Wikipedia:Reference desk/Archives/Mathematics/2015 February 10

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February 10[edit]

Is it an anomaly that the month of February 2015 contains four occurrences of each day of the week?[edit]

The month of February 2015 begins on a Sunday. And, of course, it has 28 days. As such, this month has exactly four Sundays, four Mondays, four Tuesdays, etc., etc., etc., up until four Sundays. I read somewhere that this only happens once in every 823 years! That can't be correct, can it? Wouldn't this happen every seven years? That is, whenever the month begins on a Sunday. The leap year oddity of 29 days might throw that off a little bit. But, certainly not once in every 823 years. Am I right or am I wrong? Is there something that I am not seeing here? Thanks. Joseph A. Spadaro (talk) 02:20, 10 February 2015 (UTC)[reply]

I've never seen anything that is 823 years in the calendar. There are only 14 allowable calendars (leap/non-leap and starting on each of the 7 days) and presuming that the cycle starts with a non-leap sunday(day 1) and that leapyears are surrounded by (), the starting days for the calendars in the successive years are in a 28 year cycle 123(4)671(2)456(7)234(5)712(3)567(1)345(6). Now the years like 1900 will throw this off, but only to another part of the cycle So presuming the last year in that previous entry wasn't a leap year (because it was 1900 or some other non 400 divisible year), you'd get a sort of skipped cycle of 123(4)671(2)456(7)234(5)712(3)567(1)3456712(3)567(1)345(6).Naraht (talk) 05:14, 10 February 2015 (UTC)[reply]
This 823-year thing has been going around the web for at least 5 years now. The details change, but no matter how you slice it, it contains NOT ONE GRAIN of truth. -- Jack of Oz [pleasantries] 05:19, 10 February 2015 (UTC)[reply]
Thanks. Yeah, the 823-year spread didn't seem to make any sense. Thanks. Joseph A. Spadaro (talk) 05:27, 10 February 2015 (UTC)[reply]
I've looked at the calendars for February 2014, 2013, 2011 and they all contained exactly four Sundays, four Mondays, four Tuesdays etc. Surely this happens every February that has 28 days?Widneymanor (talk) 11:19, 10 February 2015 (UTC)[reply]
Yes. Every 28-day February has 4 of each weekday. -- Meni Rosenfeld (talk) 13:29, 10 February 2015 (UTC)[reply]
As Jack says, this is a myth/hoax that pops up from time to time, though it's usually about a 31-day month having five Fridays, Saturdays and Sundays (or some other days, adjusted to suit the month and year in question). See Snopes. AndrewWTaylor (talk) 11:47, 10 February 2015 (UTC)[reply]
But they've failed to account for leap weeks!  :-D Sławomir Biały (talk) 13:08, 10 February 2015 (UTC)[reply]
Of course they haven't accounted for leap weeks. I proposed them in the 1990s, not expecting to be taken entirely seriously. They haven't been implemented. If leap weeks were implemented, the same day of the month would always fall on the same day of the week, but the number of days in the existing months would be changed. Robert McClenon (talk) 22:29, 10 February 2015 (UTC)[reply]
Even simpler than any of the above explanations - starting on a Sunday has nothing to do with it. Any 28-day February will contain exactly 4 Sundays, 4 Mondays, etc. simply by the fact that 28 days is exactly 4 weeks. So, it's true for any non-leap year. Organics talk 13:29, 10 February 2015 (UTC)[reply]
Yes, you are correct. Clearly, 28 divided by 4 always equals 7. Or 28 divided by 7 always equals 4. For some reason, I thought that the "interesting" selling point about the version I mentioned in my original question was that the days of the week were actually successive, from the beginning to the end of the week (i.e., starting on a Sunday ... going all the way through Saturday). Joseph A. Spadaro (talk) 17:51, 10 February 2015 (UTC)[reply]
I had not come across this nonsense before but the first tried Google search every 823 years confirms that the version about 5 Fridays, Saturdays and Sundays in a month is a common myth. PrimeHunter (talk) 13:36, 10 February 2015 (UTC)[reply]
That would require a month with 31 days, which starts on a Friday. Since 7/12 months have 31 days, and they should start on Friday 1/7th of the time, that means 1 of every 12 months, or one month a year, on average, should qualify. StuRat (talk) 16:32, 10 February 2015 (UTC)[reply]
Indeed. I've disabused at least half a dozen people about this over the years. The last straw was when my local newspaper published it as an item of interest for readers. The journo had obviously read it somewhere or been told about it, and just republished it without one second's thought or analysis. That earned them a letter from me, pointing out that this "amazing" calendric oddity happens almost every month of every year. To their credit, they published my letter. That was in 2010, from memory. -- Jack of Oz [pleasantries] 18:08, 10 February 2015 (UTC)[reply]
The calendar repeats every 400 years (303 X 365 + 97 X 366 = 0 mod 7). What is true is that the 13th of a month occurs oftener on Friday than on any other day, the frequencies in the 400-year cycle for Monday, Tuesday, ... Sunday being 685, 685, 687, 684, 688, 684 and 687 (the total being 4800 = 12 X 400).→31.54.246.96 (talk) 16:01, 11 February 2015 (UTC)[reply]

Thanks, all. Joseph A. Spadaro (talk) 23:39, 12 February 2015 (UTC)[reply]

User 156.61.250.250 provided a good explanation for the origin of 823 years at WP:Reference desk/Archives/Mathematics/2015 February 17#Is it an anomaly that the month of February 2015 contains four occurrences of each day of the week? (February 10). They pointed out that this is roughly the time between two full moons falling on a particularly day of the week on February 29. -- ToE 20:07, 17 February 2015 (UTC)[reply]

Queuing theory: number of servers needed to meet QoS target[edit]

I need help with a queuing problem. It should be quite elementary and well-understood to those versed in queuing theory. A system has N servers. Customer arrival is a constant-rate process with an exponentially-distributed interarrival time. Customers are served on a first-come, first-served basis. Queue length is unbounded.

