Bill Gosper <rwmgosper@yahoo.com> wrote: >rwg>This my first recollection of an algebraic series for (1/3)!. Foo, the nth difference of a monic nth degree polynomial anywhere is n!, so just forget n is an integer and run the sum to oo. For oo=9, (c540) ('sum((-1)^k*(x-k)^n*binomial(n,k),k,0,9),%% = apply_nouns(%%)) 9 ==== \ k n (d540) > (- 1) binomial(n, k) (x - k) = / ==== k = 0 n n n n (n - 1) n (x - 2) (n - 2) (n - 1) n (x - 3) x - n (x - 1) + ------------------ - -------------------------- 2 6 n (n - 3) (n - 2) (n - 1) n (x - 4) + ---------------------------------- 24 n (n - 4) (n - 3) (n - 2) (n - 1) n (x - 5) - ------------------------------------------ 120 n (n - 5) (n - 4) (n - 3) (n - 2) (n - 1) n (x - 6) + -------------------------------------------------- 720 n (n - 6) (n - 5) (n - 4) (n - 3) (n - 2) (n - 1) n (x - 7) - ---------------------------------------------------------- 5040 n (n - 7) (n - 6) (n - 5) (n - 4) (n - 3) (n - 2) (n - 1) n (x - 8) + ------------------------------------------------------------------ 40320 n (n - 8) (n - 7) (n - 6) (n - 5) (n - 4) (n - 3) (n - 2) (n - 1) n (x - 9) - -------------------------------------------------------------------------- 362880 (c541) block([ratfac : true],makelist(ratsimp(''(rhs(%))),n,0,10)) (d541) [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 10 9 8 7 6 5 4 - x + 100 x - 4500 x + 120000 x - 2100000 x + 25200000 x - 210000000 x 3 2 + 1200000000 x - 4500000000 x + 10000000000 x - 9996371200] (Why not to rely *too* heavily on EIS.) (c543) subst([x = 9,n = 9/2],n! = rhs(d540)) 945 sqrt(%pi) 151263 sqrt(7) 196875 sqrt(5) (d543) ------------- = -------------- - 8505 sqrt(6) + -------------- 32 8 128 1701 sqrt(3) 4718601 sqrt(2) 640843731 - ------------ - --------------- + --------- 1024 128 32768 (c544) dfloat(%) (d544) 52.3427777845535d0 = 52.3427798776902d0 Even easier: Just construct a telescoping series from a product limit for x!: inf x + 1 x x - 1 x x ==== (----- + n) - (----- + n) (- + 1) \ 2 2 n x + 1 x (d488) > ----------------------------------- + (-----) / binomial(x + n, n) 2 ==== n = 1 (c489) apply_nouns(subst([x = 1/2.0d0,inf = 333],d488))^2*4 (d489) 3.14159177222301d0 Finally, a really simple formula for eulerian(n,j): | n + 1 n | (d547) eulerian(n, j) = delta (max (k - j, 0))| k | |k = n | n + 1 p | = delta (max (n - j, 0))|, n | |p = n where delta^j := the jth backward difference operator. (The 'at notation just is to avoid differencing the n in the exponent. There's a subtle difference between the evaluation semantics of the two forms which necessitates quotes in one or 'eval in the other, depending on your implementation of 'delta.) With nth difference vs n+1, you get the eulerian running row sum. --rwg --------------------------------- Looking for last minute shopping deals? Find them fast with Yahoo! Search.