content_image/307838.jpg content_image/307837.jpg content_image/307836.jpg content_image/307835.jpg
Further, using the definition of g$_{j}$ ( z$_{ρ}$$_{(}$$_{j}$$_{)0}$ ) , the fact that ∑ $_{ρ}$sgn ρz j − 1 ρ ( j )0 equals the Vandermonde determinant ∏ r i>j ( z$_{i}$$_{0}$ − z$_{j}$$_{0}$ ) , we get r r 

<img src='content_image/307838.jpg'>

Substituting (52) and (53) into (51), and using (49) and (50) together with the fact that the branch of the square root in f − 1 / 2 i 2 is chosen so that √ − 1 = − i , we get after some algebra 

<img src='content_image/307837.jpg'>

Now, using the fact that S − p = O$_{p}$ (1) , we get ln ( S p ) = S − p p + O$_{p}$ ( p − $^{2}$) and 

Further, the Stirling approximations give 

<img src='content_image/307836.jpg'>

<img src='content_image/307835.jpg'>

