%% Garbage In, Garbage Out, Sigplan Notices 33,6 (June 1998), 27-31.
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\ColumnTitle{You Could Learn a Lot from a Quadratic:\\
Newton Squares the Circle
%Newton's Revenge\\
%Newton's Imagination
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{http://home.pipeline.com/\char'176hbaker1/home.html}{hbaker1@pipeline.com}
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\ColumnAuthor{Henry G. Baker}{%
Henry Baker 
has been diddling bits for 35 years, with time off for
good behavior at MIT and Symbolics.  In his spare time, he collects
garbage and tilts at windbags.  This column appeared in ACM {\it Sigplan Notices} 33,6 (Jun~1998), 27-31.
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``Newton's Method'' for the iterative approximation of square
roots---not just on the real line, but also in the complex
plane---provides one of the most beautiful and elegant applications of
`linear fractional transformations,' also known as `M\"obius
transformations.'

Briefly, Newton's Method for the square root of the real number $N$
consists of guessing an initial approximation $z_0$, and then
successively computing the members of the sequence
$z_1=(z_0+N/z_0)/2$, $z_2=(z_1+N/z_1)/2$, ...,
$z_{n+1}=(z_n+N/z_n)/2$.  As we have discussed in previous columns,
this iteration converges rapidly to $\pm\sqrt{N}$ so long as $z_0\not=
0$, for positive $N$, real $z_0$.

But the real beauty and understanding of Newton's method becomes
evident by looking at its operation in the complex plane.  In other
words, we will consider the action of the Newton iteration for {\it
complex} numbers $z_n=x_n+iy_n$.  We will see that Newton's square
root method consists of a M\"obius transformation of the initial guess
to a new ``coordinate system'' for the complex plane, followed by a
number of simple {\it squarings} of the transformed initial guess,
followed by the inverse M\"obius transformation back to the original
coordinate system.

We must first review a few properties of complex numbers.  A
complex number $z$ may be expressed in either {\it rectangular} form
as $z=x+iy$, or in {\it polar} form as
$z=|z|e^{i\theta}=|z|(\cos\theta+i\sin\theta)$, where $|z|$ is the
{\it absolute value} of $z$, i.e., $|z|=|x+iy|=\sqrt{x^2+y^2}$, and
$\theta$ is the {\it arctangent} of $y/x$, i.e.,
$\theta=\atan\left(y/x\right)=\atanii(y,x)$.

If we square a complex number, we get $z^2=(x+iy)^2=(x^2-y^2)+i(2xy)$
in rectangular form, and $z^2=(|z|e^{i\theta})^2=|z|^2 e^{i2\theta}$
in polar form.  Thus, when squaring a complex number in polar form, we
note that its (real) absolute value is {\it squared}, and its angle
$\theta$ is {\it doubled}.

Consider, for example, those complex numbers whose absolute value is
exactly {\it one}, i.e., numbers $z$ such that $|z|=1$.  These numbers
form the {\it unit circle} in the complex plane.  In this case, $z=|z|
e^{i\theta}=e^{i\theta}$, so $z$ is exactly representable as a pure
imaginary exponential.  For these complex numbers,
$z=e^{i\theta}=\cos\theta+i\sin\theta$, so that
$$\eqalign{z^2&=e^{i2\theta}\cr
&=\cos 2\theta+i\sin 2\theta\cr
&=(\cos\theta+i\sin\theta)^2\cr
&=(\cos^2\theta-\sin^2\theta)+i(2\cos\theta\sin\theta)\cr}$$
This equation trivially shows the classic ``double angle'' formulas so
laboriously memorized by high school trigonometry students.

Squaring a complex number $z$ has the following property---if the
number $|z|>1$, then $|z^2|>|z|>1$, and additional squarings will grow
the absolute value towards infinity---i.e., $|z^{2^n}|$ grows very
fast.  Similarly, if $|z|<1$, then $|z^2|<|z|<1$, and additional
squarings will shrink the absolute value towards zero---i.e.,
$|z^{2^n}|$ shrinks very fast.  Finally, if $|z|$ is {\it exactly}
one, then squaring $z$ will not change this, and $z^{2^n}$ will remain
on the unit circle for all $n>1$.  Newton's square root method makes
essential use of this growing/shrinking property of squaring to
converge quickly to the correct value.

