Wikipedia:Reference desk/Archives/Mathematics/2015 May 24

From Wikipedia, the free encyclopedia
Mathematics desk
< May 23 << Apr | May | Jun >> May 25 >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.


May 24[edit]

here's another one; it belongs to wikipedia[edit]

they (the powers to be) won't let me enhance yet another wikipedia page with a formula that a very amateurish person would understand. it goes something like this, and again no diagnosis, no advice, no prediction, etc. etc., and it certainly doesn't qualify as a homework assignment. it just follows careful observation. it could be labeled as a very, very good guess-- nothing more, nothing less. it would attract people to the talk page of Van Der Waerden Numbers!

(my) Van Der Waerden Numbers formula-- presented! before I had even understood what Van Der Waerden Numbers were about, I looked at the diagonal of the chart on wikipedia's webpage and saw a 9 and 293, and I wondered... how could those numbers be generated from 2 and 3??? then, I spotted the answer in less than 5 minutes: v(k)= (2*k^2 -1)^(k -1) +2^(k -1). it turned out that I had to back up a little to verify that for k= 1, v(k)= 2. i later read the description of a very interesting problem-- imagine that we have 9 positions in a row with only the first 8 "colored" positions:

R B B R R B B R X
1 2 3 4 5 6 7 8 9

Van Der Waerden noticed that if he listed all 16 arithmetic progressions: 1, 2, 3; 2, 3, 4; 3, 4, 5; 4, 5, 6; 5, 6, 7; 6, 7, 8; 7, 8, 9; 1, 3, 5; 2, 4, 6; 3, 5, 7; 4, 6, 8; 5, 7, 9; 1, 4, 7, 2, 5, 8, 3, 6, 9; 1, 5, 9; then the last 2 sequences containing the 9th position would either be ALL red or ALL blue iff the last position were to be "colored". the situation is forced at several levels! when i showed (my) formula to a professor, he angrily replied that "you can't just examine the data, produce a formula, and call it MATH!!! i was sorry, but sometimes a discovery of a general formula for a pattern could lead to a very humble beginning of a math proof. it looked as though arithmetic progressions must be admitted after reaching the 2nd, 9th, 293rd, and the 29,799th positions, and so on... of each Ramsey-like-coloring of a "(k+1)-length" progression. i couldn't just ignore the fact that i uncovered the answer to a phenomenon. he was just mad 'cause i saw it first! the iterations must be 2, 9, 293, 29799, since the next iteration had to be greater than 17,000. unknowingly, the 2^(to some power) described the number of "glue" variables that were present in the positions as they were described in Jerome Paul's and Michal Kouril's formal paper. ideas from their paper could be included on the wikipage also, easily!! 04/26/2015 Bill Bouris — Preceding unsigned comment added by 108.242.169.13 (talk) 04:49, 24 May 2015 (UTC)[reply]

I reformatted slightly for easier reading. Hope you don't mind. —Tamfang (talk) 05:35, 24 May 2015 (UTC)[reply]
Hi. I've been thinking about how the Reference Desk can best help you; I hope you don't mind if I make a couple of comments...
First, you need to know that I've studied mathematics to a fairly high level, but I'm not qualified to comment on Van der Waerden numbers, as they are outside my area of expertise. However, I do understand how both Wikipedia and mathematics work, so I can see where your difficulties are arising.
I can see from your question above, and from some other things you've said elsewhere in Wikipedia, that you are very interested in mathematics and have some new things to say. However, what you need to understand is that Wikipedia isn't the right place for posting new ideas, however interesting they may be. Even if Einstein came along with a new theory, we wouldn't say anything about it here until he had shown the theory to lots of other people, and they'd checked it and agreed that it was correct. It's the same for you: even if your ideas are right, we don't just rely on your saying so, we wait until other people have checked your results. In mathematics, there is a special way of checking, which consists of two parts: first mathematicians require proof, then they require peer review.
In mathematics it isn't enough just to write down a formula that gives the answer to a particular problem. You also have to prove that the result is correct. Most of the advanced training that mathematicians get at university is really training in how to prove things. These proofs can be quite short, or very, very long; it depends on lots of different factors. A proof is simply a list of steps that any other mathematician can follow, at the end of which they will be as sure as you are that your result is correct. Since some people are very sceptical, they are not convinced just by how pretty the formula is, or how it happens to fit all the cases you've examined so far: they need a way of being sure that your result is always true, in every possible case that the result is about, not just for the cases that we already know about.
Once you have a proof, you write it down and start on the second part of the process, peer review. For this, you send your result and its proof to a refereed journal for professional mathematicians; there are lots of these, and you pick the one that has readers who are interested in your particular area of mathematics. The editor of the journal sends your result and its proof to several other mathematicians, who check your work (it's just like submitting homework at school, only with a very strict teacher!). If these other mathematicians don't find any mistakes in your work, and are convinced by your reasoning, then they recommend to the journal editor that your work is published. And once it's published, we here at Wikipedia can immediately include the result in the appropriate Wikipedia article, because we can be sure that your work has now been checked.
I hope you can see that this process of proving and checking is what matters to Wikipedia. It's also what matters most to mathematicians, and if you hope to make a real contribution to mathematics you have to learn how to do it.
I'm not sure how old you are, or whether you're still at school or college, but I think what will help you best is to study more mathematics. If you're at school or college, that's easy to do - you just work harder in class, and try to learn from your mistakes every day. If you're outside formal education, a very good thing to do would be to sign up for more mathematical education - perhaps do a degree part-time or by distance learning. You will gradually learn how to prove things as well as work out formulas, and this is a good way to become a useful member of the mathematical community.
Almost finally, I want to say one thing which may give you a little encouragement: my professional view is that the professor who you mentioned earlier was not very helpful to you, because his advice was incomplete. What I think he meant to say was something like this: "Examining the data and producing a formula is math, but it's not all that you need. You also need the proof that your result is true." If you can learn how to prove your results, you'll be able to make some very good contributions to the subject. Getting to this stage isn't easy, but I hope you'll work hard to get there.
And finally, I'd like to recommend a book that will help you to understand a little about proofs and how they work: "Proofs and Refutations" by Imre Lakatos. Reading this book will help you understand how mathematics works, and how you can best make contributions to the subject. Good luck! RomanSpa (talk) 10:08, 25 May 2015 (UTC)[reply]
Einstein would be the exception though. Notable whether right or wrong — Preceding unsigned comment added by YohanN7 (talkcontribs) 10:31, 25 May 2015 (UTC)[reply]

