges
to the expression defined by the infinite ``continued radical.''
We claim next the sequence is bounded below by $0$ (clear) and above by $3$.
The second part requires some proof, but we can argue that $x_n < 3$ for all 
$n$ by induction as follows.  The base case is 
$x_1 = \sqrt{3} < 3$.  Assuming $x_k < 3$, then we have 
$x_{k+1} = \sqrt{3 + x_k} < \sqrt{3 + 3} = \sqrt{6} < 3$.  Hence $x_n < 3$ 
for all $n$.  Finally, we show by induction again that $x_n$ is monotone increasing.  
Note that $x_2 = \sqrt{3 + \sqrt{3}} > \sqrt{3} = x_1$ since $x_2^2 = 3 + \sqrt{3} > 3 = x_1^2$.  Assuming $x_{k+1} > x_k$, then 
$x_{k+2} = \sqrt{3 + x_{k+1}} > \sqrt{3 + x_k} = x_{k+1}$ also.  Hence the monotone 
convergence theorem (part A) implies that $x_n \to a$ for some real number $a < 3$.  
In fact, taking the limit on both sides of $x_{n+1} = \sqrt{3 + x_n}$, 
we can see that $a$ must satisfy 
$$a = \sqrt{3 + a} \Rightarrow a^2 - a - 3 = 0$$
By the quadratic formula, 
$$a = {1 + \sqrt{13} \over 2} \doteq 2.302775638.$$  
\bigskip
\noindent
IV.  
\item{A)} (15) Let 
$$f(x) = \cases{ x + 3 & if $x$ is a rational number\cr
                -x^2 + 3 & if $x$ is an irrational number\cr}$$
Is $f$ continuous at $x = 0$?  Why or why not? 
\bigskip
\item{} {\it Solution:}
  Yes, $f(x)$ is continuous at $0$.  To see this, let $\varepsilon > 0$,
and $\delta < {\rm min}(\varepsilon,\sqrt{\varepsilon})$.  If 
$|x| < \delta$ and $x$ is rational then we have 
$$|f(x) - f(0)| = |x + 3 - 3| = |x| < \varepsilon,$$
while if $x$ is irrational we have
$$|f(x) - f(0)| = |-x^2 + 3 - 3| = |x|^2 < (\sqrt{\varepsilon})^2 = \varepsilon.$$
Therefore $f$ is continuous at $0$.
\bigskip 
\item{B)} (25) State and prove the Intermediate Value Theorem.  (You may assume
as known the theorem that if $f$ is continuous at $c$ and 
$x_n \to c$ is a sequence contained in the domain of $f$, then $f(x_n) \to f(c)$). 
\bigskip
\item{} {\it Solution:}  The IVT in the form we proved first says that if
$f$ is continuous on $[a,b]$ with $f(a) \ne f(b)$,
and $k$ is any number between $f(a)$ and $f(b)$, then there exists a $c\in [a,b]$
such that $f(c) = k$.  We proved this as follows.  If $k = f(a)$ or $k = f(b)$, then there is nothing to prove. Hence we must consider the case that $k$
is strictly between the two values $f(a)$ and $f(b)$.  Suppose, for instance, that
$f(a) < k < f(b)$.  (The other case $f(b) < k < f(a)$ can be proved
similarly.)  Consider 
$$S = \{x \in [a,b] \mid f(x) \le k\} $$
Then $S$ is not empty since for instance $a \in S$.  $S$ is also bounded
since it is a subset of $[a,b]$.  Hence the LUB axiom implies that 
$c = {\rm lub}(S)$ exists in ${\bf R}$.  We claim that $f(c) = k$.  First
note that $c < b$ since $f(b) > k$.  Hence if we let $x_n \to c$ with
$c \le x_n \le b$, we will have $f(x_n) > k$ for all $n$, and hence (since
$f$ is continuous at $c$, $\lim_{n\to\infty} f(x_n) = f(c) \ge k$.   On 
the other hand, since $c = {\rm lub}(S)$, if $\varepsilon > 0$, then 
there will exist $x \in S$ with $c-\varepsilon < x \le c$.  Apply this
for each $\varepsilon = {1\over n}$ for $n \in {\bf N}$.  We get a
a sequence $x_n' \in S$ with $x_n' \to c$.  Hence 
$\lim_{n\to\infty} f(x_n') = f(c) \le k$ since $f(x_n') \le k$ for all $n$.  
