Wikipedia:Reference desk/Archives/Miscellaneous/2014 March 3

= March 3 =

Expression for pi
The square of the Gaussian integral is pi, so my logic follows from expanding and integrating the integrand using a multivariate Maclaurin series, with use of the binomial theorem.

Am I then correct in claiming that, briefly, $$\pi = \iint_{\mathbf{R}^2} e^{-(x^2+y^2)} dx dy = \lim_{(a,b)\to(\infty,\infty)} 4 \sum_{n=0}^\infty (-1)^n (\sum_{j=0}^n \frac{a^{(2j+1)} b^{2(n-j)+1}}{(n-j)!j!(2j+1)(2n-2j+1)})$$ ?--Jasper Deng (talk) 06:48, 3 March 2014 (UTC)


 * I think you wanted this at WP:RD/MA and not WP:RD/M. Dismas |(talk) 06:55, 3 March 2014 (UTC)
 * Indeed. I forgot my shortcuts.--Jasper Deng (talk) 07:31, 3 March 2014 (UTC)

Surrender
Russia has had troops in Ukraine for some time now, and yet there was no formal declaration of war. Ukraine has put its troops on full military alert, buy thankfully no shots have been fired. However, today, Russia demanded the surrender of the Ukrainian military by 03:00 GMT, or face out an all-out Crimea-wide assault. Would this demand (not the actual assault) be considered to be a declaration of war? Is there any protocol to decide this? KägeTorä - (影虎) ( TALK )  16:33, 3 March 2014 (UTC)


 * Whatever you call it, it's a standard formulation. In WWI, Austria-Hungary had made several demands on Serbia in response to serbian nationals assassinating the archduke. The had until a certain time to meet those demands, and when they didn't Austria began firing across he border.  Likewise, in the West, when it became clear Germany was mobilized with troops about to cross into the neutral Low Countries, Britain notified Germany they had until a certain time to cease troop movements, or a state of war would exist, which was then announced to the public when the Germans continued to advance.  In the Iraq war we gave Saddam until a certain time to step down.  When he didn't, Bush declared hostilities had commenced. Russian acts already amount to an act of war.


 * Putin's basically acting from a position of strength so he can say the Ukrainians were given a chance to back down, but didn't. Kind of like already having grabbed a woman and put your hand on her mouth telling he if he removes it and she screams he'll have to kill her. A brutal act in the form of a pretend courtesy. μηδείς (talk) 18:10, 3 March 2014 (UTC)


 * The Hague Convention of 1907 set up protocol: http://avalon.law.yale.edu/20th_century/hague03.asp, but it is now considered obsolete according to Declaration of war. Star Lord -  星王 (talk) 18:27, 3 March 2014 (UTC)
 * On top of that, there are various definitions of war. The definition in the "Encyclopia of War, latest edition", if I recommend correctly, require a certain level casualties to be included there. I am hoping this conflict will not meet that criteria, but I admit to an optimistic view. Star Lord -  星王 (talk) 18:30, 3 March 2014 (UTC)
 * In 1982, the UK and Argentina spent several months in full-blown combat without either side declaring war on the other - it wasn't even called "The Falklands War" until after the shooting had stopped, it was a "conflict" while it was in progress. Alansplodge (talk) 11:36, 4 March 2014 (UTC)

Same applies in America. The US Constitution gives Congress only ability to make war which has been declared only five times, last WW2. Normally the President just tells the military what to do, and then within a few weeks or months has to ask Congress for the money. In Ireland, parliament, government and the UN Security Council all have to agree to a war declaration.--92.25.228.93 (talk) 16:29, 5 March 2014 (UTC)

Will 1 in 28 citizens of New Orleans be murdered, based on a rate of 53.2 per 100k?
If I said 1 in 28 of New Orleans residents will be murdered in his lifetime, is that valid? What we do know is in 2012 New Orleans murder rate was 53.2 per 100,000. All I did was use a life expectancy of 68 years, then calculate 68 * 53.2/100,000 = 0.036 (3.6%) which roughly is one in twenty-eight.

This assumes the life expectancy and murder rate will hold steady of the next 68 years.

However, I notice that nobody, not even the most strident antigun activist ever uses this cumulative risk calculation for murder expectancy over a lifetime, even though it is an attention-getting statistic. Are there other problems with this calculation that I am missing? Thanks. (Please just dont repeat the life expectancy and murder rate realistically cannot be constant over 68 years, I realise that.) There is probably more that I am missing. Nobody uses this metric. Thanks Jojo Fiver (talk) 19:59, 3 March 2014 (UTC)


 * This question also needs to be posted on Reference desk/Mathematics. --   Jack of Oz   [pleasantries]  20:02, 3 March 2014 (UTC)
 * Yes, they'd be able to discuss any issues with the math better over there. Even if the math works, the whole concept seems a bit overly simplified.  It assumes crime remains constant and the population stable.  Ian.thomson (talk) 20:08, 3 March 2014 (UTC)
 * Thank you. I am not familiar with how that works really.  Can one of you cross post it for me?  Thanks ..Jojo Fiver (talk) 20:18, 3 March 2014 (UTC)


 * The numbers this was derived from is that New Orleans saw 193 murders over the course of a single year and a population of 362,874 people. So one person was murdered for every 1880 people living there.  According to List of U.S. states by life expectancy, the people of Louisiana have a life expectancy of 75.7 years.  So if that year was statistically typical - we'd expect a one in 25 chance (4%) of being murdered.  According to List of causes of death by rate, the worldwide chance of dying from "violence" (broadly construed) is 0.9% - so this 4% rate of murder as cause of death is certainly fairly alarming.


 * However, this is a very tricky statistical matter. Your life expectancy varies dramatically depending on how old you already are - and I'm sure that murders are not evenly distributed throughout the age range.


 * But no matter how you slice it, this is a horrifyingly large number. SteveBaker (talk) 20:43, 3 March 2014 (UTC)


 * Any prediction of how many will die from murder who are now living has to be based on a model predicting the future rate 80 years out or so. That takes into account a huge number of arbitrary assumptions, people who will move to and away from NO, future internal population growth and death rates from other causes, political, economic, and social and cultural changes.  In the future most people will probably be killed by poisonous drones and weapons of mass destruction, think the opening and closing scenes of Dune.


 * Calculating that rate and the change in it over time up until now is a question of brute summation of historical data. That info may be available somewhere. μηδείς (talk) 20:50, 3 March 2014 (UTC)


 * "The Better Angels of our Nature" by Harvard psychologist Steven Pinker (2011) has voluminous data on historical murder rates from the paleolithic times to the present in its endnotes and a comprehensive bibliography to find more. Sevastopol Dude (talk) 21:28, 3 March 2014 (UTC)