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\centerline{\Largebf ON CONTINUED FRACTIONS IN QUATERNIONS}
\vskip12pt
\centerline{\Largebf By}
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\centerline{\Largebf William Rowan Hamilton}
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\centerline{\vbox{\halign{\largerm #\hfil\cr
(Philosophical Magazine (4th series):\cr
\qquad vol.~iii (1852), pp.\ 371--373,\cr
\qquad vol.~iv (1852), p.~303,\cr
\qquad vol.~v (1853), pp.\ 117--118, 236--238, 321--326.)\cr}}}
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\centerline{\largerm Edited by David R. Wilkins}
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\centerline{\largerm 2000}
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\noindent
{\largeit On Continued Fractions in Quaternions. By\/}
{\largerm Sir}
{\largesc William Rowan Hamilton,} {\largeit LL.D., M.R.I.A.,
F.R.A.S., \&c., Andrews' Professor of Astronomy in the
University of Dublin, and Royal Astronomer of
Ireland\/}\footnote*{Communicated by the Author.}.
\bigbreak
\vskip 12pt
\centerline{[{\it The London, Edinburgh and Dublin Philosophical
Magazine and Journal of Science,}}
\centerline{4th series, vol.~iii (1852), pp.\ 371--373,
vol.~iv (1852), p.~303,}
\centerline{vol.~v (1853), pp.\ 117--118, 236--238, 321--326.]}
\bigskip
% VOLUME III, pp. 371-373
1.
It is required to integrate the equation in differences,
$$u_{x+1} (u_x + a) = b,$$
where $x$ is a variable whole number, but $a$, $b$, $u$ are
quaternions\footnote\dag{The advertisement, that the present
writer's Lectures on Quaternions were to be ready in January
last, was inserted, contrary to his wishes, through the over-zeal
of an agent: but the work in question is now nearly all in
type.}.
Let $q_1$ and $q_2$ be any two assumed quaternions; then
$$u_{x+1} + q_1
= b (a + u_x)^{-1} + q_1
= (b + q_1 a + q_1 u_x) (a + u_x)^{-1},$$
$$u_{x+1} + q_2
= b (a + u_x)^{-1} + q_2
= (b + q_2 a + q_2 u_x) (a + u_x)^{-1},$$
$${u_{x+1} + q_2 \over u_{x+1} + q_1}
= {b + q_2 a + q_2 u_x \over b + q_1 a + q_1 u_x}
= q_2 {q_2^{-1} b + a + u_x \over q_1^{-1} b + a + u_x} q_1^{-1}.$$
If therefore we suppose that $q_1$, $q_2$ are roots of the
quadratic equation
$$q^2 = qa + b,$$
which gives
$$q^{-1} b + a = q,$$
we shall have
$${u_{x+1} + q_2 \over u_{x+1} + q_1}
= q_2 {u_x + q_2 \over u_x + q_1} q_1^{-1},$$
and finally,
$${u_x + q_2 \over u_x + q_1}
= q_2^x {u_0 + q_2 \over u_0 + q_1} q_1^{-x}.$$
\bigbreak
2.
It was in a less simple way that I was led to the last written
result. I assumed
$$u_x = \left( {b \over a +} \right)^x c,$$
and treated this continued fraction as a particular case of the
following,
$$u_x
= {b_1 \over a_1 +} \, {b_2 \over a_2 +} \, \cdots \,
{b_x \over a_x + c}
= {N_x \over D_x}
= {N_x' (a_x + c) + N_x'' b_x \over D_x' (a_x + c) + D_x'' b_x}.$$
By changing $c$ to
$\displaystyle {b_{x+1} \over a_{x+1} + c}$,
I obtained the equations,
$$N_{x+1}' = N_x' a_x + N_x'' b_x,\quad N_{x+1}'' = N_x',$$
$$D_{x+1}' = D_x' a_x + D_x'' b_x,\quad D_{x+1}'' = D_x',$$
with the initial conditions
$$N_1' = 0,\quad N_1'' = 1,\quad D_1' = 1,\quad D_1'' = 0,$$
which allowed me to assume
$$N_0' = 1,\quad D_0' = 0.$$
Making next
$$a_x = a,\quad b_x = b,$$
there resulted
$$N_x = N_x' (a + c) + N_{x-1}' b,\quad
D_x = D_x' (a + c) + D_{x-1}' b,$$
$$N_{x+1}' = N_x' a + N_{x-1}' b,\quad
D_{x+1}' = D_x' a + D_{x-1}' b.$$
This led me to assume
$$N_x' = l q_1^x + m q_2^x,\quad
D_x' = l' q_1^x + m' q_2^x,$$
