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August 31

volume

What is the volume (ounces or mls) in a fifth of whiskey? — Preceding unsigned comment added by 71.182.193.144 (talk) 00:33, 31 August 2012 (UTC)

It depends on the mixture, the temperature, the altitude of the sample and your position on the Earth. Moreover, it depends on the orbit of the Earth around the Sun, and on the Sun's motion through our galaxy. All of these things affect the volume. — Fly by Night (talk) 00:57, 31 August 2012 (UTC)

How density varies would only matter if a fifth was a unit mass. It is not, it's a unit of volume. See below. Therefore, if the number of fifths goes up for a given mass, so does the number of ounces and milliliters, in proportion, such that the number of ounces or milliliters in a fifth remains constant. (Note that I didn't engage in a personal attack just because you gave a poor answer.) StuRat (talk) 02:20, 31 August 2012 (UTC)

Not a personal attack, but a plain statement of fact - the first response was inappropriate, being patronising, belittling and worst of all, plain wrong. If you can't give an accurate, polite and helpful answer then don't give any.←86.139.64.77 (talk) 20:46, 1 September 2012 (UTC)

See Fifth (unit). hydnjo (talk) 01:00, 31 August 2012 (UTC)

Resolved

Solving a summation

I don't have the math background to know where to start with this. I have a summation of the usual sort with n on the top and i=0 on the bottom... and then the actual summation formula itself is a constant with i as an exponent. The problem is, I don't want to know what the answer is at a given n, I want to know what n is when I reach a given constant. I want to solve for the n. How do I go about figuring this out? Broba (talk) 07:42, 31 August 2012 (UTC)

Here's a simpler example of what I'm trying to do to illustrate the basics of what I can't figure out: Summation with i=0, the formula inside the summation is 3 * i. I want to know at what value of n the summation equals 18. What's the method you use to arrive at the value of n? Is it possible to do the same if the summation equals say 17? Broba (talk) 07:47, 31 August 2012 (UTC)

Quite simple to do with a computer program, although your example is easy enough to do by hand:

i  3i   ∑
---------
0   0   0
1   3   3
2   6   9
3   9  18

So we don't get 17 for the summation, assuming n is an integer. I don't think non-integer solutions for n (like n = 8/3) make sense in a summation, since, if you add in i = 8/3, how can you justify not adding in i = 7/3, as well, along with the infinite number of other non-integers ? StuRat (talk) 07:58, 31 August 2012 (UTC)

Well, another way to do something like this would be to close form solution of the summation as a function of n (if possible) and then setting it equal to whatever number you want (17 in this case) and solving for n. Like StuRat said, there is no guarantee if n would be a natural integer or not. Are you looking only for natural integers? To illustrate with your example, let $f(n)=\sum _{i=0}^{n}3i={\frac {3n(n+1)}{2}}$. Then you want to solve for n where ${\frac {3n(n+1)}{2}}=17$ and in this case we have a simple quadratic and the answer is $n={\frac {-3+{\sqrt {417}}}{6}}=2.90343...$. There is another solution too which is negative which we ignore. As you can see this is close to n=3 which would give us the sum equals 18. Is this what you are kind of looking for? Someone here might be able to help you furthermore if you give us the sum and how/where this problem arises.68.121.32.26 (talk) 10:32, 31 August 2012 (UTC)

But are non-integer solutions actually allowed in a summation ? To me it doesn't make any sense. As I noted before, if you sum in one non-integer value (the final one), then you would have to sum in all non-integer values, wouldn't you ? This would typically mean the solution is either positive infinity or negative infinity, regardless of n, which isn't very useful. StuRat (talk) 20:22, 31 August 2012 (UTC)

For an exponential function, use $\sum _{i=0}^{n-1}a^{i}={\frac {1-a^{n}}{1-a}}$ Ssscienccce (talk) 15:27, 31 August 2012 (UTC)

Thank you both 68.121.32.26 and Ssscienccce. That is the approach I'm trying to take, although I think my problem has additional complexities. What is the name of this sort of math... I had a lot of trouble knowing what to search for for this.

