tea.mathoverflow.net - Discussion Feed (Zeta functions)2019-10-21T14:17:36-07:00http://mathoverflow.tqft.net/
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Scott Morrison comments on "Zeta functions" (18537)http://mathoverflow.tqft.net/discussion/1313/zeta-functions/?Focus=18537#Comment_185372012-02-18T09:05:46-08:002019-10-21T14:17:36-07:00Scott Morrisonhttp://mathoverflow.tqft.net/account/3/
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 ...
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.
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Vassilis Parassidis comments on "Zeta functions" (18536)http://mathoverflow.tqft.net/discussion/1313/zeta-functions/?Focus=18536#Comment_185362012-02-18T08:42:17-08:002019-10-21T14:17:36-07:00Vassilis Parassidishttp://mathoverflow.tqft.net/account/644/
@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 ...
Angelo comments on "Zeta functions" (18533)http://mathoverflow.tqft.net/discussion/1313/zeta-functions/?Focus=18533#Comment_185332012-02-18T00:56:21-08:002019-10-21T14:17:36-07:00Angelohttp://mathoverflow.tqft.net/account/483/
Yes, closing this thread seems like an excellent idea.
Mariano comments on "Zeta functions" (18532)http://mathoverflow.tqft.net/discussion/1313/zeta-functions/?Focus=18532#Comment_185322012-02-18T00:32:30-08:002019-10-21T14:17:36-07:00Marianohttp://mathoverflow.tqft.net/account/61/
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...
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...
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Angelo comments on "Zeta functions" (18531)http://mathoverflow.tqft.net/discussion/1313/zeta-functions/?Focus=18531#Comment_185312012-02-17T23:48:02-08:002019-10-21T14:17:36-07:00Angelohttp://mathoverflow.tqft.net/account/483/
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)http://mathoverflow.tqft.net/discussion/1313/zeta-functions/?Focus=18530#Comment_185302012-02-17T23:36:36-08:002019-10-21T14:17:36-07:00Ryan Budneyhttp://mathoverflow.tqft.net/account/107/
Like your previous questions, this one isn't appropriate for the MO forum.
Like your previous questions, this one isn't appropriate for the MO forum.
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Vassilis Parassidis comments on "Zeta functions" (18529)http://mathoverflow.tqft.net/discussion/1313/zeta-functions/?Focus=18529#Comment_185292012-02-17T23:00:29-08:002019-10-21T14:17:36-07:00Vassilis Parassidishttp://mathoverflow.tqft.net/account/644/
@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)http://mathoverflow.tqft.net/discussion/1313/zeta-functions/?Focus=18528#Comment_185282012-02-17T22:55:01-08:002019-10-21T14:17:36-07:00Vassilis Parassidishttp://mathoverflow.tqft.net/account/644/
@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)http://mathoverflow.tqft.net/discussion/1313/zeta-functions/?Focus=18527#Comment_185272012-02-17T22:30:28-08:002019-10-21T14:17:36-07:00Angelohttp://mathoverflow.tqft.net/account/483/
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 ...
Andy Putman comments on "Zeta functions" (18526)http://mathoverflow.tqft.net/discussion/1313/zeta-functions/?Focus=18526#Comment_185262012-02-17T22:26:27-08:002019-10-21T14:17:36-07:00Andy Putmanhttp://mathoverflow.tqft.net/account/113/
@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 ...
Vassilis Parassidis comments on "Zeta functions" (18525)http://mathoverflow.tqft.net/discussion/1313/zeta-functions/?Focus=18525#Comment_185252012-02-17T22:12:14-08:002019-10-21T14:17:36-07:00Vassilis Parassidishttp://mathoverflow.tqft.net/account/644/
@ 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 ...
Angelo comments on "Zeta functions" (18524)http://mathoverflow.tqft.net/discussion/1313/zeta-functions/?Focus=18524#Comment_185242012-02-17T21:54:31-08:002019-10-21T14:17:36-07:00Angelohttp://mathoverflow.tqft.net/account/483/
$1/n^4-1/(n^2+n)^4$ 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 ...
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)http://mathoverflow.tqft.net/discussion/1313/zeta-functions/?Focus=18523#Comment_185232012-02-17T21:04:02-08:002019-10-21T14:17:36-07:00Vassilis Parassidishttp://mathoverflow.tqft.net/account/644/
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 ...
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?]]>