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From: Valeri Astanoff on 20 Jul 2010 07:55
Good day,

How can one prove this function

1/sqrt(x-1) - gamma(x) / gamma(x+1/2)

to be positive for x > 1 ?


Any help gratefully appreciated.

--
Valeri Astanoff
From: Raymond Manzoni on 20 Jul 2010 08:12
Valeri Astanoff a �crit :
> Good day,
>
> How can one prove this function
>
> 1/sqrt(x-1) - gamma(x) / gamma(x+1/2)
>
> to be positive for x > 1 ?
>
>
> Any help gratefully appreciated.
>
> --
> Valeri Astanoff


This 2007 thread could be helpful too :
<http://groups.google.fr/group/sci.math/browse_frm/thread/17a8cd3370028d72>

Raymond
From: Valeri Astanoff on 20 Jul 2010 09:35
On 20 juil, 14:12, Raymond Manzoni <raym...(a)free.fr> wrote:
> Valeri Astanoff a écrit :
>
> > Good day,
>
> > How can one prove this function
>
> > 1/sqrt(x-1) - gamma(x) / gamma(x+1/2)
>
> > to be positive for x > 1 ?
>
> > Any help gratefully appreciated.
>
> > --
> > Valeri Astanoff
>
>    This 2007 thread could be helpful too :
> <http://groups.google.fr/group/sci.math/browse_frm/thread/17a8cd337002...>
>
>         Raymond

Ciao Raymond,

Eventually, the reference you gave me
[J. G. Wendel, Note on the gamma function,
Amer. Math. Monthly 55 (1948), no. 9, 563-564.]
is just fine for what I needed.

Many thanks

--
Valeri Astanoff
From: Rob Johnson on 20 Jul 2010 14:48
In article <bca7a674-6195-4fa5-a823-f0c55f22e77e(a)t10g2000yqg.googlegroups.com>,
Valeri Astanoff <astanoff(a)gmail.com> wrote:
>How can one prove this function
>
>1/sqrt(x-1) - gamma(x) / gamma(x+1/2)
>
>to be positive for x > 1 ?
>
>
>Any help gratefully appreciated.

Remember that log(gamma(x)) is convex, and that by the recursive
property of gamma

log(gamma(x-1)) + log(x-1) = log(gamma(x))

By convexity of log(gamma(x)), we get

log(gamma(x+1/2) >= log(gamma(x-1)) + 3/2 log(x-1)

= log(gamma(x)) + 1/2 log(x-1)

Getting rid of the (monotonic increasing) logs, we get

gamma(x+1/2) >= gamma(x) sqrt(x-1)

which leads to the inequality you mention above since all values are
positive.

Rob Johnson <rob(a)trash.whim.org>
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From: Rob Johnson on 20 Jul 2010 17:28
In article <20100720.112636(a)whim.org>,
Rob Johnson <rob(a)trash.whim.org> wrote:
>In article <bca7a674-6195-4fa5-a823-f0c55f22e77e(a)t10g2000yqg.googlegroups.com>,
>Valeri Astanoff <astanoff(a)gmail.com> wrote:
>>How can one prove this function
>>
>>1/sqrt(x-1) - gamma(x) / gamma(x+1/2)
>>
>>to be positive for x > 1 ?
>>
>>
>>Any help gratefully appreciated.
>
>Remember that log(gamma(x)) is convex, and that by the recursive
>property of gamma
>
> log(gamma(x-1)) + log(x-1) = log(gamma(x))
>
>By convexity of log(gamma(x)), we get
>
> log(gamma(x+1/2) >= log(gamma(x-1)) + 3/2 log(x-1)
>
> = log(gamma(x)) + 1/2 log(x-1)
>
>Getting rid of the (monotonic increasing) logs, we get
>
> gamma(x+1/2) >= gamma(x) sqrt(x-1)
>
>which leads to the inequality you mention above since all values are
>positive.

I just read the thread that Raymond Manzoni cited, and in the same
vein, we can also show that the function

gamma(x)/gamma(x+1/2) - 1/sqrt(x)

is positive for x > 0. By the recursive property of gamma

log(gamma(x)) + log(x) = log(gamma(x+1))

By convexity of log(gamma(x)), we get

log(gamma(x+1/2)) <= log(gamma(x)) + 1/2 log(x)

Getting rid of the (monotonic increasing) logs, we get

gamma(x+1/2) <= gamma(x) sqrt(x)

which leads to the result.

Thus, we can show, without using Stirling's formula, both

gamma(x)/gamma(x+1/2) >= 1/sqrt(x) for x > 0

and

gamma(x)/gamma(x+1/2) <= 1/sqrt(x-1) for x > 1

We could simply state that for x > 1,

1/sqrt(x) <= gamma(x)/gamma(x+1/2) <= 1/sqrt(x-1)

but that is slightly weaker since it loses the left-side inequality
when 0 < x <= 1.

Rob Johnson <rob(a)trash.whim.org>
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