\magnification=\magstep1
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Solutions for Problem Set 9 -- {\bf Due:}  Friday, April 25}
\bigskip
\noindent
{\it `A' Section}
\bigskip
\noindent
1.  Let $f(x) = x^2 + 4x - 1$ on $[2,3]$.
\bigskip
\item{(a)}  Show that $f$ is integrable on $[2,3]$ directly using
the definition (that is, do not use any general theorems giving criteria for
integrability).  Hints:  Use regular partitions of $[2,3]$, and the summation
rules
$$\sum_{i=1}^n i = {n(n+1)\over 2}\ {\rm and}\ \sum_{i=1}^n i^2 = {n(n+1)(2n+1)\over 6}.$$
\item{} {\it Solution:}  Consider the regular partition ${\cal P}_n$
of the interval $[2,3]$, so $x_i = 2 + {i\over n}$ for $i = 0,1,\ldots,n$.  We have 
$f'(x) = 2x + 4 > 0$ for all $x \in [2,3]$.  Hence $f$ is increasing on the interval
and thus $m_i$ comes from evaluating at the left endpoints and $M_i$ comes from 
evaluating at the right endpoints:
$$\eqalign{
L(f,{\cal P}_n) &= 
\sum_{i = 0}^{n-1} \left( \left(2 + {i\over n}\right)^2 + 4\left(2 + {i\over n}\right) - 1 \right) {1\over n}\cr
&= {1\over n^3} \sum_{i=0}^{n-1} i^2 + {8 \over n^2} \sum_{i=0}^{n-1} i + {11 \over n} \sum_{i=0}^{n-1} 1\cr
&= {(n-1)n(2n - 1)\over 6n^3} + {8(n-1)n\over 2n^2} + 11.\cr}$$
Similarly, evaluating at the right endpoints,
$$\eqalign{
U(f,{\cal P}_n) &= \sum_{i = 1}^{n} \left( \left(2 + {i\over n}\right)^2 + 4\left(2 + {i\over n}\right) - 1 \right) {1 \over n}\cr
&= {1\over n^3} \sum_{i=1}^{n} i^2 + {8\over n^2} \sum_{i=1}^{n} i + {11\over n} \sum_{i=1}^{n} 1\cr
&= {n(n+1)(2n + 1)\over 6n^3} + {8 n(n+1)\over 2n^2} + 11.\cr}$$
Hence 
$$U(f,{\cal P}_n) - L(f,{\cal P}_n) = {9 \over n}.$$
We can make this $< \varepsilon$ for any $\varepsilon > 0$ by letting $n > {9\over \varepsilon}$. Hence $f$ is integrable by our definition. 
\bigskip
\item{(b)}  Determine the value of $\int_2^3 x^2 + 4x - 1\ dx$.  
\bigskip
\item{} {\it Solution:}  The value of the integral is
$$\lim_{n\to \infty} U(f,{\cal P}_n) = \lim_{n\to\infty} {n(n+1)(2n + 1)\over 6n^3} + {8 n(n+1)\over 2n^2} + 11 = {1\over 3} + {8\over 2} + 11 = {46\over 3}.$$
\noindent
(This can be checked by the FTC:
$$\eqalign{
\int_2^3 x^2 + 4x - 1\ dx &= \left. {x^3\over 3} + {4x^2 \over 2} - x\right|_1^2\cr 
&= \left({27\over 3} - {8\over 3}\right) + (18 - 8) + (3 - 2)\cr
&= {46\over 3}.)\cr}$$
\bigskip
\noindent
2.   Let $F(x) = \int_0^x f(t)\ dt$ where
$$f(t) = \cases{1 - t & if $0 \le t \le 1$ \cr
                1 + t & if $1 < t \le 2$\cr}$$
\item{(a)} Find an explicit formula for $F(x)$ valid for all $0 \le x \le 2$.
(Use the FTC in an appropriate way for this.)
\bigskip
\item{} {\it Solution:}  Using the FTC, we have 
$$F(x) = \left. t - {t^2\over 2}\right|_0^x = x - {x^2\over 2}$$
if $0 \le x \le 1$.  Note that the value at $x = 1$ is $F(1) = {1\over 2}$.  Then
for $x$ between $1$ and $2$, we want an antiderivative that takes the value
${1\over 2}$ at $x = 1$, so 
$$F(x) = x + {x^2\over 2} - 1$$
for $1 \le x \le 2$.  Therefore, putting these together:
$$F(x) = \cases{x - {x^2\over 2} & if $0 \le x \le 1$\cr
                {x^2\over 2} + x - 1 & if $1 < x \le 2$\cr}$$
(The constant term in the formula on 
the second half of the interval ensures that $F$ is continuous on $[0,2]$.)