The two inequalities show that $f(c) = k$.  
\bigskip
\noindent
V.  
\item{A)} (15)  Using the limit definition of the derivative, compute $f'(c)$ for 
$f(x) = {1\over (x+3)^2}$ at a general $c \ne -3$.  
\bigskip
\item{} {\it Solution:}  We have
$$\eqalign{
f'(c) &= \lim_{x\to c} {f(x) - f(c)\over x - c}\cr
&= \lim_{x\to c} {{1\over (x+3)^2} - {1\over (c + 3)^2} \over x - c}\cr
&= \lim_{x\to c} {(c + 3)^2 - (x + 3)^2\over (x+3)^2 (c+3)^2 (x-c)}\cr
&= \lim_{x\to c} {c^2 - x^2 + 6(c - x)\over (x+3)^2(c + 3)^2(x - c)}\cr
&= \lim_{x\to c} {-c - x - 6\over (x + 3)^2 (c + 3)^2}\cr
&= {-2\over (c + 3)^3}\cr}$$
({\it Note:}  It is not permissible to use L'Hopital's Rule to compute this
limit because that requires you to use a derivative formula to derive 
$f'(x) = {-2\over (x + 3)^3}$.  Using the derivative rule to compute the
derivative by the definition is an example of circular reasoning!)
\item{B)} (10)  What theorem guarantees that 
$$F(x) = \int_1^x {1\over (t+3)^2}\ dt$$
is differentiable at $x = 2$?  Exactly why does it apply here?
What does it say about $F'(2)$?
\bigskip
\item{} {\it Solution:}  The theorem is the first part of the Fundamental Theorem of Calculus.
It applies because the function $f(x) = {1\over (x + 3)^2}$ is continuous
on the closed interval $[1,a]$ for all $a > 2$.  It implies $F'(2) = f(2) = 
{1\over 25}$.  
\bigskip  
\item{C)} (10) Show that if $f(x) = {e^x + e^{-x}\over 2}$, then 
for every real $k$, there exists a solution $c$
of the equation $f'(c) = {e^c - e^{-c}\over 2} = k$.
\bigskip 
\item{} {\it Solution:}  Note that $f'(x) = {e^x - e^{-x} \over 2}$ 
exists and is continuous at all $x \in {\bf R}$.  This function satisfies
$\lim_{x\to +\infty} f'(x) = +\infty$ and $\lim_{x\to -\infty} f'(x) = -\infty$.  
So whatever the value $k$, we can find an interval $[-a,a]$ 
such that $f'(-a) < k < f'(a)$.  On the interval $[-a,a]$, 
$f'$ is continuous, and then the Intermediate Value Theorem
implies there is some $c \in [-a,a]$ such that $f'(c) = k$.
\bigskip
\item{}   
Alternatively, it would also be possible to solve the 
equation ${e^c - e^{-c} \over 2} = k$ explicitly like this:
$$\eqalign{ {e^c - e^{-c} \over 2} = k &\Leftrightarrow e^{2c} - 2k e^c - 1 = 0\cr
&\Leftrightarrow e^c = {2k + \sqrt{4k^2 + 4}\over 2}\quad {\rm (by\ the\ quadratic\ formula)}\cr
&\Leftrightarrow e^c = k + \sqrt{k^2 + 1}\cr
&\Leftrightarrow c = \ln(k + \sqrt{k^2 + 1})\cr}$$
The function $g(x) = \ln(x + \sqrt{x^2 + 1})$ is also known as the inverse hyperbolic sine function, since $f'(x) = {e^x - e^{-x}\over 2}$ is the hyperbolic sine.  
\bigskip
\noindent
VI.  (20) In this question, you may use the summation
formulas:
$$\sum_{i=1}^n 1 = n \quad \sum_{i=1}^n i = {n(n+1) \over 2} \quad
\sum_{i=1}^n i^2 = {n(n+1)(2n+1) \over 6}.$$ 
Show that $f(x) = x^2 + x$ is integrable on $[a,b] = [0,2]$ by 
considering upper and lower sums for $f$ and determine the value of
$\int_0^2 x^2 + x\, dx$.
