The next theorem gives a sketch of irrationality proof to a constant that \ may not yet have an irrationality proofs ----------------------------------------------------------------------------\ --- Sketch of an Irrationality Proof of the constant 1 1 / / | | 1 | | ---------------------- dx dy | | (2/3) (3/2) / / x (-x y + 1) 0 0 divided by 1 1 / / | | 1 | | -------------------- dx dy | | (2/3) 1/2 / / x (-x y + 1) 0 0 By Shalosh B. Ekhad Theorem: The constant of the title 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.374181890, yielding an irrationality measure that is approximately , 150.9578309 We hope that the reader can find nu exactly. According to Maple, this constant equals (3/2) 1/2 Pi 3 + 9 GAMMA(2/3) GAMMA(5/6) 1/3 -------------------------------------- (3/2) 1/2 Pi 3 - 3 GAMMA(2/3) GAMMA(5/6) and in Maple format 1/3*(Pi^(3/2)*3^(1/2)+9*GAMMA(2/3)*GAMMA(5/6))/(Pi^(3/2)*3^(1/2)-3*GAMMA(2/3)* GAMMA(5/6)) We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 2 36 (n + 2) (30 n + 99 n + 80) (1 + 2 n) (1 + n) (1 + 3 n) X(n) - --------------------------------------------------------------------- + 6 2 (30 n + 39 n + 11) (5 + 2 n) (8 + 3 n) (3 + 2 n) (5 + 6 n) (4 + 3 n) 5 4 3 2 (n + 2) (5940 n + 33462 n + 71724 n + 72567 n + 34259 n + 5920) / 2 X(1 + n) / ((30 n + 39 n + 11) (5 + 2 n) (8 + 3 n) (5 + 6 n) (4 + 3 n)) / + X(n + 2) = 0 Subject to the initial conditions A(0) = 0, A(1) = -11/5 B(0) = 1, B(1) = -7/5 A(n) Then, ----, approximates the constant of the title, c B(n) / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof, consider the Beukers type-integral 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / F(n) = | | ---------------------- dx dy | | 2/3 3/2 | | x (-x y + 1) / / 0 0 normalized by dividing by the following constant, (independent of n) 1 1 / / | | 1 | | -------------------- dx dy | | (2/3) 1/2 / / x (-x y + 1) 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.358146767, 2.362411568, 2.357418323, 2.352643839, 2.358383963, 2.365242755, 2.370195635, 2.359287324, 2.366348222, 2.372576349, 2.371294724, 2.374133279, 2.371522929, 2.371413077, 2.369793606, 2.373509790, 2.371393618, 2.372558683, 2.369469565, 2.374181890, 2.368875619] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.01005673539, 0.009153365892, 0.01021119603, 0.01122475733, 0.01000645010, 0.008554556183, 0.007508705419, 0.009814984388, 0.008320937446, 0.007006765045, 0.007276915652, 0.006678778447, 0.007228802271, 0.007251962238, 0.007593517020, 0.006810098479, 0.007256064867, 0.007010487833, 0.007661886751, 0.006668541377, 0.007787228097] As you can see, they are all positive We leave it to the reader to fill-in the details. The next two theorems seem go give new irrationality proofs of log(2), but t\ hat's not very exciting! ----------------------------------------------------------------------------\ --- Sketch of an Irrationality Proof of the constant 1 1 / / | | 1 | | --------------- dx dy | | (3/2) / / (-x y + 1) 0 0 divided by 1 1 / / | | 1 | | ------------- dx dy | | 1/2 / / (-x y + 1) 0 0 By Shalosh B. Ekhad Theorem: The constant of the title 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.000563405, yielding an irrationality measure that is approximately , 11.86724795 We hope that the reader can find nu exactly. According to Maple, this constant equals 4 ln(2) ----------- 4 - 4 ln(2) and in Maple format 4*ln(2)/(4-4*ln(2)) We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 4 (1 + n) (1 + 2 n) (10 n + 17) X(n) - ------------------------------------- 2 (10 n + 7) (5 + 2 n) (3 + 2 n) 3 2 2 (220 n + 814 n + 950 n + 341) X(1 + n) + ------------------------------------------ + X(2 + n) = 0 (10 n + 7) (5 + 2 n) (3 + 2 n) Subject to the initial conditions A(0) = 0, A(1) = -7/3 B(0) = 1, B(1) = -1 A(n) Then, ----, approximates the constant of the title, c B(n) / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof, consider the Beukers type-integral 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / F(n) = | | ---------------------- dx dy | | 3/2 | | (-x y + 1) / / 0 0 normalized by dividing by the following constant, (independent of n) 1 1 / / | | 1 | | ------------- dx dy | | 1/2 / / (-x y + 1) 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.000563405, 1.996646589, 1.994093631, 1.991691859, 1.997405289, 1.993490208, 1.992276135, 1.998375923, 1.997370830, 1.989174258, 1.992727278, 1.991705844, 1.990342866, 1.991260295, 1.987938293, 1.988738279, 1.987751515, 1.987516394, 1.992038635, 1.984942617, 1.980600718] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.09201961757, 0.09299112014, 0.09362527095, 0.09422253955, 0.09280280197, 0.09377526782, 0.09407718274, 0.09256197391, 0.09281135369, 0.09484931303, 0.09396497332, 0.09421905989, 0.09455829104, 0.09432992930, 0.09515727758, 0.09495792585, 0.09520383200, 0.09526244153, 0.09413626358, 0.09590442745, 0.09698914943] As you can see, they are all positive We leave it to the reader to fill-in the details. ----------------------------------------------------------------------------\ --- Sketch of an Irrationality Proof of the constant 1 1 / / | | 1 | | ------------------------- dx dy | | 1/2 1/2 (3/2) / / x y (-x y + 1) 0 0 divided by 1 1 / / | | 1 | | ----------------------- dx dy | | 1/2 1/2 1/2 / / x y (-x y + 1) 0 0 By Shalosh B. Ekhad Theorem: The constant of the title 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 1.995519913, yielding an irrationality measure that is approximately , 11.72145829 We hope that the reader can find nu exactly. 1 Comment: Note that this constant appears to be , ----- ln(2) Prove it! According to Maple, this constant equals 1 ----- ln(2) and in Maple format 1/ln(2) We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 (1 + 2 n) (5 n + 8) (2 n + 3) X(n) -1/4 ----------------------------------- (n + 1) (5 n + 3) (5 + 2 n) (2 + n) 3 2 (2 n + 3) (110 n + 396 n + 445 n + 150) X(n + 1) + 1/2 -------------------------------------------------- + X(2 + n) = 0 (n + 1) (5 n + 3) (5 + 2 n) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -1 B(0) = 1, B(1) = -2/3 A(n) Then, ----, approximates the constant of the title, c B(n) / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof, consider the Beukers type-integral 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / F(n) = | | ----------------------- dx dy | | 1/2 1/2 3/2 | | x y (-x y + 1) / / 0 0 normalized by dividing by the following constant, (independent of n) 1 1 / / | | 1 | | ----------------------- dx dy | | 1/2 1/2 1/2 / / x y (-x y + 1) 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [1.990664263, 1.987474748, 1.991207831, 1.995519913, 1.990702628, 1.991588878, 1.995301549, 1.991756366, 1.992195325, 1.994691090, 1.992594970, 1.990713247, 1.984005816, 1.982432578, 1.991068796, 1.989709712, 1.983892867, 1.982686896, 1.986647805, 1.984817646, 1.986526559] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.09447827970, 0.09527282345, 0.09434298580, 0.09327089403, 0.09446872956, 0.09424816327, 0.09332513430, 0.09420648953, 0.09409728444, 0.09347679711, 0.09399787891, 0.09446608622, 0.09613828375, 0.09653124027, 0.09437758838, 0.09471594809, 0.09616648628, 0.09646769874, 0.09547901253, 0.09593561855, 0.09550925040] As you can see, they are all positive We leave it to the reader to fill-in the details.