<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">% This paper has been transcribed in Plain TeX by
% David R. Wilkins
% School of Mathematics, Trinity College, Dublin 2, Ireland
% (dwilkins@maths.tcd.ie)
%
% Trinity College, 2000.

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\centerline{\Largebf ON A THEOREM IN THE}

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\centerline{\Largebf CALCULUS OF DIFFERENCES}

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\centerline{\Largebf By}

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\centerline{\Largebf William Rowan Hamilton}

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\centerline{\largerm (British Association Report, 1843,
   Part~II, pp.\ 2--3.)}

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\centerline{\largerm Edited by David R. Wilkins}

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\centerline{\largerm 2000}

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{\largeit\noindent
On a Theorem in the Calculus of Differences.  By\/}
{\largerm Sir} {\largesc William Rowan Hamilton}.

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\centerline{[{\it Report of the Thirteenth Meeting of the British
Association for the Advancement of}}
\centerline{{\it Science; held at Cork in August 1843}.}
\centerline{(John Murray, London, 1844), Part~II, pp.\ 2--3.]}

\bigbreak

It is a curious and may be considered as an important problem in
the Calculus of Differences, to assign an expression for the sum
of the series
$${\rm X} = u_n (x + n)^n
       - u_{n-1} \mathbin{.} {n \over 1} \mathbin{.} (x + n - 1)^n
       + u_{n-2} \mathbin{.} {n (n - 1) \over 1 \mathbin{.} 2}
               \mathbin{.} (x + n - 2)^n
       - \hbox{\&amp;c.};
   \eqno (1.)$$
which differs from the series for $\Delta^n x^n$ only by its
introducing teh coefficients~$u$, determined by the conditions
that
$$u_i = +1,\enspace 0, \hbox{ or } -1,
   \hbox{ according as } x + i &gt; 0,\enspace = 0, \hbox{ or } &lt; 0.
   \eqno (2.)$$
These conditions may be expressed by the formula
$$u_i = {2 \over \pi} \int_0^\infty {dt \over t} \sin (xt + it);
   \eqno (3.)$$
and if we observe that
$$\eqalign{
{d \over dt} \sin (at + b)
   &amp;= a \sin \left( at + b + {\pi \over 2} \right),\cr
\left( {d \over dt} \right)^n \sin (at + b)
   &amp;= a^n \sin \left( at + b + {n \pi \over 2} \right),\cr}$$
we shall see that the series (1.) may be put under the form
$${\rm X} = {2 \over \pi} \int_0^\infty {dt \over t}
         \left( {d \over dt} \right)^n \Delta^n
         \sin \left( xt - {n\pi \over 2} \right);
   \eqno (4.)$$
the characteristic~$\Delta$ of difference being referred to $x$.
But
$$\eqalign{
\Delta \sin (2 \alpha x + \beta)
   &amp;= 2 \sin \alpha
         \sin \left(
            2 \alpha x + \beta + \alpha + {\pi \over 2}
         \right),\cr
\Delta^n \sin (2 \alpha x + \beta)
   &amp;= (2 \sin \alpha)^n
         \sin \left(
            2 \alpha x + \beta + n \alpha + {n \pi \over 2}
         \right);\cr}$$
therefore, changing $t$, in (4.) to $2 \alpha$, we find
$${\rm X} = \int_0^\infty {d\alpha \over \alpha} \,
         {d^n {\rm A} \over d\alpha^n},
   \eqno (5.)$$
if we make, for abridgment,
$${\rm A} = {2 \over \pi} \sin \alpha^n \sin (2x\alpha + n\alpha).
   \eqno (6.)$$
Again, the process of integration by parts gives
$$\int_0^\infty {d\alpha \over \alpha^i} \,
         {d^{n - i + 1} {\rm A} \over d \alpha^{n - i + 1}}
   = i \int_0^\infty {d\alpha \over \alpha^{i+1}}
         {d^{n - i} {\rm A} \over d \alpha^{n - i}},$$
provided that the function
$${1 \over \alpha^i} {d^{n-i} {\rm A} \over d \alpha^{n-i}}$$
vanishes both when $\alpha = 0$ and when $\alpha = \infty$, and
does not become infinite for any intermediate value of $\alpha$,
conditions which are satisfied here; we have, therefore, finally,
$${\rm X}
   =  1 \mathbin{.} 2 \mathbin{.} 3 \, \ldots \, n
         \int_0^\infty d\alpha \, {{\rm A} \over \alpha^{n+1}}.
   \eqno (7.)$$
Hence, if we make
$${\rm P}
   =  {{\rm X} \over 1 \mathbin{.} 2 \mathbin{.} 3 \, \ldots \, n},
   \quad\hbox{and}\quad
  c = 2x + n,
   \eqno (8.)$$
we shall have the expression
$${\rm P}
   =  {2 \over \pi} \int_0^\infty d\alpha \,
         \left( {\sin \alpha \over \alpha} \right)^n
         {\sin c \alpha \over \alpha},
   \eqno (9.)$$
as a transformation of the formula
$$\left. \eqalign{
{\rm P}
   &amp;= {1 \over 1 \mathbin{.} 2 \mathbin{.} 3 \, \ldots \, n
            \mathbin{.} 2^n}
      \biggl\{
         (n + c)^n - {n \over 1} (n + c - 2)^n
       + {n (n - 1) \over 1 \mathbin{.} 2} (n + c - 4)^n
       - \hbox{\&amp;c.} \cr
   &amp;\qquad\qquad
       - (n - c)^n
       + {n \over 1} (n - c - 2)^n
       - {n (n - 1) \over 1 \mathbin{.} 2} (n - c - 4)^n
       + \hbox{\&amp;c.}
      \biggr\};\cr}
   \right\}
   \eqno (10.)$$
each partial series being continued only as far as the quantities
raised to the $n$th power are positive.  Laplace has arrived at
an equivalent transformation, but by a much less simple analysis.

\bye
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