The complications I have is that I tried to use the exponential function that Ssscienccce gave but I have some coefficients that I believe are screwing it up. Non integers are fine (expected probably). This is specifically the form of the problem I'm trying to solve (and yes I could estimate it with a program, I'm not interested in that... I'm looking for the method to arrive at the answer more than any specific answer):

$\sum _{i=0}^{n}100e^{0.05n}=5000$

I'm trying to solve for $n$ here of course. As an aside, I have a hunch there's some analogy to this kind of problem in physics but I haven't grasped exactly which yet. Broba (talk) 17:47, 31 August 2012 (UTC)

Oh I got it now. I just found another version of Ssscienccce's explanation with coefficients. Thanks for you help. Broba (talk) 21:56, 31 August 2012 (UTC)

Resolved

Didn't you mean $\sum _{i=0}^{n}100e^{0.05i}=5000$ ? Bo Jacoby (talk) 06:56, 1 September 2012 (UTC).

Late to the party but whatever... If you can still use some advice, I try to walk you step by step to the solution. Some things, like the 100 make your equation look far more formidable than it really is. I'll compute the "a" for you but only outline the later steps, it's quite straightforward once you see what to compute, and last but not least, I'm doing the stuff mentally atm.

Let's look at your sum of 100exp(0.05i) = 5000. You can cancel out a factor of 100 to get "Sum of exp(0.05i) = 50", and now assume that a = exp(0.05) = 1.051..., so this is a fancy way to write "Sum of a^i = 50".

Ssscienccce's formula gives you a closed form for that kind of summation. Multiply by (1-a) to get something like "1 - foo^bar = 50 (1-a)", add foo^bar, and subtract the constants on the right-hand side to get something like "constant = foo^bar,"

something you can attack with a log operation to get "bar = _foolog(constant)". Sorry for the sloppy terminology.

- ¡Ouch! (^{hurt me} / _{more pain}) 07:23, 3 September 2012 (UTC)

compare all to all

I have a set S and I want to compare every item in S to every other item in S. The comparison is a simple function like f(S1,S2). I claim that it is impossible to do this without S*(S-1) comparisons. I've been told it can be done with S*log(S) comparisons, but not how it can be done that way. Can anyone explain to me how that is possible? — Preceding unsigned comment added by 128.23.113.249 (talk) 14:29, 31 August 2012 (UTC)

Use any of the worst-case $n\log n$ comparison sorts from the table in sorting algorithm#Comparison of algorithms, such as merge sort. Once you sort the set, you can compare any two items by comparing their index in the sorted set without invoking f. This all assumes that f is a bona fide total order.—Emil J. 14:41, 31 August 2012 (UTC)

Please correct me if I'm wrong, but that implies that S is sortable. I'm trying to come up with an example that shows how they are not sortable based on f. What if you have a set of people S. Then f(S1,S2) is a rating of how much S1 likes S2. We can make it simple and assume f(S1,S2)=f(S2,S1) if that is important. Then, you have a weighted graph where each node is an element in S and the edges have a weight, the value of f for the two nodes. If I use f to sort S with, say, merge sort, I won't compare every element in S to every other element and, therefore, will be missing edges in my graph. Does that make sense or am I just rambling nonsense? — Preceding unsigned comment added by 128.23.113.249 (talk) 14:59, 31 August 2012 (UTC)

Your f is not a comparison function (i.e., an indicator function of a total (pre)order). If you allow a completely arbitrary function as f, then there is of course no way of finding all its values for all pairs of elements other than checking all the |S|² pairs.—Emil J. 15:37, 31 August 2012 (UTC)

Thank you. I believe our class argument is actually an argument about how f is defined, not about the set S. It all depends on if f defines total order or just some arbitrary relationship between two items. — Preceding unsigned comment added by 128.23.113.249 (talk) 15:42, 31 August 2012 (UTC)

You may want to check out Closest pair of points problem Ssscienccce (talk) 16:19, 31 August 2012 (UTC)

solutions to an assignment

Hello I am an independent learner looking through material here: http://www.hutter1.net/ethz/uaiethz.htm and in particular trying the assignment here: http://www.hutter1.net/ethz/assign1.pdf but sadly there are no solutions to this assignment on that website. Some questions are easy math questions but some are hard. Could someone kind create solutions or refer to elsewhere? — Preceding unsigned comment added by Bulkc (talk • contribs) 18:05, 31 August 2012 (UTC)

That's a bit much to ask of us, I'm afraid. We also won't do your homework for you. However, if you do it, and post your answers here, we might be willing to check it for you and point out any mistakes. StuRat (talk) 20:28, 31 August 2012 (UTC)