$$q_1 = a + q_1^{-1} b,\quad
q_2 = a + q_2^{-1} b,$$
$$l + m = 1,\quad
l q_1 + m q_2 = 0,\quad
l' + m' = 0,\quad
l' q_1 + m' q_2 = 1;$$
whence there followed,
$$\eqalign{
l &= (q_1^{-1} - q_2^{-1})^{-1} q_1^{-1}
= - q_2 (q_1 - q_2)^{-1},\cr
m &= - (q_1^{-1} - q_2^{-1})^{-1} q_2^{-1}
= + q_1 (q_1 - q_2)^{-1},\cr
l' &= - m' = (q_1 - q_2)^{-1}.\cr}$$
Hence
$$N_x = l q_1^x (q_1 + c) + m q_2^x (q_2 + c),\quad
D_x = l' q_1^x (q_1 + c) + m' q_2^x (q_2 + c);$$
and making, for conciseness,
$$v_x
= {q_2^x (q_2 + c) \over q_1^x (q_1 + c)}
= q_2^x {q_2 + c \over q_1 + c} q_1^{-x},$$
it was found that
$$u_x
= \left( {b \over a +} \right)^x c
= {N_x \over D_x} = {l + m v_x \over l' + m' v_x}
= {-q_2 (q_1 - q_2)^{-1} + q_1 (q_1 - q_2)^{-1} v_x
\over (q_1 - q_2)^{-1} (1 - v_x)}.$$
Thus
$$\eqalign{
u_x + q_1
&= {1 \over (q_1 - q_2)^{-1} (1 - v_x)}
= (1 - v_x)^{-1} (q_1 - q_2);\cr
u_x + q_2
&= v_x (1 - v_x)^{-1} (q_1 - q_2);\cr}$$
and finally,
$${u_x + q_2 \over u_x + q_1} = v_x,$$
as before.
And because in no one stage of the foregoing process has the
commutative principle of multiplication been employed, the
results hold good for quaternions, and admit of interesting
interpretations.
\nobreak\bigskip
Observatory, March~20, 1852.
\bigbreak
% VOLUME IV, p. 303
3.
It results from what has been shown in the two former articles
of this paper, that, whether in quaternions\footnote*{The
writer has again to regret that unforeseen causes of delay have
occurred to retard the publication of his Volume of Lectures on
Quaternions, of which, however, he hopes that the value will be
found to have been increased, by the additions which he has
inserted.}
or in ordinary algebra, the value of the continued fraction,
$$u_x = \left( {b \over a +} \right)^x c,
\eqno (1)$$
may be found from the equation
$${u_x - u'' \over u_x - u'} = v_x,
\eqno (2)$$
where
$$v_x = u''^x {c - u'' \over c - u'} u'^{-x},
\eqno (3)$$
or from the expression
$$u_x = (1 - v_x)^{-1} (u'' - v_x u'),
\eqno (4)$$
if $u'$ and $u''$ be two unequal roots of the quadratic,
$$u^2 + ua = b.
\eqno (5)$$
If, then, $a \, b \, c \, u' \, u''$ be five real quaternions, of
which the three last are unequal among themselves, and the two
latter have unequal tensors,
$$T u' > T u'',
\eqno (6)$$
we shall have the following limiting values:
$$T v_\infty = 0,\quad v_\infty = 0,\quad u_\infty = u''.
\eqno (7)$$
We may then enunciate this Theorem:---If the real quaternion~$c$
be {\it not a root\/} of the quadratic equation (5) in $u$,
{\it the value of the continued fraction\/} (1) {\it will
converge indefinitely towards that one of the real quaternion
roots of that quadratic, which has the lesser tensor}. If the
quaternion $c$ or $u_0$ be a root of that equation, it is clear
that the fraction will be constant.
\nobreak\bigskip
September~21, 1852.
\bigbreak
% VOLUME V, p. 117--118.
4.
Those who have acquired some familiarity with the interpretation
of the results obtained by the Calculus of Quaternions, will have
now little difficulty in seeing that the following geometrical
theorems\footnote*{These theorems are taken from art.~665 of the
author's (as yet unpublished) Lectures on Quaternions.}
are obtained from the consideration of the continued fraction
$\displaystyle \rho_x
= \left( {\beta \over \alpha +} \right)^x \rho_0$,
where $\alpha$, $\beta$, $\rho_0$, $\rho_x$ are real vectors,
$\beta$ being perpendicular to the other three, and the condition
$\alpha^4 + 4 \beta^2 > 0$ being satisfied.