\bigskip  
\item{(b)} Is $F$ differentiable at $x = 1$?  Why or why not?  Does this contradict
the first part of the FTC?
\bigskip
\item{} {\it Solution:}  No, $F$ is not
differentiable at $x = 1$, since
$$\lim_{x \to 1^-} {F(x) - F(1) \over x - 1} = \lim_{x \to 1^-} {{x^2\over 2} - x + {1\over 2}\over x-1} = \lim_{x\to 1^-} {1\over 2}x - {1\over 2} = 0,$$
but 
$$\lim_{x \to 1^+} {F(x) - F(1) \over x - 1} = \lim_{x \to 1^+} {{x^2\over 2} + x - {3\over 2} \over x - 1} = 
\lim_{x\to 1^+} {1\over 2}(x + 3) = 2.$$
This does not contradict the first part of the FTC since $f$ is not continuous on $[0,2]$.
It has a jump discontinuity at $x = 1$ since $\lim_{t\to 1^-} f(t) = 0$, but
$\lim_{t \to 1^+} f(t) = 2$.  
\bigskip
\noindent
3.  Find the following derivatives using the FTC (and other derivative rules as needed):
\bigskip
\item{(a)}  ${d\over dx} \int_1^x {e^t \over t}\ dt$  (Note that the integrand is 
not defined at $t = 0$, so $0$ should not be included in the interval of integration, 
or else the FTC does not apply.)
\bigskip
\item{} {\it Solution:}  The first part of the FTC implies 
the derivative is ${e^x \over x}$.  
\bigskip
\item{(b)}  ${d\over dx} \int_{-x^2}^{x^4} \cos(e^u) \ du$ 
\bigskip
\item{} {\it Solution:} By the first part of the FTC and the chain rule, the derivative
is 
$$4x^3 \cos(e^{x^4}) + 2x \cos(e^{-x^2}).$$
\bigskip
\noindent
{\it `B' Section}
\bigskip
\noindent
1.  Let $f$ be integrable on the interval $[a,b]$, and assume 
$\int_a^b f(x)\ dx < 0$.  Show that there exist $k < 0$ and an 
interval $[c,d] \subseteq [a,b]$ such that $f(x) < k < 0$ for
all $x \in [c,d]$.  
\bigskip
\noindent
{\it Solution:}  If the function is integrable,
then the definition of integrability 
implies that for every $\varepsilon > 0$, there exists 
a partition ${\cal P}$ such that 
$$U(f,{\cal P}) - L(f,{\cal P}) < \varepsilon. \leqno(1)$$
Moreover, it is always true that
$$L(f,{\cal P}) \le \int_a^b f(x)\ dx \le U(f,{\cal P}).\leqno(2) $$
Since the integral is strictly negative, this means that if we
take $0 < \varepsilon < \left|\int_a^b f(x)\ dx\right|$, it follows from (1) and (2) that 
$$U(f,{\cal P}) < L(f,{\cal P}) + \varepsilon < 
\int_a^b f(x)\ dx + \left|\int_a^b f(x)\ dx\right| = 0.$$
In other words, the lower sum $U(f,{\cal P}) < 0$ for some partition.  This 
can only happen if $M_i =  {\rm glb}\{f(x)\mid x\in [x_{i-1},x_i]\} < 0$ 
over some interval $[x_{i-1},x_i] \subset [a,b]$ from the partition 
$\cal P$.  Hence $f(x) \le M_i < M_i/2 < 0$ for all 
$x \in [c,d] = [x_{i-1},x_i]$.  We can 
take $k = m/2$ to get the strict inequality desired.   
\bigskip
\noindent
2.  Let $f$ be continuous on $[a,b]$.  
Show that there exists $c \in [a,b]$ such that 
$$\int_a^b f(x)\ dx = f(c)(b - a).$$
(Hint: Look at Theorem 5.2.5 (b).)
\bigskip 
\noindent
{\it Solution:}  Since $f$ is continous on $[a,b]$, by the Extreme
Value Theorem, it attains a maximum $M$ and a minimum $m$ on the interval
By the theorem indicated, 
$$m (b - a) \le \int_a^b f(x)\ dx \le M(b - a),$$
so
$$m \le {1\over b - a} \int_a^b f(x)\ dx \le M.$$
But then by the Intermediate Value Theorem (in the second version
we discussed), it follows that 
$${1\over b - a} \int_a^b f(x)\ dx = f(c)$$
for some $c\in [a,b]$.  Hence after multiplying through again by $b - a$, we
get:
$$\int_a^b f(x)\ dx = f(c)(b - a)$$
for some $c \in [a,b]$.  (Comment: this result is sometimes called the 
Mean Value Theorem for Integrals.  It is related to the method
of computing the {\it average value} of a function $f$ on $[a,b]$ that you 
may have seen as an application of integration in Calculus 2.)