Ok. The first two questions and the probability questions are easy any way. I think I can do too the question MDL-ML, part (i) of MDL-Ber and some of KC-KC myself. What about the rest? For example can you show me how to do KC-KC part (iv) which does not follow from the rest I think. Or inequality K(xy) < K(x,y) — Preceding unsigned comment added by Bulkc (talk • contribs) 20:51, 31 August 2012 (UTC)

For (iv), let $T$ be the machine that, on input $a$, first runs $U(a)$. Then, if the output is a pair $(x,f)$, runs $f$ on $x$ and outputs $f(x)$. Now, if $U(a)=(x,f)$, then $T(a)=f(x)$. This shows that $K_{T}(f(x))\leq K(x,f)$, which is less than $K(x)+K(f)$ by part (iii). Since $K\leq ^{+}K_{T}$, the result follows.

For $K(xy)\leq ^{+}K(x,y)$, let $T$ be the machine that, on input $a$, first runs $U(a)$. Then, if the output is a pair $(x,y)$, $T$ outputs $xy$. Then follow the same reasoning as above.--121.73.35.181 (talk) 00:08, 1 September 2012 (UTC)

Thank you! I can do now these questions, and some other ones, due to your advice about Turing machines. But, I am still having some trouble. What about AP-CM (ii) and (iii)? Or MDL-Ber (iii), (v) and (vi)?--Bulkc (talk) 06:14, 1 September 2012 (UTC)

Actually, I just did MDL-Ber (v) using (iv)! — Preceding unsigned comment added by Bulkc (talk • contribs) 06:32, 1 September 2012 (UTC)

Woops, had one of my inequalities backwards. Fixed.--121.73.35.181 (talk) 10:28, 1 September 2012 (UTC)

• After much work and reading through the material, I think I can solve all questions except for AP-CM. I think this is hardest. — Preceding unsigned comment added by Bulkc (talk • contribs) 02:58, 2 September 2012 (UTC)

Here's (i): Fix $\epsilon >0$. Let $A_{t}$ be the set of infinite sequences $x$ with $M(0|x_{<t})-\mu (0|x_{<t})>\epsilon$. Note that $\sum _{t}\mu (A_{t})\epsilon ^{2}$ is less than the sum in theorem 4.5, so in particular is finite. So $\sum _{t}\mu (A_{t})$ is finite, and thus $\limsup _{t}A_{t}$ has $\mu$-measure 0. This contains all sequences where the limit is infinitely often above $\epsilon$. Union over all rational $\epsilon$ to get the set of all sequences where the limit is non-zero, and it still has $\mu$-measure 0.--121.73.35.181 (talk) 22:40, 2 September 2012 (UTC)

Thank you very much for your help. Let me just ask for solutions to the ones that I'm still not wholly sure about:

KC-Cmp (i)

AP-RC, especially inequality $K(x|l(x)){\stackrel {+}{<}}KM(x)$

AP-CM (ii), (iii), and maybe (iv) but I think I can figure to do that one.

I think I would like solutions to all of KC-KC too because I find Kolmogorov complexity confusing still. --Bulkc (talk) 22:56, 4 September 2012 (UTC)

September 1

Konig's lemma

In the proof of Konig's lemma it is mentioned that:

König's lemma may be considered to be a choice principle; the first proof above illustrates the relationship between the lemma and the axiom of dependent choice. At each step of the induction, a vertex with a particular property must be selected. Although it is proved that at least one appropriate vertex exists, if there is more than one suitable vertex there may be no canonical choice.

In the proof mentioned, it is clear that there are only finitely many such vertices to consider at any given stage.(Cf: "Every one of the infinitely many vertices of G can be reached from v1 with a simple path, and each such path must start with one of the finitely many vertices adjacent to v1."..."We may thus pick one of these vertices and call it v2." That is, the choice is being made among finitely many vertices. Why then do we need the axiom of choice or some weaker version of it to prove the lemma? Thanks--Shahab (talk) 13:07, 1 September 2012 (UTC)

I am not at all well up on this, but I believe it is because we are making infinitely many such choices. I do know that (because there is no uniform way of telling a left sock from a right sock, unlike for shoes) you need choice to get one sock from each of infinitely many pairs (apart, of course, from the fact that they are in a nice Euclidean space...) Straightontillmorning (talk) 13:39, 1 September 2012 (UTC)