Let ${\sc c}$ and ${\sc d}$ be two given points, and ${\sc p}$ an
assumed point. Join ${\sc d} {\sc p}$, and draw ${\sc c} {\sc
q}$ perpendicular thereto, and towards a given hand, in the
assumed plane ${\sc c} {\sc d} {\sc p}$, so that the rectangle
${\sc c} {\sc q} \mathbin{.} {\sc d} {\sc p}$
may be equal to a given area. From the derived point~${\sc q}$,
as from a new assumed point, derive a new point~${\sc r}$, by the
same rule of construction. Again conceive that ${\sc s}$ is
derived from ${\sc r}$, and ${\sc t}$ from ${\sc s}$, \&c., by an
indefinite repetition of the process. Then, {\it if the given
area be less than half the square of the given line\/}
${\sc c} {\sc d}$, and if a semicircle (towards the proper hand)
be constructed on that line as diameter, it will be possible to
inscribe a parallel chord ${\sc a} {\sc b}$, such that the given
area shall be represented by the product of the diameter
${\sc c} {\sc d}$, and the distance of this chord therefrom. We
may also conceive that ${\sc b}$ is nearer than ${\sc a}$ to
${\sc c}$, so that ${\sc a} {\sc b} {\sc c} {\sc d}$ is an
uncrossed trapezium inscribed in a circle, and the angle
${\sc a} {\sc b} {\sc c}$ is obtuse. This construction being
clearly understood, it becomes obvious, Ist, that because the
given area is equal to each of the two rectangles,
${\sc c} {\sc a} \mathbin{.} {\sc d} {\sc a}$
and
${\sc c} {\sc b} \mathbin{.} {\sc d} {\sc b}$
while the angles in the semicircle are right, then, whether we
begin by assuming the position of the point~${\sc p}$ to be at
the corner~${\sc a}$, or at the corner~${\sc b}$, of the
trapezium, every one of the derived points,
${\sc q}$, ${\sc r}$, ${\sc s}$, ${\sc t}$, \&c., will {\it
coincide\/} with the position so assumed for ${\sc p}$, however
far the process of derivation may be continued. But I also say,
IInd, that if {\it any other point\/} in the plane, {\it except
these two fixed points}, ${\sc a}$, ${\sc b}$, be assumed for
${\sc p}$, then not only will its {\it successive derivatives},
${\sc q}$, ${\sc r}$, ${\sc s}$, ${\sc t},\ldots$ be all
{\it distinct\/} from it, and from each other, but they will
{\it tend\/} successively and indefinitely to {\it coincide with
that one of the two fixed points\/} which has been above named
${\sc b}$. I add, IIIrd, that if, from any point~${\sc t}$,
distinct from ${\sc a}$ and from ${\sc b}$, we go {\it back}, by
an {\it inverse\/} process of derivation, to the {\it next
preceding point\/}~${\sc s}$ of the recently constructed series,
and thence, by the same inverse law, to ${\sc r}$, ${\sc q}$,
${\sc p}$, \&c., {\it this\/} process will produce an
{\it indefinite tendency\/} to, and an {\it ultimate
coincidence\/} with, the {\it other\/} of the two fixed points,
namely,~${\sc a}$. IVth. The common {\it law\/} of these two
tendencies, direct and inverse, is contained in the formula
$${{\sc q} {\sc b} \mathbin{.} {\sc p} {\sc a}
\over {\sc q} {\sc a} \mathbin{.} {\sc p} {\sc b}}
= {{\sc c} {\sc b} \over {\sc c} {\sc a}}
= \hbox{constant};$$
which may be variously transformed, and in which the constant is
independent of the position of ${\sc p}$. Vth. the {\it alternate
points}, ${\sc p}$, ${\sc r}$, ${\sc t}$, \&c., are all contained
on one {\it common circular segment\/} ${\sc a} {\sc p} {\sc b}$;
and the {\it other system of alternate points},
${\sc q}$, ${\sc s}$, \&c., has for its locus {\it another
circular segment}, ${\sc a} {\sc q} {\sc b}$, on the {\it same
fixed base}, ${\sc a} {\sc b}$. VIth. The {\it relation\/}
between these {\it two segments\/} is expressed by this other
formula, connecting the {\it angles\/} in them,
$${\sc a} {\sc p} {\sc b} + {\sc a} {\sc q} {\sc b}
= {\sc a} {\sc c} {\sc b};$$
the angles being here supposed to {\it change signs}, when their
vertices cross the fixed line ${\sc a} {\sc b}$.
\nobreak\bigskip
Observatory, December~30, 1852.
\bigbreak
% VOLUME V, p. 236--238.
5.