\bigskip
\noindent
{\it Solution:}  $f'(x) = 2x + 1 > 0$ for all $x \in [0,2]$.  Therefore,
$f$ is increasing on $[0,2]$ and as in the proof of the integrability
of all monotone functions, if ${\cal P}_n$ is the regular partition 
of $[0,2]$ into $n$ equal subintervals,
$$U(f,{\cal P}_n) - L(f,{\cal P}_n) = (f(2) - f(0)){(2 - 0)\over n} = 
{12\over n}.$$
If $\varepsilon > 0$ is given and $n > {12\over \varepsilon}$, then 
$$U(f,{\cal P}_n) - L(f,{\cal P}_n) < \varepsilon.$$
The limit of the upper sum for ${\cal P}_n$ as $n \to \infty$ is
$$\eqalign{\int_0^2 x^2 + x\ dx &=
\lim_{n\to\infty} \sum_{i=1}^n \left(\left({2i\over n}\right)^2 
+ {2i\over n}\right) {2\over n}\cr
&= \lim_{n\to\infty} \left({8\over n^3} \sum_{i=1}^n i^2 + {4\over n^2} \sum_{i=1}^n i\right)\cr
&=  \lim_{n\to\infty} \left({8 \over n^3} \cdot {n(n+1)(2n+1)\over 6} 
+ {4\over n^2} \cdot {n(n+1)\over 2}\right)\cr
&= {8\over 3} + 2 = {14\over 3}.\cr}$$
Check:  By Part 2 of the FTC:
$$\int_0^2 x^2 + x\ dx = \left. {x^3\over 3} + {x^2 \over 2}\right|_0^2 = 
{8\over 3} + 2 = {14\over 3}.$$
\bigskip
\noindent
VII.  
\item{A)} (5) State the definition of convergence for an infinite
series $\sum_{n=1}^\infty a_n$. 
\bigskip
\item{} {\it Solution:}  The infinite series $\sum_{n=1}^\infty a_n$
converges to $S \in {\bf R}$ 
if the sequence of partial sums $s_k = \sum_{n=1}^k a_n$
converges to $S$ (as a sequence).
\bigskip   
\item{B)} (15)  For which $x \in {\bf R}$ does the power series
$$\sum_{n=1}^\infty {3^n x^n\over n 7^n}$$
converge?  (Apply the Ratio Test and test the series at the endpoints
of your interval separately.)
\bigskip 
\item{} {\it Solution:}  We have 
$$\lim_{n\to \infty} 
\left|{3^{n+1} x^{n+1}\over (n+1) 7^{n+1}}\cdot {n 7^n\over 3^n x^n}\right|
=\lim_{n\to\infty} {3n\over 7(n+1)}|x|  = {3|x|\over 7}.$$
By the Ratio Test, the series converges absolutely whenever
${3|x|\over 7} < 1$, or $|x| < {7\over 3}$, and it diverges
when $|x| > {7 \over 3}$.  At $x = \pm {7\over 3}$, we have
the following.  If $x = {7\over 3}$, we substitute and simplify to 
obtain $\sum_{n=1}^\infty {1\over n}$.  This is the harmonic series, 
which diverges.    If $x = - {7\over 3}$, we substitute and simplify to 
obtain $\sum_{n=1}^\infty {(-1)^n \over n}$.  This series converges 
by the Alternating Series Test.  Therefore the series converges 
at all $x \in \left[-{7\over 3}, {7\over 3}\right)$ (and only for those $x$).
\bigskip
\noindent
{\it Extra Credit}  (20)  Prove the result mentioned in part B of question 
IV above:  if $f$ is continuous at $c$ and 
$x_n \to c$ is a sequence contained in the domain of $f$, then $f(x_n) \to f(c)$.
\bigskip
\noindent
{\it Solution:}  Since $f$ is continuous at $c$, given any $\varepsilon > 0$, 
there is some $\delta > 0$ such that 
$$|f(x) - f(c)| < \varepsilon\leqno(1)$$
for all $x$ with $|x - c| < \delta$.  Then, since $x_n \to c$, given that $\delta$,
there is an $n_0$ such that $|x_n - c| < \delta$ for all $n \ge n_0$.  But
then substituting $x = x_n$ from the sequence into the 
inequality (1) we get $|f(x_n) - f(c)| < \varepsilon$ for all $n \ge n_0$.
This implies that the sequence $f(x_n) \to f(c)$.  
\bye            
