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# G(x+1)/G(x+0.5),G=Gamma

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### alex.lupas

Aug 14, 2006, 6:06:15 AM8/14/06
to
Denote e(x,a)=exp( 1/( 64(x+a)^2) ) ,
E(x,a):=e(x,a)(x+0.25)^{0.5}
Prove or disprove

E(x,0.75)< G(x+1)/G(x+0.5)}< E(x,-0.25) , (for x> 1),

G being Gamma function.

### David W. Cantrell

Aug 14, 2006, 3:06:38 PM8/14/06
to

In giving bounds for G(x + 1)/G(x + 1/2) in the form E(x, a), we can do
much better than using a = 3/4 and -1/4, resp., for the lower and upper
bounds.

With a = 1/4 (rather than -1/4), we get the tightest possible upper bound.
It is valid for x > -1/4.

Using any a > 1/4, E(x, a) provides a lower bound if x is sufficiently
large. But if, say, we wish the lower bound to be valid for all x > -1/4,
then a must be at least 0.36995...

Combining the above, we may say

E(x, 0.37) < G(x + 1)/G(x + 1/2) < E(x, 1/4) for x > -1/4.

David Cantrell

Message has been deleted

### alex.lupas

Aug 15, 2006, 4:02:17 AM8/15/06
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David W. Cantrell wrote:
> In giving bounds for G(x + 1)/G(x + 1/2) in the form E(x, a), we can do
> much better than using a = 3/4 and -1/4, resp., for the lower and upper
> bounds.
> With a = 1/4 (rather than -1/4), we get the tightest possible upper bound.
> It is valid for x > -1/4. Using any a > 1/4, E(x, a) provides a lower
>bound if x is sufficiently> large. But if, say, we wish the lower bound
> to be valid for all x > -1/4, then a must be at least 0.36995...
> Combining the above, we may say
> E(x, 0.37) < G(x + 1)/G(x + 1/2) < E(x, 1/4) for x > -1/4.
> David Cantrell
Hi David, thanks for these nice remarks.If possible,please inform
me regarding the proofs.Alex

### Peter Luschny

Aug 15, 2006, 5:15:28 AM8/15/06
to

Nice bounds. I hope you will show us the proof.
If we look for simpler bounds we might use

S(x, 0.185) < G(x + 1)/G(x + 1/2) < S(x, 1/2) for x > -1/2.

Here S(x,a) = (x+a)^{1/2}.

Regards Peter

### David W. Cantrell

Aug 15, 2006, 3:10:47 PM8/15/06
to
Peter Luschny <spam...@luschny.de> wrote:
> > David W. Cantrell wrote:
> >> "alex.lupas" <alex....@gmail.com> wrote:
>
> >> Denote e(x,a)=exp( 1/( 64(x+a)^2) ) ,
> >> E(x,a):=e(x,a)(x+0.25)^{0.5}
> >> Prove or disprove
> >> E(x,0.75)< G(x+1)/G(x+0.5)}< E(x,-0.25) , (for x> 1),
> >> G being Gamma function.
>
> > In giving bounds for G(x + 1)/G(x + 1/2) in the form E(x, a), we can do
> > much better than using a = 3/4 and -1/4, resp., for the lower and upper
> > bounds.
> > With a = 1/4 (rather than -1/4), we get the tightest possible upper
> > bound. It is valid for x > -1/4.
> > Using any a > 1/4, E(x, a) provides a lower bound if x is sufficiently
> > large. But if, say, we wish the lower bound to be valid for
> > all x > -1/4, then a must be at least 0.36995...
> > Combining the above, we may say
> > E(x, 0.37) < G(x + 1)/G(x + 1/2) < E(x, 1/4) for x > -1/4.
>
> Nice bounds. I hope you will show us the proof.

Hello, Alex and Peter. There's really nothing interesting. I simply
compared the series expansions for G(x + 1)/G(x + 1/2) and E(x, a) at
x = +oo, for which a computer algebra system was useful. (And the constant
0.36995... was obtained numerically.)

> If we look for simpler bounds we might use
>
> S(x, 0.185) < G(x + 1)/G(x + 1/2) < S(x, 1/2) for x > -1/2.

I suppose you meant to say that the first inequality holds for x > -0.185.

> Here S(x,a) = (x+a)^{1/2}.

I'd thought about mentioning simpler bounds earlier, but had decided to
comment on just Alex's E(x, a). Anyway, now that you've brought up the
topic of simpler bounds:

If we wish to approximate G(x + 1)/G(x + 1/2) as well as possible by S(x,
a) for large x, then we must use a = 1/4. [Presumably, that is why Alex had
a factor of S(x, 1/4) in his E(x, a). Hmm. Careful: Alex's a and your a
represent different things.] That can be shown by comparing the series at
x = +oo. For large x, the error is approximately -1/64 x^(-3/2).

A slightly more complicated, but much more accurate, approximation for
G(x + 1)/G(x + 1/2) is

(x + 1/2)/Sqrt(x + 3/4 + 1/(32(x + 3/4))).

It is a lower bound for x > -1/2, and is precise at x = -1/2.

For large x, the error is approximately -9/4096 x^(-7/2).

Regards,
David W. Cantrell

### Peter Luschny

Aug 16, 2006, 6:06:10 AM8/16/06
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David W. Cantrell wrote:
> Peter Luschny <spam...@luschny.de> wrote:
>> > David W. Cantrell wrote:
>>>> "alex.lupas" <alex....@gmail.com> wrote:

>>>> Denote e(x,a)=exp( 1/( 64(x+a)^2) ) ,
>>>> E(x,a):=e(x,a)(x+0.25)^{0.5}
>>>> Prove or disprove
>>>> E(x,0.75)< G(x+1)/G(x+0.5)}< E(x,-0.25) , (for x> 1),
>>>> G being Gamma function.
>>> In giving bounds for G(x + 1)/G(x + 1/2) in the form E(x, a), we can do
>>> much better than using a = 3/4 and -1/4, resp., for the lower and upper
>>> bounds.
>>> With a = 1/4 (rather than -1/4), we get the tightest possible upper
>>> bound. It is valid for x > -1/4.
>>> Using any a > 1/4, E(x, a) provides a lower bound if x is sufficiently
>>> large. But if, say, we wish the lower bound to be valid for
>>> all x > -1/4, then a must be at least 0.36995...
>>> Combining the above, we may say
>>> E(x, 0.37) < G(x + 1)/G(x + 1/2) < E(x, 1/4) for x > -1/4.
>> Nice bounds. I hope you will show us the proof.
>
> Hello, Alex and Peter. There's really nothing interesting.

For me your observations are useful and interesting,
especially the upper bound.

Let me rephrase the relation. For x > -1/4:

Gamma(x + 1)
[*] ------------ < (x+1/4)^{1/2} * exp((8x+2)^{-2})
Gamma(x+1/2)

I simplified this to:

Gamma(x + 1)
------------ < (x+1/2)^{1/2}
Gamma(x+1/2)

Of course this bound is not as sharp as yours because

(x+1/4)^{1/2} * exp((8x+2)^{-2}) < (x+1/2)^{1/2}.

This relation was proved in an accompanying thread in
de.sci.mathematik. It remains to prove [*].

> I simply
> compared the series expansions for G(x + 1)/G(x + 1/2) and E(x, a) at
> x = +oo, for which a computer algebra system was useful. (And the constant
> 0.36995... was obtained numerically.)
>> If we look for simpler bounds we might use
>> S(x, 0.185) < G(x + 1)/G(x + 1/2) < S(x, 1/2) for x > -1/2.
> I suppose you meant to say that the first inequality holds for x > -0.185.

Right. Sorry.

>> Here S(x,a) = (x+a)^{1/2}.
> I'd thought about mentioning simpler bounds earlier, but had decided to
> comment on just Alex's E(x, a). Anyway, now that you've brought up the
> topic of simpler bounds:
> If we wish to approximate G(x + 1)/G(x + 1/2) as well as possible by S(x,
> a) for large x, then we must use a = 1/4. [Presumably, that is why Alex had
> a factor of S(x, 1/4) in his E(x, a). Hmm. Careful: Alex's a and your a
> represent different things.] That can be shown by comparing the series at
> x = +oo. For large x, the error is approximately -1/64 x^(-3/2).
> A slightly more complicated, but much more accurate, approximation for
> G(x + 1)/G(x + 1/2) is
> (x + 1/2)/Sqrt(x + 3/4 + 1/(32(x + 3/4))).
> It is a lower bound for x > -1/2, and is precise at x = -1/2.

2x + 1 Gamma(x + 1)
[**] ------------------------- < ------------
Sqrt(4x + 3 + 1/(8x + 6)) Gamma(x+1/2)

> For large x, the error is approximately -9/4096 x^(-7/2).

Again, a useful and interesting bound for me, which I have not seen before.

I still hope someone will show us a proof of [*].

Regards Peter

### Peter Luschny

Aug 16, 2006, 6:05:43 PM8/16/06
to
David W. Cantrell schrieb:

> Peter Luschny <spam...@luschny.de> wrote:
>>> David W. Cantrell wrote:
>>>> "alex.lupas" <alex....@gmail.com> wrote:

Hallo Alex, hallo David!

This time I will look at the lower bounds
and summarize some statements.

Alex suggested to prove the lower bound:

Gamma(x + 1)
(x+1/4)^(1/2)*exp(1/(64(x+3/4)^2)) < ------------
Gamma(x+1/2)

David suggested as a better lower bound:

Gamma(x + 1)
(x+1/4)^(1/2)*exp(1/(64(x+0.37)^2)) < ------------
Gamma(x+1/2)

Next David suggested in the context of simpler bounds:

2x + 1 Gamma(x + 1)

------------------------- < ------------
Sqrt(4x + 3 + 1/(8x + 6)) Gamma(x+1/2)

However, none of these bounds have in my opinion
the elegance of David's upper bound:

Gamma(x + 1)
------------ < (x+1/4)^{1/2}*exp((8x+2)^{-2})
Gamma(x+1/2)

So now my contribution, a lower bound, designed in
the spirit of David's upper bound and more powerful
then all the other ones cited.

Gamma(x + 1)
(x+1/4)^(1/2)*exp((8x+2)^(-4)(64x^2+32x-6)) < ------------
Gamma(x+1/2)

Regards Peter

### David W. Cantrell

Aug 24, 2006, 11:47:10 AM8/24/06
to

That's a fine bound, Peter. However...

The only reason I ever gave any bounds using an exponential function
was that Alex began this thread considering that type of bound. But I
think that such bounds for Gamma(x + 1)/Gamma(x + 1/2) are not really
what we should be considering. That ratio of Gamma functions can be
bounded very well using _algebraic_ functions. I think doing that is
better -- aesthetically, computationally, etc. -- than involving
exponential functions.

For example, consider the simple algebraic lower bound

Sqrt(u) (1 + 1/(64 u^2) - 19/(8192 u^4)), where u = x + 1/4,

which, for x > 0, is of comparable accuracy to the lower bound you
gave.

My algebraic bound was obtained merely by truncating an expansion for
large x:

Gamma(x + 1)/Gamma(x + 1/2) = Sqrt(u) f(u), where u = x + 1/4 and

f(u) = 1 + 1/(64 u^2) - 19/(8192 u^4) + 631/(524288 u^6) -
174317/(134217728 u^8) + 689888044259/(482178798452736 u^10) -+...

Of course, by suitable truncation, we can get, as desired, either
upper or lower bounds.

Furthermore, if we want an expression which works for smaller x, we
may simply shift things using

Gamma(x + 1) x + 1/2 Gamma(x + 2)
-------------- = ------- --------------
Gamma(x + 1/2) x + 1 Gamma(x + 3/2)

(and repeatedly, if desired).

For example, shifting twice, then with u = x + 9/4, we could use

x + 1/2 x + 3/2
------- ------- Sqrt(u) f(u)
x + 1 x + 2

to get an approximation which works reasonably well for x > -1.

Best regards,
David Cantrell

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