\magnification=\magstep1
\overfullrule=0pt
\centerline{MATH 242 -- Principles of Analysis}
\centerline{Solutions for Problem Set 2 -- due:  Feb. 8}
\baselineskip=15pt
\bigskip
\noindent
{\it `A` Section}
\bigskip
\item{1.}  Let $x \in [-1,2]$.  Determine the largest and smallest
values of $|x - 5|$, $|x + 5|$.
\item{} {\it Solution:}  If $x \in [-1,2]$, then $x - 5 \in [-6,-3]$, so the largest
and smallest values of $|x - 5|$ are $6$ and $3$ respectively.  (You can also see these
as the distances from $-1$ to $5$ and $2$ to $5$ along the number line, thinking geometrically.)
 Next, $x + 5 \in [4,7]$,
so the largest and smallest values of $|x + 5|$ are $7$ and $4$ respectively. 
\item{2.} Use the binomial theorem (Theorem 1.4.1) for all parts of this problem.
\itemitem{a.}  Expand using the binomial theorem and simplify as much as possible:
$$(a^2 + 3b)^5.$$
\itemitem{} {\it Solution:}  We have
$$(a^2 + 3b)^5 = a^{10} + 15a^{8}b + 90a^6b^2 + 135 a^4 b^3 + 405 a^2 b^4 + 243b^5.$$
\itemitem{b.}  What is the coefficient of $x^3$ in the expansion of
$$\left({x^5 + 3\over x^2}\right)^3.$$
\itemitem{} {\it Solution:}  This coefficient is zero.  The powers of $x$ actually 
appearing in the binomial expansion will be $(x^5)^3 = x^{15}$, 
$(x^5)^2 x^{-2} = x^8$, $x^5 (x^{-2})^2 = x$ and $(x^{-2})^3 = x^{-6}$.  
\itemitem{c.}  What is $\sum_{k=0}^n {n\choose k} 2^k$?  Explain.
\itemitem{} {\it Solution:}  From the binomial theorem this sum is what we obtain from
$$(1 + 2)^n = 3^n.$$
\itemitem{d.}  What is $\sum_{k=0}^n (-1)^k {n \choose k}$?  Explain.
 \itemitem{} {\it Solution:}  From the binomial theorem this sum is what we obtain from
$$(1 - 1)^n = 0^n = 0.$$
\item{3.}  For each of the following statements, say whether the statement is true or false.
If it is false, give a counterexample; if it is true, give a short reason.
\itemitem{a.}  A set $A \subset {\bf R}$ is bounded if there exists some $B > 0$ such
that $|x| \le B$ for all $x \in A$.
\itemitem{} {\it Solution:}  TRUE -- the set is bounded above by $B$ and below by $-B$.
\itemitem{b.}  If $A, B\subset {\bf R}$ are bounded, then $A \cap B$ is also bounded.
\itemitem{} {\it Solution:}  TRUE -- If $A$ is bounded above by $M_A$ and $B$ is bounded
above by $M_B$, then $A \cap B$ is bounded above by $\min(M_A,M_B)$.  Similarly,
if $A$ is bounded below by $m_A$ and $B$ is bounded
below by $m_B$, then $A \cup B$ is bounded above by $\max(m_A,m_B)$.
\itemitem{c.}  If $A, B\subset {\bf R}$ are bounded, then $D = \{x + y \mid x \in A, y \in B\}$
is also bounded.
\itemitem{} {\it Solution:}  TRUE -- Suppose $A$ is bounded above by $M_A$ and $B$ is bounded
above by $M_B$.  Similarly, suppose $A$ is bounded below by $m_A$ and $B$ is bounded
below by $m_B$.  Then for all $x \in A$ and $y \in B$ we have $m_A \le x \le M_A$ and
$m_B \le y \le M_B$.  It follows that $m_A + m_B \le x + y \le M_A + M_B$.  Therefore
$D$ is bounded.
\itemitem{d.}  If $A, B\subset {\bf R}_{>0}$ are bounded, then $Q = \{y/x \mid x \in A, y \in B\}$
is also bounded.
\itemitem{} {\it Solution:}  FALSE -- Let $A = (0,1)$ and let $B = \{1\}$.  The set $Q$
is the set $\{1/y \mid y \in (0,1)\}$ which is not bounded above.
\item{4.}
\itemitem{a.}  Let $A = [0,3) \cap (2,5]$.  What is $a = {\rm lub}\ A$?  What is $b = {\rm glb}\ A$?
Are $a,b \in A$?
\itemitem{} {\it Solution:}  We have $A = (2,3)$.  Therefore $a = 3$, which is not in $A$. Similarly,
$b = 2$ is not in $A$ either.
\itemitem{b.}  Let $B = \{x \in {\bf R} \mid 0 \le x^2 - 2x + 1 \le 1\}$.  What is $a = {\rm lub}\ B$?  What is $b = {\rm glb}\ B$?   Are $a,b \in B$?
\itemitem{} {\it Solution:}  $B$ is the set where $(x - 1)^2 \le 1$, or $x \in [0,2]$, so $a =2$ and $b = 0$ are both
in $B$.   
\bigskip
\noindent
{\it `B` Section}
\bigskip
\item{1.}  Let $x,y$ be any real numbers.
\itemitem{a.}  Show that $|x| - |y| \le |x - y|$ and deduce that $||x| - |y|| \le |x - y|$.
