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% School of Mathematics, Trinity College, Dublin 2, Ireland
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% Trinity College, 2000.
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\centerline{\Largebf ON THE SOLUTION OF THE EQUATION}
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\centerline{\Largebf OF LAPLACE'S FUNCTIONS}
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\centerline{\Largebf By}
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\centerline{\Largebf William Rowan Hamilton}
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\centerline{\largerm (Proceedings of the Royal Irish Academy,
6 (1858), pp.\ 181--185.)}
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\centerline{\largerm Edited by David R. Wilkins}
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\centerline{\largerm 2000}
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\centerline{\sc On the Solution of the Equation of Laplace's Functions.}
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\centerline{Sir William Rowan Hamilton.}
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\centerline{Communicated February 26th, 1855.}
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\centerline{[{\it Proceedings of the Royal Irish Academy},
vol.~vi (1858), pp.\ 181--185.]}
\bigbreak
Rev.\ Professor Graves communicated the following extract from a
letter addressed to him (under date of January~26th, 1855) by Sir
William R. Hamilton:---
``{\sc My dear Graves},---You may like, perhaps, to see a way in
which I have to-day, for my own satisfaction, confirmed (not that
they required confirmation) some of the results announced by you
to the Academy on Monday evening last.
``Let us then consider the function (suggested by you),
$$\Sigma i^l j^m k^n
= (l, m, n) i^l h^m k^n;
\eqno (1)$$
where $l$, $m$, $n$ are positive and integer exponents ($0$
included); the summation~$\Sigma$ refers to all the possible
arrangements of the $l + m + n$ factors, whereof the number is
$$N_{l,m,n} = {(l + m + n)! \over l! \, m! \, n!};
\eqno (2)$$
each of these $N$ arrangements gives (by the rules of
$i \, j \, k$) a product $= \pm 1 \mathbin{.} i^l j^m k^n$; and
the sum of these positive or negative unit-coefficients, $\pm 1$,
thus obtained, is the numerical coefficient denoted by
$(l, m, n)$.
``Since each arrangement must have $i$ or $j$ or $k$ to the left,
we may write,
$$\Sigma i^l j^m k^n
= i \, \Sigma i^{l-1} j^m k^n
+ j \, \Sigma i^l j^{m-1} k^n
+ k \, \Sigma i^l j^m k^{n-1};
\eqno (3)$$
and it is easy to see that the coefficient $(l,m,n)$, or the sum
$\Sigma (\pm 1)$, vanishes, if {\it more than one\/} of the
exponents, $l$, $m$, $n$, be {\it odd}. Assume, therefore, as a
new notation,
$$(2 \lambda, 2 \mu, 2 \nu) = \{ \lambda, \mu, \nu \};
\eqno (4)$$
which will give, by (3), and by the principle last mentioned
respecting odd exponents,
$$\eqalignno{
(2 \lambda + 1, 2 \mu, 2 \nu)
&= \{ \lambda, \mu, \nu \};\cr
(2 \lambda - 1, 2 \mu, 2 \nu)
&= \{ \lambda - 1, \mu, \nu \}.
&(5)\cr}$$
We shall then have, by the mere notation,
$$\Sigma i^{2 \lambda} j^{2 \mu} k^{2 \nu}
= \{ \lambda, \mu, \nu \} i^{2 \lambda} j^{2 \mu} k^{2 \nu};
\eqno (6)$$
and, by treating this equation on the plan of (3),
$$\{ \lambda, \mu, \nu \}
= \{ \lambda - 1, \mu, \nu \}
+ \{ \lambda, \mu - 1, \nu \}
+ \{ \lambda, \mu, \nu - 1 \}.
\eqno (7)$$
By a precisely similar reasoning, attending only to $j$ and $k$,
or making $\lambda = 0$, we have an expresssion of the form,
$$\Sigma j^{2 \mu} k^{2 \nu}
= \{ \mu, \nu \} j^{2\mu} k^{2\nu},
\eqno (8)$$
where the coefficients $\{ \mu, \nu \}$ must satisfy the analogous
equation in differences,
$$\{ \mu, \nu \} = \{ \mu - 1, \nu \} + \{ \mu, \nu - 1 \},
\eqno (9)$$
together with the initial conditions,
$$\{ \mu, 0 \} = 1,\quad
\{ 0, \nu \} = 1.
