 \documentstyle{article}
\begin{document}
\title{The Miracles with Exponents}
\author{Kevin Iga}
\date{June 16, 1990}
\maketitle

\section{Preliminaries}

\subsection{Integer powers}
The way you were probably introduced to exponents was by repeated
multiplication, $a^5=a \times a \times a \times a \times a$.  It is,
in fact, a little reminiscent of the way that multiplication is
repeated addition.  We can, in fact, use this analogy to think about
exponents.  The nth term in an arithmetic series is $a+d \times
(n-1)$.  The nth term in a geometric series is $a \times r^{n-1}$.
There are other comparisons, but we'll leave this aside for now.

One thing you might have noticed was that this definition was rather
limited.  It only let you raise numbers to positive integer powers.
To raise numbers to negative integer powers, we have to look at what
happens to positive powers, then try reversing that.

We know that 

\[a \times a^n = \overbrace{a \times a \times \cdots \times a}^{n+1}
= a^{n+1} \]

That is, incrementing the exponent power by one is equivalent to
multiplying one more of the base.  In fact, we can think of
incrementing the exponent one by one, multiplying the bases one at a
time.

To undo this, of course, means to divide.  So

\begin{eqnarray*}
a^3 \div a & = & a^2 \\
a^2 \div a & = & a^1 = a \\
1 = a \div a & = & a^0 \\
\frac{1}{a} = a^0 \div a & = & a^{-1} \\
\end{eqnarray*}

In other words, $a^0 = 1$ and $a^{-1} = \frac{1}{a}$.  Likewise, we
can keep dividing to get $a^{-2}$ or any negative integer power.  But
we have forgotten one case: zero.  The way we derived negative integer
powers requires dividing by the base.  Indeed, if we believe our
result for negative powers, zero to any negative integer power would
be undefined.  Zero to the zero power is a little unclear.  It is
supposed to be zero divided by zero.  Can the zeroes cancel, or not?
In itself, zero divided by zero is undefined.  This is, however useful
when the numerator and denominator approach zero at different speeds,
and one may win over the other.

In summary, zero to a positive power is, of course zero, but zero to
the zeroth power, or to a negative power, is undefined.  All other
powers to the zeroth power result in one, while anything to a negative
power is the reciprocal to the equivalent positive power.

We can now understand the following formulas:

\begin{equation}
x^a x^b = x^{a+b}
\end{equation}
\begin{equation}
x^a \div x^b = x^{a-b}
\end{equation}
\begin{equation}
(x^a)^b = x^{ab}
\end{equation}

These are known as the ``laws of exponents'' and, although intuitive
now, when we are dealing with integer exponents, we will have to
depend on them blindly when dealing with more complicated exponents.

\subsection{Rational powers}

We have therefore explored exponentiation with integer exponents.
What about rational exponents?  We haven't defined that yet, and we
would like to define it in the most convenient way possible.  And what
would really be convenient is if we could still use the laws of
exponents.  Take a look at equation (3) above.  If we try to make this
apply to $a=q$ and $b=\frac{1}{q}$, the result is

\begin{equation}
(x^q)^{\frac{1}{q}} = x
\end {equation}

In other words, $x=y^{1/q}$ is the same thing as $y=x^q$.  We are
therefore justified in ``raising both sides to the q power'' and using
the third law of exponents, {\em even when the law was meant
originally for integers!\/}

As you know, this is also known as taking the qth root.  As you are
also aware, this raises many difficulties.  For one, there are two
square roots of a number.  If you allow yourself to wander in the
complex plane, there are n nth roots of a complex number.
Furthermore, if we are taking things to rational roots, we have to
reduce the exponent to lowest terms before deciding how many roots
there are.  In other words, it's a mess.

We choose to avoid all of that for now, and leave it to a complex
analysis course.  Instead, we will investigate logarithms.

\subsection{Logarithms}

Roots correspond to one of the ways of inverting exponentiation.  If
you have an equation like $a^b=c$ and want to solve for a, we can
write $a=c^{\frac{1}{b}}$.  There is another way of inverting
exponentiation: logarithms.  If we want to solve for b in the equation
above, we can write $b=\log_a c$, read, ``b is the logarithm of c,
base a.''

