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December 10

Roulette

The numbers on a roulette wheel are not sequential. They run (on the French wheel):

0-32-15-19-4-21-2-25-17-34-6-27-13-36-11-30-8-23-10-5-24-16-33-1-20-14-31-9-22-18-29-7-28-12-35-3-26

and (on the American wheel):

0-28-9-26-30-11-7-20-32-17-5-22-34-15-3-24-36-13-1-00-27-10-25-29-12-8-19-31-18-6-21-33-16-4-23-35-14-2.

These are not random sequences - black and red numbers alternate according to the rule that even numbers between 2 and 10 and 20 and 28 (inclusive) are black. The French sequence was not standardised originally, so how did these sequences originate? Is there some underlying pattern and did the arrangement of the dominant manufacturer prevail as happened with the typewriter? How many different sequences can be generated within the constraint of the red/black rule? 92.27.49.50 (talk) 13:29, 10 December 2017 (UTC)

One of your questions is easy to answer. If you already have a rule that assigns red to half of the numbers and black to the other half, and have already constructed and painted your wheel, then there are 18! ways of numbering the black pockets and 18! ways of numbering the red pockets. See factorial and permutation. Thus the total number of different sequences which can be generated within the constraint of the red/black rule is the product of those two numbers, or (18!)² ≈ 4.1×10³¹ (a very big number). Without following a predetermined red/black rule (but still keeping the zeros where they are), then the total number of different sequences is 36! ≈ 3.7×10⁴¹ (an even bigger big number). -- ToE 16:26, 10 December 2017 (UTC)

The relevant section of our article, Roulette#Roulette wheel number sequence, gives the sequences but not their history. -- ToE 16:33, 10 December 2017 (UTC)

There is a pattern in the placement of high (19-36) and low (1-18). The French wheel alternates high, low, high, low, ... in either direction from the zero, meeting half-way across the wheel at two adjacent lows, 10 & 5. This, combined with the alternating red and black pockets, ensures that half of either color is high and half is low. The American wheel alternates once, then alternates in pairs, that is, H,L,H,H,L,L,H,H,L,L,..., clockwise around the wheel, starting from either zero. Since the pockets adjacent to a zero (either zero) are the same color on an American wheel, this pattern also ensures that half of either color is high and half is low. Since 18 is not divisible by 4, each half of the American wheel could not alternate in pairs while keeping the same number of high and lows on each half, hence the initial singleton alternation. -- ToE 17:03, 10 December 2017 (UTC)

Additionally, for the half of the American wheel clockwise from 0 to 00, the highs are all even and the lows are all odd, with the opposite the case for the other half of the wheel, and with the consecutive numbers 1&2, 3&4, etc. diametrically opposed on the wheel. -- ToE 17:47, 10 December 2017 (UTC)

Looking at the sequences, there appears to be an effort to make it so that the bets listed at Roulette#Outside bets don't cluster on one half of the wheel. I presume that this is to avoid the possibility of a skilled player being able to hit one side more often than the other. I wonder if either of the traditional sequences are actually optimized for this. Sounds like an interesting math puzzle. --Guy Macon (talk) 19:58, 10 December 2017 (UTC)

Detail correction: it's not the people betting who launch the ball, it's the croupier. But it doesn't matter, since no honest casino wants their croupiers to be able to influence who wins. --69.159.60.147 (talk) 05:34, 11 December 2017 (UTC)