\eqno (10)$$
Hence, it is easy to infer that
$$\{ \mu, \nu \} = {(\mu + \nu)! \over \mu! \, \nu!};
\eqno (11)$$
one way of obtaining which result is, to observe that the
generating function has the form,
$$\Sigma \{ \mu, \nu \} u^\mu v^\nu = (1 - u - v)^{-1}.
\eqno (12)$$
In like manner, if we combine the equation in differences (7),
with the initial conditions derived from the foregoing solution
of a less complex problem, namely, with
$$\{ 0, \mu, \nu \} = \{ \mu, \nu \},\quad
\{ \lambda, 0, \nu \} = \{ \lambda, \nu \},\quad
\{ \lambda, \mu, 0 \} = \{ \lambda, \mu \},
\eqno (13)$$
when the second members are interpreted as in (11), we find that
the (slightly) more complex generating function sought is,
$$\Sigma \{ \lambda, \mu, \nu \} t^\lambda u^\mu v^\nu
= (1 - t - u - v)^{-1};
\eqno (14)$$
and therefore that the required form of the coefficient is,
$$\{ \lambda, \mu, \nu \}
= {(\lambda + \mu + \nu)! \over \lambda! \, \mu! \, \nu!};
\eqno (15)$$
as, I have no doubt, you had determined it to be.
``With the same signification of $\{ \, \}$, we have, by (2),
$$N_{l,m,n} = \{ l, m, n \};
\eqno (16)$$
therefore, dividing $\Sigma$ by $N$, or the {\it sum\/} by the
{\it number}, we obtain, as an expression for what you happily
call the {\sc mean value} of the product
$i^{2 \lambda} j^{2 \mu} k^{2 \nu}$ the following:
$$M i^{2 \lambda} j^{2 \mu} k^{2 \nu}
= { \{ \lambda, \mu, \nu \}
\over \{ 2 \lambda, 2 \mu, 2 \nu \} }
i^{2 \lambda} j^{2 \mu} k^{2 \nu};
\eqno (17)$$
or, substituting for $\{ \, \}$ its value~(15), and writing for
abridgement
$$\kappa = \lambda + \mu + \nu,
\eqno (18)$$
$$M i^{2 \lambda} j^{2 \mu} k^{2 \nu}
= { (-1)^\kappa \kappa! (2 \lambda)! \, (2 \mu)! \, (2 \nu)!
\over (2 \kappa)! \, \lambda! \, \mu! \, \nu! }.
\eqno (19)$$
In like manner,
$$M i^{2 \lambda + 1} j^{2 \mu} k^{2 \nu}
= { i (-1)^\kappa \kappa! \, (2 \lambda + 1)! \, (2 \mu)! \, (2 \nu)!
\over (2 \kappa + 1)! \, \lambda! \, \mu! \, \nu! }.
\eqno (20)$$
``The whole theory of what you call the {\it mean values}, of
{\it products\/} of positive and integer {\it powers\/} of
$i \, j \, k$, being contained in the foregoing remarks, let us
next apply it to the determination of the mean value of a
{\it function\/} of $x + iw$,~$y + jw$,~$z + kw$; or, in other
words, let us investigate the equivalent for your
$$M f(x + iw, \, y + jw, \, z + kw):
\eqno (21)$$
by developing this function~$f$ according to ascending powers of
$w$, and by substituting, for every product of powers of
$i \, j \, k$, its {\it mean\/} value determined as above.
Writing, as you propose,
$${d \over dw} = D,\quad
{d \over dx} = D_1,\quad
{d \over dy} = D_2,\quad
{d \over dz} = D_3,
\eqno (22)$$
we are to calculate and to sum the general term of (21), namely,
$$M i^l j^m k^n \times { w^{l + m + n} \over l! \, m! \, n! }
D_1^l D_2^m D_3^n f(x, y, z).
