ON A THEOREM IN THE CALCULUS OF DIFFERENCES

By

William Rowan Hamilton

(British Association Report, 1843, Part II, pp. 2–3.)

Edited by David R. Wilkins On a Theorem in the Calculus of Differences. By Sir William Rowan Hamilton.

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

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

X = u_{n}(x + n)^{n} − u_{n−1} . \frac{n}{1} . (x + n − 1)^{n} + u_{n−2} . \frac{n(n − 1)}{1 . 2} . (x + n − 2)^{n} − &c.; (1.)

which differs from the series for Δ^{n}x^{n} only by its introducing teh coefficients u, determined by the conditions that

u_{i} = +1, 0, or − 1, according as x + i > 0, = 0, or < 0.  (2.)

These conditions may be expressed by the formula

u_{i} = \frac{2}{π} ∫_{0}^{∞} \frac{dt}{t} sin(xt + it);  (3.)

and if we observe that

\frac{d}{dt} sin(at + b) = a sin (at + b + \frac{π}{2} ) ,
(\frac{d}{dt})^{n} sin(at + b) = a^{n} sin (at + b + \frac{nπ}{2} ) ,

we shall see that the series (1.) may be put under the form

X = \frac{2}{π} ∫_{0}^{∞} \frac{dt}{t} (\frac{d}{dt})^{n} Δ^{n} sin (xt − \frac{nπ}{2} );  (4.)

the characteristic Δ of difference being referred to x. But

Δ sin(2αx + β) = 2 sin α sin (2αx + β + α + \frac{π}{2} ) ,
Δ^{n} sin(2αx + β) = (2 sin α)^{n} sin (2αx + β + nα + \frac{nπ}{2} );

therefore, changing t, in (4.) to 2α, we find

X = ∫_{0}^{∞} \frac{dα}{α} \frac{d^{n}A}{dα^{n}},  (5.)

if we make, for abridgment,

A = \frac{2}{π} sin α^{n} sin(2xα + nα).  (6.)

Again, the process of integration by parts gives

∫_{0}^{∞} \frac{dα}{α^{i}} \frac{d^{n−i+1}A}{dα^{n−i+1}} = i ∫_{0}^{∞} \frac{dα}{α^{i+1}} \frac{d^{n−i}A}{dα^{n−i}},

provided that the function

\frac{1}{α^{i}} \frac{d^{n−i}A}{dα^{n−i}}

vanishes both when α = 0 and when α = ∞, and does not become infinite for any intermediate value of α, conditions which are satisfied here; we have, therefore, finally,

X = 1 . 2 . 3 … n ∫_{0}^{∞} dα \frac{A}{α^{n+1}} .  (7.)

Hence, if we make

P = \frac{X}{1 . 2 . 3 … n}, and c = 2x + n,  (8.)

we shall have the expression

P = \frac{2}{π} ∫_{0}^{∞} dα (\frac{sin α}{α})^{n} \frac{sin cα}{α},  (9.)

as a transformation of the formula

\begin{matrix}P & = \frac{1}{1 . 2 . 3 … n . 2^{n}} \{(n + c)^{n} − \frac{n}{1}(n + c − 2)^{n} + \frac{n(n − 1)}{1 . 2}(n + c − 4)^{n} − &c. \\ & − (n − c)^{n} + \frac{n}{1}(n − c − 2)^{n} − \frac{n(n − 1)}{1 . 2}(n − c − 4)^{n} + &c.\}; \end{matrix} \} (10.)

each partial series being continued only as far as the quantities raised to the nth power are positive. Laplace has arrived at an equivalent transformation, but by a much less simple analysis.