- Thread starter
- #1

- Mar 22, 2013

- 573

Any comment/question regarding the thread should be noted here rather than the tutorial thread.

- Thread starter mathbalarka
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- Thread starter
- #1

- Mar 22, 2013

- 573

Any comment/question regarding the thread should be noted here rather than the tutorial thread.

- Thread starter
- #2

- Mar 22, 2013

- 573

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

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- Jan 17, 2013

- 1,667

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

Keep it up , you are doing well.

- 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

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

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

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

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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- Admin
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Yes, thank you for creating the commentary thread. I added the link to this thread.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...

- Sep 16, 2013

- 337

@ 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!

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...

- Thread starter
- #7

- Mar 22, 2013

- 573

Your understanding is correct. Let me clarify my points here :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.

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.

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:More explicitly, as I understand it, the Weierstrauss infinite product for the Gamma function is used to define the Gamma function for $z \in \mathbb{C}, Re(z) \neq \mathbb{Z}^-$

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.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...

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- Jan 31, 2012

- 253

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.

- Thread starter
- #9

- Mar 22, 2013

- 573

Yes, that is a typo.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.

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

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

- Jan 31, 2012

- 253

- Sep 16, 2013

- 337

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...

All the best!

Gethin

- Sep 16, 2013

- 337

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!

- Jan 31, 2012

- 253

The only thing I assumed was the infinite product representation.

- Sep 16, 2013

- 337

Nice!!!

The only thing I assumed was the infinite product representation.

Would you mind showing how, as and when you have time...?

You're much,

[Sorry to be a pest! lol ]

- Jan 31, 2012

- 253

Ummm... didn't you click on the link?

- Sep 16, 2013

- 337

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

[runs and hides]

- Thread starter
- #17

- Mar 22, 2013

- 573

Sure. I will let you know when the pre-print releases on arXiv.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...

- Sep 16, 2013

- 337

Thank you!Sure. I will let you know when the pre-print releases on arXiv.

Can't wait...

- Sep 16, 2013

- 337

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 --> "].

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

- Jan 31, 2012

- 253

- Sep 16, 2013

- 337

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...

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.