FIG. 8: Left: y = n ( p − 1) versus z = n ( q − 1) for C$_{n}$ , with n = 8 , 16 , 32 , 64 , 128 with larger n extending farther to the left. Right: K giving the maximum y vs 1 /n for C$_{n}$ , n = 8, 12, 16, 24, 32, 48, 64, 96, 128, 192, 256. $^{GLYPH<Minus>}$15 $^{GLYPH<Minus>}$10 $^{GLYPH<Minus>}$5 z 0.2 0.4 0.6 0.8 1.0 y 8 12 16 24 32 64 256 1 GLYPH<Slash1> n 0.105 0.110 0.115 0.120 0.125 0.130 0.135 K Though we can not exactly solve what y and z should be at K = 2 / 15 we can at least see FIG. 9: Left: Maximum value of y = n ( p − 1) versus 1 /n for C$_{n}$ . Right: value of z ( q − 1) versus 1 /n for C$_{n}$ when y is at its maximum. In both cases n = 8, 12, 16, 24, 32, 48, 64, 96, 128, 192, 256. how y and z relate at this point. For an infinite 1-dimensional lattice we have that χ = e 2 $^{K}$, see e.g. $^{13}$. The second moment then should behave as Let ℓ = k − n/ 2 and σ = √ σ$_{2}$ . For high temperatures we expect ℓ/σ to be normally distributed and thus The probability ratio is then 37 R ( n, n/ 2 , ℓ ) = P ( ℓ ) P (0) = exp ( − ℓ $^{2}$/ 2 σ $^{2}$) (146) P ( ℓ ) ∼ exp ( − ( ℓ/σ ) $^{2}$/ 2) σ √$_{2}$ π (145) σ$_{2}$ ∼ n χ 4 = n e 2 K 4 (144) 8 12 16 24 32 64 256 1 GLYPH<Slash1> n 1.00 1.02 1.04 1.06 1.08 y 8 12 16 24 32 64 256 1 GLYPH<Slash1> n $^{GLYPH<Minus>}$3.6 $^{GLYPH<Minus>}$3.5 $^{GLYPH<Minus>}$3.4 $^{GLYPH<Minus>}$3.3 $^{GLYPH<Minus>}$3.2 $^{GLYPH<Minus>}$3.1 z