\magnification=\magstep1
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Final Examination Solutions}
\centerline{May 17, 2011}
\bigskip
\noindent
I.  
\item{A)} State the $\varepsilon, n_0$ definition of convergence for
a sequence of real numbers.
\medskip
\item{} {\it Solution:}  The sequence $x_n$ converges to $c \in {\bf R}$
if for every $\varepsilon > 0$, there exist $n_0 \in {\bf N}$ such
that $|x_n - c| < \varepsilon$ for all $n \ge n_0$.  
\medskip
\item{B)}  Identify $\lim_{n\to\infty} {2n^2 - 2\over n^2 + n}$.
\medskip
\item{} {\it Solution:}  We have
$$\eqalign{
\lim_{n\to\infty} {2n^2 - 2\over n^2 + n} &= \lim_{n\to\infty} {2n^2 - 2\over n^2 + n}\cdot {{1\over n^2}\over {1\over n^2}}\cr
&= \lim_{n\to\infty} {2 - {2\over n^2}\over 1 + {1\over n}}\cr
&= 2.\cr}$$
\item{C)}  Show that your result in part B is correct using the definition.
\medskip
\item{} {\it Solution:}  Let $\varepsilon > 0$ and take 
$n_0 > 2/\varepsilon$.   Then for all $n \ge n_0$, 
$2/n < \varepsilon$, so
$$\eqalign{
\left|{2n^2 - 2 \over n^2 + n} - 2\right| 
&= \left|{-2n - 2\over n^2 + n}\right|\cr
&= {2n + 2 \over n^2 + n}\cr
&= {2\over n}\cr
&< \varepsilon.\cr}$$
Therefore, by definition, ${2n^2 - 2\over n^2 + n}$ converges to $2$. 
\bigskip
\noindent
II.  Show using mathematical induction that the sequence $\{x_n\}$ defined
by $x_1 = 1$, and 
$$x_n = \sqrt{x_{n-1} + 1}, \qquad {\rm if\ } n\ge 2$$
is bounded above by 2 and monotone increasing.  Does this imply that
$\{x_n\}$ is convergent?  If so, say why and find the limit.  If not, say why not.
\medskip
\noindent
{\it Solution:}  By assumption $x_1 = 1 < 2$.  Then if $x_k < 2$, we have
$x_{k+1} = \sqrt{x_k + 1} < \sqrt{2 + 1} < \sqrt{3} < 2$.  Hence $x_n < 2$
for all $n$ by induction.

Similarly, $x_1 = 1$ and $x_2 = \sqrt{2}$ satisfy $x_1 < x_2$.  Assume that
$x_k < x_{k+1}$.  Then by definition
$$x_{k+1} = \sqrt{x_k + 1} < \sqrt{x_{k+1} + 1} = x_{k+2}.$$
Hence the sequence is monotone strictly increasing by induction.  

Since the sequence is monotone increasing and bounded above, the monotone
convergence theorem implies that $\{x_n\}$ converges to some $c$ in ${\bf R}$
(indeed, $c = {\rm lub}\{x_n \mid n \ge 1\}$).  From the recurrence relation,
taking $\lim_{n\to\infty}$, and using the continuity of the square root function,
$$c=\lim_{n\to \infty} x_n = \lim_{n\to\infty} \sqrt{x_{n-1} + 1} = \sqrt{c + 1}.$$
This implies $c^2 = c+1$, so by the quadratic formula, 
$c = {1 \pm \sqrt{5}\over 2}$.  Since all
the terms $x_n > 0$, this implies 
$$c = {1 + \sqrt{5}\over 2} \doteq 1.618.$$ 

\noindent
III.  The Mean Value Theorem states:  If $f$ is continuous on $[a,b]$ and
differentiable on $(a,b)$, then there exists $c \in (a,b)$ such that
$$f'(c)= {f(b) - f(a)\over b - a}.$$
\bigskip
\noindent
{\it Proof:}  First, consider the case $f(b) = f(a)$.  If $f$ is
constant on $[a,b]$, then we can take $c \in (a,b)$ arbitrary.  Otherwise,
by the Extreme Value Theorem, $f$ must attain a maximum or minimum not equal
to $f(a) = f(b)$ at some $c \in (a,b)$.  Hence $f$ has a local maximum
or minimum at $c$ so it follows from the definition of the derivative that
$f'(c) = 0$.  This establishes the special case where $f(b) = f(a)$.  

