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**Extra resources for Communications In Mathematical Physics - Volume 286**

**Example text**

1 ∂ (69) (b) Note that ∂ bk = c − δ k . So, setting all auxiliary parameters t, s ( ) , β equal to zero, one ultimately shows that the p equations (68) can be transformed into p equations, purely in terms of the boundary points of E and the final points bi , also taking into account the value18 τn (R) of the integral (67) over the full range R, since Pn (E) = τn (E)/τn (R) for t = s = β = 0. The only remaining undesirable partials, to be eliminated, are the partials with regard to the β ’s, which appear on the right hand side of the final equation for Pn , ∂ (β) ∂ ln Pn , F ∂ = H (1) ,F ∂ − H (2) ,F ∂ (b) , 1≤ ≤ p.

P−1 ) 1 p 1 n ! ⎛ = p EN n n ( ) i ∞ i=1 ti (x j ) e =1 j=1 p ⎝ n (1) ( p) ) N (x , . . , x (x ( ) ) =1 e ( )2 − 21 x j +b ( ) x j +β ⎞ ( )2 xj − ∞ ( ) ( )i i=1 si x j ( ) dx j ⎠ j=1 ⎞ (µi(1) j )0≤i≤n 1 −1, 0≤ j≤N −1 ⎟ ⎜. ⎟, = det ⎜ ⎠ ⎝ .. ( p) (µi j )0≤i≤n p −1, 0≤ j≤N −1 ⎛ p 1 ci bi =0 p 1 ci βi =0 (67) x2 where F1 (x) := e− 2 in the inner-product below and where one only indicates the dependency on the auxiliary variables, ( ) µi j (t − s ( ) , β ) = E˜ x i+ j e− = x i e− for x2 2 +b ∞ ( ) k 1 sk x f |g 1 x+β x 2 eb = e x+β x 2 E˜ ( ) k ∞ 1 (tk −sk )x x je dx ∞ i 1 tk x 1 , f (x)g(x)F1 (x)d x.

4 of [21]. Now, we give some details concerning the term IV. Recall first that IV = ∂x N, where N = f (R + Rc + W ) − f (R + Rc ) − f (R)W . 2), thus the contribution of these terms to Fk , , G k , are in Y. For k = p, the term f (k−1) (R(x)) is bounded p and the term of lower order in ((Rc + W ) p − Rc ) which is not in Y comes from B1,0 = B 1,0 + b1,0 ϕ. Thus, the lowest order term not localized in the y variable is p−1 pb1,0 (Q c p−1 Q c ϕ)x = pb1,0 (Q c p−2 Q c + ( p − 1)Q c p−1 (Q c )2 ) + pb1,0 Q c Qcϕ .