We are particularly interested in sets of complex numbers whose
absolute values are all the same---these form concentric {\it circles}
about the origin.  The squaring function maps one such circle into a
second such circle, with the mapping ``covering'' the image circle
twice.  If we square again, we map the original circle into a final
circle {\it four} times, and if we square the original circle $n$
times, we map the original circle into a final circle $2^n$ times.

If we {\it invert} a general complex number---i.e., perform the
operation $z\rightarrow 1/z$, then in polar form this operation
becomes $|z| e^{i\theta}\rightarrow (1/|z|) e^{-i\theta}$.  For the
case where $|z|=1$, $1/z=(1/|z|) e^{-i\theta}=e^{-i\theta}$, so that
$1/z$ is identical to the {\it conjugate} $\bar{z}=x-iy$ of $z=x+iy$.

The next complex operation we will need is that of `linear fractional
transformation' or `M\"obius transformation,' for short.  A M\"obius
transformation transforms $z\rightarrow(Az+B)/(Cz+D)$, where $A$, $B$,
$C$, and $D$ are complex numbers, and we will need the additional
condition that $AD-BC\not= 0$.  These transformations are elegant and
important because they are one-to-one and onto functions of the
complex plane extended with the additional point $\infty=1/0$.  These
transformations have inverses---e.g., the inverse transformation to
$z\rightarrow(Az+B)/(Cz+D)$ is $z\rightarrow(Dz-B)/(-Cz+A)$:
$$\eqalign{&{A(Dz-B)/(-Cz+A)+B\over C(Dz-B)/(-Cz+A)+D}\cr
&={A(Dz-B)+B(-Cz+A)\over C(Dz-B)+D(-Cz+A)}\cr
&={ADz-AB-BCz+AB\over CDz-BC-CDz+AD}\cr
&={(AD-BC)z\over AD-BC}\cr
&=z\cr}$$

Working with M\"obius transforms as ratios works, but we can see the
properties much more clearly if we write these transforms as $2\times
2$ matrices:
$$\left(\matrix{A&B\cr C&D\cr}\right)\left(\matrix{z\cr 1\cr}\right)=
\left(\matrix{Az+B\cr Cz+D\cr}\right)$$

Once we have this bidirectional mapping between M\"obius
transformations and matrices, the other properties of M\"obius
transformations become obvious.  Thus, the requirement that
$AD-BC\not= 0$ is simply the requirement that the {\it determinant} of
the $2\times 2$ matrix $\left(\matrix{A&B\cr C&D\cr}\right)$ be
non-zero, so that the matrix has an inverse.  Thus, the inverse of
this matrix is
$${1\over AD-BC} \left(\matrix{D&-B\cr -C&A\cr}\right)$$
However, since we intend to form the {\it ratio} of the two elements
of the final column matrix, we can safely ignore any ``scale
factors''---so long as any such scale factor (in this case $AD-BC$) is
non-zero.

One of the truly remarkable properties of M\"obius transforms is that
they {\it preserve circles}---i.e., given {\it any} circle in the
complex plane, a M\"obius transform will transform that circle into
another circle.  For this property to work, however, we must define
``circles'' as including the limiting case of {\it straight lines},
which are circles having an ``infinite radius.''

For example, the transform $z\rightarrow (z-1)/(z+1)$ transforms the
imaginary (or ``$y$'') axis into the unit circle---i.e., $0$
transforms into $-1$, $\pm i$ transforms into $\pm i$, and $\infty$
transforms into $1$.

Proving that M\"obius transforms preserve circles is a bit tricky,
since M\"obius transforms do {\it not} preserve the {\it centers} of
the circles.  Thus, the center of a circle does {\it not} usually map
into the the center of the M\"obius-transformed circle.  As a result,
one cannot simply transform the origin to the center of the circle and
show that a transformed rotating clock hand sweeps out the transformed
circle.

There is another characterization of a circle, however, that was known
to the ancient Greeks, but not to most modern high school students.
This characterization describes a circle as the locus of points the
ratio of whose distance to two given fixed points is a constant.
Thus, we start with two arbitrary points $P$ and $Q$ in the plane, and
then find those points $Z$ such that $d(Z,P)/d(Z,Q)=k$, where $k$ is a
real (positive) constant and $d(Z,P)$ means the (Euclidean) distance
between the point $Z$ and the point $P$.  The locus of these points
$Z$ will be a circle or a straight line.

We will now prove this fact twice---once in traditional $x,y$
coordinates, and once in complex numbers.  Hopefully, you will agree
that the complex number formulation is much shorter and sweeter.

