Dr. Katz DEq Homework Solutions 94

# Dr. Katz DEq Homework Solutions 94 - p~oo p pd = I(N In...

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94 N.M. Katz: Hence (6.6.0.20) holds, and for all sufficiently large p with pA- 1 (N), we have (6.6.0.21) 1Rp(-a)>lRv(-c)>l Rp(-b). P P P Substituting via (6.5.2.1), (6.6.0.21) may be rewritten (6.6.0.22) (aA)-~-->(cA) C >(bA) b. O P P Because c-aeZ, (aA)= (cA), and hence --a --r (6.6.0.23) >-- P P Because (6.6.0.16) holds for A, (6.6.0.15) must hold for -A. Using the limit formula, (6.6.0.15) may be rewritten 1> lim 1Rp(-b)> lim 1Rp(-c) p~ p p~ p (6.6.0.24) pn ----- -- I(N) pA =- - 1 (N) 1 = lim
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Unformatted text preview: p~oo p pd =-- -- I(N} In particular, for all sufficiently large p with p A - - 1 mod (N), we have (6.6.0.25) 1__ Rv ( _b)> 1__ R~,(-c). P P As this is incompatible with (6.6.0.20), (6.6.0.19) holds for such p, and in particular, (6.6.0) gives (6.6.0.26) 1Rp(-c)>__l Rp(-a). Substituting via (6.5.2.1), this gives (6.6.0.27) (_cA)__c_>__(_aA) a P P As c-aeZ, (-cA)=(-aA), hence --r --a (6.6.0.28) --=>_ , P P which contradicts (6.6.0.23), and concludes the proof that c-aCZ. By symmetry we conclude that c-bCZ....
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## This note was uploaded on 12/21/2011 for the course MAP 4341 taught by Professor Normankatz during the Fall '11 term at UNF.

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