Yep. You need choice to make infinitely many choices; it doesn't matter whether each of those choices is from finitely or infinitely many possibilities.--121.73.35.181 (talk) 21:02, 1 September 2012 (UTC)

This is a complete guess: if there's only one suitable vertex, you can identify it by trying out paths of increasing length for every vertex. If only one is suitable, this search will end eventually. If more then one is suitable, it won't. Ssscienccce (talk) 14:59, 1 September 2012 (UTC)

September 2

Graph theory

How many trees are there with n vertices such that each vertex has degree at most 4, if ones that are the same other than the naming of the vertices are excluded? --168.7.237.248 (talk) 04:08, 2 September 2012 (UTC)

By working out the first few terms by hand and searching the OEIS, I found A000602. -- BenRG (talk) 06:06, 2 September 2012 (UTC)

What's the closed form for that sequence? --128.42.223.234 (talk) 16:03, 2 September 2012 (UTC)

I don't know. The OEIS entry links to some unreadable Maple code. -- BenRG (talk) 21:00, 3 September 2012 (UTC)

The answer would depend on the meaning of a tree—whether you consider a tree as defined in a graph theory (i.e. undirected connected graph with no cycles) or as defined in computer science (a connected directed graph with a specific root and a transitive is-a-child relation, in which every node is a child of exectly one parent, except a root which has no parent). In the latter case also decide whether to consider (or not) specific ordering of children (is a root with a single left child equivalent to the root with a single right child etc.).
Anyway I do not know the answer in any case. --CiaPan (talk) 05:35, 5 September 2012 (UTC)

PS. Is this problem related to classifying carbon chain structures of organic compounds in organic chemistry? (e.g. IUPAC nomenclature of organic chemistry) CiaPan (talk)

Maths and music

Hi. First of all, I want to acknowledge that the following is homework. However, I've come here because I actually don't understand the problems:

1. I'm asked to find the continued fraction for $\log _{2}(3/2)$, calculate convergents ${\frac {p_{n}}{q_{n}}}$, and then find the first $q_{n}>12$. That's all fine: $q_{n}=29$. But then "Compute the relative frequency of the fifth harmonic in equal temperament in this scale with $q_{n}$ semitones and compare to the error to 3/2 to that of our usual scale with 12 semitones." I don't understand any of this. I've tried reading a few Wikipedia articles, but it still makes no sense to me; I have no musical background.
2. And I have no idea where to begin with this one: "When building a guitar it is useful to be able to determine where to put fret i in relation to fret i + 1. Find the numerical value of the ratio of the length of the string remaining when pressing down at fret i to the distance of fret i to fret i + 1, assuming the usual 12 semitone scale and equal temperament."

Can anyone give some advice on these questions? As I said, I've tried reading a few articles, but the musical terminology is lost on me. I'd appreciate it if someone could distill these problems into a more pure mathematical form, or – just as helpfully – explain in very simple terms the music theory I need to know to comprehend them. PS: I have quoted these questions verbatim. Feeling like they're grammatically incorrect? You and me both. —Anonymous Dissident^Talk 13:36, 2 September 2012 (UTC)

What does "the first $q_{n}>12$" mean? Please, write down something like log₂ 3/2 =(some fraction with p and q). Incnis Mrsi (talk) 14:50, 2 September 2012 (UTC)

Surely that means "the value of $q_{n}$ for the least n such that this is more than 12" (I don't know continued fractions.) However, I do know what 12-semitone equal temperament is so I'll try to be helpful. Musical intervals, in terms of frequencies, are ratios. An octave corresponds to doubling the frequency. In an octave under the standard system there are 12 semitones (A-A#-B-C-C#-D-D#-E-F-F#-G-G#-A), and equal temperament divides them equally (in a logarithmic sense), so that going up a semitone is multiplies the frequency by the twelfth root of two. Presumably you are considering a twenty-nine semitone equally tempered scale so going up a semitone is multiplying by the twenty-ninth root of two. Then for the fifth harmonic, I think we're looking at perfect fifths. I can't make out harmonic series (music), but the former says that the "just" interval for a perfect fifth is 3/2 (ie the frequency goes up by that factor). In equal temperament, it's a bit wrong (because the twelfth root of two is ugly) - I don't see where the fifth harmonic comes in. For the second question, Guitar#Frets says that changing by one fret changes by a semitone. By virtue of the assumption of twelve-tone equal temperament, that means the frequency changes by (a factor of) the twelfth root of two. If you (like I was) are having trouble with parsing the sentence we want the ratio of the (remaining string when you press at the i'th fret) [i.e. that bit that vibrates to give you that note] to the (distance between the i'th and (i+1)'st). So you need to assume frequency f for the ith fret, work out how long the string needs to be, work out how long it needs to be for the next note, and hope f cancels.