Now consider a general $f$ satisfying the conditions of the theorem, 
and consider 
$$g(x) = f(x) - {f(b) - f(a) \over b - a}(x - a).$$
The function $g$ is a linear combination of $f(x)$ and the linear function 
$x - a$. Hence $g$ is also continuous on $[a,b]$ and differentiable on $(a,b)$.  
Moreover, $g(b) = f(b) - {f(b) - f(a)\over b - a}(b - a) = f(a)$ and 
$g(a) = f(a) - 0 = f(a)$.  Therefore, $g$ satisfies the extra condition
of the special case.  It follows that there exists a $c\in (a,b)$ where
$g'(c) = 0$.  But $g'(c) = f'(c) - {f(b) - f(a) \over b - a} = 0$, so 
$f'(c) = {f(b) - f(a)\over b - a}$, and the theorem is proved.
\bigskip
\noindent
IV.  True/False.  For each true statement, give a short proof.  For each false
statement, give a counterexample.
\medskip
\item{A)}  Let $f$ be differentiable and $f'$ be continuous on $[a,b]$.
If $f'(x) \ne 0$ for all $x \in [a,b]$, then $f(x)$ is either increasing
on the whole interval $[a,b]$, or decreasing on the whole interval $[a,b]$.  
\medskip
\item{} {\it Solution:}  This is TRUE.  Since $f'$ is continuous on $[a,b]$,
the Extreme and Intermediate Value Theorems can be applied to $f'$.  Since 
$f'$ never takes the value 0 on $[a,b]$, either the minimum value of $f'$
is strictly greater than zero, or else the maximum value of $f'$ is
strictly less than zero.  In the first case, by one of our consequences
of the Mean Value Theorem, $f$ is increasing on $[a,b]$.  In the second
case, $f$ is decreasing on $[a,b]$.
\medskip
\item{B)}  If $f$ is uniformly continuous on $[a,b]$, then $f$ is
differentiable at every point in $[a,b]$.
\medskip
\item{} {\it Solution:}  This is FALSE.  By a theorem we proved
in class, any continuous function on $[a,b]$ is uniformly continuous.
But there are continuous functions that are not differentiable 
at some $c \in [a,b]$.  An example is $f(x) = |x|$ on $[a,b] = [-1,1]$.
$f$ is continuous, but is not differentiable at $x = 0$.
\medskip
\item{C)}  Let $f : [a,b] \to {\bf R}$ be continuous and let
$\{x_n\}$ be a sequence with $x_n \in [a,b]$ for all $n$.  Then
The sequence $\{f(x_n)\}$ has a convergent subsequence. 
\medskip
\item{} {\it Solution:}  This is TRUE.  By the Extreme Value Theorem,
the sequence $\{f(x_n)\}$ is bounded (it lies in the interval $[m,M]$
where $m,M$ are the minimum and maximum values of $f$ on the interval).  
Then the Bolzano-Weierstrass theorem implies that it has a convergent 
subsequence.  
\bigskip
\noindent
V.  Show that 
$$f(x) = \cases{1 & if $0 \le x < 1/2$\cr
                2 & if $1/2 \le x \le 1$\cr}$$
is integrable on $[a,b] = [0,1]$ by considering the upper and lower sums for $f$.  Also, determine the value of $\int_0^1 f$.
\medskip
\noindent
{\it Solution:}  Let $\delta < 1/2$ and let ${\cal P}_\delta$ be the partition
$\{0,1/2 - \delta, 1/2 + \delta, 1\}$.  Then
$$\eqalign{
L_{{\cal P}_\delta}(f) &= 1\cdot (1/2 - \delta) + 1\cdot(2 \delta) + 2\cdot(1/2 - \delta),\cr
U_{{\cal P}_\delta}(f) &= 1\cdot (1/2 - \delta) + 2\cdot(2 \delta) + 2\cdot(1/2 - \delta),\cr
U_{{\cal P}_\delta}(f) - L_{{\cal P}_\delta}(f) &= 2\delta.\cr}$$
Given $\varepsilon > 0$, let $\delta < \varepsilon/2$.  Then for any such $\delta$,
the difference between the upper and lower sums is $< \varepsilon$.  This implies
that $f$ is integrable on $[0,1]$.  This also shows
$$\int_0^1 f = {\rm glb}\{U_{{\cal P}_\delta}(f) \mid \delta > 0\} = {3\over 2},$$
since 
$$U_{{\cal P}_\delta}(f) = 1\cdot (1/2 - \delta) + 2\cdot(2 \delta) + 2\cdot(1/2 - \delta) = {3\over 2} + \delta.$$
\noindent
VI.
\item{A)}  Noting that 
$$\lim_{x\to 1} {\ln(x)\over x - 1} = \lim_{x\to 1} {\ln(x) - \ln(1)\over x - 1},$$
evaluate the limit.  (You may use any calculus facts you need here without proof.)
\medskip
\item{} {\it Solution:}  Let $f(x) = \ln(x)$.  By definition, the limit above
computes the derivative $f'(1)$.  Since $f'(x) = 1/x$ by calculus facts, 
$f'(1) = 1$, so 
$$\lim_{x\to 1} {\ln(x)\over x - 1} = 1.$$
\item{B)}  Let $x = 1 + {1\over n}$ in the limit from part A and use that result
and the sequential criterion for continuity applied to $f(x) = \exp(x) = e^x$ to 
compute
$$\lim_{n\to\infty} \left(1 + {1\over n}\right)^n.$$
\item{} {\it Solution:}  With $x = 1 + {1\over n} \to 1$, we have, using
properties of logarithms and continuity of the exponential function,
$$\eqalign{
1 &= \lim_{n\to\infty} {\ln\left(1 + {1\over n}\right)\over {1\over n}}\cr
&= \lim_{n\to\infty} \ln\left(\left(1 + {1\over n}\right)^n\right)\cr
\Rightarrow e^1 &= \lim_{n\to\infty} \exp\left(
\ln\left( 1 + {1\over n} \right)^n\right).\cr}$$
Hence
$$\lim_{n\to\infty} \left(1 + {1\over n}\right)^n = e.$$

