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72 Chapter 2. On a conjecture about addition modulo 2 k − 1valid for |z| < 1. We obtain the following expression where we shift the summation index j by k:P [X d = k] =2k ∑∞ ( )( )d − 1 + k + j 2k + 2j3 2d+2k 2 j 3 −2jd − 1 jj=0∑∞ ( )( )d − 1 + j 2j= 3 −2d 2 j 3 −2j .j k + jj=kHere is an elementary proof. We know thatso, by Parseval’s theorem,[∞∑ d∑]P G i = j e ijθ 1=(2 − e iθ ) d ,j=0 i=1[∞∑ d∑] [ d∑]P G i = j P G i = j + ki=1i=1j=0= 12π∫ 2π0= 1 ∫ 2π2π= 12π0∫ 2π0e ikθ ∣ ∣∣∣ 1(2 − e iθ ) d ∣ ∣∣∣2dθcos(kθ)(5 − 4 cos θ) d dθcos(kθ)(9 − 8 cos 2 (θ/2)) d dθ .Moreoverandso that1 19 d (1 −89 cos2 (θ/2) ) d = ∑∞3−2dj=0( d − 1 + j( ecos 2j iθ/2 + e −iθ/2 ) 2j( (2j )(θ/2) == 2 −2j +2j12πHence, we have the identity∫ 2π0P [X d = k] = 3 −2dj)cos 2j (θ/2)2 3j 3 −2j ,j∑m=1( ) 2jcos(kθ) cos 2j (θ/2)dθ = 2 −2j k + j∞∑j=k( d − 1 + jj( ) )2j2 cos(mθ)j + m)( ) 2j2 j 3 −2j .k + j.,This expression is interesting because it can be used to strengthen Proposition 2.6.7.Corollary 2.6.11. For d ≥ 1, X d follows a unimodal distribution centered at 0, i.e. P [X d = k]increases for k ≤ 0 and decreases for k ≥ 0.Proof. Indeed, P [X d = k] is an even function of k and for fixed j ≥ 0 and k ≥ 0 each summandof the expression given in the previous proposition decreases as k increases.Moreover, specializing this expression at k = 0 yields an expression for P d where d appearsonly twice.