May 10
Is it possible for two different continuously defined functions to intersect over a range of values (all points from a to b) while diverging for other values? Or if I haven't defined the functions enough, under what criteria can this take place? 184.98.183.164 (talk) 11:27, 10 May 2012 (UTC)[reply]
- Depends what you think you mean by "continuously defined". For example, the functions f(x) = x and g(x) = |x| take the same values for all x ≥ 0, but not for x < 0. Do you consider |x| to be "continuously defined" ? In the complex plane, defining a function over a small region uniquely determines its values over a larger domain as long as we restrict ourselves to holonomic functions and connected domains - see the identity theorem and analytic continuation. Gandalf61 (talk) 12:16, 10 May 2012 (UTC)[reply]
- If you meant the very stringent condition that both functions must be smooth (having [continuous] derivatives of all orders) everywhere, it is still possible to have this matching-but-diverging behaviour. For an example, follow the link. — Quondum☏ 13:38, 10 May 2012 (UTC)[reply]
- Taylor series, see the ln(x+1) example. Ssscienccce (talk) 15:28, 10 May 2012 (UTC)[reply]
In science, we hear about scientists committing scientific fraud by publishing fraudulent papers where the data was made up. Are there any known fraudulent Mathematical Papers? 202.177.218.59 (talk) 14:22, 10 May 2012 (UTC)[reply]
- It seems like it would be more difficult in math, as any mathematician should be able to reproduce the results. In science, many experiments require a great deal of equipment, money, and time to duplicate, so you tend to get scientists taking the words of others, rather than trying it for themselves.
- I suppose a mathematician could claim they solved some difficult math problem, and refuse to provide the solution, but they would tend to be ignored as a crackpot if they did that. (An exception might be if they are dead, as in Fermat's Last Theorem, in which case they seem to have a valid reason for not providing the solution when asked.) StuRat (talk) 15:42, 10 May 2012 (UTC)[reply]
- There's the Bogdanov affair, although that's mathematical physics. Sławomir Biały (talk) 15:59, 10 May 2012 (UTC)[reply]
- Plagiarism of research mathematics is more common than most people know. My first publication was plagiarized 5 years later by people at a university in Iran. New mathematics research is so specialized that it's easy for this sort of thing to go unnoticed- I only discovered my case by chance. Here's two links about the issue:
- Staecker (talk) 16:40, 10 May 2012 (UTC)[reply]
- I have a friend in Minsk, who has a friend in Pinsk.... --Trovatore (talk) 23:15, 10 May 2012 (UTC) [reply]
- For all issues of academic malpractice, I heartily recommend Retraction Watch [1] where two journalists track journal retractions and attempt to dig the dirt. Most of the maths section looks like plagiarism (or, self-plagiarism) but there's at least a few that were deemed ultimately nonsensical. 129.234.186.45 (talk) 17:23, 10 May 2012 (UTC)[reply]
Chaos, Solitons and Fractals is a fascinating case. There is only a little information in the wiki article but Google will find a lot more. Tinfoilcat (talk) 08:06, 11 May 2012 (UTC)[reply]
Making up data would be possible in computer-assisted proofs I suppose.. Did anyone verify the 1976 proof of the Four color theorem completely? Ssscienccce (talk) 12:01, 11 May 2012 (UTC)[reply]
- Computer-assisted proofs are typically verified by (a) checking that the algorithm is logically sound and (b) writing and running independent software implementations of the algorithm. In the case of the four-colour theorem, the proof has also been verified by implementing at least two different algorithms - see Four color theorem#Simplification and verification. Gandalf61 (talk) 12:43, 11 May 2012 (UTC)[reply]
- Actually, (like Gandalf's link says) the four color theorem has been fully verified in Coq, which right now is about as good as it gets for verification. Staecker (talk) 12:51, 11 May 2012 (UTC)[reply]
- I'm not doubting the theorem, I was just wondering if the first (version of the) proof was independently confirmed. The article mentions rumors of a flaw in the Appel-Haken proof, and there's no mention of mathematicians other than Appel and Haken responding to these claims, suggesting that they might have been the only persons who really understood all the details of the original proof. My point was that using fake data in mathematics could only be succesful if the proof was so complex that no one would want to completely check it. Ssscienccce (talk) 16:05, 11 May 2012 (UTC)[reply]
- Or if the paper is obsure or unimportant enough. I have published a paper with computer-checked steps (complicated algebraic rewriting) which were probably not independently checked by the referee. Eventually somebody would notice an error if the paper ended up being widely cited, but it probably wouldn't be hard to fudge a step or two citing "computer calculations", and get it through peer review for an ordinary (i.e. not super-prestigious) journal. Staecker (talk) 16:19, 11 May 2012 (UTC)[reply]
Hello,
I came across this symbol in an article on graph theory: [2]
It doesn't seem to be so well known, it's not in List of mathematical symbols
I can sort of deduce from the context that it's supposed to mean that the graph has as vertex set the disjoint union of two sets, but I do not know for sure if it also means that there are no edges between those two graphs?
Many thanks,
Evilbu (talk) 16:33, 10 May 2012 (UTC)[reply]
- I don't know for sure, but yes, that would be my guess. That's presumably the coproduct in the category of graphs. If you want more than a sense that yes-this-is-what-that-should-mean, I'm afraid I don't have any refs to give you. --Trovatore (talk) 07:27, 11 May 2012 (UTC)[reply]
- Yes, it means the disjoint union of two graphs. More generally it is used to indicate the disjoint union of two sets. Often the dot is on top rather than inside the U. It isn't very common and I would recommend defining it if you want to use it. McKay (talk) 07:41, 11 May 2012 (UTC)[reply]
Hi all, please help me. I'm trying to work out the weight of a peice of card and my maths skills are non existent. The card is 210mm by 198mm (two thirds the size of a sheet of A4) and it is made of card that weighs 160 grams per square metre. So it's possible to work out how much it weighs, just not possible by me. thanks. — Preceding unsigned comment added by 31.185.42.46 (talk) 21:19, 10 May 2012 (UTC)[reply]
- 210 mm is 210/1000 = 0.21 meters, 198 mm is 198/1000 = 0.198 meters. So the area is 0.21*0.198 = 0.04158 square meters. At 160 grams per square meter the mass is 160*0.04158 = 6.6528 grams. -- Meni Rosenfeld (talk) 22:17, 10 May 2012 (UTC)[reply]
Many thanks.31.185.42.46 (talk) 22:20, 10 May 2012 (UTC)[reply]
- Alternatively, use the fact that by definition, A4 is half of A3 is half of A2 is half of A1 is half of A0. A0 has an area of 1.000000 square metre, therefore a mass of 160 g. Therefore one sheet of A4 (one-sixteenth) weighs 10 g, and your two-thirds A4 weighs 6.6666 grams. Sussexonian (talk) 05:51, 11 May 2012 (UTC)[reply]
Thanks again. Out of interest, why the small difference in figures after the decimal point?87.115.25.177 (talk) 07:43, 11 May 2012 (UTC)[reply]
- This is because the ISO standard dimensions of A4 (210mm x 297 mm) gives an area that is not exactly 1/16 of a square metre - it is a little under. Area of A4 is 0.210 x 0.297 = 0.06237 m2, whereas 1/16 m2 would be 0.0625 m2. ISO standard paper sizes are rounded to nearest mm for convenience - see here for more details. Gandalf61 (talk) 09:00, 11 May 2012 (UTC)[reply]
- Beware that the actual weight may vary, depending on things like humidity. StuRat (talk) 03:02, 12 May 2012 (UTC)[reply]