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By Daniel J. Velleman

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Extra info for American Mathematical Monthly, volume 117, August September 2010

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Unfortunately, this hides the fact that this change of variables introduces a transcendental function with a multivalued inverse. One has to proceed with care. 3) of G n gives G n+1 = 2 π π/2 (cos θ)2n dθ. 1) is obtained by writing (cos θ)2n = (cos θ)2n−2 + sin θ d (cos θ)2n−1 2n − 1 dθ and then integrating by parts. Yet another recurrence for G n is obtained by a doubleangle substitution yielding G n+1 = August–September 2010] 2 π π/2 0 1 + cos 2θ 2 n dθ, WALLIS-RAMANUJAN-SCHUR-FEYNMAN 619 and a binomial expansion (observe that the odd powers of cosine integrate to zero).

2) cancel out. This may be checked directly by combining the summands corresponding to j and k. 2) is symmetric while the critical factors q j − qk in the denominator combine to form the antisymmetric Vandermonde determinant. Accordingly, they have to cancel. 1A 622 minor correction from [2]. 4. The identities ∞ n+1 2 π 2 π 0 j =1 ∞ n+1 0 j =1 2 π x2 ∞ 0 x2 1 1 , dx = 2 +j (2n + 1)n! (n + 1)! 2. Here, A(n, k) are the Eulerian numbers which count the number of permutations of n objects with exactly k descents.

Moreover, a routine calculation confirms that (6) is true for 0 ≤ k ≤ li + 1. Finally, we have to treat the case i = 5, which is less involved than the cases i ∈ I . Here we directly show that v2k = ti 2k−2 + 2k−2 and v2k+1 = ti 2k−1 + 2k for k ≥ 1, so that we do not have to bother about initial conditions. ) Summing up, we have that the intervals [ξi,1 , ξi,2 ) are disjoint for i = 1, 2, . . , 8 √ √ 2 2 and completely cover [1 − 2 , 2 ). 2. ACKNOWLEDGMENTS. The author is a recipient of an APART-fellowship of the Austrian Academy of Sciences at the University of Waterloo, Canada.

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