We can translate the three laws of exponents into the three laws of
logarithms.  Take the first one: $x^a x^b = x^(a+b)$.  This is the
same thing as $\log_x (x^a x^b) = a+b$.  Now if we let $u=x^a$
and $v=x^b$, which is the same thing as saying that $a=\log_x u$ and
$b=\log_x v$, we have $\log_x (uv) = \log_x u + \log_x v$.  We can do
the same thing for the other laws, yielding the following results:

\[
\begin{array}{ll}
Exponents & Logarithms \\
x^a x^b = x^{a+b} & \log_x{uv} = \log_x u + \log_x v \\
x^a \div x^b = x^{a-b} &\log_x{u \div v} = \log_x u - log_x v\\
(x^a)^b = x^{ab} & \log_x(u^b) = b\log_x u \\
\end{array}
\]

Note that exponents allow you to write addition as multiplication,
subtraction as division, and multiplication as exponentiation.
Likewise, logarithms allow you to write multiplication as addition,
division as subtraction, and exponentiation as multiplicaiton.

\section{Miracle number one}

After Newton and Leibnitz invented calculus, and derived that 

\begin{equation}
\frac{d}{dx}x^{n}=nx^{n-1}
\end{equation}

it was clear that all you had to do to integrate was reverse the
process.  Instead of first taking down the exponent then subtracting
one, we should first add one to the exponent, then divide by that new
exponent.

\begin{equation}
\int x^{n} = \frac{1}{n+1} x^{n+1}
\end{equation}

This seems to work fine until you try to integrate $x^{-1}$.  You end
up dividing by zero!  It obviously doesn't work.  But the integral of
$x^{-1}$ must {\em mean} something.  We intend to find that out now.

First, for easy reference, we should give it a name.  For now, we will
use the letter F.  An indefinite integral has an arbitrary constant.
It will not do to have this in a single-valued function.  So we need
to integrate over some bounds.  We might not want to integrate from
zero, since $1/x$ is not defined there.  So we will integrate from 1,
as the ``next nicest number'' to zero.  It doesn't really matter, but we're
making the definition here.

\begin{equation}
F(x)=\int_1^x \frac{dt}{t}
\end{equation}

It would take us a long time to find out what F(x) really is.  There's
no way you would get this off the top of your head.  It took a lot of
people a lot of trial and error.  I'll quickly give you many years of
effort, displaying in the important pieces of progress, hiding the
humiliating dead ends.

First, we try to find $F(x)+F(y)$.

\[F(x)+F(y)= \int_1^x \frac{dt}{t} + \int_1^y \frac{dt}{t}\]

Now we let $z=xt$ and substitute it in the second integral, where x is
a constant.  Note that we must change the limits of integration.  When
$t$ ranged from $1$ to $y$, $z$ ranges from $x$ to $xy$.  Note also
that we are substituting $z/x$ for $t$ in the second integral.

\begin{eqnarray*}
 & = & \int_1^x \frac{dt}{t} + \int_x^{xy} \frac{dz/x}{z/x}\\
 & = & \int_1^x \frac{dt}{t} + \int_x^{xy} \frac{dz}{z}\\
\end{eqnarray*}

Now since the $z$ in the second integral is just a dummy variable we
integrate over, and has nothing to do with the $t$ in the first
integral, we are free to rename it $t$.

\begin{eqnarray*}
 & = & \int_1^x \frac{dt}{t} + \int_x^{xy} \frac{dt}{t} \\
 & = & \int_1^{xy} \frac{dt}{t} \\
\end{eqnarray*}

\begin{eqnarray}
F(x)+F(y) & = & F(xy)\\
\end{eqnarray}

Now letting $z=xy$,

\[F(x)+F(z/x) = F(z)\]

or

\begin{equation}
F(x)-F(z) = F(z/x)
\end{equation}

By iterating either of these, we can show that

\[\overbrace{F(x)+F(x)+\cdots+F(x)}^n = F(\overbrace{x \times x \times
\cdots \times x}^n)\]

or

\begin{equation}
nF(x)=F(x^n)
\end{equation}

The last three numbered equations look suspiciously like the laws of
logarithms.  In fact, the laws of logarithms were derived from the
laws of exponents which completely characterize exponents.  What this
means is that our friend $F(x)$ is really a logarithm to some base.

This is the first miracle we will explore.  There's no real reason why
this should work, other than it does.  The integral of $1/x$ is a
logarithm.  Or the derivative of a particular logarithm is $1/x$.
It's strange.  But it will blossom into a bigger miracle in the next
section.

