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March 30

Deciding whether a pentagon is cyclic

I'm reposting here this question from Georgia guy (talk) from Talk:Pentagon:

A triangle ABC is always cyclic.

A quadrilateral ABCD is cyclic if and only if A+C = 180 degrees (B+D will also always be 180 degrees.)

A pentagon ABCDE is cyclic if and only if....

Duoduoduo (talk) 19:26, 29 March 2012 (UTC)[reply]

A pentagon ABCDE is cyclic if and only if quadrilaterals ABCD and BCDE are both cyclic. Bo Jacoby (talk) 20:28, 29 March 2012 (UTC).[reply]

And the next case, if you're interested: a hexagon ABCDEF is cyclic if and only if the sum of both sets of alternate angles (A + C + E and B + D + F) is 360 degrees.→31.53.233.105 (talk) 11:39, 30 March 2012 (UTC)[reply]

So is it true whenever the number of sides n is even that a polygon is cyclic iff each set of alternate angles sums to 90°×(n–2)? And, is there any comparable rule when n is odd? Duoduoduo (talk) 14:40, 30 March 2012 (UTC)[reply]

This condition is necessary but not sufficient. For a counterexample, start with a regular hexagon, then take a pair of opposite parallel sides and increase their length by a factor of 100. Angles are all still 120 degrees so this polygon satisfies the alternate angle condition, but it is clearly not cyclic. Gandalf61 (talk) 07:38, 31 March 2012 (UTC)[reply]

March 29

math onto graphmatica

How to insert an equation like y=2-3 with the exponent 2 above 2, and y=5 to the power of x+3 and y=2 to the power of 2x-1 onto the program Graphmatica? — Preceding unsigned comment added by 70.31.20.138 (talk) 19:50, 29 March 2012 (UTC)[reply]

I haven't used graphmatica, but the usual way is to add parentheses, like this: y=5^(x+3). That works on Wolfram alpha, for instance. 130.76.64.118 (talk) 21:40, 29 March 2012 (UTC)[reply]

A notation for Hadamard product

I was curious if anyone knew of a common notation for the Hadamard product akin to capitol sigma notation for summation. That is, I need to denote the Hadamard product of $P_{i}$ over all i. Anyone seen something like this? Is there a common convention? Thanks, --TeaDrinker (talk) 19:54, 29 March 2012 (UTC)[reply]

Q_{jk}=\Pi _{i}P_{ijk}

. Bo Jacoby (talk) 20:34, 29 March 2012 (UTC).[reply]

Ah sure, just describe the product pointwise; that would certainly work. Thanks, --TeaDrinker (talk) 21:32, 29 March 2012 (UTC)[reply]

World Population

Roughly how do various agencies and organizations around the world estimate the world's population? I'm guessing that they mainly use censuses, but what calculations do they use to estimate uncounted areas or non-responders? Thanks, 207.6.208.66 (talk) 01:53, 30 March 2012 (UTC)[reply]

I don't have much knowledge in this field, but you may find it of interest to read through this page on the US Census Bureau estimates and projects population. From a quick skim-through, it seems that a lot of it for areas without a census is basically informed guessing, looking at things like immigration/emigration, any natural or health disasters, and data from public health efforts. 151.163.2.8 (talk) 18:22, 30 March 2012 (UTC)[reply]

Births, deaths, school enrollments, and consumption of energy and food figure in to many calculations. Dru of Id (talk) 09:47, 31 March 2012 (UTC)[reply]

You have to do sampling and a thorough check of the sampled areas. And when they say that hut in the back garden is just there so they can accommodate their aged grandmother on visits you have to reassure them you're not going to hand in illegal aliens or persecute them for renting it out. Dmcq (talk) 13:59, 31 March 2012 (UTC)[reply]

Lottery odds

Can someone check my calculation of the odds of winning the lottery described here? The article says one in 176 million (which would give a positive expectation, which is very usual for a lottery, even with a rollover), but I calculate one in 21 billion (which would give a very negative expectation). I suspect that the 176 million figure is the chance of winning anything at all, rather than just the jackpot, but the article is only talking about the jackpot. --Tango (talk) 22:25, 30 March 2012 (UTC)[reply]

Here is a more detailed article with the same odds. --Tango (talk) 22:27, 30 March 2012 (UTC)[reply]

(56*55*54*53*52)/(5*4*3*2*1)*46 = 175,711,536 combinations. The 1 in 176 million chance of winning the jackpot is correct. Dragons flight (talk) 22:33, 30 March 2012 (UTC)[reply]

Yeah, I forgot the 5 don't need to be in order - careless of me! In that case, it's a rather good lottery! Once you allow for diminishing marginal utility of money, there probably isn't a positive expected utility, but that's still the highest expectation I've seen in a lottery. Thanks for your help! --Tango (talk) 00:02, 31 March 2012 (UTC)[reply]

Also, the expected return is not "the entire jackpot;" it's "the jackpot, divided by number of winners." If it were possible to win the entire jackpot, everybody with $176,000,000 would purchase one of each possible numeric tickets, and would win the $500,000,000 jackpot, netting some three hundred million dollars. Nimur (talk) 00:35, 31 March 2012 (UTC)[reply]