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August 11

Mandala problem

I'm using the word in the sense of a number of equally-spaced points in a circular pattern, with each connected by a straight line to all of the others. For n points, then, the total number of lines will be n(n-1)/2. I'm trying to determine the greatest number of these which can remain so that no triangle is formed by the original points and the lines. For n being 2,...,8 I got the values 1,2,4,5,9,10,16 by actually drawing the patterns. However, this sequence is unknown to OEIS, so maybe it's wrong. Does anyone have an explicit formula for a general value of n?–81.153.219.75 (talk) 09:50, 11 August 2008 (UTC)[reply]

Do you only count triangles with vertices among the n points? If so, the sequence is A002620. Otherwise, I don't know. -- BenRG (talk) 11:23, 11 August 2008 (UTC)[reply]

Thanks. Yes, that's what I meant by original points. As I suspected, there are errors in my sequence, in that it is possible to have 6 lines from 5 points and 12 from 7. It's surprisingly difficult to draw the maximal pattern with certainty. Of course, the equal-spacing and circularity conditions are unnecessary, they just make it easier to visualise the problem.—81.153.219.75 (talk) 12:06, 11 August 2008 (UTC)[reply]

The maximal pattern is actually rather easy to visualize if you don't use a circle. It's a complete bipartite graph, with either the same number of vertices in each part (if n is even) or a difference of one (if n is odd). Algebraist 12:29, 11 August 2008 (UTC)[reply]

If n is even the optimal solution makes a nice circular pattern in which you connect each point to the other points that are an odd distance away (e.g. for n = 6 connect the outside of a hexagon and its long diagonals; for n = 8 connect the outside of an octagon and an eight-point star; etc.) —David Eppstein (talk) 16:55, 11 August 2008 (UTC)[reply]

Can two asymptotic functions touch or cross?

This is currently being discussed on Talk:Asymptote. If you define f as being asymptotic to g if the limit of f is g, then f is asymptotic to f, which is not the way the word is usually imagined, and contrary to its etymology. --Slashme (talk) 14:26, 11 August 2008 (UTC)[reply]

Why do you say this is contrary to the way the term is ordinarily used? I have always viewed asymptotic as a formal way of saying that two functions behave similarly in the limit; I would expect that with the right definition of "asymptotic" the relation "f is asymptotic to g at

\infty

" will be an equivalence relation. Typically, though, the definition is only made for functions that are asymptotic to lines. — Carl (CBM · talk) 14:57, 11 August 2008 (UTC)[reply]

I can't see anything wrong with saying f is asymptotic to f. It's completely trivial, but it's true by any reasonable definition. --Tango (talk) 17:04, 11 August 2008 (UTC)[reply]

Question - at the talk page y=sin(x)/x is given as an obvious example approaching y=0 as x becomes infinity - is this asymptopic or not? or is it a matter of opinion on the meaing of the word..?87.102.45.156 (talk) 19:20, 11 August 2008 (UTC)[reply]

Yes, that would be considered asymptotic. It doesn't matter that y = sin(x)/x actually equals 0 at numerous points. To answer the title question: Yes, two asymptotic functions may touch or cross. 76.224.123.69 (talk) 19:38, 11 August 2008 (UTC)[reply]

Thanks, that's what I thought.87.102.45.156 (talk) 22:30, 11 August 2008 (UTC)[reply]

The above discussion leads me to ask:

Is y = sin(x)/x asymptotic to y = sin(x)/(x+420) ?

Thanks. Wanderer57 (talk) 23:20, 11 August 2008 (UTC)[reply]

Yes, y = sin(x)/x is asymptotic to y = sin(x)/(x+420). Both functions are asymptotic to y = 0, and the asymptotic property in this sense is transitive. 76.224.123.69 (talk) 23:26, 11 August 2008 (UTC)[reply]

Yes. They both tend to zero, which is all you need. --Tango (talk) 23:24, 11 August 2008 (UTC)[reply]

Conversion

Okay, hopefully someone here can help with this. Basically, I need a rough estimation of the following:

If 250,000 roubles in 1907 has the present day value of 3.4 million roubles, then what would the present day value be for 20 roubles in 1899?

I need this for the Joseph Stalin article. Thanks in advance for any help. Jennavecia ^(Talk) 17:19, 11 August 2008 (UTC)[reply]

Insufficient information. The best we could do would be to assume inflation was constant over the 109 year period, which is almost certainly far from the truth. The internet should be able to provide you with the information you need - historic inflation indices for Russia/USSR. Once you find those, I can show you how to use them if you're not sure. --Tango (talk) 17:24, 11 August 2008 (UTC)[reply]

I think the OP wanted a rough estimate. Barring a major economic catastrophe in turn-of-the-century Russia, 20 1899 rubles would be somewhere around 275-ish rubles today. However, if you want to put it in an article, you're going to need a better source than me. Paragon ¹²³²¹ 17:33, 11 August 2008 (UTC)[reply]

It's also important to keep in mind that there has been a series of redenominations of the ruble, so you need to know which ruble the numbers you have are measured in. It looks like the 250,000 figure is in the original ruble (the first redenomination was in 1922), is the same true of the 20 figure? If you have a contemporary source for the figure, then it will be, if it's a modern source, however, you'll need to check the source carefully. Chances are, common sense will tell you if you have it right - the total redenomination between the turn of the century and now is a factor of 5 quadrillion, if I've done the sums right. When you find inflation figures you'll need to check if they take into account the redonominations or not, and if they don't, you'll need to do so. --Tango (talk) 18:07, 11 August 2008 (UTC)[reply]

Jennavecia, I would strongly suggest that you stay away from doing such conversions and estimates yourself and putting them in the article, unless you find a direct source for this. Anything else is likely to be WP:OR. In the case of Russia and the Soviet Union such conversions/estimations are particularly problematic since during the Soviet period the meaning of money was rather different and much of the economy was non-monetary. Nsk92 (talk) 14:00, 12 August 2008 (UTC)[reply]

linear regression

Hi, our article on linear regression says: A linear regression model need not be a linear function of the independent variable: linear in this context means that the conditional mean of y is linear in the parameters $\beta$ .

Can someone confirm this for me, and does that mean quadratic regression is just a special case of linear regression? If so, how do you deal with the fact that the quadratic term, $x^{2}$ and the linear term, $x$ , cannot both be normally distributed, and are not independent? It's been emotional (talk) 19:23, 11 August 2008 (UTC)[reply]

This is correct: the term "linear" refers to the dependent variable as a function of the parameters, not the independent variables. The model $y=a+bx^{2}+u$ is the same as $y=a+bz+u$ where $z=x^{2}$ . I don't follow your question regarding x being normally distributed. OLS imposes no requirement on the distribution of x. The imposition is on the distribution of u. Wikiant (talk) 19:36, 11 August 2008 (UTC)[reply]

Thanks for the help, and from someone in the actual profession too :). Much appreciated. I had misunderstood the condition you referred to, which is mentioned in the textbook, and which I had come across; I had simply forgotten it. If you actually lecture in statistics, I would remind you that, for students like me, it's easy to completely forget stuff like that, even when you've already understood it clearly the first time round: new info goes in, and the old stuff gets scrambled around somewhat. It's been emotional (talk) 15:19, 13 August 2008 (UTC)[reply]

It might be worth noting that the terminology differs in some fields. In statistical machine learning, for example, regression with models that aren't actually linear functions of the independent variables often go by other terms, like "polynomial regression" in the example with an x^2. --Delirium (talk) 18:26, 13 August 2008 (UTC)[reply]