By Todhunter, I. (Isaac)
The beneficial reception which has been granted to my heritage of the Calculus of diversifications in the course of the 19th Century has inspired me to adopt one other paintings of an analogous sort. the topic to which I now invite consciousness has excessive claims to attention as a result of the delicate difficulties which it includes, the dear contributions to research which it has produced, its vital useful purposes, and the eminence of these who've cultivated it.
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Extra info for A history of the mathematical theory of probability : from the time of Pascal to that of Laplace
Then S(x) can be expressed in the form S(x) = a0 x2 ∆2 a0 x∆a0 + + ··· . 247) This result is known as Montmort’s theorem on infinite summation. Note that if ak is a polynomial in k of degree n, then ∆m a0 will be zero for all m > n and thus a finite number of terms for the series S(x) will occur. 32 Difference Equations Proof. We have S(x) = a0 + a1 x + a2 x2 + · · · + ak xk + · · · = (1 + xE + x2 E 2 + · · · + xk E k + · · · )a0 = (1 − xE)−1 a0 = [1 − x(1 + ∆)]−1 a0 −1 1 x 1− ∆ a0 1−x 1−x x∆ x2 ∆2 1 1+ + · · · a0 + = 1−x 1 − x (1 − x)2 a0 x2 ∆2 a0 x∆a0 = + + ··· .
6. For λ = 1, λk Pk = λk λ−1 1− λ∆ λ2 ∆2 − ··· + λ − 1 (∆ − 1)2 Pk . 249) Proof. Let Fk be a function of k. 250) k = λ (λE − 1)Fk . Now, set (λE − 1)Fk = Pk ; consequently Fk = (λE − 1)−1 Pk . 253) Pk . 2 a short listing of the antidifferences and definite sums of selected functions. In each case, the particular item is calculated using the definition of ∆−1 yk and the fundamental theorem of the sum calculus. For example, k−1 ∆−1 1 = 1 = k + constant. 1. Likewise, from the fundamental theorem of calculus, we have n 1 = ∆−1 1|n+1 = n + 1.
Difference of a Product Apply the difference operator to the product xk yk : ∆(xk yk ) = xk+1 yk+1 − xk yk = xk+1 yk+1 − xk+1 yk + yk xk+1 − xk yk = xk+1 (yk+1 − yk ) + yk (xk+1 − xk ) = xk+1 ∆yk + yk ∆xk . 98) Leibnitz’s Theorem for Differences We now prove that ∆n (xk yk ) = xk ∆n yk + n (∆xk )(∆n−1 yk+1 ) 1 n (∆2 xk )(∆n−2 yk+2 ) 2 n + ···+ (∆n xk )(yk+n ). 99) THE DIFFERENCE CALCULUS 15 Proof. Define operators E1 and E2 which operate, respectively, only on xk and yk . Therefore, E1 (xk yk ) = xk+1 yk , E2 (xk yk ) = xk yk+1 .