Re: not so easy

O/H Raymond Manzoni ??????:
glenn a écrit :
O/H Raymond Manzoni ??????:

An equivalent formulation is :
Gamma(n/2+1/2)/Gamma(n/2+1)>(2/(n+1))^(1/2)>Gamma(n/2+1)/Gamma(n/2+3/2)

And the proof :
True because the geometric mean of Gamma(n/2+1/2)/Gamma(n/2+1) and Gamma(n/2+1)/Gamma(n/2+3/2) is sqrt(Gamma(n/2+1/2)/Gamma(n/2+3/2)) =sqrt(2/(n+1)) QED


But wait. How you got this "equivalent formulation"? Assuming induction this is what exactly you are trying to prove.

?? No.

Gamma(n/2+1/2)/Gamma(n/2+1) > Gamma(n/2+1)/Gamma(n/2+3/2)
is trivial since log(Gamma(x)) is convex nearly 'by definition' : http://en.wikipedia.org/wiki/Bohr-Mollerup_theorem so that
log(Gamma(n/2+1)) <= (log(Gamma(n/2+1/2))+log(Gamma(n/2+3/2)))/2
Gamma(n/2+1)^2 <= Gamma(n/2+1/2)*Gamma(n/2+3/2)
and Gamma(n/2+1/2)/Gamma(n/2+1) >= Gamma(n/2+1)/Gamma(n/2+3/2)

The geometric mean value of these values is sqrt(2/(n+1)) and must be between these values so that for all n>0 we have :
Gamma(n/2+1/2)/Gamma(n/2+1)>sqrt(2/(n+1))>Gamma(n/2+1)/Gamma(n/2+3/2)

Take the righ part and replace n+1 by n to get
(2/n)^(1/2) > Gamma(n/2+1/2)/Gamma(n/2+1) as wished

Hoping it clarified things!
Raymond

Now I've got it. So I guess that my friend either _proved_ in his mail the Bohr?Mollerup theorem, either he didn't new it so he tried with a different method.
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