Timeline for Combinatorial equation
Current License: CC BY-SA 2.5
29 events
| when toggle format | what | by | license | comment | |
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| May 3, 2012 at 23:31 | answer | added | Scott Snow | timeline score: 0 | |
| Oct 25, 2010 at 1:01 | comment | added | Theo Johnson-Freyd | @Mark: By my request for more motivation, I mean only that I'm always curious why people want to know things. "The following identity came up in my research on ...." is always a fun tidbit. Note that I'm one of the people who did post a solution. | |
| Oct 24, 2010 at 19:01 | comment | added | Gjergji Zaimi | Downvoted because I would like to see less questions like this in the future. On the other hand this question is perfect for artofproblemsolving. | |
| Oct 24, 2010 at 15:32 | comment | added | Thierry Zell | @Mark: $0^0=1$ is a much more reasonable definition in the current context of combinatorics (because the number of ways of doing something that can only be done trivially should be 1) than in the context of calculus (where limits that evaluate as $0^0$ can be anything). Which is why when I was in high school, we were always told that $0^0$ was undefined (even though we were never told why). | |
| Oct 24, 2010 at 11:14 | history | edited | Harald Hanche-Olsen | CC BY-SA 2.5 |
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| Oct 24, 2010 at 7:38 | comment | added | user6976 | @Gerry: You are correct. I also think that little can be gained by trying to kick a good problem out of MO without any justifiable reason. Anyway, it turned out nicely after all. I do not know about Will, but I learned some non-trivial math by reading the answers. | |
| Oct 24, 2010 at 5:39 | answer | added | Todd Trimble | timeline score: 6 | |
| Oct 24, 2010 at 5:34 | comment | added | Gerry Myerson | Mark, there are many reasons why one might not post a solution to a problem, and being unable to solve the problem is only one of them. I'm of the opinion that little is to be gained in public speculation about the mathematical abilities of fellow members of MO. | |
| Oct 24, 2010 at 3:55 | answer | added | Theo Johnson-Freyd | timeline score: 16 | |
| Oct 24, 2010 at 3:17 | vote | accept | Amir | ||
| S Oct 24, 2010 at 3:17 | vote | accept | Amir | ||
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| Oct 24, 2010 at 3:16 | vote | accept | Amir | ||
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| Oct 24, 2010 at 3:16 | vote | accept | Amir | ||
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| Oct 24, 2010 at 3:15 | vote | accept | Amir | ||
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| Oct 24, 2010 at 3:15 | comment | added | user6976 | @Theo: It is a good looking problem, and most mathematicians would consider that as a motivation. What is a motivation for solving $x^n+y^n=z^n$? As for the title, I agree with you more, but you did not give any suggestions. Anyway, the problem has been solved several times already. | |
| Oct 24, 2010 at 3:01 | answer | added | Faisal | timeline score: 11 | |
| Oct 24, 2010 at 3:00 | comment | added | Theo Johnson-Freyd | I would prefer if this question had (a) more motivation, (b) a better title. I think that (b) is the more important one: there are many "combinatorial equations", and so you should try to rewrite the title to include a focused version of your question. (Remember that titles can be 240 characters --- if it fits in a tweet, it fits as a title on mathoverflow.) | |
| Oct 24, 2010 at 2:42 | comment | added | user6976 | @Will: The problem is certainly given by somebody. But I do not think it is a homework problem. Perhaps I am wrong. Since there is no definition of a research problem, I guess it would be correct to say that a research problem is a problem which is liked by a research mathematician. I am a research mathematician and I like this problem (and the solution below). Of course you may have a different opinion. My conjecture that you cannot solve the problem is based solely on the fact that you have not posted a solution yet. | |
| Oct 24, 2010 at 2:31 | comment | added | user6976 | @Will: I do not know, but the formula seems to work with that assumption. To avoid the ambiguity, just replace the last term by $n^{n-1}$. When I was a calculus student, I was taught that $0^0=1$ is a usual assumption, otherwise the function $x^x$ would not be continuous at $0$. | |
| Oct 24, 2010 at 2:25 | comment | added | user6976 | @Will: It is a fine problem and the answer below is nice. Why do you want to kick it somewhere else? Just because you cannot solve it? | |
| Oct 24, 2010 at 2:25 | comment | added | Maximiliano Valle | @Will: You get $4=4$ if you accept that $0^0=1$ | |
| Oct 24, 2010 at 2:19 | comment | added | user6976 | @Thierry: $0^0=1$ here. | |
| Oct 24, 2010 at 2:17 | answer | added | Matti Åstrand | timeline score: 6 | |
| Oct 24, 2010 at 2:15 | comment | added | Thierry Zell | For $i=n$, the last term in your sum contains $0^0$. How are we supposed to lift the ambiguity? | |
| Oct 24, 2010 at 2:08 | comment | added | user6976 | @Yemon: It looks like an Olympiad-type problem. Do you think it is a homework problem? For what kind of class? | |
| Oct 24, 2010 at 2:03 | comment | added | Yemon Choi | Will: are you sure it doesn't work for $n=2$? | |
| Oct 24, 2010 at 2:02 | comment | added | user6976 | @Will: It is correct, $4=4$. | |
| Oct 24, 2010 at 2:02 | comment | added | Yemon Choi | Why do you know this equality is true? is this something you've been asked to work out? | |
| Oct 24, 2010 at 1:10 | history | asked | Amir | CC BY-SA 2.5 |