1/2 arctan(a ) Irrationality Proofs and Measures for, ------------, 1/2 for ALL positive integers a that are Congruent to 3 Modulo 4 By Shalosh B. Ekhad Theorem: Let a be a positive integer such that a (mod 4)=3 then arctan(sqrt(a))/sqrt(a) is an irrational number and an irrationality measure is, | 1/2 | | 1/2 | | -a + (a (a + 1)) | | a + (a (a + 1)) | 2 ln(| ------------------- |) - 2 ln(| ------------------ |) | a | | a | ------------------------------------------------------------ | 1/2 | | -a + (a (a + 1)) | 2 ln(| ------------------- |) + ln(a) + 2 | a | In Maple format the irrationality measure is (2*ln(abs((-a+(a*(a+1))^(1/2))/a))-2*ln(abs((a+(a*(a+1))^(1/2))/a)))/(2*ln(abs( (-a+(a*(a+1))^(1/2))/a))+ln(a)+2) Proof: Consider the integral 1 / n n | x (1 - x) E(n) = | --------------- dx | 2 (n + 1) / (x + a) 0 Let c be the constant 1/2 arctan(a ) ------------ 1/2 a It is readily seen that E(n) can be written as E(n)=A(n)+B(n)*c for some sequences A(n) and B(n) of RATIONAL NUMBERS. It follows from the Almkvist-Zeilberger algorithm that E(n), and hence A(n) and B(n), all satisfy the following second-order linear\ recurrence with polynomial coefficients (n + 1) F(n) - 2 a (2 n + 3) F(n + 1) - 4 a (n + 2) F(n + 2) = 0 but of course with different initial conditions. It follows from the Poincarre lemma that A(n),B(n)=OMEGA(C1(a)^n), E(n)=OMEGA( C2(a)^n ) where C1(a) is the larger root and C2(a) the smaller root (in absolute value\ ) of the indicial equation of the constant-coefficient recurrence approximating the above recurrence that indicial polynomial, in N, is 2 -4 N a - 4 N a + 1 and in Maple format -4*N^2*a-4*N*a+1 Solving this quadratic, we get that the absolute value of the root that has \ the larger absolute value, let's call is C1(a) is | 2 1/2 | | a + (a + a) | 1/2 | --------------- | | a | and the smaller root (in absolute value), let's call it, C2(a), is | 2 1/2 | | -a + (a + a) | 1/2 | ---------------- | | a | and in Maple format,they are respectively 1/2*(a+(a^2+a)^(1/2))/a, 1/2*(-a+(a^2+a)^(1/2))/a We now need the following divisibility lemma that is left as an exercise to \ the reader. Lemma: A1(n)=lcm(1..n)*2^n*a^(trunc(n/2))*A(n) and B1(n)=lcm(1..n)*2^n*a^(t\ runc(n/2))*B(n) are always integers. Let E1(n)=lcm(1..n)*2^n*a^(trunc(n/2))*E(n) , then E1(n)=A1(n)+B1(n)*c where now A1(n) and B1(n) are INTEGERS. Since lcm(1..n)=OMEGA(e^n), we have A1(n),B1(n)=OMEGA((2*sqrt(a)*e*C1(a))^n ), E(n)=OMEGA( (2*sqrt(a)*e*C2(a))^n\ ) Hence E(n)=OMEGA(1/max(A(n),B(n))^delta ), where delta is -log((2*sqrt(a)*e*C2(a))/log(b^2*e*C1(a,b)) | 2 1/2 | | -2 a + 2 (a + a) | 1/2 ln(| -------------------- |) + ln(2 a ) + 1 | 4 a | that equals , - --------------------------------------------- | 2 1/2 | | 2 a + 2 (a + a) | 1/2 ln(| ------------------- |) + ln(2 a ) + 1 | 4 a | and in Maple format it is -(ln(abs(1/4*(-2*a+2*(a^2+a)^(1/2))/a))+ln(2*a^(1/2))+1)/(ln(abs(1/4*(2*a+2*(a^ 2+a)^(1/2))/a))+ln(2*a^(1/2))+1) It is readily seen that this is always positive and an irrationality measure can be taken to be 1+1/de that equals | 1/2 | | 1/2 | | -a + (a (a + 1)) | | a + (a (a + 1)) | 2 ln(| ------------------- |) - 2 ln(| ------------------ |) | a | | a | ------------------------------------------------------------ | 1/2 | | -a + (a (a + 1)) | 2 ln(| ------------------- |) + ln(a) + 2 | a | as claimed in the statement. Finally, here is a plot ot this irrationality measure from a=3 to 1000. Recall that it is ONLY valid for integers a that are 3(mod 8) or 7 (mod 8) * 8 * * * * 7 * * * 6 * * * * 5 * * *H 4 +H +H +HH + HHHH 3 + HHHHHHHHHHHHHHHH +--+--+--+--+--+--+--+****************************************************** 200 400 600 800 1000