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December 5

How can you tell when you have correct digits of pi in equations with limits?

The specific equation is:
$\lim _{a\to 0}{\sqrt {.5-(.5\cos a)}}\times {\frac {360}{a}}=\pi$
By the way, I found this equation myself, but its underlying principle is so simple that someone else must have already found it first.

If I read your equation properly, you appear to be working with cos in degrees rather than radians. The cosine function in radians can be simply expressed as an Taylor series expansion as

\cos x=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+\cdots

. However, when working with degrees, you have to first convert to radians using

a={\frac {\pi }{180}}x

. So, what you effectively have is a way to calculate pi that requires that you know pi ahead of time. -- Tcncv (talk) 03:46, 5 December 2008 (UTC)[reply]

A trigonometric identity tells you that

{\sqrt {0.5-(0.5\cos a)}}={\sqrt {\frac {1-\cos a}{2}}}=\left|\sin {\frac {a}{2}}\right|.

Now recall from calculus that

\lim _{a\to 0}\left|\sin {\frac {a}{2}}\right|\times {\frac {1}{|a/2|}}=1

provided you use radians rather than degrees. Since you're using degrees, you then just make the necessary conversions. Michael Hardy (talk) 04:25, 5 December 2008 (UTC)[reply]

In general to know you have gone far enough with an approximation that a particular digit doesn't change you need to limit the possible value om either side with a remainder term, and if the two limits don't change he digit then the digit is correct. For Taylor's series Taylor's theorem gives an easy to compute maximum remainder within which any addition will fall. A variant for less easy calculations and just using numbers is to use interval arithmetic, this would be able to deal with your problem above. 13:18, 5 December 2008 (UTC)

statistical significance and Wikipedia:Arbitration Committee Elections December 2008/Results

see Wikipedia:Arbitration Committee Elections December 2008/Results. Only seven Arbs will be elected. Looking at the results as they stand today, many folks seem to have no hope, and some folks should be in like Flynn. The problem will choosing that seventh slot (and perhaps even the sixth). I could do the necessary googling etc. for the formulas and crank them out, but just on casual inspection I'm not sure that the difference between these candidates is statistically significant (particularly the four in the middle) : Vassyana, Cool Hand Luke, Jayvdb, WJBscribe, Wizardman, and Carcharoth.

So the challenge is: anyone wanna track (track == "ongoing"; just doing it today and then forgetting about it doesn't help much) the candidates on the cusp and determine the statistical significance of the difference between their vote percentages? I know the short answer is "It doesn't matter, Jimbo decides." But.. come on.. aren't you stats folks up for the challenge?

Thanks Ling.Nut ^{(talk—WP:3IAR)} 01:16, 5 December 2008 (UTC)[reply]

By Monte Carlo means, assuming that votes cast reflect a random statistical sampling of the potential voters and that the underlying voter preferences doesn't change over time but may get clarified with additional sampling, I get that:

Casliber, Risker, Roger Davies, and Rlevse are essentially certain to finish in the top 7.
Cool Hand Luke and Vassyana are likely to each get a spot (94% and 83% of trials, respectively).
Jayvdb and WJBscribe are essentially a coin toss for the last spot (63% and 58%, respectively).
Wizardman is a long-shot with 2% chance to defeat both of the above and get to the seventh spot.
Everyone else, including Carcharoth, has essentially no real chance to finish in the top 7.

Of course, voter preferences don't necessarily have to be stable if some new fact is revealed about one of the candidates, their level of support could swing either way. Dragons flight (talk) 07:55, 5 December 2008 (UTC)[reply]

Much more importantly, it's a very dubious assumption that the votes will reflect a random sampling. Taemyr (talk) 15:11, 5 December 2008 (UTC)[reply]

Why? You think early voters will have significantly different tendencies to late voters in the absence of new information coming to light during the vote? I see no reason for that. (It's possible, sure, but I think it's reasonable to assume the opposite.)--Tango (talk) 15:29, 5 December 2008 (UTC)[reply]

If early voting in the US election was any indicator, I believe there were significant demographic differences between early and late voters. I wouldn't hazard to guess what the differences between early and late voters in ArbCom elections might be like, but it doesn't seem at all bizarre that there might be some, particularly as regards interest level, experience and comfort level with Wikipedia politics, etc.RayAYang (talk) 18:12, 5 December 2008 (UTC)[reply]

Early voting in the US election was rather different - the election took place on a single day, early voting referred to voting before that day. In this case the vote takes place over a period of time and early voting just means voting nearer the beginning of that period. It's a far less significant fact. There could well be all kinds of differences between people that vote early and those that vote late, but in the absence of any actual reason to believe there are, believing there are not is a good assumption. --Tango (talk) 18:19, 5 December 2008 (UTC)[reply]

Why not? RayAYang (talk) 18:44, 5 December 2008 (UTC)[reply]

Why not what? --Tango (talk) 19:09, 5 December 2008 (UTC)[reply]

Why isn't it a good working assumption, to believe that later voters are likely to be substantially different from early ones? The assumption makes modelling harder (I can see that), but I can also think of very plausible reasons for it to be correct. RayAYang (talk) 19:20, 5 December 2008 (UTC)[reply]

It makes modelling impossible. You can't assume they are different without saying how they are different. We have no reason to assume any particular difference over any other, so we have no choice but to assume there is no difference. --Tango (talk) 21:36, 5 December 2008 (UTC)[reply]

Question for Dragons flight: what were your specific modelling assumptions? Specifically, did you assume that votes on different arbitrators were independent, proceeding at different rates, or something else, etc, etc. I'm asking not so much for nitpicking, but because while I know a bit of the probability theory, I'm rather unfamiliar with the assumptions and practices that go into actual modelling. Thanks, RayAYang (talk) 18:47, 5 December 2008 (UTC)[reply]

I assumed that votes on different arbs are independent. Not strictly true, but the alternative is hard to quantify/model. For any true preference p, there is a finite probability of obtaining the observed percentages with the observed number of total votes. Given the observed votes so far, that gives a probability density function for p for each arb, with peaks at the observed percentages. I then choose instantiations of those p's in a randomly consistent way and looked at the resulting rankings to determine the frequency that someone would have a p in the top 7. So, in essence, I am looking in the limit of an infinite number of votes. Dragons flight (talk) 22:25, 5 December 2008 (UTC)[reply]

The Fabulous number i

I am looking for an article written extolling how the creation of the imaginary number 'i' is as important as one of the ten wonders of the world. The article is one I use in my math classes when students complain how useless this concept is. The article's author is unknown to me but was published some time ago (10-15 years). It is a great article and I would like to find it again.

Thanks75.141.56.16 (talk) 23:43, 5 December 2008 (UTC)[reply]

Not sure if this is the one you mean, but there was an essay written by Isaac Asimov called "The Imaginary That Isn't," which was along those lines. It was originally published in The Magazine of Fantasy & Science Fiction in March 1961, and re-printed in Asimov's book Asimov on Numbers (ISBN 978-0517371459) published in 1977. It is very entertaining and enlightening. — Michael J 01:33, 6 December 2008 (UTC)[reply]