\noindent
VII.
\item{A)}  State the definition of convergence for an infinite series
$\sum_{n=1}^\infty a_n$.
\medskip
\item{} {\it Solution:}  The series $\sum_{n=1}^\infty a_n$ converges
if the sequence $\{s_k\}$ of partial sums, defined by  
$$s_k = \sum_{n=1}^k a_n,$$
converges to some $S \in {\bf R}$ (as a sequence).
\medskip
\item{B)}  Does the series $\sum_{n=1}^\infty {2^n\over 5^n}$ converge?  If so, what is
the sum of the series?
\medskip
\item{} {\it Solution:}  The answer is yes -- this is a geometric series 
with ratio $r = 2/5$.  Since $|r| < 1$, the geometric series converges to
$${2/5\over 1 - 2/5} = {2\over 3}.$$
\item{C)}  Does the series $\sum_{n=1}^\infty n e^{-n}$
converge?  (Use the Integral Test.)  
\medskip
\item{} {\it Solution:}  The terms of the series are $f(n)$ for $f(x) =  x e^{-x}$.
(Note that $f'(x) = e^{-x}(x - 1) < 0$ for all $x > 1$.  So $f(x)$ is decreasing
to 0 as $x \to \infty$.)  Integrating by parts and using the Fundamental Theorem of Calculus, 
we have
$$\eqalign{
\lim_{b \to \infty} \int_1^b x e^{-x}\ dx &= \lim_{b \to \infty} \left. -x e^{-x} - e^{-x}\right|_1^b\cr
&=\lim_{b\to\infty} -b e^{-b} + e^{-b} - e^{-b} + 2e^{-1}\cr
&=2e^{-1}\cr}$$
Since this is finite, the series {\it converges}.  
\medskip
\item{D)}  Use the Ratio Test to determine the $x \in {\bf R}$ for which
the series $\sum_{n=1} {nx^n \over 2^n}$ converges absolutely.
\medskip
\item{} {\it Solution:}  We have
$$\lim_{n\to\infty} \left| {(n+1)x^{n+1}\over 2^{n+1}} \cdot {2^n\over nx^n}\right| = \lim_{n\to\infty} \left(1 + {1\over n}\right){ |x| \over 2} = {|x|\over 2}.$$
This is $< 1 $ if and only if $|x| < 2$, or $x \in (-2,2)$.  The series converges absolutely for $x \in (-2,2)$, but diverges
for $x = \pm 2$.  
\bigskip
\noindent
{\it Extra Credit}  Let 
$$f = \cases{x + x^2 \cos(\pi/x) & if $x \ne 0$\cr
                    0 & if $x = 0$.\cr}$$
Show that $f$ is differentiable at all $x \in {\bf R}$, but $f'$ 
is not continuous at $x = 0$. 
\medskip
\noindent
{\it Solution:}  For $x \ne 0$, we can apply the usual derivative
rules to see 
$$f'(x) = 1 + 2x\cos(\pi/x) + \pi \sin(\pi/x).$$
For $x = 0$, we must use the definition of the derivative:
$$\eqalign{
f'(0) &= \lim_{x \to 0} {f(x) - f(0)\over x - 0}\cr
&= \lim_{x\to 0} {x + x^2 \cos(\pi/x)\over x}\cr
&= \lim_{x \to 0} 1 + x \cos(\pi/x)\cr
&= 1\cr}$$
(note that $\lim_{x\to 0} x \cos(\pi/x) = 0$ by the squeeze theorem).
Hence $f'(x)$ exists for all $x \in {\bf R}$.  

Finally, we claim that $f'$ fails to be continuous at $x = 0$.
this is true because 
$$\lim_{x\to 0} f'(x) = \lim_{x \to 0} 1 + 2x\cos(\pi/x) + \pi \sin(\pi/x)$$
does not exist.  The last term, $\sin(\pi/x)$ has no limit as $x \to 0$,
but the other two terms do.  It follows that $f'$ has a discontinuity 
at $x = 0$.  

\bye
