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October 17

Binomial expansions

Does Wikipedia have an article(s) about binomial expansions for non-iteger and/or negative powers? asyndeton 14:26, 17 October 2007 (UTC)[reply]

Did you check out the article on the binomial theorem, in particular the section labelled "Newton's generalized binomial theorem"? And then the bit in that section that says "For a more extensive account of Newton's generalized binomial theorem, see binomial series"? That pretty much answers your question, I believe. Confusing Manifestation 14:38, 17 October 2007 (UTC)[reply]

It's not wikipedia-related, but you might want to read Knuth's Concrete Mathematics about such binomial expansions. – b_jonas 21:01, 17 October 2007 (UTC)[reply]

Sorry, perhaps this wasn't a good suggestion. An introductory complex analysis book might be better. – b_jonas 19:11, 18 October 2007 (UTC)[reply]

An infinite amount of monkeys...

Well, maybe I've been watching too many Twilight Zone episodes but my question was as such: What IS the probability that on any normal day, no one died? My reasoning leads me to believe that it is possible (or probable) but would it be sooooo small that one would realize that the likelihood of the event happening would be "impossible". So help me out here, is it possible to have a day where no one died, or is it so far away in probability that it "impossible" for it to occur? Thanks! ♥ECH3LON ♥ 22:36, 17 October 2007 (UTC)[reply]

P.S.: Some statsitics would help me out as well!

It is plausible to assume that the number of deaths per day follows the Poisson distribution. Under that assumption, if the average number of deaths per day is

\lambda

, the probability that in a given day there will be 0 deaths is

e^{-\lambda }

. Now we only need some statistics about the average death rate. I'd say

\lambda \approx 300000

, making our probability

e^{-300000}\approx 10^{-130000}

, an unimaginably small probability. It's comparable to the probability that if you buy a lottery ticket every day, you will win every single day for 40 years. -- Meni Rosenfeld (talk) 22:45, 17 October 2007 (UTC)[reply]

The probability is zero. Think about it. Car accident fatalities. Drug overdose. You might as well ask what is the probability that noone gets born on any normal day. 202.168.50.40 00:29, 18 October 2007 (UTC)[reply]

Not true. Just because something is incredibly improbable, doesn't mean it's completely impossible. What's the probability of no-one dying in a given second? Probably still small, but non-zero. So what's the probability of no-one dying in a given minute? It's the probability of no-one dying in each of the 60 seconds, multiplied together. Even smaller, but still possible. Take that up to an hour, and a day, and even further if you like (technically you can't just do straight multiplication since it's probably not reasonable to argue perfect independence, but it should make for a reasonably close approximation at this level). So, like Meni suggests above, as unlikely as winning the lottery every day for 40 years, but still, technically, possible. Confusing Manifestation 00:42, 18 October 2007 (UTC)[reply]

Just by the way :-) not even a probability of exactly zero implies that something is impossible. --Trovatore 00:45, 18 October 2007 (UTC)[reply]

202, I'm not sure what new insight did you try to convey with "Think about it. Car accident fatalities. Drug overdose.". That people tend to die? Yeah, we already knew that. That doesn't invalidate my calculations (though by no means am I saying they are correct).

Yes, you can ask what is the probability that nobody will be born in a given day, and I'd say it is about the same (give or take a factor of

10^{100000}

, of course).

As for the difference between 0 and

10^{-130000}

, see below. -- Meni Rosenfeld (talk) 09:25, 18 October 2007 (UTC)[reply]

Question: What is the probability that on any normal day, no one died?

{\begin{aligned}P(Nobody\,dies|NormalDay)&={\frac {P(Normal\,Day|Nobody\,dies)\,P(Nobody\,dies)}{P(Normal\,Day)}}\end{aligned}}

But a day in which nobody dies is not a normal day hence:

P(Normal\,Day|Nobody\,dies)=0

so

{\begin{aligned}P(Nobody\,dies|NormalDay)&={\frac {0\times P(Nobody\,dies)}{P(Normal\,Day)}}&=0\end{aligned}}

202.168.50.40 00:44, 18 October 2007 (UTC)[reply]

So you're defining a normal day as being one in which no-one dies? Sounds like begging the question, to me. I would suggest that if you want to get a meaningful answer, you could go for a definition such as "a normal day is a 24 hour period in which no major events occurred that would affect the death rate". That way, you're not implicitly assuming your answer in the definition, and the answer is, as said above, possible but immensely improbable. (It probably needs some tweaking itself, though, since it doesn't allow for a situation where something prior to that day affects that day's death rate - e.g. if everyone dies on the previous day, no-one dies that day, and yet by the definition it technically counts as a normal day). Confusing Manifestation 05:44, 18 October 2007 (UTC)[reply]

If this were asked on the Miscellaneous or Humanities Desk, a reasonable (and, in my view, correct) answer might have been "For all practical purposes, the probability can be assumed to be zero". But since it's being asked on the Mathematics Desk, it's not about practicalities but theory, and a more precise answer is called for. (This reminds me of: In theory, there's no difference between theory and practice - but in practice, there is.). -- JackofOz 05:59, 18 October 2007 (UTC)[reply]

I'll give as an example a theory I have come up with for the creation of life. I doubt it's original but I haven't yet encountered it anywhere. Opponents of evolution, atheism etc. often mention that it is extremely unlikely that a bunch of random molecules will band together to create a life form, and they're probably right. What they forget to acknowledge is that it is currently believed (last time I checked, anyway) that the universe has infinitely many planets. In this light, the probability for the creation of life on any planet can be 1 / Graham's number for all I care - as long as it's positive (and assuming independence), the probability that life will be created on at least one planet is 1 (we would not have this guarantee if the probability was 0). It is then trivial to conclude that the probability that life has been created on our planet, given that we are sentient living creatures who are capable of asking about said probability, and under the assumption that we were not transported here from elsewhere, is 1. So it is not at all surprising that life has been created here, even without assuming divine intervention.