I hope that helps slightly, but it probably won't. Straightontillmorning (talk) 16:09, 2 September 2012 (UTC)

I think when they say "fifth harmonic" they must be really mean the perfect fifth, since they ask to compare the ratio to 3/2. In that case all the question is asking is to compare 2^7/12 to 3/2 and then compare 2^17/29 to 3/2. Rckrone (talk) 23:58, 2 September 2012 (UTC)

"fifth harmonic" may really mean fifth harmonic, or rather the major third; the history of musical temperament can be seen as a series of attempts to express 4:5 with factors of 2 and 3. So here's my paraphrase of the first question: Using the method of continued fractions, find the next rational approximation to lg(3/2) after 7/12, and discuss the quality of major thirds in the equally-tempered scale implied by that ratio. (I don't get 17/29; I get 24/41 and then 31/53 – but never mind, maybe you used a slightly different form of the CF, and you ought to get 41 next!) So: what is 2^9/29 or 2^13/41? —Tamfang (talk) 04:35, 3 September 2012 (UTC)

Reading Perfect fifth#The pitch ratio of a fifth, I would interpret "compare to the error to 3/2 to that of our usual scale with 12 semitones" as: for 12 semitones you get 1200*ln(1.5)/ln(2)=701.995 or 2% of a semitone wider than semitone 7; for 29 semitones it's 2900*ln(1.5)/ln(2)= 1696.39... 3.6% of a semitone different from semitone 17. But I've got no idea what the "relative freq of the 5th harm" would mean; relative to what, a semitone? Ssscienccce (talk) 05:34, 3 September 2012 (UTC)

Relative to the tonic. —Tamfang (talk) 23:15, 3 September 2012 (UTC)

In guitar tuning, there are subtleties in frettage that I don't understand, and you're probably not expected to understand either. The simple version is that the frets are placed so that the lengths of a given string from the bridge to each fret form a geometric sequence, whose ratio is (usually) the twelfth root of two. Note that you're not asked for the length of the string, though. —Tamfang (talk) 04:45, 3 September 2012 (UTC)

Scale (string instruments) has the details; lets call a=$\scriptstyle {\sqrt[{12}]{2}}$ for short; P(i+1)=P(i)/a , you need Pi/(Pi-P(i+1)) = Pi/(Pi-Pi/a) = ... = a/(a-1) = 17.817 Ssscienccce (talk) 06:30, 3 September 2012 (UTC)

Are you sure about 29? I get: [0;1,1,2,2,3,1,5,2,23,2,2,1,1,55,..], calculate it till 23: 1/(1+1/(1+1/(2+1/(2+1/(3+1/(1+1/(5+1/(2+1/23)))))))) = 0.5849625024, and 2^0.584... gives me 1.500000002 Dooh! I should learn to read.. Ssscienccce (talk) 06:58, 3 September 2012 (UTC)

I also get q_n=41 rather than 29. How did you get 29? I also know that Harry Partch composed for a 41-note scale. My first thought (like some others) about "fifth harmonic" was that it really should have said perfect fifth, but Tamfang's explanation about the major third also makes sense. The equal-tempered major third in the 12-note scale is off by about 0.8% and you can similarly calculate the error in the 41-note scale. I'll skip the details since I think figuring them out yourself is part of the exercise. 67.119.15.30 (talk) 15:05, 7 September 2012 (UTC)

Woops, after I noticed my mistake (thinking it was about a_n > 12, instead of the resulting denominator), I didn't bother checking any further, but you're right, [0;1,1,2,2,3] becomes 24/41. Ssscienccce (talk) 17:01, 7 September 2012 (UTC)

Binomial process

I have a closed population which is subjected to a hazard which removes them from the population. I am modelling the number removed each month using a binomial distribution with a constant percentage hazard rate. What is the distribution of the cumulative number of removals after the first T months? I don't necessarily need an exact distribution - the population is large enough that a normal approximation should be sufficient for my purposes. Is it as simple as a mean of n*(1-(1-p)^T) and a variance of n*(1-(1-p)^T)*(1-p)^T? I know it is that simple for a Poisson process, but I suspect it doesn't work the same way for a Binomial. Thanks for your help. --Tango (talk) 17:25, 2 September 2012 (UTC)

