\magnification=\magstep1
\leftline{Mathematics 242 -- Principles of Analysis}
\leftline{Final Exam Solutions}
\bigskip
\noindent
I.  
\item{A)}  $A = \{\sin(x) : 0 \le x \le 3\pi/4\} = [0,1]$ and 
$B = \{x : 1 < x^2 < 4\} = (-2,-1) \cup (1,2)$, so 
$A \cup B = (-2,-1) \cup [0,2)$.
\item{B)}  $C = \{|x - 2| : x \in A\}$ equals the set $[1,2]$, since
we think of $|x - 2|$ as the distance from $x$ to 2.  Hence 
${\rm lub}(C) = 2$ (the largest distance to 2 of any element of $C$).
\bigskip
\noindent
II.  
\item{A)}  $x_n$ converges to $L$ if for every $\epsilon > 0$, there
exists some corresponding $n_0 \in {\bf N}$ such that $|x_n - L| < \epsilon$ for 
all $n \ge n_0$.
\item{B)}  By the ``Big Limit Theorem'' (for instance),
$$\eqalign{
\lim_{n\to\infty} {3n^2 + n\over n^2 + 1} &=
\lim_{n\to\infty} {(3n^2 + n)\cdot{1\over n^2}\over (n^2 + 1)\cdot{1\over n^2}}\cr
&=\lim_{n\to\infty} {3 + {1\over n}\over 1 + {1\over n^2}}\cr
&=3}$$
To prove this by the definition, let $n \ge 3$.  Then
$$\left|{3n^2 + n\over n^2 + 1} - 3\right| = \left|{n-3\over n^2 + 1}\right|
= {n-3\over n^2 + 1} < {n\over n^2} = {1\over n}$$
Hence, given $\epsilon > 0$, if $n_0 \ge \max\{3,1/\epsilon\}$ and 
$n\ge n_0$, then
$$\left|{3n^2 + n\over n^2 + 1} - 3\right| <  {1\over n} < \epsilon$$
\bigskip
\noindent
III.  
\item{A)}  Since $\sin(x)$ takes values only between -1 and 1, 
$x_n$ is a bounded sequence.  Hence, the Bolzano-Weierstrass
Theorem implies that $x_n$ has a convergent subsequence
$x_{n_k}$.
\item{B)}  The answer is {\it yes}.  Using the index sequence $n_k$
found in part A, the sequence $y_{n_k}$ is also a bounded
sequence since the cosine function is also bounded.  By the 
Bolzano-Weierstrass Theorem again, there is a convergent subsequence
$y_{n_{k_\ell}}$.  The corresponding subsequence $x_{n_{k_\ell}}$
is also convergent, since every subsequence of a convergent
sequence (like $x_{n_k}$) is also convergent.
\bigskip
\noindent
IV.  
\item{A)}  $\lim_{x\to c} f(x) = L$ if for every $\epsilon > 0$ there
exists a corresponding $\delta > 0$ such that $|f(x) - L| < \epsilon$
for all $x$ with $0 < |x - c| < \delta$.
\item{B)}  $\lim_{x\to 0} x^2\sin(1/x) = 0$ since for all $x \ne 0$,
$$|x^2 \sin(1/x)| = |x|^2 |\sin(1/x)| \le |x|^2$$
(the last inequality follows since $|\sin(\theta)| \le 1$ for all $\theta$).
Hence, given $\epsilon >0$, let $\delta = \sqrt{\epsilon}$.  Then
if $0 < |x| < \delta$, 
$$|x^2 \sin(1/x) - 0| \le |x|^2 < \delta^2 = \epsilon$$
\bigskip
\noindent
V.  
\item{A)}  The Intermediate Value Theorem:  Let $f$ be continuous
on $[a,b]$ and let $m = \min\{f(x) : x \in [a,b]\}$, 
$M = \max\{f(x) : x \in [a,b]\}$.  Then for all $y_0$, $m \le y_0 \le M$, 
there exist $x_0 \in [a,b]$ such that $f(x_0) = y_0$.
\item{B)}  For $f(x) = a_1 + a_2\sin(x)$,
$$\cases{f(-\pi/2) = a_1 - a_2 &\cr
         f(\pi/2) = a_1 + a_2 &\cr}$$
By hypothesis, $a_2 > |a_1|$.  Hence $a_1 - a_2 < 0$ and $a_1 + a_2 > 0$.