I see what is almost one more pattern in the American wheel, but I don't quite know what to make of it. Since one side of the wheel determines the other, it is sufficient to look for patterns within and between the high/even and the low/odd sequences on the first half of the wheel, separating them as 28-26-30-20-32-22-34-24-36 and 9-11-7-17-5-15-3-13-1. For relative ordering of the first sequence, subtract 18 and divide by two, and for the second sequence, add one and divide by two, giving 5-4-6-1-7-2-8-3-9 and 5-6-4-9-3-8-2-7-1. If you cast out the 5 (which is from the singleton high/low alternation before they start alternating in pairs) then you get two quite similar sequences: 4-6-1-7-2-8-3-9 and 6-4-9-3-8-2-7-1. Rotate the first sequence left by two (i.e., take the 4-6 from the start and put it on the end), and you have the the reverse of the second sequence. This seems too significant to be mere coincidence, but not significant enough to have much meaning. The individual sequence 1-7-2-8-3-9-4-6 isn't quite broken at the end, as it comes in pairs of n, (n mod 4) + 6. This reflects the pattern in the alternating high, high and low, low pairs where within each pair there is one number from the upper half of the range and one from the lower, with the difference between them 12 except for when that take the higher number beyond its range in which case the difference is 4. For example, break the even highs into the lower highs, 20, 22, 24, & 26, the middle high 28, and the upper highs, 30, 32, 34, & 36. 28, the middle high comes first, then the pair 26 & 30, namely the fourth of the lower highs and the first of the upper highs, then 20 & 32, namely the first of the lower highs and the second of the upper highs, then 22 & 34, namely the second of the lower highs and the third of the upper highs, and so on. Using the symbols Even/Odd, High/Low, Upper/Middle/Lower, & numbers 1..4, we have EH:(M--L4-U1--L1-U2--L2-U3--L3-U4). The odd lows run OL:(M--U1-L4--U4-L3--U3-L2--U2-L1). (And the two are appropriately interleaved.) So there is a pattern there, with the two running in the opposite direction, but jumbled up just a bit at the start. -- ToE 20:03, 10 December 2017 (UTC)

December 12

Cheryl’s Birthday

I came across this problem (http://abc7.com/society/are-you-smart-enough-to-solve-this-riddle/656040/) and couldn’t figure it out thinking there wasn’t enough information. So I looked at the solution but I’m still not convinced. I think the problem lies with Albert’s second statement. While if you take Albert’s statement to be true you can solve it, I believe there is no way Albert could make that deduction. Am I right? —�-Polyknot (talk) 16:06, 12 December 2017 (UTC)

This problem is old enough that the solution is available if you just google it. Here is the first such solution I found through Google. If you want to find others, try typing "cheryl's birthday solution" into Google. --Jayron32 16:09, 12 December 2017 (UTC)

Thanks Jay, but there was a solution on the web page I posted. The problem I have is that, as I mentioned, I don’t see how Albert could have enough information to make his second statement. —Polyknot (talk) 16:24, 12 December 2017 (UTC)

Wasn't this a question in a Singapore school exam which went viral? B was given the number 14, 15, 16, 17 or 18 (if he had been given the number 19 he would know it was May). A must have been given May because if B had been given 19 he would know the date. B now knows the date (it's the number given to him and the month is May). The possible birthdays in May are the 15th and 16th (19 is out). Originally 15 could have been May or August and 16 could have been May or July. That's as far as I can take it. 82.13.208.70 (talk) 17:05, 12 December 2017 (UTC)

Albert knows (from the start) that it is July.

After Albert's 1st statement, Bernard and the readers know that it's either July or August.

Based on Bernard's statement, Albert and the readers know that the day is not 14 (if it was 14, Bernard would not know if it's July 14 or August 14).

So Albert knows that it is July and not 14. So he knows it is July 16. -- Meni Rosenfeld (talk) 20:37, 12 December 2017 (UTC)

We actually have an article about this. --JBL (talk) 00:47, 13 December 2017 (UTC)

Here are some teasers to keep you entertained over Christmas. Answers are included, and Cheryl's birthday is explained well:

[1], [2] [3] [4] [5] [6] [7]. 86.171.242.205 (talk) 12:16, 13 December 2017 (UTC)

Meni says the answer is 16 July and that Albert knows from the start that it's July. We can add to that that Bernard knows it's the 16th. Before anyone speaks, A knows that the date is 14 or 16, B knows that it is July or August. Albert starts off by saying

I don't know when Cheryl's birthday is, bit I know that Bernard does not know too.

Agreed, since he knows that if B has 14 it's either July or August and if B has 16 it's either May or July. What I don't understand is why, when A speaks,

Bernard and the readers know that it's either July or August.