\itemitem{} {\it Solution:}  From the usual triangle inequality,
$$|x| = |(x-y) + y| \le |x - y| + |y|.$$
Subtracting, we obtain $|x| - |y| \le |x - y|$ as desired.  Similarly, reversing the
roles of $x,y$, we have $|y| - |x| \le |y - x| = |x - y|$.  Since either $|x| \ge |y|$
or $|y| \ge |x|$ is true, we have either $||x| - |y|| = |x| - |y|$ or $||x| - |y|| = |y| - |x|$.
Since both of those are $\le |x - y|$, it follows that $||x| - |y|| \le |x - y|$
as desired.
\itemitem{b.}  Show that if $x,y > 0$, then $x < y$ is equivalent to $x^n < y^n$ for all $n \ge 1$
\itemitem{} {\it Solution:} $\Rightarrow:$ We argue by induction on 
$n$.  The base case is the same as the hypothesis so there is nothing to prove. 
Assume $x^k < y^k$ for some positive integer $k$.  Then since $x > 0$, we can 
multiply by $x$ on both sides to get
$x^{k+1} < y^k x$.  Similarly, we can multiply the base case $x < y$ by $y^k > 0$ on
both sides to get $xy^k < y^{k+1}$.   But then transitivity of the order relation 
implies $x^{k+1} < y^{k+1}$.  This shows that $x^n < y^n$ for all $n\ge 1$ by induction.
Conversely, if $x^n < y^n$ for all $n \ge 1$, then $x < y$ directly from 
the case $n = 1$ (!)
\itemitem{c.}  Show that if $0 < x < y$, then $\sqrt{y} - \sqrt{x} < \sqrt{y - x}$.
\itemitem{} {\it Solution:}  By part b with $n = 2$, since $\sqrt{y} - \sqrt{x} > 0$ and
$\sqrt{y - x} > 0$, it suffices to show that
$$(\sqrt{y} - \sqrt{x})^2 < (\sqrt{y - x})^2.$$
But the left side here is $y - 2\sqrt{y}\sqrt{x} + x$ and the right side is $y - x$.
We have
$$(y - x) - (y - 2\sqrt{y}\sqrt{x} + x) = 2\sqrt{y}\sqrt{x} - 2x = 2\sqrt{x}(\sqrt{y} - \sqrt{x}).$$
This is $>0$ because of the assumption $y > x$ and part b.  Hence the desired inequality follows.
\item{2.}  Let $a,b$ be any real numbers.  Define $\max(a,b)$ and $\min(a,b)$
to be the larger and smaller of the two numbers, respectively.  (That is, $\max(a,b) = a$
if $a \ge b$ and $\max(a,b) = b$ if $b \ge a$.  Similarly for the minimum.)
Show that
$$\max(a,b) = {a + b\over 2} + {|a - b|\over 2}$$
and
$$\min(a,b) = {a + b\over 2} - {|a - b|\over 2}.$$
\item{} {\it Solution:}  Geometrically, the average ${a + b\over 2}$ is the midpoint
of the line segment along ${\bf R}$ between $a$ and $b$ and $|a - b|$ is the
distance between $a$ and $b$ (the length of the line segment).  So starting at
the midpoint and going $1/2$ the distance to the right gives the maximum of the
endpoints, and going $1/2$ the distance to the left gives the minimum of the endpoints.
More analytically, we can also prove these by breaking into cases.  Suppose first that
$a \ge b$ so $a$ is the right endpoint.  Then $|a - b| = a - b$, and
$${a + b\over 2} + {|a - b|\over 2} = {a + b\over 2} + {a - b\over 2} = a = \max(a,b),$$
while
$${a + b\over 2} + {|a - b|\over 2} = {a + b\over 2} - {a - b\over 2} = b = \min(a,b).$$
If $b$ is the maximum and $a$ is the minimum, then
$${a + b\over 2} + {|a - b|\over 2} = {a + b\over 2} + {b - a\over 2} = b = \max(a,b),$$
while
$${a + b\over 2} - {|a - b|\over 2} = {a + b\over 2} - {b - a\over 2} = a = \min(a,b).$$
\item{3.}  Show by mathematical induction that
$$(1 + x)^n \ge 1 + nx$$
for all $x > -1$ and all $n\ge 0$.
\item{} {\it Solution:}  With $n = 0$, we have $1 = 1$, so the base case is established.
Now assume the inequality has been proved for $n = k$ and consider the next case $n = k+1$.
We have by the induction hypothesis
$$(1 + x)^k \ge 1 + kx.$$
Since $x > -1$, $1 + x > 0$, so multiplying this factor on both sides we get
$$(1 + x)^{k+1} \ge (1 + kx)(1 + x) = 1 + (k+1)x + kx^2 \ge 1 + (k+1)x,$$
since $k \ge 0$ and $x^2 \ge 0$.  This establishes the desired inequality
in all cases.
\item{4.}  Find a suitable $n_0$ and then show by mathematical induction that $n! \ge 5^n$ for all $n \ge n_0$.
\item{} {\it Solution:}  The smallest $n_0$ such that $n_0! \ge 5^{n_0}$ is $n_0 = 12$ since $12! = 479,001,600$, while
$5^{12} = 244,1140,625$.  This gives the base case here:  $12! > 5^{12}$ is
true.  For the induction step, assume $k! > 5^k$ and consider $(k+1)!$.
We see, by the induction hypothesis (and since $k + 1 > 5$ whenever $k \ge 12$):
$$(k+1)! = (k+1)k! > (k+1) 5^k > 5\cdot 5^k = 5^{k+1}.$$
\bye