Let us now consider the continued fraction,
$$u_x = \left( {\beta \over \alpha +} \right)^x u_0,$$
where $u_0$ and $u_x$ are quaternions, and $\alpha$, $\beta$ are
two rectangular vectors, connected by the relation,
$$\alpha^4 + 4 \beta^2 = 0;$$
and, as a sufficient exemplification of the question, let it be
supposed that $\alpha$, $\beta$ have the values
$$\alpha = i - k,\quad \beta = j.$$
It may easily be shown, by the rules of the present Calculus,
that the expression,
$$u_1 = {j \over i - k + u_0},$$
gives the relations,
$$(u_1 - k)^{-1} = k + i (u_0 - k)^{-1} k,$$
$${\rm S} \mathbin{.} (i \pm k)
\{ (u_1 - k)^{-1} \pm (u_0 - k)^{-1} \} = \mp 1,$$
$$(u_2 - k)^{-1} - (u_0 - k)^{-1} = k - i;$$
and generally, by an indefinite repetition of the last process,
$$(u_{2n+x} - k)^{-1} - (u_x - k)^{-1} = n (k - i).$$
There is no difficulty in hence inferring that
$$\left( {j \over i - k +} \right)^\infty u_0 = u_\infty = k,$$
whatever arbitrary {\it quaternion\/} ($u_0$) may be assumed as
the original subject of the operation, which is thus indefinitely
repeated. By assuming for this original operand a {\it vector\/}
$\rho_0$ in the plane of $ik$, some geometrical\footnote*{Note
added during printing.---Since the foregoing communication was
forwarded, I have perceived that the theorem~VIII. of art.~6,
which presented itself to me as an interpretation of the
expression for $(u_1 - k)^{-1}$, when $k = {\sc c} {\sc e}$,
$i = {\sc d} {\sc e}$, $u_0 = {\sc c} {\sc p}$,
$u_1 = {\sc c} {\sc q}$, may be very simply proved by means of
the two similar triangles ${\sc q} {\sc e} {\sc c}$,
${\sc e} {\sc c} {\sc p}''$: and may be then employed to deduce
{\it geometrically\/} all the other theorems of that article.
(Each of these two triangles is similar to
${\sc e} {\sc p}_\prime {\sc c}$, if ${\sc p}_\prime$ be on
${\sc e} {\sc p}''$, and
${\sc p} {\sc p}_\prime \parallel {\sc d} {\sc c}$.) I see also
that the lately published results of art.~4 may all be deduced
geometrically, from the consideration of the two pairs of similar
triangles, ${\sc a} {\sc d} {\sc p}$, ${\sc q} {\sc c} {\sc a}$,
and ${\sc b} {\sc d} {\sc p}$, ${\sc q} {\sc c} {\sc b}$. These
geometrical simplifications have only recently occurred to me;
but it may have been perceived that, on the present occasion,
{\it geometry\/} has been employed merely to illustrate and
exemplify the signification and validity of certain new
symbolical expressions, and methods of calculation; some account
of which expressions and methods I hope to be permitted to
continue. (March~15, 1853.)}
theorems arise, less general indeed in their import than the
foregoing results respecting quaternions, yet perhaps not
uninteresting, as belonging to a somewhat novel class, and coming
fitly to be stated here, because they bear a sort of {\it
limiting\/} relation to the results recently published in the
Philosophical Magazine as part of the present paper.
\bigbreak
6.
Let ${\sc c}$ and ${\sc d}$ be the extremities, and ${\sc e}$ the
summit of a semicircle. Assume any point~${\sc p}$ in the same
plane, and draw ${\sc c} {\sc q}$ perpendicular to
${\sc d} {\sc p}$, so that the rectangle
${\sc c} {\sc q} \mathbin{.} {\sc d} {\sc p}$
may be equal to the given square ${\sc c} {\sc e}^2$. Then it is
clear, Ist, that if the {\it hand\/} (or direction of rotation)
be duly attended to, in thus drawing
${\sc c} {\sc q} \perp {\sc d} {\sc p}$,
the point ${\sc q}$ will {\it coincide\/} with ${\sc p}$, when
the latter point~${\sc p}$ is so assumed as to coincide with the
given summit~${\sc e}$. But I say also, IInd, that if the
point~${\sc p}$ be taken {\it anywhere else\/} in the same plane,
and if, after deriving ${\sc q}$ from it as above, we derive
${\sc r}$ from ${\sc q}$, \&c., by repeating the same process,
these {\it new\/} or derivative points ${\sc q}$, ${\sc r}$,
${\sc s}$, \&c., will {\it tend}, successively and indefinitely,
to coincide with the point~${\sc e}$. I add, IIIrd, that if,
from an arbitrarily assumed point ${\sc s}$, we go {\it back}, on
the same plan, to other points ${\sc r}$, ${\sc q}$, ${\sc p}$,
\&c., these new points, thus inversely derived, will {\it also\/}
tend indefinitely to coincide with the same fixed
summit~${\sc e}$. IVth. The {\it alternate\/} points ${\sc p}$,
${\sc r}$, ${\sc t},\ldots$ are all contained on one common
circular circumference; and the {\it other\/} alternate system of
derived points ${\sc q}$, ${\sc s}$, ${\sc u},\ldots$ are all
contained on {\it another circular locus}. Vth. These two new
circles {\it touch\/} each other and the given semicircle at the
given summit~${\sc e}$; and their {\it centres\/} are
{\it harmonic conjugates\/} with respect to the completed circle
${\sc c} {\sc e} {\sc d}$. (The same harmonic conjugation of the
centres of the two loci might easily have been derived for the
more general case considered in an earlier part of this paper,
from the last formula of art.~4; I have found that it holds good
also in another equally general case, hereafter to be considered,
when the given area of the rectangle under ${\sc c} {\sc q}$ and
${\sc d} {\sc p}$ is {\it greater\/} than the square on the
quadrantal chord ${\sc c} {\sc e}$, in which case there {\it
can\/} be {\it no convergence\/} to a limiting position, but
there {\it may} be, under certain conditions, {\it circulation}.)