I don't think it's a question of Poisson versus binomial. My intuition is that with a constant percentage hazard rate, the strong law of large numbers should give you a log-normal distribution in the limit of large T. Looie496 (talk) 17:46, 2 September 2012 (UTC)

Thanks, but I should have said, T is quite small. For a Poisson, the absolute rate is constant, so you can just scale it. In this case, the rate is reducing because n is reducing, so I doubt it is so simple. --Tango (talk) 18:17, 2 September 2012 (UTC)

Any single individual will have a probability of survival equal to (1-(1-p)^T). And this will be independent from all other individuals. So the whole T months period for all individuals will be n independent trials with sucess chance (1-(1-p)^T). This means that the number of survivors will be Binomially distributed with mean and variance as given in original post. Taemyr (talk) 21:24, 2 September 2012 (UTC)

Of course it is - I was over complicating the whole thing. Thanks! --Tango (talk) 21:36, 2 September 2012 (UTC)

Resolved

Stone Cetch Compactification using ultrafilters

Hello, In this topology, I saw in a proof that: $x\in {\overline {\cup A}}$ and $x\in {\overline {\cup B}}$ implies that $(\cup A)\cap (\cup B)\neq \emptyset$. Can somone explain why this is true? since, this claim is not true in topology in general... Thanx. — Preceding unsigned comment added by General Topology (talk • contribs) 18:32, 2 September 2012 (UTC) General Topology (talk) 18:43, 2 September 2012 (UTC)

If I understand from your mention of ultrafilters and such, you are talking about the Stone–Čech compactification of a discrete space X, is that right? In that case, the compactificaiton of X is equivalent to the Stone Space of the set of all subsets of X. For x a subset of X let b(x) be all ultrafilters of X containing x, then {b(x) : x} is a basis. It's obvious, then, that for every open set U there is a collection Y of subsets of X so that U is the collection of ultrafilters meeting Y. [I'm putting TA, TB, etc. for the union of all sets in A, B, etc.] So, in your case, given collections of ultrafilters A and B, it follows cl(A) is exactly the ultrafilters made from sets in TA, similarly for cl(B). Thus, Tcl(A) = TA and the result follows. Two quick notes: since you didn't say what A was exactly, I'm guessing that A is a set of uf's and that you meant to write union(closure A) not closure(union A), I'm sorry if I'm wrong; I haven't thought about this in a while and am just working this out in my head, so I apologize if there is any stupidity in my answer. :-) Phoenixia1177 (talk) 22:43, 2 September 2012 (UTC)

Hi:) yes, I ment the Stone–Čech compactification of a discrete space X. I have a problem with some convergence properties of this space. Let me try and ask a slightly different question. From what I understand, this space is not Frechet Urysohn (a topological space is frechet urysohn if for every $x\in {\overline {A}}\setminus A$ there is a sequence of points in A that converges to x). Can anyone tell why?? General Topology (talk) 11:31, 4 September 2012 (UTC)

I'm working this out on a pad of sketch paper at work, so I apologize for any stupidity. That said, via our above basis, if a(n) is a sequence of ultrafilters, then a(n) converges to a iff for every b in a there is an n(b) so N >= n(b) implies b is in a(N)[assume n(b) is the smallest such indice in what follows]. Now, let X be infinite and a(n) be a nontrivial sequence converging to a, an ultrafilter. Well order a and define F(N) to be the least element larger than everything in {b : n(b) <= N}. Clearly, F(n(b)) >= b for any b, thus, F has to have an unbounded subsequence. However, by Easton's Theorem, the cofinality of a must be larger than the cardinality of X [a has the cardnality of X's power set.] In other words, since X is, at least, countable, we cannot have an unbounded sequnce in a, thus, there can be no nontrivial convergent sequences; which answers your question:-) Phoenixia1177 (talk) 06:28, 5 September 2012 (UTC)

A simpler proof would be to observe that the set of principal ultrafilters is dense, but has the same cardinality as X, where as the set of ultrafilters has cardinality the power set of the power set of X; hence there should be more ultrafilters than can be gotten than through convergence, in general.Phoenixia1177 (talk) 08:15, 5 September 2012 (UTC)