Recall that the equation of a circle in the $x,y$ plane is in the form
$(x-O_x)^2+(y-O_y)^2=r^2$, where $(O_x,O_y)$ is the center of the
circle, and $r$ is the radius of the circle.  If we now form the ratio
of the distances from $(x,y)$ to $P=(P_x,P_y)$ and $Q=(Q_x,Q_y)$,
respectively, we get by brute algebraic force:
$${\sqrt{(x-P_x)^2+(y-P_y)^2}\over\sqrt{(x-Q_x)^2+(x-Q_y)^2}}=k$$
Squaring both sides, we get
$${(x-P_x)^2+(y-P_y)^2\over (x-Q_x)^2+(y-Q_y)^2}=k^2$$
Clearing fractions, we get
$$(x-P_x)^2+(y-P_y)^2=k^2(x-Q_x)^2+k^2(y-Q_y)^2.$$
Without even finishing this derivation, we notice that the
coefficients of both $x^2$ and $y^2$ are the same ($1-k^2$), so that
if the radius squared of the resulting equation is positive, we will
have a ``real'' circle.

Let us now consider the same derivation in complex numbers.  We know
that the ``length'' of a complex number $z=x+iy$ is the distance from
$z$ to the origin, or
$|z|=\sqrt{x^2+y^2}=\sqrt{z\bar{z}}=\sqrt{(x+iy)(x-iy)}$.  So the
equation of a circle in complex coordinates is $|z-O|=r$, where $O$ is
the center and $r>0$ is the radius.  This equation is more usually
written as
$$\eqalign{|z-O|^2&=(z-O)(\bar{z}-\bar{O})\cr
&=[(x-O_x)+i(y-O_y)]\cr
&\qquad[(x-O_x)-i(y-O_y)]\cr
&=(x-O_x)^2+(y-O_y)^2\cr
&=r^2\cr}$$

So the complex formulation of a circle is identical to the Cartesian
formulation.  We can now go back to our {\it ratio} formulation for
a circle:
$$\eqalign{{|z-P|^2\over |z-Q|^2}&=k^2\cr}$$
Clearing fractions, we have
$$\eqalign{|z-P|^2&=k^2|z-Q|^2\cr
%(z-P)(\bar{z}-\bar{P})-k^2(z-Q)(\bar{z}-\bar{Q})&=0\cr
%z\bar{z}-P\bar{z}-\bar{P}z+P\bar{P}-k^2(z\bar{z}-Q\bar{z}-\bar{Q}z+Q\bar{Q})&=0\cr
%(1-k^2)
{\left|{z-{P-k^2Q\over 1-k^2}}\right|}^2&=
{|P-k^2Q|^2\over (1-k^2)^2}-
{|P|^2-k^2|Q|^2\over 1-k^2}\cr
%{|P-k^2Q|^2\over 1-k^2}-(|P|^2-k^2|Q|^2)\cr
{\left|{z-{P-k^2Q\over 1-k^2}}\right|}^2&={k^2|P-Q|^2\over (1-k^2)^2}\cr
&={\left({|k||P-Q|\over 1-k^2}\right)}^2\cr
&=r^2\cr
}$$

Thus, in the general case, we have a circle of positive radius
$|P-Q||k|/(1-k^2)$ centered at $(P-k^2Q)/(1-k^2)$.  If $k=0$, then we
have a circle of zero radius centered at $P$, while if $k=\pm 1$, we
have a circle of infinite radius---i.e., a straight line perpendicularly
bisecting the line $PQ$.  Note that the sign of $k$ is irrelevant, and
we can therefore take it to be positive.

Now we can use this ratio characterization of circles to easily prove
that M\"obius transforms preserve circles.  We simply plug in the
result of the transformation and see that it is still the ratio of
distances, as before.