B certainly knows that, but how do the readers know it? Now A has spoken B says (in effect) that he knows it's July. How does he know? 86.171.242.205 (talk) 15:00, 13 December 2017 (UTC)

See Cheryl's Birthday. JBL linked it in "We actually have an article about this". Blue text means a link. PrimeHunter (talk) 21:04, 13 December 2017 (UTC)

No, before A speaks, B doesn't "know that it is July or August". He knows that it is 16, which means it is May or July.

After A speaks, the readers know that it is July or August, because otherwise, A could not have been sure that B doesn't know the answer. For example, let's assume it's May. So for all A knows, it could be May 19. In this case, B, knowing it's 19, would deduce that it is May, because that is the only possibility for 19. So it can't be May. It can't be June either (because then it could be June 18, in which case B would know the answer off the bat). This leaves July and August.

B, of course, having already known that it's May or July, can now be confident it is July. -- Meni Rosenfeld (talk) 22:14, 13 December 2017 (UTC)

December 13

Infinite Product

Where can I find a worked example of how to convert an infinite series into an infinite product, presumably through Weierstrass factorisation? All I can find are convergence properties of these products, and proofs, but no actual demonstrations of application. Plasmic Physics (talk) 09:12, 13 December 2017 (UTC)

How about $\sum _{i=0}^{\infty }a_{i}=\ln(\prod _{i=0}^{\infty }e^{a_{i}})$?--Jasper Deng (talk) 11:29, 13 December 2017 (UTC)

There are also several given at Euler product, where the resulting product is over the primes. –Deacon Vorbis (carbon • videos) 15:43, 13 December 2017 (UTC)

The Fundamental_theorem_of_algebra basicly says that $\sum _{i=0}^{N}a_{i}x^{i}=a_{N}\prod _{r\in C}(x-r)^{m_{r}}$. If $r$ is a root then $m_{r}$ is its multiplicity, otherwise $m_{r}=0$. Bo Jacoby (talk) 16:04, 13 December 2017 (UTC).

This does not extend to infinite series and products. The Taylor series of the exponential function has no roots. and so one cannot speak of defining that function as an infinite product of monomials.--Jasper Deng (talk) 18:22, 13 December 2017 (UTC)

It does extend to some infinite series and products, but not to all infinite series and products. The (taylor series of the) sine function has roots, and one can indeed speak of defining that function as an infinite product of monomials.

$\sin(z)=z\prod _{k=1}^{\infty }(1-({z \over k\pi })^{2})$

([8]). Bo Jacoby (talk) 20:01, 13 December 2017 (UTC).

As in the OP's question, Weierstrass factorization is the relevant generalization of the fundamental theorem to all entire functions (including ones defined by series of course), using "elementary factors" instead of monomials.John Z (talk) 02:41, 14 December 2017 (UTC)

My function is: $\ z=\sum _{j=0}^{\infty }(2j+1)e^{{\epsilon _{0} \over \tau }(j^{2}+j-{1 \over 2})}$. Plasmic Physics (talk) 06:33, 14 December 2017 (UTC)

What is your independent variable, $\epsilon _{0}$ or $\tau$ or ${\epsilon _{0} \over \tau }$? Why not call the independent variable $\ z$ and write $\ f(z)=\sum _{j=0}^{\infty }(2j+1)e^{z(j^{2}+j-{1 \over 2})}$ ? Bo Jacoby (talk) 10:39, 14 December 2017 (UTC).

I beg your pardon, I thought that $\tau$ was implied. Plasmic Physics (talk) 10:48, 14 December 2017 (UTC)

I assume ε₀ is a positive real number, say the permittivity of free space? In that case your series diverges whenever the real part of τ is positive. Are you hoping to find an analytic continuation of the part where it converges, similar to what is done with the Riemann zeta function? --Trovatore (talk) 11:08, 14 December 2017 (UTC)

You need to find the complex zeroes of $\ f({\epsilon _{0} \over \tau })$. You need to use analytical continuation as the sum is divergent for ${\epsilon _{0} \over \tau }\geq 0$. Bo Jacoby (talk) 11:04, 14 December 2017 (UTC).