VIth. If the chords ${\sc p} {\sc e}$, ${\sc r} {\sc e}$,
${\sc t} {\sc e},\ldots$ of the one circular locus, and also the
chords ${\sc q} {\sc e}$, ${\sc s} {\sc e}$,
${\sc u} {\sc e},\ldots$ of the second locus, be prolonged
through the point of contact~${\sc e}$, so as to render the
following rectangles equal to the given square or area,
$$ {\sc p} {\sc e} {\sc p}'
= {\sc r} {\sc e} {\sc r}'
= {\sc t} {\sc e} {\sc t}'
= \cdots
= {\sc q} {\sc e} {\sc q}'
= {\sc s} {\sc e} {\sc s}'
= {\sc u} {\sc e} {\sc u}'
= \cdots
= {\sc c} {\sc e}^2,$$
then not only will the points
${\sc p}' \, {\sc r}' \, {\sc t}',\ldots$
be ranged on {\it one straight line}, and the points
${\sc q}' \, {\sc s}' \, {\sc u}',\ldots$
on another, but also the {\it intervals\/}
${\sc p}' {\sc r}'$, ${\sc r}' {\sc t}',\ldots$
${\sc q}' {\sc s}'$, ${\sc s}' {\sc u}',\ldots$
will all be {\it equal\/} to each other and to the given diameter
${\sc c} {\sc d}$; and will have the same {\it direction\/} as
that diameter. Thus the four points
${\sc e} \, {\sc p} \, {\sc r} \, {\sc t}$,
or the four points
${\sc e} \, {\sc q} \, {\sc s} \, {\sc u}$,
form what may be called an {\it harmonic group}, on the one or on
the other circular locus: and if, as in some modern methods, the
{\it directions\/} (and not merely the {\it lengths\/}) of lines
be attended to, the chords
${\sc e} {\sc p}$, ${\sc e} {\sc r}$, ${\sc e} {\sc t},\ldots$
or
${\sc e} {\sc q}$, ${\sc e} {\sc s}$, ${\sc e} {\sc u},\ldots$
may be said to form, each set within the circle to which they
belong, a species of {\it harmonical progression}. VIIth. The
orthogonal projection of ${\sc p}' {\sc q}'$ or
${\sc q}' {\sc r}'$, \&c., on ${\sc c} {\sc d}$, is equal in
length and direction to the {\it half\/} of that given diameter.
VIIIth. If ${\sc p}' {\sc p}''$ be so drawn as to be
perpendicularly bisected by the common tangent to the three
circles, the line ${\sc p}'' {\sc q}'$ will be equal in length
and direction to the given quadrantal {\it chord}
${\sc c} {\sc e}$.
\nobreak\bigskip
Observatory, February~19, 1853.
\bigbreak
% VOLUME V, p. 321--326.
7.
The geometrical theorems stated in recent articles of this paper,
although perhaps not inelegant, cannot pretend to be important:
indeed a hint has been given (in a note) of a quite elementary
way, in which they may be geometrically demonstrated. But I
think that the analytical process, by which I was led to the
formul{\ae} of art.~5, whereof the geometrical statements of
art.~6 are in part an interpretation, may deserve to be
considered with attention, on account of the novelty of the
{\it method\/} employed: and especially for the examples which it
supplies of calculation with {\it biquaternions}.
\bigbreak
8.
After obtaining the result already published in this Magazine
(compare the number for May, 1852), for a certain continued
fraction in quaternions, namely that if\footnote*{The numbering
of the equations commences here anew.}
$$u_x = \left( {b \over a+} \right)^x c,
\eqno (1)$$
then\footnote\dag{This result holds good also in ordinary
algebra, and even in {\it arithmetic\/}: but in applying it to
quaternions, the {\it order of the factors\/} must be attended
to.}
$${u_x - u'' \over u_x - u'}
= u''^x \left( {c - u'' \over c - u'} \right) u'^{-x},
\eqno (2)$$
$u'$, $u''$ being roots of the quadratic equation
$$u^2 + ua = b;
\eqno (3)$$
and after hence deducing the theorem (given in page~303 of the
Philosophical Magazine for October, 1852 [article~3]), that for
the case of {\it real quaternions}, and of {\it unequal tensors},
$$u_\infty = u'', \hbox{ if } T u'' < T u';
\eqno (4)$$
it was obvious, as a particular application, that by changing
$a$, $b$, $c$, $u'$, $u''$, $u_x$ to
$\alpha$, $\beta$, $\rho_0$, $\rho'$, $\rho''$, $\rho_x$, and by
supposing these last to be six {\it real vectors}, among which
$\beta$ is perpendicular to all the rest, I might write
$${\rho_x - \rho'' \over \rho_x - \rho'}
= \rho''^x {\rho_0 - \rho'' \over \rho_0 - \rho'} \rho'^{-x};
\eqno (5)$$
and ultimately,
$$\rho_\infty = \rho'', \hbox{ if } T \rho'' < T \rho',
\eqno (6)$$
the vectors $\rho'$, $\rho''$ being roots of the quadratic,
$$\rho^2 + \rho \alpha = \beta.
