By Armand Borel

Comprises sections on Reductive teams, representations, Automorphic kinds and representations

**Read or Download Automorphic Forms, Representations, and L-Functions: Symposium in Pure Mathematics. Volume XXXIII Part 1 PDF**

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**Additional info for Automorphic Forms, Representations, and L-Functions: Symposium in Pure Mathematics. Volume XXXIII Part 1**

**Sample text**

Elementary Results PROOF. 1), we find that x 2 - 2xy + 2y2 - Z = o. 3) by 2 and rearranging terms, we arrive at (2y)2 - 2(2y)x + 2X2 - 2z = O. 4). 1). The previous three entries demonstrate Ramanujan's keen eye in spotting beautiful algebraic relationships. Entry 14 (p. 337). 1) where p and q are constants. 2) cp(x) Let Fl be defined by F(x) = Fl(X)X(X). Then F l(X + p) + F l(X + q) = PROOF. 1), we find that F(x + p) F(x + q) F 1(x+p)+F 1(x+q)= ( )+ ( ) xx+p xx+q = + p) + -'----'---'---''------=l/J(x)F(x + q) X(x + p) cp(x)X(x + p) F(x cp(x)F(x + p) + l/J(x)F(x + q) cp(x)X(x + p) f(x) cp(x)x(x + p)' which completes the proof.

328). 1) and a[3 = 1. 2) 24 22. 3) where y and () are positive numbers such that y{) = 1 and 125z y-{)=11+6 • x PROOF. f3 = 1, (1 + ~Y = (1. 1/5 - ((1. - f3) - 5{ (1 = = ((1. 2), + 11 ;6 = (1 + 5f33/5 - f3 +~)} + 10(1 +~) +~). y + ~Y + 5(1 + ~ + 5(1 + ~). 4) Expanding and simplifying, we deduce that z = 25xy 5 + 25x 2y 4 + 15x3 y 3 + 5x4 y2 + x 5 y. This may be rewritten in the form 11 + l~~Z = 11 + C:Y + 5c:r + 15(~y + 25C:Y + 25C:) = (1 + 5:Y + 5(1 + ~y + 5(1 + ~). 5). 1). Entry 13 (p. 328). 2) 25 22.

N aj - bnj+ 1 ) 1 ,-1 j=o ... b 1, a 1 ••• a,-l a 1 ••• aj )n. 1) Observe that the introduction of the parameter n is adventitious. We thus will assume without loss of generality that n = 1. There are two misprints in Ramanujan's formulation. PROOF. Induct on r. 1) is trivial. 1) holds. Then by induction, if a, #- 0, which completes the proof. Entry 27 (p. 356). 3) are the roots of the equation Z3 PROOF. 4) 0. 3) is apparent. 5) 38 22. Elementary Results and x2 Set x +y= - + xy + y2 + a + 1 = o.