but we do not really need it at this moment. This gives a distribution with a width of the order √$_{n}$$_{. Computing the limit ratio gives us}$ thus giving us an essentially gaussian distribution. Note that it does not depend on z in its current form. Of course, we expect other terms to depend on z but these vanish when n → ∞ . Suppose that we want the peaks located outside the middle. Defining k = n 2 (1 + µ ) for 0 < | µ | < 1 means that we move the peaks out from the middle and that we move to another isocurve. We keep q = 1 + z/n and p = 1 + y$_{1}$/n + y$_{2}$/n 2 + · · · and solve R$_{p,q}$ ( n, k, 1) = 1, i.e. which we rewrite as At this point it would be appropriate to define the function s = Ω$_{µ}$ ( x ), for 0 < µ < 1, as the maximum solution to the equation Note here that the function e x sinh( µ x ) has a minimum at ( − atanh µ ) /µ for 0 < µ < 1 and, due to symmetry, a maximum at the same point for − 1 < µ < 0. The function Ω returns a value in the interval The solution sought in our equation is thus y$_{1}$ = 2 w where 22 w = Ω$_{µ}$ ( e z/ 2 sinh µ z 2 ) (76) − atanh µ µ < s < 0 (75) x = e s sinh ( s µ ) (74) e y$_{1}$/ 2 sinh µ y$_{1}$ 2 = e z/ 2 sinh µ z 2 (73) exp ( y$_{1}$ 2 (1 − µ ) ) − exp ( y$_{1}$ 2 (1 + µ ) ) = exp ( z 2 (1 − µ ) ) − exp ( z 2 (1 + µ ) ) (72) p $^{n}$−k +1 − p k = q $^{n}$−k +1 − q k (71) With k = n 2 (1 + µ ) we use (35) on both sides and find the equation D. Two separate peaks R$_{p,q}$ ( n, n/ 2 $^{$_{, x}$√$_{n}$}$) ∼ e a x 2 (70)