\eqno (7)$$
This last equation gave, by taking separately the scalar and
vector parts,
$$\rho^2 + {\rm S} \mathbin{.} \rho \alpha = 0;
\eqno (8)$$
$${\rm V} \mathbin{.} \rho \alpha = \beta;
\eqno (9)$$
whereof the former (8) expressed that $\rho$ terminated on a {\it
spheric surface}, passing through the origin, and having the
vector $-\alpha$ for its diameter; while the latter (9) expressed
that $\rho$ terminated on a {\it right line}, which was drawn
through the extremity of the vector $\beta \alpha^{-1}$, in a
direction parallel to that diameter. Thus (9) gave, by the rules
of the present calculus,
$$\rho = \beta \alpha^{-1} + x \alpha,\quad
\rho^2 = - \beta^2 \alpha^{-2} + x^2 \alpha^2,\quad
{\rm S} \mathbin{.} \rho \alpha = x \alpha^2;
\eqno (10)$$
and therefore, by (8), I had the ordinary quadratic equation,
$$x^2 + x = \beta^2 \alpha^{-4},
\quad\hbox{or}\quad
(2x + 1)^2 \alpha^4 = \alpha^4 + 4 \beta^2 > 0,
\eqno (11)$$
as in art.~4 (Phil.\ Mag.\ for February, 1853): the two values of
the vector~$\rho$, which answer to the two values of the scalar
coefficient~$x$, being here supposed to be geometrically real and
unequal; or the right line (9) being supposed to meet the spheric
surface (8), in two distinct and real points, ${\sc a}$,
${\sc b}$. Hence by assuming
$$\rho' = {\sc c} {\sc a},\quad
\rho'' = {\sc c} {\sc b},\quad
\rho_0 = {\sc c} {\sc p},\quad
\rho_1 = {\sc c} {\sc q},\quad
\alpha = {\sc d} {\sc c},\quad
\beta = {\sc c} {\sc a} \mathbin{.} {\sc d} {\sc a},
\eqno (12)$$
I was conducted with the greatest ease to the theorems of the
last-cited article.
\bigbreak
9.
But in the case of art.~5, namely when
$$\alpha^4 + 4 \beta^2 = 0,
\eqno (13)$$
and when consequently
$$x = - {1 \over 2},\quad
\rho'' = \rho' = \beta \alpha^{-1} - {1 \over 2} \alpha,
\eqno (14)$$
the {\it equality\/} of the {\it two roots\/} of the quadratic
(11) in $x$, or of the two real and vector roots of the equation
(7) in $\rho$, appeared to reduce the formula (5) to an {\it
identity\/}: and the simple process of the article last cited did
not immediately occur to me. I therefore had recourse to
certain {\it imaginary\/} or purely {\it symbolical solutions},
of that quadratic equation (7), or rather of the following, by
which we may here conveniently replace it,
$$u^2 + u (i - k) = j;
\eqno (15)$$
the continued fraction to be studied being now,
$$u_x = \left( {j\over i - k +} \right)^x u_0,
\eqno (16)$$
where $i \, j \, k$ are the usual symbols of this calculus, and
$u_0$ may denote any arbitrarily assumed quaternion. By an
application of a general process (described in art.~649 of my
unpublished Lectures on Quaternions), I found that the quadratic
(15) might be symbolically satisfied by the two following {\it
imaginary quaternions}, or {\it biquaternion expressions\/}:
$$u' = -i - h(1 - j);\quad u'' = -i + h(1 - j);
\eqno (17)$$
where $h$ is used as a temporary and abridged symbol for the
{\it old and ordinary imaginary\/} of common algebra, denoted
usually by $\sqrt{-1}$, and regarded as being always a {\it
free\/} or {\it commutative factor\/} in any multiplication: so
that
$$h^2 = -1,\quad hi = ih,\quad hj = jh,\quad hk = kh,
\eqno (18)$$
although $ji = -ij$ \&c. In fact the first of these expressions
(17) gives,
$$\eqalignno{
u' (u' + i - k)
&= \{ i + h (1 - j) \} \{ k + h(1 - j) \} \cr
&= ik + h \{ i(1 - j) + (1 - j)k \} + h^2 (1 - j)^2 \cr
&= -j + h (i - k + k - i) + h^2 (1 - 2j -1) \cr
&= -j + 0h + 2j = j;
&(19)\cr}$$
and the second expression (17) gives, in like manner,
$$u'' (u'' + i - k) = j:
\eqno (20)$$
so that, without entering at present into any account of the
process which enabled me to {\it find\/} the biquaternions (17),
it has been now {\it proved, \`{a} posteriori}, by actual
substitution, that those expressions do in fact {\it symbolically
satisfy\/} the quadratic equation (15). And because they are
{\it unequal roots\/} of that equation, as differing by the sign
of $h$, I saw that they might be employed in the general formula
(2), without being liable to the practical objection that lay
against the employment of the two {\it real\/} by {\it equal\/}
roots $\rho'$, $\rho''$ of the equation (7).
