------------------------------ This file finds Verbose versions for the recurrences for the integral, from \ 0 to 1 of powers of the Chebyshev polynomials --------------------------------------- 1 / | For the sequence defined by, | CHEBYSHEVfirstKind[n](x) dx, we have | / 0 ----------------------------------------------------------------------------\ ---------------- A Linear Recurrence With Polynomial Coefficients Satisfied by The Inegral of\ the, 1, -th power of Certain C-finite Polynomial Sequence By Shalosh B. Ekhad Proposition: Let , P[n](x), be the sequence of polynomial definied by the following recurrence P[n](x) = 2 x P[n - 1](x) - P[n - 2](x) Subject to the initial conditions P[0] = 1, P[1] = x Equivalently, in terms of generating function infinity ----- \ n -t x + 1 ) P[n](x) t = -------------- / 2 ----- t - 2 t x + 1 n = 0 1 n Define the Umbra, U, to be equal to, -----, when applied to , x , n + 1 and extended linearly Let's define the sequence of numbers s(n), to be the action of U applied to, P[n](x) Then s(n) satisfies the following linear recurrence equation with polynomial\ coefficients (n + 2) (n - 1) s(n) n (n + 3) s(n + 1) n (n + 3) s(n + 2) - -------------------- + ------------------ - ------------------ + s(n + 3) = 0 (n + 4) (n + 1) (n + 4) (n + 1) (n + 4) (n + 1) Subject to the initial condition s(1) = 1/2, s(2) = -1/3, s(3) = -1/2 Just for fun, using this recurrence we get that -1 s(1000) = ------ 999999 This ends this article, that took, 0.128, seconds to generate. ---------------------------------------------------- --------------------------------------- 1 / | 2 For the sequence defined by, | CHEBYSHEVfirstKind[n](x) dx, we have | / 0 ----------------------------------------------------------------------------\ ---------------- A Linear Recurrence With Polynomial Coefficients Satisfied by The Inegral of\ the, 2, -th power of Certain C-finite Polynomial Sequence By Shalosh B. Ekhad Proposition: Let , P[n](x), be the sequence of polynomial definied by the following recurrence P[n](x) = 2 x P[n - 1](x) - P[n - 2](x) Subject to the initial conditions P[0] = 1, P[1] = x Equivalently, in terms of generating function infinity ----- \ n -t x + 1 ) P[n](x) t = -------------- / 2 ----- t - 2 t x + 1 n = 0 1 n Define the Umbra, U, to be equal to, -----, when applied to , x , n + 1 and extended linearly Let's define the sequence of numbers 2 s(n), to be the action of U applied to, P[n](x) Then s(n) satisfies the following linear recurrence equation with polynomial\ coefficients (2 n - 1) s(n) 4 (n + 1) s(n + 1) -------------- - ------------------ + s(n + 2) = 0 2 n + 5 2 n + 5 Subject to the initial condition s(1) = 1/3, s(2) = 7/15 Just for fun, using this recurrence we get that 1999999 s(1000) = ------- 3999999 This ends this article, that took, 0.232, seconds to generate. ---------------------------------------------------- --------------------------------------- 1 / | 3 For the sequence defined by, | CHEBYSHEVfirstKind[n](x) dx, we have | / 0 ----------------------------------------------------------------------------\ ---------------- A Linear Recurrence With Polynomial Coefficients Satisfied by The Inegral of\ the, 3, -th power of Certain C-finite Polynomial Sequence By Shalosh B. Ekhad Proposition: Let , P[n](x), be the sequence of polynomial definied by the following recurrence P[n](x) = 2 x P[n - 1](x) - P[n - 2](x) Subject to the initial conditions P[0] = 1, P[1] = x Equivalently, in terms of generating function infinity ----- \ n -t x + 1 ) P[n](x) t = -------------- / 2 ----- t - 2 t x + 1 n = 0 1 n Define the Umbra, U, to be equal to, -----, when applied to , x , n + 1 and extended linearly Let's define the sequence of numbers 3 s(n), to be the action of U applied to, P[n](x) Then s(n) satisfies the following linear recurrence equation with polynomial\ coefficients 3 2 (n - 1) (3 n - 1) (3 n + 1) (57 n + 425 n + 1006 n + 748) s(n) ---------------------------------------------------------------- (3 n + 11) (3 n + 13) (n + 5) %1 3 114 n (3 n + 4) (3 n + 2) (n + 3) s(n + 1) - ------------------------------------------- (3 n + 11) (3 n + 13) (n + 5) %1 4 3 2 2 (3 n + 5) (3 n + 7) (57 n + 456 n + 1285 n + 1492 n + 496) s(n + 2) + ------------------------------------------------------------------------ (3 n + 11) (3 n + 13) (n + 5) %1 3 114 (3 n + 10) (n + 4) (3 n + 8) (n + 1) s(n + 3) - -------------------------------------------------- + s(n + 4) = 0 (3 n + 11) (3 n + 13) (n + 5) %1 3 2 %1 := 57 n + 259 n + 342 n + 124 Subject to the initial condition -9 -7 -37 s(1) = 1/4, s(2) = --, s(3) = --, s(4) = --- 35 20 715 Just for fun, using this recurrence we get that -2333333 s(1000) = ------------- 2999996666667 This ends this article, that took, 1.561, seconds to generate. ---------------------------------------------------- --------------------------------------- 1 / | 4 For the sequence defined by, | CHEBYSHEVfirstKind[n](x) dx, we have | / 0 ----------------------------------------------------------------------------\ ---------------- A Linear Recurrence With Polynomial Coefficients Satisfied by The Inegral of\ the, 4, -th power of Certain C-finite Polynomial Sequence By Shalosh B. Ekhad Proposition: Let , P[n](x), be the sequence of polynomial definied by the following recurrence P[n](x) = 2 x P[n - 1](x) - P[n - 2](x) Subject to the initial conditions P[0] = 1, P[1] = x Equivalently, in terms of generating function infinity ----- \ n -t x + 1 ) P[n](x) t = -------------- / 2 ----- t - 2 t x + 1 n = 0 1 n Define the Umbra, U, to be equal to, -----, when applied to , x , n + 1 and extended linearly Let's define the sequence of numbers 4 s(n), to be the action of U applied to, P[n](x) Then s(n) satisfies the following linear recurrence equation with polynomial\ coefficients 2 (4 n - 1) (2 n - 1) (4 n + 1) (24 n + 73 n + 54) s(n) ------------------------------------------------------ n (2 n + 5) (4 n + 7) (4 n + 9) 3 2 2 (4 n + 5) (4 n + 3) (24 n + 14 n - 28 n + 9) s(n + 1) - --------------------------------------------------------- + s(n + 2) = 0 n (2 n + 5) (4 n + 7) (4 n + 9) Subject to the initial condition 107 s(1) = 1/5, s(2) = --- 315 Just for fun, using this recurrence we get that 2666664888889 s(1000) = ------------- 7111108888889 This ends this article, that took, 1.264, seconds to generate. ---------------------------------------------------- -------------------------------------- This took, 3.227, seconds.