\centerline{\bf How to $k$-Sect an Angle}
\centerline{And Save Your Marriage}
\centerline{Easy Method}
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{\sl Stephen Judd
Los Alamos National Laboratory / Northwestern University
judd@sgt-york.lanl.gov
7/7/93}
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{\bf Abstract}
Haven't we all had this happen? You're out on a date and your date
hands you a compass, a straightedge, and a piece of paper with an angle drawn
on it, and then asks you to divide it into $k$ equal parts, where $k$ is some
integer. $x^n + y^n \ne z^n$ was a breeze,
and squaring the circle took a little thought, but now it looks like the
gig is up: sooner or later your date is going to find out your deepest,
darkest, most secretest secret that ever reared it's black ugly head out of
the inner sanctum of your soul.
Sooner or later,
your date will find out that you're not really a mathematician but
were only masquerading as one, since (as everyone knows)
it's the only way to get a really good-looking date.
But fear no longer!
In a very famous and well-known paper, written earlier, a now
much celebrated method was presented for trisecting an angle very quickly.
This paper presents a generalization of the earlier (very famous and well-known)
paper to provide a general algorithm for $k$-secting an angle, using a
compass and a straightedge.
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{\bf General and Sufficiently Vague Idea}
So you're on a date, and you need to $k$-sect this angle $\alpha$,
where $k$ is some integer. First step: don't panic. Relax. Pick a
number $q$, such that
$$q = k+1.$$
Next, $q$-sect the angle. You might have to work backwards, and
figure out how to $q$-sect the angle, but you can always reduce it to
bi-secting the angle (then trisecting, etc.). Now you have an angle of
size
$${1\over q}\alpha.$$
It turns out that instead of doing the above, you could just start by
guessing what the $k$-section angle would be, and continue the iterations
described below, with this guess as the new angle (this is explained later).
Add this new angle to the original angle $\alpha$ and $q$-sect again,
giving a new angle of
$${q+1\over q^2}\alpha .$$
Continue this process as long as you like; the next three terms are:
$${q^2+q+1\over q^3}\alpha$$
$${q^3+q^2+q+1\over q^4}\alpha$$
$${q^4+q^3+q^2+q+1\over q^5}\alpha$$
and in general, after $N$ iterations, the factor multiplying the angle will be
$${1\over q^N}\sum_{i=0}^{N-1} q^i.\eqno(1)$$
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{\bf Hmmmmmm...}
Consider the series
$$\sum_{i=0}^{N-1} q^i.$$
If we multiply it by $q-1$, we can write an identity
$$(q-1)\sum_{i=0}^{N-1} q^i = q^{N} - 1$$
or
$$\sum_{i=0}^{N-1} q^i = {q^N - 1\over q-1}.$$
If we then substitute this expression into equation (1), we have
$$\eqalign{{1\over q^N}\sum_{i=0}^{N-1} q^i
&= {1\over q^N}\Bigl({q^N - 1\over q-1}\Bigr)\cr
&= {1\over q-1}\Bigl({1 - {1\over q^N}}\Bigr)}$$
Any child of eight can tell you that as $N$ gets larger this sequence
gets closer and closer to $1\over q-1$, so that as $N$
goes to infinity this sequence converges to
$$1\over k.$$
Thus, you have now successfully $k$-sected the angle $\alpha$. If you
want to stop after $N$ iterations, the error will be
$${1\over k}\Bigl({1\over k+1}\Bigr)^N$$
\centerline{\bf Q.E.D.}
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{\bf Conclusions and Rationalization }
This method converges very quickly. A 9-sect will gain an order of
magnitue accuracy with every iteration.
To $k$-sect an angle, you really only need to be able to prime-sect
an angle, and you can always work backwards to bisecting the angle if you
need to. For instange, to 11-sect an angle, the method requires a 12-sect.
A 12-sect consists of two bisections and a trisection. A trisect requires a
quad-sect, or two bisections. "But dost this not err thy method?" you may
say. The answer is yes, but in the addendum we find that errors can be
introduced and the method will still work. So, in fact, if speed is a concern
you only need to guess at the trisection.
An interesting thing to notice is that for any
integer $k$, the number $k^N$ can always be expressed as $c(k-1) + 1$, that
is, that any integer raised to an integer power can be expressed as a
constant times the integer below it, plus one. One case of this was
presented in the earlier paper on trisection. I just thought this was
kinda neat.
An alternative way of writing each iteration is
$$x_N = \sum_{i=1}^N {1\over q^i}$$
which means that this summation converges to
$$1\over q-1$$
in the limit, naturally as long as $|q| > 1$.
A very important question to ask is what if your date is a prude and
thinks you have
cheated, are faking, and that the answer is not exact? Well then, that really
depends on what you mean by exact.
In a Euclidian sense I
can construct a line that is ``exactly'' $\sqrt{2}$ long
(or for that matter $\sqrt{n}$, naturally for $n$ integer,
which is left as an excercise to the interested student). What is not
clear is how to find the cube root of a number, or the $n^{\rm th}$ root.
In a similar way
I can calculate exactly and in finite time the trisection of some angles
(multiples of $45^\circ$ and $108^\circ$, for instance, again left as an
excecise to the interested student). Both tasks have limitations. So if
your date balks, just ask your date to calculate the cube root of ten.