notes 13 - Special values Laurent coecients algebraic...

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( September 22, 2013 ) Special values, Laurent coefficients, algebraic relations Paul Garrett [email protected] http: / /www.math.umn.edu/ e garrett/ [This document is http://www.math.umn.edu/˜garrett/m/mfms/notes 2013-14/03c spec vals Laurent.pdf] 1. Special values ζ (2 n ) in a Laurent expansion 2. Special values L (2 n, χ ) in Laurent expansion of ( z ) 1. Special values ζ (2 n ) in a Laurent expansion Via Liouville’s theorem, by cancelling poles, and so on, f ( z ) = X n Z 1 ( z + n ) 2 (secretly f ( z ) = π 2 sin 2 πz , but we don’t use this) satisfies f 0 2 = 4 f 2 ( f - π 2 ) The Laurent coefficients of f ( z ) at 0 have direct relations to the special values ζ (2) , ζ (4) , . . . , producing algebraic relations among these values, as follows. Let g ( z ) = f ( z ) - 1 z 2 , so g ( z ) is holomorphic at z = 0, and g ( z ) = g (0) + g 0 (0) 1! z + g 00 (0) 2! z 2 + . . . = X n 6 =0 1 n 2 + X n 6 =0 - 2 1! · n 3 z + X n 6 =0 ( - 2)( - 3) 2! · n 4 z 2 + X n 6 =0 ( - 2)( - 3)( - 4) 3! · n 5 z 3 + X n 6 =0 ( - 2)( - 3)( - 4)( - 5) 4! · n 6 z 4 + . . . In the odd-degree sums the ± n terms cancel, giving f ( z ) = 1 z 2 + 2 ζ (2) + 6 ζ (4) z 2 + 10 ζ (6) z 4 + 14 ζ (8) z 6 + . . . and f 0 ( z ) = - 2 z 3 + 12 ζ (4) z + 40 ζ (6) z 3 + 84 ζ (8) z 5 + . . . The simplified relation f 0 2 = 4 f 2 ( f - π 2 ) from above gives a recursion to determine ζ (2 n ) from ζ (2) , ζ (4) , . . . , ζ (2 n - 2), for 2 n 6, since all the Laurent coefficients of 0 = f 0 2 - 4 f 2 ( f - π 2 ) vanish: namely, the first/lowest-degree term involving ζ (2 n ) is the z 2 n - 6 term 0 = 2 · - 2 z 3 · (4 n - 2)(2 n - 2) ζ (2 n ) z 2 n - 3 - 4 · 3 · 1 z 2 2 · (4 n - 2) ζ (2 n ) z 2 n - 2 + (previous) = - 4(2 n - 2) + 12 (4 n - 2) · ζ (2 n ) + (previous) = - (8 n + 4)(4 n - 2) · ζ (2 n ) + (previous) where previous is a polynomial involving ζ (2) , ζ (4) , . . . , ζ (2 n - 2). In fact, given that ζ (2) = π 2 / 6 and ζ (4) = π 4 / 90, this approach can prove [1.0.1] Claim: ζ (2 n ) 2 n is rational , for 2 n = 2 , 4 , 6 , 8 , . . . . Proof: Rewrite the relation in terms of normalizations ζ (2 m ) 2 m . From the Laurent expansion, π - 2 f ( z/π ) = 1 z 2 + 2 ζ (2) π 2 + 6 ζ (4) π 4 z 2 + 10 ζ (6) π 6 z 4 + 14 ζ (8) z 6 π 8 + . . .
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