An Irrationality measure of arctan(1/p) Doron ZEILBERGER and Wadim ZUDILIN

Abstract. We establish an irrationality measur for arctan(\frac{1}{p}) for any odd prime p. The Maple package. This article is accompanied by a shoft Maple package

• ArcTan.txt ;

It is available, along with an input and output file, from the front of this article

http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/arctan.html .

Let p be an odd prime. Consider the following integral

E_{n}(p) = ∫_{0}^{1} \frac{x^{n} (1 − x)^{n}}{(1 + \frac{x^{2}}{p^{2}} )^{n+1}} dx .

Fact 1: There exists a constant C_{1} such that

|E_{n}(p)| ≤ \frac{C}{a_{1}(p)^{n}} ,

where

a_{1}(p) = 2 \frac{1}{(−p + \sqrt{1 + p^{2}}) p} .

Proof: Differentiating f(x) = _{\frac{x(1−x)}{1+ \frac{x^{2}}{p^{2}}}} with respect to x and setting it equal to 0 gets that the maximum is at x = (−p + \sqrt{1 + p^{2}}) p. Plugging it into f(x) gives 1/a_{1}(p). Fact 2: E_{n}(p) satisfies the linear recurrence equation

(4n + 8)E_{n+2}(p) − 2p^{2}(2n + 3)E_{n+1}(p) − (n + 1)p^{2}E_{n}(p) = 0 .

Proof: Follows from the Almkvist-Zeilberger algorithm [AZ]. Fact 3: There exist rational numbers A_{n}(p) and B_{n}(p) such that

E_{n}(f) = −A_{n} + B_{n} arctan(\frac{1}{p}) .

Proof: Either by integration by parts, or using Fact 2, once it is verified for n = 0 and n = 1.

Fact 4: The sequence of rational numbers A_{n}(p), B_{n}(p) satisfy the same recurrence as of E_{n}(p), namely

(4n + 8)A_{n+2}(p) + 2p^{2}(2n + 3)A_{n+1}(p) − (n + 1)p^{2}A_{n}(p) = 0 ,
(4n + 8)B_{n+2}(p) + 2p^{2}(2n + 3)B_{n+1}(p) − (n + 1)p^{2}B_{n}(p) = 0 .

Proof: Use fact 3, comparing the coefficients of 1 and arctan(\frac{1}{p}). Fact 5: There exist constants C_{2}, C_{3} such that

A_{n}(p) ≤ C_{2}a_{2}(p)^{n} , B_{n}(p) ≤ C_{3}a_{2}(p)^{n} ,

where

a_{2}(p) = \frac{1}{2} (p + \sqrt{1 + p^{2}}) p

Proof: From Fact 4 and the Poincaré Lemma. Fact 6: There exists a constant C_{4} such that

|arctan(\frac{1}{p}) − \frac{A_{n}(p)}{B_{n}(p)}| ≤ \frac{C_{4}}{(a_{1}(p) a_{2}(p))^{n}} .

Proof: Divide Fact 3 by B_{n} and use Facts 1 and 5. Surprising Fact 7: Let d(n) := lcm(1...n). Then

Ā_{n}(p) = A_{n}(p) d(n) 2^{[3n/2]}/p^{n} , B̄_{n}(p) = B_{n}(p) d(n) 2^{[3n/2]}/p^{n}

are integers. Proof: Checked empirically for many p and many n. Formal proof coming up. Fact 8:

B̄_{n}(p) = O((2^{3/2} a_{2}(p)/p)^{n}d(n)) ,

Proof: From Fact 5 and the definition of B̄_{n}(p). From Fact 6 we have Fact 9

|arctan(\frac{1}{p}) − \frac{Ā_{n}}{B̄_{n}} | ≤ \frac{C_{4}}{(a_{1}(p) a_{2}(p))^{n}} = \frac{C_{5}}{(B̄_{n}(p))^{1+δ}} , ,

where (since d(n) = O(e^{n})),

1 + δ = \frac{log(a_{1}(p)) + log(a_{2}(p))}{\frac{3}{2} log 2 + log(a_{2}(p)) − log(p) + 1} .

As usual (see [vdP]), the irrationality measure is 1 + 1/δ, hence, we have established Theorem: If p is an odd prime, then arctan(\frac{1}{p}) has irrationality measure

−2 (ln (2 \frac{p + \sqrt{1 + p^{2}}}{p} ) + ln (1/2 (p + \sqrt{1 + p^{2}}) p)) (−2 ln (2 \frac{p + \sqrt{1 + p^{2}}}{p} ) + 3 ln (2) + 2 − 2 ln (p))^{−1} .

Here are a few sample values

p = 3 : 7.707357546 , p = 5 : 4.788327432 , p = 7 : 4.075567842 , p = 11 : 3.542000040 , p = 1009 : 2.429989300 , p = 1000000007 : 2.134188776 , p = 100000000000000000039 : 2.059322676 . References

[AZ] Gert Almkvist and Doron Zeilberger, The Method of differentiating Under The integral sign, J. Symbolic Computation 10 (1990), 571-591. Available from

http://sites.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/duis.html .

[PWZ] Marko Petkovsek, Herbert S. Wilf, and Doron Zeilberger, “A=B”, A.K. Peters, 1996. Freely

available from https://www.math.upenn.edu/~wilf/AeqB.html .

[vdP] Alfred van der Poorten, A Proof that Euler missed, Math. Intell. 1(1979), 195-203.

Wadim Zudilin, Institute for Mathematics, Astrophysics and Particle Physics Radboud Universiteit, PO Box 9010 6500 GL Nijmegen, The Netherlands Email: wzudilin [at] gmail [dot] com .

Doron Zeilberger, Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen Rd., Piscataway, NJ 08854-8019, USA.

Email: DoronZeil [at] gmail [dot] com .

Written: Oct. 8, 2019.