Chaggara

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13 years, 269 days

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These are answers submitted by Chaggara

thank you for the interst to this formula of course "i" is positive integer because we have sum from 0 to "i" Yes fir the "k", unfortunatly I have missing a factor on containing "k": (k-r-s)! Here is the correct expression J:=Sum(Sum((-1)^(r+s)*factorial(i+j-r-s)*pochhammer(mu+1/2, i+j-r-s)/(k-r-s)!*(pochhammer(mu+1/2, i-r)*pochhammer(mu+1/2, j-s)*factorial(r)*factorial(s)*factorial(i-r)*factorial(j-s)), r = 0 .. i), s = 0 .. j) Note here that J is symmetric on i and j I have tryed a great number of particular case and J is always positive So, is it possible to proove the positivity for all parametres i,j,k (positive integers and 0\leq k\leq min(2i,2j)) and mu (real positive)
of course mu is supposed to be positive conditions concerning i,j and k are given in the first message best

Yes ok  the econd summation index is "s" and not "r"

here is the corect expression with Maple notations

The question is to justify the positivity of this term

I:=Sum(Sum((-1)^(r+s)*factorial(i+j-r-s)*pochhammer(mu+1/2, i+j-r-s)/(pochhammer(mu+1/2, i-r)*pochhammer(mu+1/2, j-s)*factorial(r)*factorial(s)*factorial(i-r)*factorial(j-s)), r = 0 .. i), s = 0 .. j)

thanks

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