\bigbreak
10.
Introducing therefore into the formula (2), or into the
following, which is a transformation thereof,
$${u_x - u'' \over u_x - u'}
= {u''^x (u_0 - u'') \over u'^x (u_0 - u')},
\eqno (21)$$
the values (17), or these which are equivalent,
$$u' = -h (1 - j - hi),\quad
u'' = h (1 - j + hi);
\eqno (22)$$
and observing that
$$(j \pm hi)^2 = j^2 \pm h (ji + ij) - i^2 = 0,
\eqno (23)$$
and that therefore\footnote*{More generally, with these rules of
combination of the symbols $h \, i \, j \, k$, if $f$ be any
algebraic function, and $f'$ the derived function,
$$f(1 + tj \pm thi) = f(1) + t f'(1) (j \pm hi);$$
because $(tj \pm thi)^2 = 0$, if $t$ be any scalar coefficient.}
$$(1 - j \mp hi)^x = 1 - xj \mp xhi;
\eqno (24)$$
we see that
$${u_x + i - h(1 - j) \over u_x + i + h(1 - j)}
= (-1)^x {{\sc a}_x - h {\sc b}_x \over {\sc a}_x + h {\sc b}_x},
\eqno (25)$$
where ${\sc a}_x$, ${\sc b}_x$ are {\it two real quaternions},
namely,
$$\left. \eqalign{
{\sc a}_x &= (1 - xj) (u_0 + i) + xi (1 - j),\cr
{\sc b}_x &= (1 - xj) (1 - j) - xi (u_0 + i);\cr}
\right\}
\eqno (26)$$
or, as we may also write them,
$$\left. \eqalign{
{\sc a}_x &= (1 - xj) (u_0 - k) + i + k,\cr
{\sc b}_x &= - xi (u_0 - k) + 1 - j.\cr}
\right\}
\eqno (27)$$
In this manner I found it possible to {\it eliminate the
symbol\/}~$h$, or to return from imaginary to real quaternions,
and so perceived that
$${u_{2n} + i \over 1 - j}
= {{\sc a}_{2n} \over {\sc b}_{2n}};\quad
{u_{2n+1} + i \over 1 - j}
= - {{\sc b}_{2n+1} \over {\sc a}_{2n+1}}.
\eqno (28)$$
Both of these two last formul{\ae} agree in giving, as a limit,
$${u_\infty + i \over 1 - j}
= {j \over i} = {-i \over j} = + k;
\eqno (29)$$
and therefore (as in art.~5),
$$\left( {j \over i - k +} \right)^\infty u_0
= u_\infty = -i + k(1 - j) = k,
\eqno (30)$$
whatever {\it real\/} quaternion may be assumed for $u_0$. This
last restriction becomes {\it here\/} necessary, from the
generality of the analysis employed: because, for the very reason
that $u'$, $u''$ are admitted as being at least {\it
symbolical\/} (or imaginary) {\it roots\/} of the equation (15),
therefore we must here say that
$$\hbox{if } u_0 = u', \hbox{ then }
u_x = u',\enspace u_\infty = u';
\eqno (31)$$
and in like manner,
$$\hbox{if } u_0 = u'', \hbox{ then }
u_x = u'',\enspace u_\infty = u''.
\eqno (32)$$
\bigbreak
11.
By the first of the two real quaternion equations (28), we have,
$$u_{2n} - k = - i - k + {\sc a}_{2n} {\sc b}_{2n}^{-1} (1 - j);
\eqno (33)$$
but also, by the latter of the two values (27),
$${\sc b}_{2n}^{-1} (1 - j)
= \{ (1 - j)^{-1} {\sc b}_{2n} \}^{-1}
= \left( {1 + j \over 2} \mathbin{.} {\sc b}_{2n} \right)^{-1}
= \{ 1 + n (k - i) (u_0 - k) \}^{-1};
\eqno (34)$$
again, by the former of the same two values (27),
$${\sc a}_{2n} - (k + i) \{ 1 + n (k - i) (u_0 - k) \}
= {\sc a}_{2n} - (k + i) + 2nj (u_0 - k)
= u_0 - k;
\eqno (35)$$
therefore
$$u_{2n} - k
= (u_0 - k) \{ 1 + n (k - i) (u_0 - k) \}^{-1}
= \{ (u_0 - k)^{-1} + n (k - i) \}^{-1};
\eqno (36)$$
or more simply,
$$(u_{2n} - k)^{-1} - (u_0 - k)^{-1} = n (k - i).
