Timeline for Sum of product maximum
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
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| Aug 14, 2013 at 13:45 | comment | added | fedja | @Todd "you" in my message is an impersonal pronoun :) I'm just too lazy to write "I believe that if one...". | |
| Aug 14, 2013 at 11:17 | comment | added | Todd Trimble | @fedja: I certainly agree, and have even said the same thing on a number of occasions to people who voted to close a question on the grounds that it looked like homework, that you should be able to do the homework before dismissing it as homework! My comment you are quoting was written to tell OP what I think happened (from a speculative POV of closers), but as for my personal ethics they agree with yours. | |
| Aug 14, 2013 at 4:17 | comment | added | Yemon Choi | Downvoting, but not voting to close - see @fedja's comment | |
| Aug 14, 2013 at 4:08 | comment | added | Brendan McKay | Also $f(x)$ is the coefficient of $y^m$ in $\prod_{i=1}^n (1-y x_i)^{-1}$. Converting that to a contour integral will allow an asymptotic formula in some cases. There's a chance it might help for large $m,n$. | |
| Aug 14, 2013 at 3:28 | comment | added | Brendan McKay | @user36162: Do you know of any case where the maximum doesn't occur either for all $x_i$ equal or for all but one $x_i$ being 0? | |
| Aug 13, 2013 at 23:34 | history | reopened |
Todd Trimble fedja Tom Leinster Michael Renardy Benjamin Steinberg |
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| Aug 13, 2013 at 23:16 | comment | added | fedja | 'I think it's the elementary formulation which makes it a bit hard to evaluate as to whether it's "research level"'. Can you solve it in under 1 hour? If not, and if you are an "established researcher", it certainly is. It MAY be trivial after all but sheer politeness requires that you convince yourself about it first before trying to convince anyone else. Voting to reopen for now. :) Of course, the comments about background and motivation stand but the mere absence of those may be grounds for downvoting but not for the closure, IMHO. | |
| Aug 13, 2013 at 22:50 | comment | added | Theo Johnson-Freyd | @ToddTrimble: I will support a reopen vote just as soon as OP adds some background and motivation. I ask for various reasons. One is that I learn quite a lot from reading MO questions, provided they include such material. | |
| Aug 13, 2013 at 22:38 | review | Reopen votes | |||
| Aug 13, 2013 at 23:34 | |||||
| Aug 13, 2013 at 22:24 | history | edited | Todd Trimble | CC BY-SA 3.0 |
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| Aug 13, 2013 at 22:21 | comment | added | Todd Trimble | Just to be clear: I was not one of the downvoters, and I half-expected the problem was in fact difficult and not at all homework -- my questions were meant to draw out user36162 a bit. I think it's the elementary formulation which makes it a bit hard to evaluate as to whether it's "research level", but I'm inclined to take OP at his/her word and vote to reopen, with a word of advice to OP that it's especially questions that look possibly elementary and are unmotivated that are susceptible to being shut down. Hope he/she won't take this to heart. Voting to reopen. | |
| Aug 13, 2013 at 20:25 | comment | added | user36162 | Your computation is right. The problem is that for example, if $n=2$ and $m=5$, that value is not maximal. | |
| Aug 13, 2013 at 19:21 | comment | added | Steve Huntsman | My casual computation (possibly wrong) is that if $x_1 = \dots = x_n$ and $\lVert x \rVert = 1$, so that $x_k \equiv n^{-1/2}$, then $f(x) = n^{-m/2} \binom{n+m-1}{m}$. | |
| Aug 13, 2013 at 19:11 | comment | added | Yemon Choi | Just to add to @TheoJohnson-Freyd's comment (with which I agree entirely) -- it is often worth giving some details of what partial progress you have been able to make yourself towards answering this question, as a show of good faith. | |
| Aug 13, 2013 at 14:44 | history | closed |
Fernando Muro Noah Stein David White Bill Johnson Theo Johnson-Freyd |
Not suitable for this site | |
| Aug 13, 2013 at 14:43 | comment | added | Theo Johnson-Freyd | Hi user36162: The down votes are probably in response to the question being entirely unmotivated. Background, what you've tried, and motivation are very important at MathOverflow. Why is this an interesting research question? Some further discussion of what makes a good MO question is at meta.mathoverflow.net/questions/70/how-to-ask-page. When you edit the question to include this part, it will automatically get nominated for reopening, and we'll look it over again. | |
| Aug 13, 2013 at 14:25 | comment | added | user36162 | I do not understand votes down. The question is far from trivial and comes from a norm of Bergman projection research. | |
| Aug 13, 2013 at 14:15 | comment | added | Steve Huntsman | FYI, $f(x) = \sum_{k=1}^n x_k^{n+m-1} \cdot \prod_{\ell \ne k}(x_k - x_\ell)^{-1}$ when no two components of $x$ are equal. | |
| Aug 13, 2013 at 13:49 | comment | added | Todd Trimble | It looks as though this question will be closed soon, so you might try over at Math Stack Exchange. But where did this problem come from, and why are you interested? Is it homework? | |
| Aug 13, 2013 at 12:45 | review | Close votes | |||
| Aug 13, 2013 at 14:44 | |||||
| Aug 13, 2013 at 12:24 | history | edited | user36162 | CC BY-SA 3.0 |
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| Aug 13, 2013 at 11:40 | history | asked | user36162 | CC BY-SA 3.0 |