\bigskip
\noindent
3.  Logarithm and Exponential.  In this problem, we will construct the 
natural logarithm and and the exponential function $e^x$ ``from scratch,'' without
relying on intuition about exponentials (as you probably did in calculus). 
We start by considering the function
$$L(x) = \int_1^x {1\over t}\ dt,\leqno(1)$$
for $x > 0$.  Note that ${1\over t}$ is continuous on $(0,+\infty)$,
hence the FTC applies to show that $L$ is a differentiable function. 
From calculus you probably recognize that $L(x) = \ln(x)$.  We want
to show directly that this makes sense and use this function 
to construct the inverse function $E(x) = L^{-1}(x)$ which is callled $E(x) = e^x$.  
Why would we proceed ``backwards'' like this?  The issue is that, while
$a^{m/n} = (a^{1/n})^m$ makes immediate sense for any positive $a \in {\bf R}$
and any rational exponent, what does $a^x$ actually mean if 
$x \notin {\bf Q}$?  Instead of trying to define that directly, we
will take an end run around the question.
\bigskip
\item{(a)}  We claim that the function $L$ ``has the right property to 
be a logarithm'' -- namely that $L(x\cdot x') = L(x) + L(x')$ for all $x, x' > 0$.
Prove this by showing that for any constant $x' > 0$, the function $L(xx')$ 
also satisfies ${d\over dx} L(xx') = {1\over x}$ for all $x > 0$.  Deduce
that $L(xx') = L(x) + c$ for some constant $c$, then determine $c$ by 
substituting an appropriate value for $x$.  
\bigskip
\item{} {\it Solution:}  By the first part of the FTC and the chain rule, 
we have ${d\over dx}L(xx') = {1\over xx'}\cdot x' = {1\over x}$.  Since
$L(x)$ also satisfies $L'(x) = {1\over x}$, this implies by one of our
corollaries of the Mean Value Theorem that $L(xx') - L(x) = c$ is a constant.
By definition $L(1) = 0$, so if we substitute $x = 1$, we get $c = L(x')$. 
Therefore $L(xx') = L(x) + L(x')$.  
\bigskip
\item{(b)}  Show that $L(x)$ is strictly increasing for $x > 0$, hence
is a 1-1 function on the domain $(0,+\infty)$.   Hence $L$ has an inverse function
$E : {\bf R} \to \{x \in {\bf R} \mid x > 0\}$.    
\bigskip
\item{} {\it Solution:}  Most of this follows since $L'(x) = {1\over x} > 0$
on $(0,+\infty)$, by the first part of the FTC.  Hence $L$ is strictly increasing
by one of the corollaries of the MVT, hence 1-1.  
\bigskip
\item{} {\it Comment:}  To give every detail here, we should also 
show that the range of $L$ is ${\bf R}$ to show that the domain of the inverse
function is actually equal to ${\bf R}$.  To see this, we can ``borrow''
from the fact that the harmonic series is divergent, like this:  We want to 
show first that $\{L(x) \mid x \in (0,\infty)\}$ is not bounded above.   Let 
$x = n$ be an integer $n > 1$.   By 
considering the lower sum of $\int_1^x {dt\over t}$ for the partition
${\cal P} = \{1,2,\ldots, n\}$ for the interval $[1,n]$, we can see
that 
$$L(n) = \int_1^n {dt \over t} > {1\over 2} + \cdots + {1\over n} 
= L\left({1\over t},{\cal P}\right).$$
As $n \to \infty$, though, the right side increases without any bound, since
the harmonic series diverges.  Hence the numbers $L(n)$ have no upper
bound either.  On the other hand by part (a) we have 
$$0 = L(1) = L\left(n \cdot {1\over n}\right) = L(n) + L\left({1\over n}\right).$$
Hence the values $L\left({1\over n}\right)$ are not bounded below either
(they go to $-\infty$ as $n \to \infty$).  Since $L$ is continuous, its range
must equal $(-\infty,\infty) = {\bf R}$ (by an application of the Intermediate
Value Theorem -- do you see why?)  
\bigskip
\item{(c)}  Show that the inverse function $E$
satisfies the equation $E(x + x') = E(x)\cdot E(x')$, hence $E$ looks like an 
exponential function.  