Anti incest version of adam and eve matematical question.[edit]

I am curious about something: Imagine adam and eve, if we came from them, their childrens would need to make sex with each other or their parents to makes things work and make population be able to increase. Now, here comes the question, Imagine that no one can make sex with her sister/brother, fathers, uncles (and their sons), grandparent. Now, what is the amount of starter humans and their sex, needed to make population be able to increase under this rule.200.165.182.235 (talk) 21:54, 24 May 2015 (UTC)[reply]

You didn't exclude first cousins (unless that's what "uncles and their sons" means). One set of first cousins having kids isn't so bad, but if you repeated it generation after generation you would end up with some serious inbreeding. StuRat (talk) 23:00, 24 May 2015 (UTC)[reply]
Evolutionary theory claims that a population evolves into a new species first by being separated from the main population, by geography or some other prevention of interbreeding. I think the maths after that are covered in the article on Genetic drift. Fiddlersmouth (talk) 23:42, 24 May 2015 (UTC)[reply]
I think people are missing the point of the original question - it isn't about genetics, it is simply about the smallest population that can mathematically increase in size according to the rules given. AndyTheGrump (talk) 23:53, 24 May 2015 (UTC)[reply]

Step number one is to list all the rules explicitly. 175.45.116.105 (talk) 04:08, 25 May 2015 (UTC)[reply]

How long can a population last without incest? is a related, well-defined puzzle analyzed on Stack Exchange, which assumes an initial unrelated population and asks how many generations can be formed if parents are restricted from having any common ancestors.
It appears that our questioner intends to exclude pairings between first cousins (such as with an uncle's son), but do they intend to permit great grandparent - great grandchild pairing which have the same r = 0.125 coefficient of relationship as first cousins? What about quadruple second cousins, also with r = 0.125? Is a certain longevity and maximum number of children per individual assumed, or are they unrestricted as in the stack exchange problem? Are faithful pairings required, or may one person reproduce with several others?
The question has piqued my curiosity and I intended to work a few variations (particularly that of lock-stepped generations with unlimited pairing between members of the same generation who don't share common grandparents; an initial population of 8 is clearly necessary, but I've yet to show that it is sufficient), but don't expect to have the free time for a few weeks (and shouldn't even be typing this). I encourage the questioner to register a Wikipedia account and drops a note here so they may be contacted after this question archives. -- ToE 13:44, 25 May 2015 (UTC)[reply]
Apply rules 1, 3, & 4 from the Stack Exchange discussion, but in place of their rule 2 which prohibits reproduction between two people sharing any common ancestry, instead prohibit reproduction only between those who have a non-zero intersection of their respective sets of immediate ancestors which includes themselves, their parents, and their grandparents. This would prohibit the pairings specified by the OP, but while such a rule is simple to define, it may not be the most natural limit on consanguinity, as while it would prohibit the pairing of two people who share a single grandparent, it would allow the pairing of two people who share an identical set of great-grand parents as long as their parents and grandparents differ. Given this restriction and permitting reproduction only between members of the same generation, the minimum viable initial population is 8 (4&4). If reproduction between members of different generations is permitted, then the minimum viable initial population is only 4 (either 2&2 or 1&3). A minimum viable population of 3 is only possible if pairings such as those between half-siblings or with half-uncles are permitted.
Conclusion: Treating the question purely as a mathematical puzzle, and taking my best interpretation of its wording, the answer is four, either two of each sex or one of one sex and three of the other. -- ToE 21:16, 28 May 2015 (UTC)[reply]