I'm not sure exactly how the service time of customers is distributed. If the analysis can be done for general distributions, it would be great. If that's too complicated (to explain or calculate), a discrete uniform distribution with would seem like a reasonable approximation. If even that's too complicated, we can model the service time using a continuous uniform distribution.

What I need is a formula or algorithm for determining the number of servers needed to meet a given QoS target, which is of the form:

(wait time + service time) ≤ T, for ≥ X% of customers, for some given T and X.

Thanks in advance. --173.49.19.4 (talk) 05:37, 10 February 2015 (UTC)[reply]

Not a direct answer, but it would help dramatically if you could process some of the requests as batch jobs. You could do this up front, by offering to email them the results of a search, etc., or you could do it when servers are overloaded, by saying "We're sorry, we can't do that right now, but would be glad to email you the results later". StuRat (talk) 16:38, 10 February 2015 (UTC)[reply]
I don't know the solution to this specific problem, but it can be solved by modeling as a Continuous-time Markov chain, assuming the service time can be modelled as exponential, with some rate .
New customers arrive at rate . Then we look at a Markov chain of the total number of customers being served + waiting. From state i, there is a transition of rate to state i+1 and a transition of rate to state . You find the stationary distribution, and after finding the distribution of wait+service time for every state, you're well on your way to find the required bounds. It's possible that it's easier to solve numerically than symbolically. -- Meni Rosenfeld (talk) 20:13, 10 February 2015 (UTC)[reply]

Exponent of monster group F1?[edit]

I'm curious.2601:7:6580:5E3:DDCC:B8E0:B818:800C (talk) 10:12, 10 February 2015 (UTC)[reply]

A "meta-Pólya" conjecture?[edit]

Is there a word for counterexamples to the Pólya conjecture, such as 906,150,257 (Pólya numbers)? If so, is it known whether there are infinitely many of them? And either way, has it been conjectured/proven that the majority of integers less than N are non-counterexamples, for all N? ± Lenoxus (" *** ") 16:30, 10 February 2015 (UTC)[reply]

From reading around the links you provided, Liouville function claims that the Pólya conjecture can be rephrased as for all . It goes on to say: it has since been shown that L(n) > 0.0618672√n for infinitely many positive integers n. Which I think answers your second question. 129.234.186.11 (talk) 11:20, 11 February 2015 (UTC)[reply]

A 'random' sequence where the previous value influences the next[edit]

Is there a way to describe a sequence of numbers that appears random, but where a low value (10) is more likely to be followed by a high value (36)? I've determined that this is a pattern by looking at how the standard deviation decreases as the average of groups of subsequent numbers increases in comparison to if I generate a random sequence with the same mean and standard deviation. For this sequence, the SD decreases more rapidly than if the values are completely random.

  • individual values: 20,22,10,36,22,26
  • average of two values: 21,16,23,29,24
  • average of three values: 17.3,19.3,23.3,28

Thanks SmartSE (talk) 21:11, 10 February 2015 (UTC)[reply]

Is the question how to model such a phenomenon, or what phenomena like this is normally called. If the latter, then hysteresis has (elements of) what you are suggesting. I would say that to model it, I'm sure that there are lots of stochastic processes with this property, so probably you need more phenomenological insight to pin it down. (But there may also be some go-to example that has exhibits exactly this behavior.) FWIW Sławomir Biały (talk) 22:01, 10 February 2015 (UTC)[reply]
Broadly speaking, a random process of this simply called a stochastic process (Sławomir used the term). However having more specific behaviour would not necessarily already be named specially. A random process with a negative correlation between adjacent values could behave this way. A sequence like this could be generated, for example, by first generating a random sequence (with some distribution, say normal with fixed standard deviation), then running along the sequence, subtraction a fraction of each value from the previous value, before outputting the resulting sequence. A typical way of analyzing such a sequence might start with examining the power spectral density of a long sequence of values if the generating process is likely to be linear. —Quondum 23:59, 10 February 2015 (UTC)[reply]
To add a little more to what has already been said... This is indeed a form of stochastic process. One well-known stochastic process with the kind of behaviour you mention is the Ornstein–Uhlenbeck process, which, as you will see from our article, has some particularly helpful characteristics. The best way to understand the process is to set up this equation in a spreadsheet: . As you experiment with different values for the parameters you'll find that changing the value of will change the "speed" with which low values are pulled up and high values are pulled down. You'll also find (and you're on the right track in your question) that as you calculate averages of more and more individual values these will (almost certainly) converge to .
There are lots of other processes that also have the behaviour you mention, but they tend to be more complicated. One reasonably well-known class of such processes arises when the in some simpler forms of stochastic process is drawn not from the usual normal distribution but from transformations of some stable distributions. These models tend not to be analytically tractable, so are not widely studied. Some GARCH-type models also have the behaviour you're interested in, and so do some jump diffusion models. Some of these are relatively easy to manage in the real world, and have important applications in financial market modelling.
It will also be useful for you to read our article on autocorrelation. A useful next step beyond those you've already taken would be to examine the autocorrelation of the differences between successive members of your sequence. This will lead you naturally towards constructing a correlogram of the data, which may in turn lead you to more complex concepts, if you're feeling brave! RomanSpa (talk) 16:25, 11 February 2015 (UTC)[reply]