Notice that before, because we only had integer or sometimes rational
exponents, logarithms could only return rational numbers.  Now, the
logarithm is a continuous function that can take on all real values.
Our miracle has given us a continous logarithm where before we had
discrete points. But what base is this logarithm?  For now, we'll call
the base b.  Then, we'll differentiate it and require it to be $1/x$.

\begin{eqnarray*}
\frac{dF}{dx} & = & \lim_{\Delta x \rightarrow 0} \frac{\Delta F}{\Delta
x} \\
 & = & \lim_{\Delta x \rightarrow 0} \frac{F(x+\Delta x) -
F(x)}{\Delta x} \\
 & = & \lim_{\Delta x \rightarrow 0} \frac{\log_b(x+\Delta x) - \log_b
x}{\Delta x} \\
 & = & \lim_{\Delta x \rightarrow 0} \frac{\log_b \frac{x+\Delta
x}{x}}{\Delta x \frac{x}{x}} \\
 & = & \lim_{\Delta x \rightarrow 0} \frac{\log_b (1+\frac{\Delta
x}{x})}{\frac{\Delta x}{x} x} \\
 & = & \lim_{\Delta x \rightarrow 0} \frac{\log_b (1+\frac{\Delta
x}{x})^\frac{x}{\Delta x}}{x} \\
 & = & \frac{1}{x} \lim_{\frac{\Delta x}{x} \rightarrow 0} \log_b (1 +
\frac{\Delta x}{x})^\frac{x}{\Delta x} \\
\end{eqnarray*}

Substituting w for $\frac{\Delta x}{x}:$

\begin{eqnarray*}
 & = & \frac{1}{x} \lim_{w \rightarrow 0} \log_b (1+w)^\frac{1}{w} \\
\end{eqnarray*}

Now we wanted this to equal $\frac{1}{x}$, which already appears on
the right hand side.  All we need is for the remainder of the
expression to equal one.

\begin{eqnarray*}
1 & = & \lim_{w \rightarrow 0} \log_b (1+w)^\frac{1}{w} \\
 & = & \log_b \lim_{w \rightarrow 0} (1+w)^\frac{1}{w} \\
b^1 & = & \lim_{w \rightarrow 0} (1+w)^\frac{1}{w} \\
b & = & \lim_{w \rightarrow 0} (1+w)^\frac{1}{w} \\
\end{eqnarray*}

We have now found the base for the logarithm.  It's not a nice
expression, but it can be done.  It seems this is the base of
logarithm ``picked out'' by natural manipulations of calculus, so we
give the number b a name: $e$, after Leonhard Euler, who discovered
it.  The logarithm we write $\ln x$ and we call it the ``natural
logarithm of x''.

The number e can be calculated numerically, and it turns out to be
approximately 2.7182818284...  It is irrational, and cannot be written
in terms of nth roots.  Like pi, it seems that this number is somehow
favored by nature.  It's a part of the big miracle that we may one day
begin to appreciated more, but probably will never be able to really
comprehend.

\section{The 1st miracle in exponent form}

\subsection{Population explosion}

Suppose we assume na\"ively that a fixed fraction of the population
will have a certain number of kids.  How will the population expand?

Consider the problem in the following way: Let's say we have ten
couples in the world.  On average, say each couple will have 1.5 kids in
the next ten years.  Then, ten years later, we will have 20+1.5*10=35
people.  The next decade, we have about 17 couples, so we have about
$17 \times 1.5$ children.  So our new population is $35+17 \times
1.5$.

Now step back at what's happening here.  Each year, we're adding on a
new bunch of people.  The number we add is proportional to the number
we have.  So all we need to do is solve the problem of finding a
function that adds on to itself a chunk that is proportional to its
existing value.  In other words, the derivative (what it adds on to
itself) is proportional to its value.

\subsection{Chickens hatching}

Now suppose we have a dozen eggs.  We don't know when they will hatch.
We want to know how many are likely to be hatched at some later time.

We can think of this problem in the following way.  Suppose we start
out with a dozen unhatched eggs.  During the first second, we
(quickly!) flip a dozen coins, each weighted the same, say, with 25\%
probability of heads.  Each coin corresponds to an egg.  If the coin
comes up heads, we ``hatch'' the egg.  Otherwise we leave it alone.

