By Percival Frost Joseph Wolstenholme

The Authors of the next Treatise have endeavoured to provide prior to scholars as entire a view of the topic as definite barriers have allowed them to do. the need of those obstacles has built itself during getting ready the paintings in the course of a interval of 4 years. The examine of innumerable papers, via the main celebrated mathematicians of all international locations, has confident the authors that the topic is sort of inexhaustible, and that, to have handled all components of it with something forthcoming to the fulness with which they've got taken care of the 1st component, may have swelled their paintings in a apprehensive percentage to what it has already attained.

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205) i=−∞ ∞ nsni k i . 170). 173) are used. 208) k=−∞ and k n+1 = kk n . 209) gives ∞ Sin+1 k (i) = k ∞ Sin kk (i) . 211) (i) + ik . 172). 8 OPERATOR ∆−1 AND THE SUM CALCULUS We define ∆−1 yk to be such that ∆(∆−1 yk ) = yk . 215) 28 Difference Equations and we have zk+1 − zk = yk , zk − zk−1 = yk−1 , zk−1 − zk−2 = yk−2 , .. .. . z2 − z1 = y1 . 218) r=1 and k−1 zk = z1 + yr . 220) r=1 since z1 is an arbitrary constant. In summary, ∆−1 yk is a function whose difference is yk . In fact, for any positive integer n, we define ∆−n yk as a function whose nth difference is yk .

N) (a) Prove that Cnk = kn! (b) For fixed n, show that k ∆Cnk = Cn−1 , and thus k Cnk+1 = Cnk + Cn−1 . 16. Sin = 0, n i=r sir Sin = δrn .

2 Example B Prove the following (Newton’s theorem): If fk is a polynomial of the nth degree, then it can be written in the form ∆f0 (1) ∆2 f0 (2) ∆n f0 (n) k + k + ···+ k . 1! 2! n! 187) where a0 , a1 , . . , an are constants. Differencing fk n times gives ∆fk = a1 + 2a2 k (1) + 3a3 k (2) + · · · + nan k (n−1) , ∆2 fk = 2 · 1 · a2 + 3 · 2 · a3 k (1) + · · · + n(n − 1)an k (n−2) , .. .. . ∆n fk = an n(n − 1) · · · (1). 188) THE DIFFERENCE CALCULUS 25 Setting k = 0 in the original function and its differences allows us to conclude that ∆m f0 am = , m = 0, 1, .