content_image/329320.jpg content_image/329317.jpg content_image/329316.jpg content_image/329315.jpg content_image/329314.jpg content_image/329313.jpg content_image/329312.jpg
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 → ∞ . 

## D. Two separate peaks 

<img src='content_image/329320.jpg'>

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. 

With k = n 2 (1 + µ ) we use (35) on both sides and find the equation 

<img src='content_image/329317.jpg'>

<img src='content_image/329316.jpg'>

which we rewrite as 

<img src='content_image/329315.jpg'>

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 

<img src='content_image/329314.jpg'>

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 

<img src='content_image/329313.jpg'>

<img src='content_image/329312.jpg'>