Suppose 3 eggs hatched this time.  That means there are 9 left to
hatch.  During the next second, we flip 9 coins, and see which come up
heads.  Suppose we then find 3 coins flipped heads.  That means there
are 6 more.  And so forth.

Note that the number that hatched each time is approximately
proportional to the number of unhatched eggs.  Each time, the number
of unhatched eggs decreases by a chunk that is proportional to the
number of unhatched eggs.  In other words, the rate of change of the
number of unhatched eggs is proportional to itself.

\subsection{The answer}

The last two problems don't even scratch the surface of applications
of this selfsame problem.  The hatching of eggs can be thought of as
radioactive nuclei that haven't decayed yet.  Or molecules that
haven't reacted yet.  The population explosion not only works for
humans and bacteria, but for money in compound interest.  The list
goes on.

Finding a function whose derivative is proportional to itself is
important.  Here, we will find the answer!

Let f(x) be the function we are looking for, and let k be the constant
of proportionality.

\begin{eqnarray*}
\frac{df}{dx} & = & k f(x) \\
\frac{1}{f(x)} \frac{df}{dx} & = & k \\
\int \frac{1}{f(x)} \frac{df}{dx} dx & = & \int k \, dx \\
\int \frac{df}{f(x)} & = & kx + C \\
\log_e f(x) & = & kx + C \\
f(x) & = & e^{kx+C}\\
f(x) & = & e^C(e^k)^x
\end{eqnarray*}

Note that we used the first miracle in this proof, when we integrated
$\frac{df}{f(x)}$.  We can now let $A=e^C$ and let $B=e^k$.  The
result is:

\[f(x) = A \, B^x\]

In terms of our problem,

\begin{equation}
\frac{d}{dx} A\,B^x = Ake^x = A \ln B \, e^x
\end{equation}

In particular, if $A=1$ and $k=1$,

\begin{equation}
\frac{d}{dx} e^x = e^x
\end{equation}

In other words, $e^x$ is a function whose derivative is itself.
Exponents like $b^x$ are functions whose derivatives are proportional
to themselves.  So these are the solutions of the population problem
and the chicken hatching problem, as well as the others mentioned
earlier.  This is one of the many things that the first miracle has
brought us.

Not only this, $e^x$ allows us to take e to any real power, as the
inverse of $\ln x = \int_1^x\frac{dt}{t}$.  We can then use the third
law of exponents to write $a^x=(e^{\log_e a})^x=e^{x \ln a}$,
so we can take any positive real number to any real power.

\section{The first miracle in differential equations form}

The last section was a special case of differential equations.
In algebraic equations, we have functions in an equation, and we wish
to solve for a variable.  In differential equations, we have functions
{\em and derivatives\/} in an equation, and we wish to solve for a
function.  Here we will investigate a very special case: linear
homogeneous ordinary (one variable) differential equations with constant
coefficients.

A linear differential equation is written in the form

\[f(x) \frac{d^ny}{dx^n} + g(x) \frac{d^{n-1}y}{dx^{n-1}} + \cdots + 
u(x) \frac{dy}{dx} + v(x) y = 0 \]

and has many interesting properties.  It is interesting to note that
mathematicians have invented a plethora of strategies of dealing with
linear differential equations, but are notoriously stumped by
the nonlinear ones.  Even most linear differential equations cannot be
solved in closed form.  For that we need f(x), g(x), and so forth to
be constants.  The equation then looks like:

\[c_n \frac{d^ny}{dx^n} + c_{n-1} \frac{d^{n-1}y}{dx^{n-1}} + \cdots +
c_1 \frac{dy}{dx} + c_0 y = 0 \]

Linear differential equations have the nice property that if you have
two solutions $y_1(x)$ and $y_2(x)$, any linear combination of them
(like $y_1(x)+y_2(x)$ or $y_1(x)-4y_2(x)$ or $-49.3y_1(x)+32.1y_2(x)$)
is also a solution.  Furthermore, the order---the highest derivative
in the equation (in the equation above, the order is n)---is equal to
the number of independent solutions needed.  To understand this
requires a little bit of linear algebra, but let it suffice to say
that if we have an nth order differential equation, we only need to
find n solutions, provided that one of them is not really a linear
combination of the others.

\subsection{First order linear differential equations}

If the order is one, we have a simpler version of the equation above.

\[c_1 \frac{dy}{dx} + c_0 y = 0\]

Since this is first order, all we need to find is ONE solution, and
all solutions will be linear combinations of that one (which in this
case just means it is multiplied by an arbitrary constant).