(This is my last response, I swear) If you're looking for something more concrete using free ultrafilters, let's take X to be the naturals. Put p(k) for the kth prime, A(k) for all numbers divisible by p(k), and B(k) for all numbers congruent 1 mod p(k). Then, for any collection of infinite sets with the finite intersection property there is a free ultrafilter contatining each of them, thus we have free ultrafilters U(n) corresponding to the collection {A(k) : k >= n} union {B(k) : k < n}. Clearly, there is an ultrafilter U for {A(n) : n} in {U(n) : n}'s closure since A(n) is in U(n) for all n, however, there is no sequence in {U(n): n} that converges to it. Since A(k) cannot be in any U(n > k) because A(k) is disjoint B(k), we cannot have A(k) in a tail of any sequence in {U(n) : n}, thus the result:-)Phoenixia1177 (talk) 09:46, 5 September 2012 (UTC)

September 5

Project Euler question

There is a Project Euler problem (whose number I am not quoting so the answer is not given away directly here), where one needs integer solutions to 2b^2 - 2b - n^2 + n = 0. b=3 and n=4 is one solution to this equation. The equations b' = 3b + 2n - 2 and n' = 4b +3n - 3 generate b' and n' which are also solutions to the equation. Two questions please: (1) How do I prove that the b' and n' equations are valid? (2) How does one do Diophantine magic and generate those equations in the first place? -- SGBailey (talk) 08:46, 5 September 2012 (UTC)

The equation is equivalent to 2b(b − 1) = n(n − 1). Now it becomes obvious b = n = 1 is another solution... --CiaPan (talk) 09:22, 5 September 2012 (UTC)

I know the equations work, I've got a dozen solutions and could have a dozen more if I wanted them (they get quite big...). But I'm interested in the proof and creation of the "next set of solutions" equations. Yes 1,1 is a solution. So is 0,1. But so is 3,4. -- SGBailey (talk) 09:36, 5 September 2012 (UTC)

I don't know where they come from, but it's simple enough to show that they work: just plug your formulas for b' and n' into the original equation and simplify. You'll get 2b^2 - 2b - n^2 + n = 0. Under the assumption that you started with a solution, this shows that b' and n' are a solution.--121.73.35.181 (talk) 10:25, 5 September 2012 (UTC)

(1) solved. You are right thanks - when I tried that previously, I went wrong; but it worked ok second time. So (2) where did the generating equations come from??? -- SGBailey (talk) 11:26, 5 September 2012 (UTC)

Convergents (p,q) to the continued fraction expansion of sqrt(2) are (1,1), (3,2), (7,5), (17,12), (41,29) etc (see Pell number). Alternate convergents (1,1), (7,5), (41,29) etc. satisfy

$p^{2}=2q^{2}-1$

Successive convergents in this sequence are related by the recurrence relations

$p'=4q+3p\quad q'=3q+2p$

Use the transform

$n={\frac {p+1}{2}}\quad b={\frac {q+1}{2}}$

and you get the (n,b) sequence (1,1), (4,3), (21, 15) etc. which satisfies

$(2n-1)^{2}=2(2b-1)^{2}-1\Rightarrow 4n^{2}-4n+1=8b^{2}-8b+1\Rightarrow n^{2}-n=2b^{2}-2b$

with the recurrence relations

$n'=4b+3n-3\quad b'=3b+2n-2$

Gandalf61 (talk) 12:08, 5 September 2012 (UTC)

Thank you (I think). I didn't understand most of that, but I'll study it for a few days and see if it makes any sense then. -- SGBailey (talk) 13:11, 5 September 2012 (UTC)

Complex variables inequality

Hi. I'm working in a book, and looking at a claim that goes as follows: For $|e^{z}-1|\leq {\frac {1}{2}}$, we have ${\frac {1}{2}}|z|\leq |e^{z}-1|\leq 2|z|$. I've checked this by doing a change of variables $\zeta =e^{z}-1$, parameterizing the circle $|\zeta |={\frac {1}{2}}$, and checking that ${\frac {1}{2}}|\log(\zeta +1)|\leq |\zeta |={\frac {1}{2}}\leq 2|\log(\zeta +1)|$ all around the circle (for the principal branch). I checked just by graphing the left and right as functions of a real variable, and sure enough, they stayed away from ${\frac {1}{2}}$. This is terribly awkward, though. Does anyone know an easier way to see this? