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Commentary for "The Barnes' G-Function, and related higher functions"

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Uh, I know it's not my job to make commentary topics for tutorial threads, but couldn't resist (or wait) to note some issued regarding the thread. I do hope the author or any moderator will put a link to this commentary in the original thread.

Now, first things first, excellent thread DW. As functional analysis is one of my interests, I was able to digest the first post in the thread. Not sure about the others though...

In the first post, DW introduced the multiple gamma and indicated that it can be defined for \(\displaystyle z \in \mathbb{C}\) too. I don't contradict it, but the 3 conditions are not sufficient to do it.

In the same post, DW also indicated the property of log-convexity. Now, at this point, I would like to note that something similar can be defined for gamma function, also called Bohr-Mollerups theorem, which is pretty well-known. Now this reminds me of an exercise I posted on MMF a long time ago, less than a year, anyway, which is on finding another analytic continuation of Gamma that is not log-convex.

Balarka
.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Thanks Balara for starting the commentary thread. I wanted to thank DW. Further I think there is a typo under the integral representation EQ(09) where you said

We prove integral (07) somewhat indirectly ...
I think you meant (09) .

Keep it up , you are doing well.
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
I just wanted to mention another way to derive the first reflection formula.


In a previous thread I showed that you can express $\log G(1+z)$ as

$$ \log G(1+z) = \frac{z}{2} \log(2 \pi) - \frac{1}{2} z(z+1) - \frac{1}{2}\gamma z^{2} +\sum_{k=2}^{\infty} \frac{(-1)^{k}}{k+1} \zeta(k) z^{k+1}$$


Then

$$\log G(1-z) = -\frac{z}{2} \log(2 \pi) + \frac{1}{2} z(1-z) - \frac{1}{2}\gamma z^{2} - \sum_{k=2}^{\infty} \frac{(-1)^{k}}{k+1} \zeta(k) z^{k+1}$$

So

$$ \log \frac{G(1+z)}{G(1-z)} = z \log(2 \pi) -z + 2 \sum_{k=1}^{\infty} \frac{\zeta(2k)}{2k+1} z^{2k+1} = z \log(2 \pi) + z - 2 \zeta(0) z + 2 \sum_{k=0}^{\infty} \frac{\zeta(2k)}{2k+1} z^{2k+1} $$

$$= z \log(2 \pi) + 2 \sum_{k=0}^{\infty} \zeta(2k) \int_{0}^{z} x^{2k} \ dx = z \log(2 \pi) + 2 \int_{0}^{z} \sum_{k=0}^{\infty} \zeta(2k) x^{2k} \ dx $$

$$ = z \log(2 \pi) - \pi \int_{0}^{z} x \cot (\pi x) \ dx $$
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Uh, I know it's not my job to make commentary topics for tutorial threads, but couldn't resist (or wait) to note some issued regarding the thread. I do hope the author or any moderator will put a link to this commentary in the original thread...
Yes, thank you for creating the commentary thread. I added the link to this thread. :D
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Thanks for the feedback, everyone... And a special thanks to Mathbalarka for setting up this thread. :D


@ Random Variable...

That's a very nice proof of the reflection formula - you are a series Jedi... In my notes, some of which will be added shortly, I've done the same thing but in reverse, namely proved the reflection formula from the loggamma integral, and then converted that into the Zeta series.


@ MathBalarka...

Once again, your help and feedback is very much appreciated! :D

Re logarithmic convexity, there is an extension to the Bohr-Mollerup theorem that can be applied to the Multiple Gamma function, but it's a bit beyond me.

In terms of the Multiple Gamma function being extended to \(\displaystyle \mathbb{C}\), I might be mistaken here, but it's my understanding that, since the double Gamma function is essentially defined as a function of the Gamma function, which itself can be extended to \(\displaystyle \mathbb{C}\), the the Double Gamma function inherits a \(\displaystyle \mathbb{C}\)-extension from the Gamma function.

More explicitly, as I understand it, the Weierstrauss infinite product for the Gamma function is used to define the Gamma function for \(\displaystyle z \in \mathbb{C}, Re(z) \ne \mathbb{Z}^{-}\), and the same holds for the Infinite product for the Barnes G-function...???

As I say though, my understanding is a little sketchy here, so if you'd like to write a bit about the uniqueness, analyticity, etc of the Barnes' G-function, I could add it to the tutorial (with full and unequivocal credit to you, of course). Totally up to you, mind, but I think it would make a great addition... (Hug)
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
DreamWeaver said:
In terms of the Multiple Gamma function being extended to $\mathbb{C}$, I might be mistaken here, but it's my understanding that, since the double Gamma function is essentially defined as a function of the Gamma function, which itself can be extended to $\mathbb{C}$, the the Double Gamma function inherits a $\mathbb{C}$-extension from the Gamma function.
Your understanding is correct. Let me clarify my points here :

What we are basically referring to extensions are really analytic continuations, otherwise it doesn't make much sense. For example, there are lots of extensions of Riemann $\zeta$, even some beth-extensions, but that is completely useless.

Now, this analytic continuation is a strange beast. What I tried to point out in a couple of posts back is that the conditions to define a certain analytic function might not be sufficient to define an continuation, especially unique ones (i.e., analytic), in certain domain. This applies to both double/multiple or the usual gamma function. Also, bear in mind that gamma cannot be extended to whole $\mathbb{C}$ since it attains branch points around the negative integers.

DreamWeaver said:
More explicitly, as I understand it, the Weierstrauss infinite product for the Gamma function is used to define the Gamma function for $z \in \mathbb{C}, Re(z) \neq \mathbb{Z}^-$
In general, yes, but there are other continuations too. For example, take the functional equation of zeta function. In fact, cobbling up a bunch of reflection formulas, alongside with the basic properties might have some luck at specific points.

DreamWeaver said:
As I say though, my understanding is a little sketchy here, so if you'd like to write a bit about the uniqueness, analyticity, etc of the Barnes' G-function, I could add it to the tutorial (with full and unequivocal credit to you, of course). Totally up to you, mind, but I think it would make a great addition... (Hug)
I can try, but I don't see I can have time anywhere soon. I can't even complete my zeta tutorial (I am collaborating a paper with a few people, that is the main reason to it.) I'll definitely try, though.
 
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Random Variable

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MHB Math Helper
Jan 31, 2012
253
@ mathbalarka

I'm sure you meant to say that the gamma function has poles at the negative integers (and the origin), not branch points.

And fortunately an analytic continuation is unique. So it's technically incorrect to speak of analytically continuing the gamma function in a different way.

I think what you're referring to is extending the factorials to values besides positive integers in a different way. But that's not a process of analytic continuation, and is thus not unique.


And also the Barnes G function is an entire function, and that infinite product representation defines it everywhere.
 

mathbalarka

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MHB Math Helper
Mar 22, 2013
573
Random Variable" said:
I'm sure you meant to say that the gamma function has poles at the negative integers (and the origin), not branch points.
Yes, that is a typo.

Random Variable said:
And fortunately an analytic continuation is unique.
I never contradicted that.

Random Variable said:
And also the Barnes G function is an entire function, and that infinite product representation defines it everywhere.
Yes, but as I said, there are other elegant ways of continuing it.
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
In your post from a couple of days ago, you misspoke and mentioned finding an analytic continuation of the gamma function that is not log convex. Clearly what you meant is finding an extension of the factorial that is not the gamma function, similar to how you can find an extension of the super factorial that is not the Barnes G-function.
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Thanks MathBalarka! That helps... A lot :D

And please DON'T worry about my suggestion of writing a few bits about the Barnes' function for this 'ere tutorial... You clearly have bigger fish to fry, and not much time to fry them in.

If I can be a tad cheeky though, would you mind PMing me a link when you get your paper published??? Sounds right up my street... :eek::eek::eek:

All the best!

Gethin
 

DreamWeaver

Well-known member
Sep 16, 2013
337
@ Random Variable...

I hope you don't mind, but it (belatedly) strikes me that, in your Zeta-series proof of the Barnes' function reflection formula, you must (???) have either assumed or used something equivalent to the reflection formula or the parametric Loggamma integral evaluation... Perhaps?

Just wanted to check that you weren't using a modicum of circular reasoning, is all...

Would you mind explaining how you connected the series

\(\displaystyle \sum_{k=2}^{\infty}(-1)^k\frac{\zeta(k)}{k+1}z^{k+1}\)

with the Barnes' function...???


Thanks! :D
 

DreamWeaver

Well-known member
Sep 16, 2013
337

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
Ummm... didn't you click on the link?
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Doh! Massive face-palm moment... (Sleepy)

Thanks fella, but I just thought you were using rainbow text...

[runs and hides]
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Random Variable said:
If I can be a tad cheeky though, would you mind PMing me a link when you get your paper published??? Sounds right up my street...
Sure. I will let you know when the pre-print releases on arXiv.
 

DreamWeaver

Well-known member
Sep 16, 2013
337

DreamWeaver

Well-known member
Sep 16, 2013
337
I'm wondering... Have any of you either managed to prove the multiplication formula for the Barnes' function, or even come across a not-insanely-difficult proof for it? I've had no luck with either.

And I stubbornly refuse to use it, because I can't prove it. I know, I know, it's silly, but... [ "I'm with stupid --> (Headbang) "].

Anyhoo, here it is:


\(\displaystyle G(n\, z) = \)


\(\displaystyle n^{n^2z^2/2-nz+5/12}(2\pi)^{(n-1)(1-nz)/2} \text{exp} \Bigg((1-n^2)\, \zeta ' (-1)\Bigg)\, \times\)


\(\displaystyle \prod_{i=0}^{n-1}\, \prod_{j=0}^{n-1} G\left(z+\frac{i+j}{n}\right)\)


For \(\displaystyle n\in \mathbb{N}\)
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
I haven't thought about it much, but perhaps it can be derived by using the multiplication formula for the Hurwitz zeta function.
 

DreamWeaver

Well-known member
Sep 16, 2013
337
I haven't thought about it much, but perhaps it can be derived by using the multiplication formula for the Hurwitz zeta function.

I think you might be right there, RV... There's nothing scientific about it, obviously, but I 'feel' like I'm not far off it. At least that's the hope: just a bit more gruntwork and I'll accidentally stumble across it when least expected.

It's quite a formidable formula, though... (Headbang)


I suspect the relation \(\displaystyle \log G(1+z)-z\log \Gamma(z)= \zeta ' (-1)-\zeta ' (-1,z)\) might also offer a decent starting point.