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 ( holds, and for all sufficiently large p with pA- 1 (N), we have ( 1Rp(-a)>lRv(-c)>l Rp(-b). P P P Substituting via (, ( may be rewritten ( (aA)-~-->(cA) C >(bA) b. O P P Because c-aeZ, (aA)= (cA), and hence --a --r ( >-- P P Because ( holds for A, ( must hold for -A. Using the limit formula, ( may be rewritten 1> lim 1Rp(-b)> lim 1Rp(-c) p~ p p~ p ( 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 ( 1__ Rv ( _b)> 1__ R~,(-c). P P As this is incompatible with (, ( holds for such p, and in particular, (6.6.0) gives ( 1Rp(-c)>__l Rp(-a). Substituting via (, this gives ( (_cA)__c_>__(_aA) a P P As c-aeZ, (-cA)=(-aA), hence --r --a ( --=>_ , P P which contradicts (, 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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