A sequence of Diophantine Approximations to Pi that implies its irrationality With a fully rigorous crude upper bound on the irrationality measure of, 10.747747465671804677 and a not-yet-rigorous smaller upper bound around, 7.1956394230370108057 By Shalosh B. Ekhad Consider the following Salikhov-type integral, let's call it E(n) 4 + 2 I / | (2 n) (2 n) (2 n) -I | (x - 4 + 2 I) (x - 4 - 2 I) (x - 5) | / 4 - 2 I (2 n) (2 n) / (3 n + 1) (3 n + 1) (x - 6 + 2 I) (x - 6 - 2 I) / (x (x - 10) ) dx / It is readily seen that E(n) can be written as E(n)=C(n) + D(n)*Pi For certain sequences of RATIONAL numbers C(n), D(n) It follows from the amazing Almkvist-Zeilberger algorithm that E(n), and hen\ ce C(n) and D(n) satisfy the following third-order linear recurrence equation with polynomial coefficients. -1024 (2 n + 5) (2 n + 3) (2 n + 1) (n + 2) (n + 1) 4 3 2 (559455 n + 5637736 n + 21207869 n + 35294652 n + 21926016) F(n) + 64 6 5 (2 n + 5) (2 n + 3) (n + 2) (1208982255 n + 14601112006 n 4 3 2 + 71554462078 n + 181444545414 n + 250049018747 n + 176898555884 n 8 7 + 50159702880) F(n + 1) - 6 (2 n + 5) (440102548665 n + 7075613606958 n 6 5 4 + 49182254436312 n + 192978557313766 n + 467319046780891 n 3 2 + 714892923466956 n + 674376299494052 n + 358497976451216 n + 82191375084672) F(n + 2) + 3 (4 n + 11) (3 n + 8) (3 n + 7) (4 n + 9) 4 3 2 (n + 3) (559455 n + 3399916 n + 7651391 n + 7554302 n + 2760952) F(n + 3) = 0 and in Maple format -1024*(2*n+5)*(2*n+3)*(2*n+1)*(n+2)*(n+1)*(559455*n^4+5637736*n^3+21207869*n^2+ 35294652*n+21926016)*F(n)+64*(2*n+5)*(2*n+3)*(n+2)*(1208982255*n^6+14601112006* n^5+71554462078*n^4+181444545414*n^3+250049018747*n^2+176898555884*n+ 50159702880)*F(n+1)-6*(2*n+5)*(440102548665*n^8+7075613606958*n^7+ 49182254436312*n^6+192978557313766*n^5+467319046780891*n^4+714892923466956*n^3+ 674376299494052*n^2+358497976451216*n+82191375084672)*F(n+2)+3*(4*n+11)*(3*n+8) *(3*n+7)*(4*n+9)*(n+3)*(559455*n^4+3399916*n^3+7651391*n^2+7554302*n+2760952)*F (n+3) = 0 The initial conditions are as follows Pi 1196 Pi 11272 18662336 Pi 6156093056 E(0) = - ----, E(1) = - ------- + -----, E(2) = - ----------- + ---------- 20 5 15 5 525 and in Maple notation E(0) = -1/20*Pi, E(1) = -1196/5*Pi+11272/15, E(2) = -18662336/5*Pi+6156093056/ 525 Using this recurrence, we can compute many values, and use them to ESTIMATE \ the implied bound irrationality measure. Using the values from n=, 990, to M=, 1000, and taking the largest value yields the following estimate 7.0702336785252052513 We can use the Poincare lemma to find the asymptotic behavior of the exponti\ ally growing C(n), D(n) and the exponentially decaying E(n). The constant-coefficient approximation to the above recurrence has indicial \ polynomial 3 2 -108 N + 2359989 N - 138304 N + 2048 and in Maple format -108*N^3+2359989*N^2-138304*N+2048 whose largest root let's call it a, is, 21851.691396219561697, and absolute value of the two smaller roots, let's call it b is, 0.029458495928116876368 It follows that, up to polynomial correction that will get out in the wash, \ C(n) and D(n) are OMEGA(a^n) and E(n) is O(b^n) We now need a divisibility lemma that is left to the reader Let d(n)=lcm(1..n) Lemma: d(4 n) C(n) d(4 n) D(n) C1(n) = ------------, D1(n) = ------------ trunc(n/2) trunc(n/2) 32 32 are integers Defining d(4 n) E(n) E1(n) = ----------- 1/2 n (4 2 ) We have E1(n) = C1(n) + Pi D1(n) where C1(n) and D1(n) are INTEGERS Since, famously, d(n)=OMEGA(e^n), we have C1(n) and D1(n) are OMEGA of n exp(4 n) a ----------- 1/2 n (4 2 ) and E1(n) is Big O of n exp(4 n) b ----------- 1/2 n (4 2 ) It follows that E1(n)=max(C1(n),D1(n))^(-delta) where delta equals 1/2 2 ln(b) + 4 + ln(----) 8 - -------------------- 1/2 2 ln(a) + 4 + ln(----) 8 recall that a is , 21851.691396219561697, and b is , 0.029458495928116876368 and delta is .10258780333832551066 implying the rigorous, but crude upper bound for the irrationality measure, \ 1+1/delta that equals 10.747747465671804677 confirming the first part of the statement in the title We now need a harder lemma, that we confess that we can't do, but you the re\ ader may be able to Harder Lemma: There exists a rigorously proved constant K1 such that gcd(C1(n),D1(n))=O(exp(K1*n)) Once we find such a constant K1 the delta can be improved to By looking at the smallest value of log(gcd(C1(n),D1(n))/n for n between n=, 500, and , 1000, we estimate that K1 can be taken to be .62083103467201405134 and the better delta is .16140384094686627049 that implies an irrationality measure, 1+1/delta that equals 7.1956394230370108060 ----------------------------------------- This ends this paper that took, 37.623, to generate.