Since $f$ is continuous everywhere, the Intermediate Value Theorem
implies that there is some $x_0 \in [-\pi/2,\pi/2]$ such 
that $f(x_0) = 0$.
\bigskip
\noindent
VI.  The Fundamental Theorem of Calculus:  Let $f$ be continuous
on $[a,b]$, and let $F(x) = \int_a^x f(t)\ dt$.  Then:
\item{1)}  $F$ is differentiable on $[a,b]$, and $F'(x) = f(x)$.
\item{2)}  If $G$ is any antiderivative of $f$, then
$\int_a^b f(x)\ dx = G(b) - G(a)$.

\noindent
See the class notes for the proof.
\bigskip
\noindent
VII.
\item{A)}  {\bf False} -- $f$ has a jump discontinuity at $x = 1$.
If $f$ is differentiable at $c$ it is also continuous at $c$.
\item{B)}  {\bf True} -- the sequence of partial sums is bounded
above and monotone increasing since $a_n \ge 0$ for all $n$.  Hence
the sequence of partial sums converges to some real number.
\item{C)}  {\bf True} -- the conclusion of the MVT is that
there exists some $c\in (a,b)$ such that $f'(c) = {f(b) - f(a) \over b - a}$. 
This is true here (even though the hypotheses of the MVT are not 
satisfied!!) since
$$f'\left({1\over 8}\right) = 2\cdot {1\over 8} = {1\over 4} = {f(2) - f(0)\over 2 - 0}$$
\item{D)}  {\bf False} -- this series is absolutely convergent, not 
conditionally convergent.  The series $\sum_{n=1}^\infty {(-1)^n\over n^3}$
itself converges by the 
Alternating Series Test.  Then, the series of absolute values:
$$\sum_{n=1}^\infty \left|{(-1)^n\over n^3}\right| = 
\sum_{n=1}^\infty {1\over n^3}$$
also converges since it is a $p$ series with $p = 3 > 1$.
\item{E)} {\bf True} -- We proved in class that if $f$ is 
continuous on $[a,b]$ except for a finite number of jump or
removable discontinuities, then $f$ is integrable.  That is 
the case for $f$ here.  
\item{F)} {\bf False} -- By the Ratio Test, 
$$\lim_{n\to\infty}\left|{x^{n+1}\over (n+1)2^{n+1}}{n2^n\over x^n}\right|
=\lim_{n\to\infty} {n\over n+1} {|x|\over 2} = {|x|\over 2}$$
This is $> 1$ if $|x| > 2$.  Hence the series diverges for all 
$x$ with $|x| > 2$.  (It converges absolutely by the Ratio Test
for all $|x| < 2$.  The series also converges conditionally for
$x = -2$, but diverges for $x = 2$.)
\item{G)} {\bf True} -- This basically comes from the Integral
Test calculation that shows the $p$-series with $p > 1$
is convergent.  By thinking of the partial sum
$\sum_{n=1}^N {1\over n^p}$ as the upper sum (left-hand Riemann sum) for the 
decreasing function $f(x) = {1\over x^p}$ on $[1,N+1]$ corresponding 
to the partition $P = \{1,2,\ldots, N+1\}$, we
get
$$\sum_{n=1}^N {1\over n^p} \ge \int_1^{N+1} {1\over x^p}\ dx 
= {(N+1)^{1-p} - 1\over 1-p}.$$
As $N\to\infty$, we get
$$\sum_{n=1}^\infty {1\over n^p} \ge \lim_{n\to\infty}{(N+1)^{1-p} - 1\over 1-p} = {1\over p - 1}$$
On the other hand, omitting the first term in the series gives
$\sum_{n=2}^N {1\over n^p}$ as the lower sum (right-hand Riemann sum) for the 
decreasing function $f(x) = {1\over x^p}$ on $[1,N]$ corresponding 
to the partition $P = \{1,2,\ldots, N\}$.  So
$$\sum_{n=2}^N {1\over n^p} = \sum_{n=1}^N {1\over n^p} -1 \le \int_1^{N} {1\over x^p}\ dx 
= {N^{1-p} - 1\over 1-p}$$
Again letting $N \to \infty$, we get
$$\sum_{n=1}^\infty {1\over n^p} \le 1 + \lim_{N\to\infty} {N^{1-p} - 1\over 1-p} = 1 +{1\over p-1}
={p\over p-1}$$
\bye