You can rewrite you expression as

$\ z=e^{-{\frac {3\epsilon _{0}}{4\tau }}}\sum _{n=1,3,..}^{\infty }n\exp \left({{\frac {\epsilon _{0}}{4\tau }}n^{2}}\right)$

and then try to find its sum using a standard method when $\epsilon _{0}<0$. Ruslik_Zero 20:59, 14 December 2017 (UTC)

Apologies - I made an mistake. The actual function is: $z=\sum _{j=0}^{\infty }(2j+1)e^{-{\epsilon _{0} \over \tau }(j^{2}+j)}$, which rearranges to $z=2e^{\epsilon _{0} \over 4\tau }\sum _{j=0}^{\infty }(j+{1 \over 2})e^{-{\epsilon _{0} \over \tau }(j+{1 \over 2})^{2}}$. It should converge for $\tau >0$, $\epsilon _{0}$ is a positive constant. Plasmic Physics (talk) 05:57, 15 December 2017 (UTC)

Let $w=e^{-{\epsilon _{0} \over \tau }}$ to get the simpler sum $\sum _{j=0}^{\infty }(j+{1 \over 2})w^{(j+{1 \over 2})^{2}}$. The exponential function has disappeared from the problem. Bo Jacoby (talk) 09:35, 15 December 2017 (UTC).

Ok, so what then? Plasmic Physics (talk) 11:21, 15 December 2017 (UTC)

Then solve the equation $\sum _{j=0}^{\infty }(j+{1 \over 2})x^{(j+{1 \over 2})^{2}}=0$. Sorry, I don't know how. Bo Jacoby (talk) 12:11, 15 December 2017 (UTC).

Do you mean find the general solution, or just evaluate the zeroes? $\lim _{k\to 0^{+}}z=0$.

The equation is $f(x)=0$ where $f(x)=\sum _{j=0}^{\infty }(2j+1)x^{j^{2}+j}$. The truncated equation $1+3x^{2}+5x^{6}+7x^{12}=0$ has the 12 complex solutions (x₁, . . . , x₁₂) = (–0.54122i, 0.54122i, –0.5983–0.616984i, –0.5983+0.616984i, 0.5983–0.616984i, 0.5983+0.616984i, –0.898509–0.371691i, –0.898509+0.371691i, 0.898509–0.371691i, 0.898509+0.371691i, –i, i) and the product representation is $f(x)\approx 7\prod _{j=1}^{12}(x-x_{j})$. Bo Jacoby (talk) 17:41, 15 December 2017 (UTC).

In general, one cannot approximate the roots of an analytic function using roots of its Taylor polynomials, so this will not suffice for an exact infinite product representation.--Jasper Deng (talk) 18:38, 15 December 2017 (UTC)

May be not in general, but in this case it works fine! Bo Jacoby (talk) 23:03, 15 December 2017 (UTC).

All you did was demonstrate that the fundamental theorem of algebra holds for Taylor polynomials (like any polynomials). Whether the sequence of roots converges to that of the function remains to be shown.--Jasper Deng (talk) 00:26, 16 December 2017 (UTC)

Criticism without improvement is not helpful. Show it yourself! (Note that small roots, such as |x₁|<0.6, are unaffected by terms of degree ≥20) Bo Jacoby (talk) 08:22, 16 December 2017 (UTC).

This problem seemed familiar to me. I eventually remembered that it is rotational partition function. There is certainly rich literature discussing its properties like this. Ruslik_Zero 17:57, 15 December 2017 (UTC)

I think that the question is meaningless as the function $z(\epsilon _{0}/kT)$ is not an entire function and can not be factorized: it has infinite number of poles located at $\epsilon _{0}/kT=2\pi im,\;m\in Z$. Ruslik_Zero 08:35, 16 December 2017 (UTC)

I'm not quite sure what a pole is, but from the introduction in our article on the matter, a pole exists where the limit of a function tends to infinity as the variable tends to some constant. If this is true of all poles, then this function has at most only one pole for tau approaching zero from the left. Plasmic Physics (talk) 10:39, 16 December 2017 (UTC)

