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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Problem Set 9 -- {\bf Due:}  Friday, April 25}
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{\it `A' Section}
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\noindent
1.  Let $f(x) = x^2 + 4x - 1$ on $[2,3]$.
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\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 $[1,2]$, 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{(b)}  Determine the value of $\int_2^3 x^2 + 4x - 1\ dx$.  
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\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.)
\item{(b)} Is $F$ differentiable at $x = 1$?  Why or why not?  Does this contradict
the first part of the FTC?
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3.  Find the following derivatives using the FTC (and other derivative rules, as needed):
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\item{(a)}  ${d\over dx} \int_0^x {e^t \over t}\ dt$
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\item{(b)}  ${d\over dx} \int_{-x^2}^{x^4} \cos(e^u) \ du$ 
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{\it 'B` Section}
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\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 $x \in [c,d]$.  
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\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.)
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\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.
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\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 $x$.  
\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
$E : {\bf R} \to \{x \in {\bf R} \mid x > 0\}$.    
\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.  
\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 $x = m/n$.
\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 $L(E(x)) = x$?
\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  a logarithm and use L'Hopital's
Rule.  
\bye