\bigskip
\item{} {\it Solution:}  By the definition of an inverse function we have
$$y = E(x) \Leftrightarrow x = L(y).$$
Let $y,y'$ be any two positive real numbers and let $x = L(y)$, $x' = L(y')$, 
so $y = E(x)$ and $y' = E(x')$.   
By part (a) we know that 
$$L(yy') = L(y) + L(y').$$
Hence 
$$E(x)E(x') = yy' = E(L(y) + L(y')) = E(x + x').$$
\item{(d)}  We define $x = e$ as the unique solution of the equation $L(x) = 1$.
Show that $x = e^{m/n}$ satisfies $L(x) = m/n$ for all rational numbers $x = m/n$.
\bigskip
\item{} {\it Solution:}  The given information says $L(e) = 1$.  
First  we claim $L(e^m) = m$ for all $m \in {\bf N}$.  This follows 
by induction since $L(e^1) = 1$ by definition, and if $L(e^k) = k$, then
$L(e^{k+1}) = L(e^k\cdot e) = L(e^k) + L(e) = k + 1$ by part (a) and 
the induction hypothesis.

\item{}  Next,  
since $L(e^m) + L(e^{-m}) = L(e^m\cdot e^{-m}) = L(1) = 0$, we have $L(e^{-m}) = -m$, 
which shows that $L(e^m) = m$ for all integers $m$.  Next, by part (a)
again, if $n$ is a positive integer,  $1 = L(e) = L((e^{1/n})^n) = n L(e^{1/n})$.
So $L(e^{1\over n}) = {1\over n}$ whenever $n$ is a positive integer.  Finally,
combining all of these steps we can write any rational number $m/n$ as a fraction 
in which $n > 0$ and then
$$L(e^{m/n}) = L((e^{1\over n})^m) = m L(e^{1/n}) = {m\over n}L(e) = {m\over n}.$$
\item{(e)}  Show that the inverse function $E(x)$ of $L(x)$ is 
differentiable and satisfies $E'(x) = E(x)$.   Hint: Look at Section 4.4.
What happens if you differentiate on both sides of the equation $L(E(x)) = x$?
\bigskip
\item{} {\it Solution:}  Theorem 4.4.4 implies that $E$ is differentiable at all
real $x$, since $L'(x) = {1\over x}$ is never equal to zero.  Hence if we
differentiate the equation $L(E(x)) = x$ on both sides using the chain rule we get
$L'(E(x)) E'(x) = 1$.  But $L'(u) = {1\over u}$, so this says ${1\over E(x)}\cdot E(x) = 1$
for all $x$, so $E'(x) = E(x)$.  This is the familiar differentiation rule for the 
exponential function: ${d\over dx} e^x = e^x$.  
\bigskip
\item{(f)}  Show that 
$$E(x) = e^x = \lim_{n\to\infty} \left(1 + {x\over n}\right)^n$$ 
for all $x \in {\bf R}$.  Hint:  Take the logarithm and use L'Hopital's
Rule.  
\bigskip
\item{} {\it Solution:}  We have 
$$L\left(\left(1 + {x\over n}\right)^n\right) = n L\left(1 + {x\over n}\right) = 
{L\left(1 + {x\over n}\right)\over {1\over n}}.$$
This is the value of a ratio $g(u)/h(u)$ at $u = {1\over n}$.  The top 
is $g(u) = L(1 + ux)$ and the bottom is $h(u) = u$.  Both of these are
``nice functions'' differentiable at $u = 0$.   Note that 
$\lim_{u \to 0} g(u) = 0 = \lim_{u\to 0} h(u)$.  So we can apply the version 
of L'Hopital's Rule developed on Problem Set 8.  We have 
$g'(u) = {x\over 1 + ux}$ and $h'(u) = 1$.  The ratio $g'(u)/h'(u)$
does have a limit as $u \to 0$:
$$\lim_{u\to 0} {g'(u)\over h'(u)} = \lim_{u\to 0} {x\over 1 + ux} = x.$$
Therefore by L'Hopital's Rule we know 
$$\lim_{u\to 0} {g(u)\over h(u)} = x$$
as well.  Since $u = {1\over n} \to 0$ as $n\to \infty$, this implies
$$\lim_{n\to \infty} {L\left(1 + {x\over n}\right)\over {1\over n}} = x.$$
Since $L$ is differentiable function at all $x > 0$, it also continuous
everywhere.  Hence 
$$L\left(\lim_{n\to\infty}\left(1 + {x\over n}\right)^n\right) 
= \lim_{n\to\infty} L\left(\left(1 + {x\over n}\right)^n\right) = x$$
This implies 
$$\lim_{n\to\infty}\left(1 + {x\over n}\right)^n = E(x) = e^x.$$

\bye