The title of the question probably alludes to the infinite monkey theorem, which is another example. -- Meni Rosenfeld (talk) 09:25, 18 October 2007 (UTC)[reply]

It is a common method of reasoning - though not everyone thinks that there are infinite planets..(see below)87.102.3.9 13:30, 18 October 2007 (UTC)[reply]

The more difficult way of realising this is that any probabilty is irrelevent - since if p(life)<>1 then we can't even think about it.(this has nothing to do with the original question - sorry)87.102.3.9 13:32, 18 October 2007 (UTC)[reply]

Where did you get the idea that astronomers predict an infinite number of planets? Not so. We have a good estimate of age and expansion rate, and thus a good estimate of a finite size, implying a finite number of planets. See Drake equation. --KSmrq^T 10:16, 18 October 2007 (UTC)[reply]

All observations are consistent both with a large but finite universe and an infinite one. An infinite universe can also be subject to metric expansion of space. See also Shape of the Universe#Hyperbolic universe. --Lambiam 12:28, 18 October 2007 (UTC)[reply]

Current predictions are that the universe will expand forever, and that the rate of expansion is increasing. Nevertheless, if we assume that signals reach us at the speed of light, we can't see "to infinity and beyond". If you can find a mainstream claim that the universe has an infinite number of potentially visible planets I'll be surprised. --KSmrq^T 13:16, 18 October 2007 (UTC)[reply]

There is another view that the universe is currently infinite in size (and infinitely old) - so would I expect include infinite planets (not sure what "potentially visible" really means here of if it's relevent) - there's always a problem with people expecting current popular theories to be absolutely true - especially when they have been printed in textbooks for general use.. Surely we know by now that current knowledge in the scientific field is never the final word.87.102.3.9 13:30, 18 October 2007 (UTC)[reply]

Did I say "potentially visible"? It is clear that the observable universe is finite in size and contains finitely many planets. I was referring to the entire universe, which may very well be infinite. -- Meni Rosenfeld (talk) 14:35, 18 October 2007 (UTC)[reply]

We have a tangle of issues. It is commonly said that intelligent life exists on planet Earth; some days I wonder, but let's take that as true. If we ask, "How likely is that?", one answer is that it is certain — otherwise we would not be here to ask the question. Such an idea has been proposed in physics as the anthropic principle; it makes many physicists uncomfortable. (Compare with attitudes about renormalization.) One reason it is unsatisfying is because we stop looking for more powerful answers, ones that explain more. The Drake equation concerns the likelihood that we will encounter extra-terrestrial intelligent life, which is one reason we'd like a better explanation of our existence. Within the finite universe we can observe (because of the speed of light), we'd like to estimate the chances of another "accident" like us, and that requires more than the anthropic principle. --KSmrq^T 04:52, 19 October 2007 (UTC)[reply]

Here's another way to look at this. The average human lifespan is, let's say, around 30,000 days (an overestimate for the global population, but a good enough ballpark). Let's assume that life/death is a Bernoulli process, so that the probability that a random person who is alive a midnight tonight will die before midnight tomorrow is 1/30,000. If we have a group of n people then the probability that no-one will die tomorrow is

P({\mbox{no-one dies tomorrow}})=\left(1-{\frac {1}{30000}}\right)^{n}

If n is 24,000 then P(no-one dies tomorrow) is just under 45%. The town where I live has a population of about 24,000. So on more days than not, at least one person dies in my town.
About an hour's drive away is a city with a population of 118,000. The probability that no-one dies on a given day in this city is less than 2%. But still, on a few days each year you would expect there to be a day with no deaths in this city.
For a larger city, with a population of say 320,000, then the probability of a day in which no-one dies is less than 1/42,000 - so you only would expect a day with no deaths to happen about once a century.
Once n gets up to 7 million, roughly the population of a major city like London, then the probability of a day in which no-one dies is less than 10^-100, or 1 in a googol. Gandalf61 09:35, 18 October 2007 (UTC)[reply]

Using data from the The World Factbook [1] (see also List of countries by death rate and World), the world death rate is about 8.37 deaths/1,000 population (2007 est.), while the world population is about 6,602,224,175 (July 2007 est.). This means the average number of deaths in a single day is about 151,300. Assuming a Poisson distribution, the probability of 0 deaths is then about e^−151300, or roughly one in 3×10⁶⁵⁷⁰⁸. The assumption of a Poisson distribution is justified if individual deaths are independent occurrences. This is not fully true – think of disasters with multiple deaths – but close enough for giving an indication of how small the chances are. This can be compared to the probability of getting Hamlet out without any errors in one trial, if 26 keys are hit uniformly randomly, of about 3.4×10¹⁸³⁹⁴⁶. --Lambiam 12:28, 18 October 2007 (UTC)[reply]