Consider the circle defined by the two points $P$, $Q$ and the ratio
$k$.  Then $|z-P|^2=k^2|z-Q|^2$ is a circle.  Consider transforming
$z\rightarrow (Az+B)/(Cz+D)$.  Its inverse is
$z\rightarrow(Dz-B)/(-Cz+A)$, as we found above.  Plugging this into
the equation for the circle, we have:
$$\eqalign{{\left|{\hbox{${Dz-B\over -Cz+A}$}-P}\right|}^2&=k^2{\left|{\hbox{${Dz-B\over -Cz+A}$}-Q}\right|}^2\cr
%{\left|{(Dz-B)-P(-Cz+A)}\right|}^2&=k^2{\left|{(Dz-B)-Q(-Cz+A)}\right|}^2\cr
{\left|\hbox{\small ${Dz-B+PCz-PA}$}\right|}^2&=k^2{\left|\hbox{\small ${Dz-B+QCz-QA}$}\right|}^2\cr
{\left|\hbox{\small ${(CP+D)z-(AP+B)}$}\right|}^2&=k^2{\left|\hbox{\small ${(CQ+D)z-(AQ+B)}$}\right|}^2\cr
%|CP+D|^2\left|z-\hbox{${AP+B\over CP+D}$}\right|^2&=k^2|CQ+D|^2\left|z-\hbox{${AQ+B\over CQ+D}$}\right|^2\cr
\left|z-\hbox{${AP+B\over CP+D}$}\right|^2&={\left|\hbox{${k{CQ+D\over CP+D}}$}\right|}^2\left|z-\hbox{${AQ+B\over CQ+D}$}\right|^2\cr
|z-P'|^2&=k'^2|z-Q'|^2\cr}$$
But we now recognize this as the equation of another circle defined by
the points $P'$, $Q'$ and the ratio $k'$, where $P'$ is the
M\"obius-transformed $P$, $Q'$ is the M\"obius-transformed $Q$, and
$k'=k|CQ+D|/|CP+D|$.  Thus, we have proved that {\it M\"obius
transforms preserve circles} (suitably defined).

We are finally ready for the {\it pi\`ece de r\'esistance}---we show
how M\"obius transforms factor the Newton square root iteration into
its essential parts.

Consider a circle defined by the two points $P=\sqrt{N}$,
$Q=-\sqrt{N}$ and a ratio $k$.  We would like to 1) map this circle
onto a circle centered on the origin; 2) {\it square} the resulting
quantity; and then 3) apply the mapping inverse to that in \#1.  In
functional form, this sequence consists of a mapping $z\rightarrow
f(z)$, following by a squaring, followed by the inverse mapping
$z\rightarrow f^{-1}(z)$, or more succinctly, $z\rightarrow
f^{-1}(f(z)^2)$.  If we now apply this mapping twice---i.e., we {\it
compose} the mapping with itself---we get the mapping
$f^{-1}(f(f^{-1}(f(z)^2))^2)=f^{-1}(f(z)^4)$.  If we continue
iterating, then the $n$-th iterate will produce the mapping
$z\rightarrow f^{-1}(f(z)^{2^n})$.

Consider now the function $f(z)=(z-\sqrt{N})/(z+\sqrt{N})$.  $f(z)$ is
a M\"obius mapping which maps
$$\eqalign{
0 & \rightarrow -1\cr
\infty & \rightarrow 1\cr
i\sqrt{N} & \rightarrow i\cr
-i\sqrt{N} & \rightarrow -i\cr
\sqrt{N} & \rightarrow 0\cr
-\sqrt{N} & \rightarrow \infty\cr
}$$
In short, $f(z)$ maps $\sqrt{N}\rightarrow 0$, $-\sqrt{N}\rightarrow
\infty$, and the $y$-axis onto the unit circle.  If we now square the
result of this mapping, the unit circle will map to the unit circle,
while circles smaller than the unit circle will get smaller, and
circles larger than the unit circle will get larger.

But in the ``transformed coordinate system'' produced by the mapping
$f(z)$, in which the origin stands for $\sqrt{N}$ and $\infty$ stands
for $-\sqrt{N}$, we see that the squaring achieves the effect of
mapping circles which were originally close to $\sqrt{N}$ even closer
to $\sqrt{N}$, and circles which were originally close to $-\sqrt{N}$
even closer to $-\sqrt{N}$.

Let us now prove that $f^{-1}(f(z)^2)=(z+N/z)/2$:
$$\eqalign{f^{-1}(f(z)^2)&=f^{-1}\left({\left({z-\sqrt{N}\over z+\sqrt{N}}\right)}^2\right)\cr
&=\sqrt{N}{{\left({z-\sqrt{N}\over z+\sqrt{N}}\right)}^2+1\over -{\left({z-\sqrt{N}\over z+\sqrt{N}}\right)}^2+1}\cr
&=\sqrt{N}{(z-\sqrt{N})^2+(z+\sqrt{N})^2\over -(z-\sqrt{N})^2+(z+\sqrt{N})^2}\cr
&=\sqrt{N}{2z^2+2N\over 4\sqrt{N}z}\cr
&={z^2+N\over 2z}\cr
&={z+N/z\over 2}\cr
}$$