\eqno (37)$$
It was in this way that I was originally led to the formula of
art.~5, namely,
$$(u_{2n + x} - k)^{-1} - (u_x - k)^{-1} = n (k - i);
\eqno (38)$$
but having once come to {\it see\/} that this result held good,
it was easy then to pass to a much more simple {\it proof}, such
as that given in the last-cited article, which was entirely
independent of the imaginary symbol here called $h$, and employed
only {\it real quaternions}.
\bigbreak
12.
It may be regarded as still more remarkable, that the {\it same
real results\/} are obtained, when we {\it combine a a real root
with an imaginary one}, instead of combining {\it two\/} real
roots, or {\it two\/} imaginary ones. Thus the
quadratic\footnote*{An equation of the $n$th dimension in
quaternions has generally $n^4$ {\it roots}, real or imaginary;
because it may be generally resolved into a system of {\it
four\/} ordinary and algebraical equations, which are {\it
each\/} of the $n$th degree. However, it is shown in my Lectures
that for the particular form (3), $u^2 + ua = b$ (or
$q^2 = qa + b$), which occurs in the present investigation,
{\it only six\/} (out of the sixteen) roots are {\it finite\/};
and that of these six, {\it two\/} are generally {\it real}, and
{\it four imaginary}. In the particular case of the equation
(15), $k$ is by this theory a {\it quadruple root}, representing
at once two real and two imaginary solutions, which have all
become {\it equal\/} to each other, by the vanishing of certain
radicals. Thus there remain in this case {\it only three
distinct roots\/} of the quadratic (15), namely the one real
root~$k$, and the two imaginary roots (17): and what appears to
me remarkable in the analysis of the present article~12, although
otherwise exemplified in my Lectures, is the {\it mixture of these
two classes\/} of solution of an equation in quaternions, a root
of one kind being {\it combined\/} with a root of the other kind,
so as to conduct to a correct determination of the value of a
certain continued fraction, regarded as a real quaternion, which
admits (as in art.~6) of being geometrically interpreted.}
equation (15) has {\it one\/} root, namely $k$, which must be
considered as {\it real\/} in this theory, whether by contrast to
the symbol~$h$ (or to the old imaginary of algebra), or because
in the geometrical interpretation it is constructed by a
{\it real line}, namely by the {\it chord\/} ${\sc c} {\sc e}$
drawn to the point of {\it contact\/}~${\sc e}$ of the spheric
surface (8) with the right line (9), under the condition (13);
$\alpha$ and $\beta$ being then for convenience replaced, as in
art.~5, by the more special symbols $i - k$ and $j$. Now if we
adopt this {\it real root\/} $k$ as the value of $u'$, but retain
the second of the two imaginary or biquaternion roots (17), as
being still the expression for $u''$, the numerators of the
formula (21) will remain unchanged, but the denominators will be
altered; and instead of (25) we shall have this other formula,
$${u_x + i - h (1 - j) \over u_x - k}
= {h^x ({\sc a}_x - h {\sc b}_x) \over k^x (u_0 - k)},
\eqno (39)$$
with the significations (26) or (27) of ${\sc a}_x$, ${\sc b}_x$,
and therefore with the relations (34) (35). Observing that
$$h^{2n} = (-1)^n = k^{2n},
\eqno (40)$$
we find that the formula (39), by comparing separately the real
and imaginary parts, in the two cases of $x$ even and $x$ odd,
gives these four others, not involving the symbol~$h$:
$${u_{2n} + i \over u_{2n} - k}
= {{\sc a}_{2n} \over u_0 - k};\quad
{1 - j \over u_{2n} - k}
= {{\sc b}_{2n} \over u_0 - k};
\eqno (41)$$
$${u_{2n+1} + i \over u_{2n+1} - k}
= {{\sc b}_{2n+1} \over k (u_0 - k)};\quad
{1 - j \over u_{2n+1} - k}
= {- {\sc a}_{2n+1} \over k (u_0 - k)};
\eqno (42)$$
of which the consistency with (28) is evident, and which are
found to agree in all other respects with conclusions otherwise
obtained. Thus all these different processes of calculation
conduct to consistent and interpretable results, although the
method of the present article appears to depart even more than
those of former ones from the ordinary analogies of algebra.
\nobreak\bigskip
Observatory of T.~C.~D.
\quad March~17, 1853.
\centerline{[To be continued.]}
\bye