\eqno (23)$$
{\it One\/} only of the exponents $l$,~$m$,~$n$, can usefully be
{\it odd}, by properties of the {\it mean\/} function, which have
been already stated. If {\it all\/} be {\it even}, and if we
make
$$l = 2 \lambda,\quad
m = 2 \mu,\quad
n = 2 \nu,
\eqno (24)$$
the corresponding part of the general term of $M f$, namely, the
part independent of $i \, j \, k$, is by (15), (18), (19),
$${ (-w^2)^\kappa \over (2 \kappa)! } \{ \lambda, \mu, \nu \}
D_1^{2 \lambda} D_2^{2 \mu} D_3^{2 \nu} f(x, y, z);
\eqno (25)$$
whereof the sum, relatively to $\lambda$,~$\mu$,~$\nu$, when
{\it their\/} sum~$\kappa$ is given, is,
$${ (-w^2)^\kappa \over (2 \kappa)! }
(D_1^2 + D_2^2 + D_3^2)^\kappa f(x, y, z)
= { (w \vecd )^{2 \kappa} \over (2 \kappa) ! }
f(x, y, z),
\eqno (26)$$
if my signification of $\vecd$ be adopted, so that
$$\vecd = i D_1 + j D_2 + k D_3;
\eqno (27)$$
and another summation, performed on (26), with respect to
$\kappa$, gives, for the part of $M f$ which is independent of
$i \, j \, k$, the expression,
$${\textstyle {1 \over 2}} ( e^{w \vecd} + e^{-w \vecd} )
f(x, y, z).
\eqno (28)$$
``Again, by supposing, in (23),
$$l = 2 \lambda + 1,\quad
m = 2 \mu,\quad
n = 2 \nu,
\eqno (29)$$
and by attending to (20), we obtain the term,
$${ w i D_1 (-w^2)^\kappa \over (2 \kappa + 1)! } \{ \lambda, \mu, \nu \}
D_1^{2 \lambda} D_2^{2 \mu} D_3^{2 \nu} f(x, y, z).
\eqno (30)$$
Adding the two other general terms correspondent, in which
$i D_1$ is replaced by $j D_2$ and by $k D_3$, we change $iD_1$
to $\vecd$; and obtain, by a first summation, the term
$${ (w \vecd)^{2 \kappa + 1} \over (2 \kappa + 1)! }
f(x, y, z);
\eqno (31)$$
and, by a second summation, we obtain
$${\textstyle {1 \over 2}} ( e^{w \vecd} - e^{-w \vecd} )
f(x, y, z),
\eqno (32)$$
as the {\it part\/} of the mean function $Mf$, which involves
expressly $i \, j \, k$. Adding the two parts, (28) and (32), we
are conducted finally to the very simple and remarkable
transformation of the {\sc mean function}~$Mf$, of which the
discovery is due to you:
$$Mf (x + iw, y + jw, z + kw)
= e^{w \vecd} f(x, y, z).
\eqno (33)$$
In like manner,
$$M\phi (x - iw, y - jw, z - kw)
= e^{- w \vecd} \phi(x, y, z).
\eqno (34)$$
Each of these two means of arbitrary functions, and therefore
also their sum, is thus a value of the expression
$$(D^2 - \vecd^2)^{-1} 0;
\eqno (35)$$
that is, the partial differential equation,
$$(D^2 + D_1^2 + D_2^2 + D_3^2) V = 0,
\eqno (36)$$
has its general integral, with two arbitary functions, $f$ and
$\phi$, expressible as follows:
$$V = Mf (x + iw, y + jw, z + kw)
+ M\phi (x - iw, y - jw, z - kw);
\eqno (37)$$
which is another of your important results. You remarked that if
the second member of the equation~(36) had been $U$, the
expression for $V$ would contain the additional term,
$$e^{w \vecd} D^{-1} e^{-2 w \vecd} D^{-1} e^{w \vecd} U.
\eqno (38)$$
In fact,
$$D + \vecd = e^{-w \vecd} d e^{w \vecd},\quad
D - \vecd = e^{w \vecd} d e^{-w \vecd},
\eqno (39)$$
and therefore,
$$(D - \vecd)^{-1} (D + \vecd)^{-1}
= e^{w \vecd} D^{-1} e^{-2 w \vecd} D^{-1} e^{w \vecd}.
\eqno (40)$$
``Most of this letter is merely a repetition of your remarks, but
the analysis employed may perhaps not be in all respects
identical with yours: a point on which I shall be glad to be
informed.
\nobreak\bigskip
\line{\hfil\vbox{\halign{#\hfil\cr
``I remain faithfully yours,\cr
\quad\quad ``{\sc William Rowan Hamilton}\cr}}}
\nobreak\bigskip
{\it ``The Rev.~Charles Graves, D.~D.''}
\bye