Instead of transforming to the new coordinate system for each
iteration, we can conceptually think of performing a number of
iterations in the transformed coordinate system before transforming
back.  In fact, we can perform sufficient squarings in the transformed
coordinate system so that when we transform back, we will be
sufficiently close to either $\sqrt{N}$ or $-\sqrt{N}$. 
These insights have thus given us a {\it closed form} solution for the
$n$-th Newton iteration:
$$z\rightarrow \sqrt{N}{(z+\sqrt{N})^{2^n}+(z-\sqrt{N})^{2^n}\over (z+\sqrt{N})^{2^n}-(z-\sqrt{N})^{2^n}}$$

Our explorations have shown us that instead of focussing upon the
sequence of {\it points} $z_0$, $z_1$, ..., $z_n$, we should instead
focus upon the sequence of {\it circles} $C_0$, $C_1$, ..., $C_n$ on
which these points lie, since all of the points of the circles will be
treated essentially alike by the Newton iterations.  In fact, unless
$z_0$ lies exactly {\it on} the $y$-axis, then these circles will very
quickly converge to zero-radius circles about either $\sqrt{N}$ or
$-\sqrt{N}$.

Given an initial guess $z_0$, how do we find the appropriate circle on
which it lies?  The answer should now be obvious: these circles are
defined by the two points $\sqrt{N}$, $-\sqrt{N}$, and the ratio
$k=|z-\sqrt{N}|/|z+\sqrt{N}|$, and the $n$-th converging circle is
defined by the same two points $\sqrt{N}$, $-\sqrt{N}$, and the ratio
$k^{2^n}$.

If we plot a circle of points having the same ratio of distances to
$\sqrt{N}$, $-\sqrt{N}$, respectively, and then plot the images of
those points under the Newton square root iteration, we find that a)
these image points form a circle; and b) that the intersections $I_1$,
$I_2$ of this image circle with the $x$-axis have some interesting
properties: 1) the intersection point $I_2$ further from the origin is
the {\it center} of the first circle; and 2) the intersection point
$I_1$ lies on every line which connects a point on the first circle
with its image point under the Newton square root mapping.
\begin{center}
\leavevmode
\hbox{%
\epsfxsize=3in\epsfclipon
\epsffile{sigcol08-fig1.ps}}
\end{center}
In the figure, we have plotted a sequence of circles generated by
successive Newton iterations, and we see that these circles converge
very quickly to $\sqrt{N}$.  We have also plotted a number of points
on the initial circle and drawn lines to their images under a single
Newton iteration.  These points are 10 degrees apart on the circle in
the coordinate system in which the squarings take place, and we
readily see that they are not equally spaced in the original
coordinate system.

%\vfil\pagebreak[4]
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Beeler, M, Gosper, R.W., and Schroeppel, R.  {\it HAKMEM}.  MIT AI
Memo No. 239, Feb. 29, 1972, items \#3, \#126.
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%
\bibitem[Cayley79]{cayleyii}
Cayley, A.  ``Application of the Newton-Fourier Method to an Imaginary
Root of an Equation.''  {\it Quart. J. Pure \& Appl. Math.}, xvi (1879), 179-185.
{\tt\tiny http://home.pipeline.com/\char'176hbaker1/quaternion/cayley/Newton.ps.gz}
%
\bibitem[Corless98]{corless}
Corless, Robert M.  ``Variations on a Theme of Newton.''  {\it
Math. Mag.}, 71, 1 (Feb. 1998), 34-41.
%
\bibitem[Curry83]{curry}
Curry, J., Garnett, L., and Sullivan, D.  ``On the iteration of
rational funtions: Computer experiments with Newton's method.''  {\it
Commun. Math. Phys. 91} (1983), 267-277.
%
\bibitem[Friedman86]{friedman}
Friedman, J.  ``On Newton's Method for Polynomials.''  {\it IEEE 27th
Symp. on FOCS}, Oct. 1986, 153-161.
%
\bibitem[Klein56]{klein}
Klein, Felix.  {\it Lectures on the Icosahedron and the Solution of
Equations of the Fifth Degree}.  Morrice, G.G., transl. from 1888,
1913 manuscript.  Dover Publs. Inc., New York, 1956.
%
\bibitem[Schwerdtfeger79]{schwerdtfeger}
Schwerdtfeger, H.  {\it Geometry of Complex Numbers}.  Dover Publ.,
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%
\end{thebibliography}

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