tea.mathoverflow.net - Discussion Feed (Zeta functions)

Scott Morrison comments on "Zeta functions" (18537)

2012-02-18T09:05:46-08:00

As per the previous thread, Vassilis Parassidis is not welcome either here or at the main site. In future, could I request of everyone to simply leave any threads he starts here (or posts at the main site) unanswered, and simply contact the moderators? It's easier on everyone.

Vassilis Parassidis comments on "Zeta functions" (18536)

2012-02-18T08:42:17-08:00

@Angelo.To support what I say I will ask you to present a technique with which you calculate the constant quantity of the pi polynomial which we obtain from the Zeta binomials (1/n-1/n+1)^2k when n takes values from one to infinity for all k .If anyone of you presents such a technique then I accept all of you know what you are talking about and you do not possess only textbook and web based knowledge.

Angelo comments on "Zeta functions" (18533)

2012-02-18T00:56:21-08:00

Yes, closing this thread seems like an excellent idea.

Mariano comments on "Zeta functions" (18532)

2012-02-18T00:32:30-08:00

Well, as Vassilis seems to have concluded that no one here can be of help, we can simply close this thread before it escalates yet again...

Angelo comments on "Zeta functions" (18531)

2012-02-17T23:48:02-08:00

I have no idea what "solving 1/n^2k-1/(n^2+n)^2k" means. But I agree with Andy and Ryan, and if the question gets posted I'll vote to close it as spam.

Ryan Budney comments on "Zeta functions" (18530)

2012-02-17T23:36:36-08:00

Like your previous questions, this one isn't appropriate for the MO forum.

Vassilis Parassidis comments on "Zeta functions" (18529)

2012-02-17T23:00:29-08:00

@Angelo.My opinion is you do not know to solve 1/n^2k-1/(n^2+n)^2k .As a matter of fact none of you knows that.

Vassilis Parassidis comments on "Zeta functions" (18528)

2012-02-17T22:55:01-08:00

@Andy.If you think can force your opinion you are wrong.As for the question you mentioning and so unfairly is received the shame is on you.My claim still stands.

Angelo comments on "Zeta functions" (18527)

2012-02-17T22:30:28-08:00

No, it is not easier, because of the well known problem of the values of the zeta function at odd positive integers. In any case my opinion stays the same: your question is not acceptable, and I doubt there is anything you can do to fix it.

Andy Putman comments on "Zeta functions" (18526)

2012-02-17T22:26:27-08:00

@Vassilis : I would have thought after the ugly display on the thread http://tea.mathoverflow.net/discussion/1287/displayed-question-incomplete/ you would have figured out that your questions are not appropriate here. I am not a moderator and thus can only speak for myself, but I would strongly urge you to find some other website to pester.

Vassilis Parassidis comments on "Zeta functions" (18525)

2012-02-17T22:12:14-08:00

@ Angelo If you think the above is easy then it should easier to "sum" the following over $n$: $1/n^3-1/(n^2+n)^3. I added the formula for clarification but if you think it is not appropriate we can remove it.

Angelo comments on "Zeta functions" (18524)

2012-02-17T21:54:31-08:00

$1/n^4-1/(n^2+n)^4$ looks close enough to me. It can be manipulated in various ways using elementary algebra. Or do you want to *sum* it over $n$? That's a little harder, but I suggest you try Wolfram Alpha, which gives a very neat form to the answer. Also, whatever the question is, suggesting the use of the formula $a^4-b^4=( a-b)(a^3+a^2 b+a b^2+b^3)$ would not really impress the users of this website.

I don't think this question would be welcome here, and I doubt you can massage it into acceptability.

Vassilis Parassidis comments on "Zeta functions" (18523)

2012-02-17T21:04:02-08:00

Is this a good MO question?
We know a^4-b^4=( a-b)(a^3+a^2 b+a b^2+b^3). Having this in mind, how do we express in closed form
1/n^4-1/(n^2+n)^4, where n takes values from one to infinity?
If it is not acceptable in this form, how can I improve it?