See pole (complex analysis). Weierstrass factorization won't work in the presence of them, even if you only intend real values for the variables here.--Jasper Deng (talk) 10:47, 16 December 2017 (UTC)

Yes, that's the article. What if I use absolute tau instead, then I'll have no poles? Plasmic Physics (talk) 10:58, 16 December 2017 (UTC)

I'm not sure what you mean, but an entire function must have a globally convergent Taylor series. In this case, we can then rule out the function being entire because the Taylor series clearly does not have an infinite radius of convergence.--Jasper Deng (talk) 11:15, 16 December 2017 (UTC)

Ok, I see. Well, if I substitute τ with |τ| in the function, then, the series should be absolutely convergent. Plasmic Physics (talk) 11:43, 16 December 2017 (UTC)

Then it fails to be differentiable as a function of |τ| and hence can't be entire. This is not to say you can't find an infinite product representation - just not one (directly) using the Weierstrass factorization theorem.--Jasper Deng (talk) 20:19, 16 December 2017 (UTC)

December 14

Notation for applying function to series

After seeing the previous entry I wondered if there was a generalised notation similar to the sigma and pi notation for sum and product for applying a general function to each element in a sequence in turn? Would this differ if the operation /function is not commutative (for example raising each element to the power of the next)? I know that this could be done with lambda expressions if the sequence is written in full, but this is not like the simplicity of the sigma and pi notations. (Caveat, my maths is not much past high-school level except from some computer-related topics, apologies if this is a silly question). -- Q Chris (talk) 09:51, 14 December 2017 (UTC)

The standard mathematical notation does not generalize the sigma and pi notations. The J programming language uses + for addition and * for multiplication and +/ for sigma and */ for pi and f/ for the generalization to any function f.

  2+3+4
9
  +/ 2 3 4
9
  2*3*4
24
  */ 2 3 4
24
  2-(3-4)
3
  -/ 2 3 4
3

Bo Jacoby (talk) 10:46, 14 December 2017 (UTC).

It's interesting that it treats operations as right-associative, I would have assumend

(2-3)-4
-5

-- Q Chris (talk) 11:35, 14 December 2017 (UTC)

J inherited its right-associativity from APL. -- ToE 17:39, 14 December 2017 (UTC)

The usual thing for repeated operations in math, sum and product being exceptions, is to use a big version of whatever the operator is. For example $\bigcap _{\alpha }S_{\alpha }$ for intersections, $\bigcup _{\alpha }S_{\alpha }$ for unions. Without some sort of convention on the order of operations, the resulting expressions are ambiguous unless you assume the operator is associative, and the operands would have to be ordered in some way unless the operation is commutative. I don't think there is standard notation for repeated exponentiation unless you count tetration. --RDBury (talk) 13:10, 14 December 2017 (UTC)

$\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n}}$ can have any value you like if you reorder the sum.. Dmcq (talk) 22:57, 15 December 2017 (UTC)

Drawing ellipse, from angled cylinder

I need to cut an elliptical hole into a (very thin) flat piece of metal, so that a cylinder (a pipe) tilted at some angle (relative to the flat piece of metal) will insert into it perfectly. How do I calculate, and more importantly, draw, that ellipse? Ariel. (talk) 15:26, 14 December 2017 (UTC)

It is clear that the minor axis of the ellipse will have the same length as the diameter of the cylinder. The major axis length is found to be $D\sec(\theta )$ where D is the cylinder diameter, $\theta$ is the angle your plane makes from the base of your cylinder (equivalently, 90 degrees minus the angle with the side of the cylinder), and $\sec()$ denotes the secant function. Geometric constructions of an ellipse can be found at the article on the ellipse.--Jasper Deng (talk) 18:28, 14 December 2017 (UTC)

Thank you! Ariel. (talk) 20:38, 14 December 2017 (UTC)

... or, if you have a spare piece of pipe of the correct size, then you could cut the spare pipe at the appropriate angle and then use the cut end of the pipe as template to mark the required ellipse on the metal. Gandalf61 (talk) 10:48, 15 December 2017 (UTC)