If you look at it carefully, you will remember that we already solved
this in the last section.  It is the same as

\[\frac{dy}{dx} = -\frac{c_0}{c_1} y \]

and the solution is:

\[y=A e^{-\frac{c_0}{c_1} x} \]

\subsection{Second order differential equations}

Let's try something harder:

\[c_2 \frac{d^2y}{dx^2} + c_1 \frac{dy}{dx} + c_0 y = 0 \]

To be specific, let $c_2$ be 1, $c_1$ be -1, and $c_0$ be $-6$.  The
equation now looks like:

\[\frac{d^2y}{dx^2} - \frac{dy}{dx} - 6y = 0 \]
 
Now we need two solutions, since this is a second order linear differential
equation.  

I'm now going to go through a trick which would be hard to think of
until you've memorized the answer.  Don't pretend there's a motivation
to every step because some parts were pure luck.

\begin{eqnarray*}
\frac{d^2y}{dx^2} - \frac{dy}{dx} - 6 y & = & \frac{dy}{dx}
\frac{dy}{dx} - \frac{dy}{dx} - 6 y \\
& = & \frac{dy}{dx} \frac{dy}{dx} + 2 \frac{dy}{dx} - 3 \frac{dy}{dx} -6y \\
& = & \frac{d}{dx} (\frac{dy}{dx} + 2y) - 3 (\frac{dy}{dx} + 2 y)
\end{eqnarray*}

We now let u be the function $\frac{dy}{dx} + 2y$ and continue:

\begin{eqnarray*}
& = & \frac{du}{dx} -3u = 0
\end{eqnarray*}

\[u=Ae^{3x}=\frac{dy}{dx} + 2y\]
\[0 = \frac{dy}{dx} + 2y - Ae^{3x}\]

We now let $v=y-Ae^{3x}$, which means $y=v+Ae^{3x}$.

\begin{eqnarray*}
0 & = & \frac{dv}{dx} + 3Ae^{3x} +2v - 2Ae^{3x}-Ae^{3x} \\
  & = & \frac{dv}{dx}+2v \\
v & = & B e^{-2x} \\
y & = & A e^{3x} + B e^{-2x} \\
\end{eqnarray*}

If this all looks a bit mysterious, notice that in the first part I
merely factored the equation as if it were an algebraic equation.  In
fact, we can use the quadratic formula to factor

\[ax^2+bx+c=a(x-\frac{-b+\sqrt{b^2-4ac}}{2a})(x-\frac{-b-\sqrt{b^2-4ac}}{2a})\]

In this way, we can follow the same logic to tell us that the
solution to the differential equation

\[a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + c y = 0 \]

is the following:

\[y=Ae^{r_1x}+Be^{r_2x}\]

where $r_1$ and $r_2$ are roots of $ax^2+bx+c=0$.

By using this factoring method, we can generalize this to the
following: if we have a linear homogeneous ordinary differential
equation with constant coefficients, we write down an associated
algebraic polynomial with the same coefficients, then try to solve
THAT equation.  Take the roots and put them in the exponents as in the
case with the second order differential equations, and multiply by an
arbitrary constant and add the solutions.

This is all a result of the fact tha $e^x$ is the same no matter how
many times you differentiate it, allowing it to cancel itself within
the differential equation---a result derived from the first
miracle of exponentiation.

\subsection{Complex roots in the second order differential equation}

Look at the following differential equation:
\[\frac{d^2y}{dx^2}+1=0\].

If we try to apply the same method as last time, we get $\pm i$ as
roots.  Our result, then, is

\[y=Ae^{ix} + Be^{-ix}\]

What can it mean to raise e to an imaginary power?  We will answer
this question later.  First I'll drop the suspense and reveal the
solution to the differential equation.

You may recall that the derivative of $sin$ is $cos$ and that the
derivative of $cos$ is $-sin$.  The result of this is that the {\em
second derivative} of $sin$ is $-sin$ and the second derivative of
$cos$ is $-cos$.  If you look at the equation at hand, you will notice
that both $sin$ and $cos$ satisfy the differential equation.  $Sin$
and $cos$ are also independent solutions, so we have all the solutions
we need.

\[y=A \sin x + B \cos x\]

\end{document}