Don't you dare bring simple, practical solutions to a nice math problem! --Stephan Schulz (talk) 12:50, 15 December 2017 (UTC)

I was assuming that option was out of the question if he wanted the pipe to pass through uncut.--Jasper Deng (talk) 19:34, 15 December 2017 (UTC)

This method from a couple of bow thruster installation articles shows another simple, practical, non-mathematical solution which extends to a cylinder intersecting a possibly non-planar surface. For the illustrated application, the rod passes through a pilot hole in either side of the hull, but it could be adapted for a single surface via a jig which maintains the rod's desired angle to the surface. -- ToE 02:20, 16 December 2017 (UTC)

December 15

colorize equations for better comprehension

I am new, so if this is not the right place, please let me know. I think colorizing equations would enhance users' comprehension. Here are some examples --Backinstadiums (talk) 17:56, 15 December 2017 (UTC)

I think WT:WPM might be a better place to bring this up, but since we're here, it's hard to see this catching on. You'd run into all sorts of accessibility issues with highly colored text (see WP:COLOR for things to keep in mind); and such text could hide links that we'd want to keep. And personally, color me skeptical (sorry) that something like that is really all that useful. –Deacon Vorbis (carbon • videos) 18:45, 15 December 2017 (UTC)

A similar argument is common in computer programming. Some people are adamant that code editors must have context-highlighting. Others don't care and often don't use any form of context highlighting. The argument for it tends to be that it makes programming easier. The argument against it is that the color choices are arbitrary and don't lend any real information to the content. Does green mean a variable or a function? Then, someone mentions vi or emacs and the entire discussion turns into a nuclear war. 71.85.51.150 (talk) 00:59, 16 December 2017 (UTC)

December 16

4 sided die procedure.

I want to roll a 4-sided die, but only have a six sided one. Is the following procedure "fair": Roll die, if 1,2,3,4 use that value. If 5,6 remember the value and re-roll. If 1,2,3,4 use that value. If 5,6 remember the value. If the remembered values are 55 use 1, 56 use 2, 65 use 3, 66 use 4. ? -- SGBailey (talk) 18:29, 16 December 2017 (UTC)

For any a in 1,2,3,4, the probability of getting a using your procedure is 1/6 + 2/6*1/6 + 1/6*1/6 = 9/36 = 1/4 (assuming the d6 is fair). So yes, fair.--108.52.27.203 (talk) 18:38, 16 December 2017 (UTC)

You can also see this by the fact that each of 1,2,3,4 is treated equivalently, and that after two throws one of the mentioned criteria will be met, so they must have equal probability, i.e., 1/4. Gap9551 (talk) 02:46, 17 December 2017 (UTC)

Why remember {5,6}, just ignore them. Simple enough! Ockham's Razor: "Entities are not to be multiplied without necessity" or do not make things (more) complicated for no reason. 110.22.20.252 (talk) 02:15, 17 December 2017 (UTC)

Maybe they only want to roll twice. Gap9551 (talk) 02:44, 17 December 2017 (UTC)

What is the expected ratio of long "stick" to short "stick"?

A unit rod is broken into two at a uniformly random position. What is the expected ratio of the long length to the short length?

My calculations is to imagine a random variable x such that x is from 0.5 to 1

Calculate Integral[x/(1-x),0.5,1] but my calculator kept on giving me "MATH ERROR".

Please help. 110.22.20.252 (talk) 02:21, 17 December 2017 (UTC)

You need to use an open interval, else you have a problem at x=1. Bubba73 ^{You talkin' to me?} 02:43, 17 December 2017 (UTC)

Your integral is basically correct, but you cannot simply use the 1 as upper limit, instead you need a limit to 1. Then, the result will approach infinity as the upper boundary of your interval approaches 1 (from below). Gap9551 (talk) 02:56, 17 December 2017 (UTC)

Which would mean that the expected value is infinite. It would still be infinite for the expected value of ratio of the left and right ends. On the other hand if you wanted the expected ratio of the short to long lengths then take the reciprocal of the integrand and you'd get an answer. --RDBury (talk) 03:01, 17 December 2017 (UTC)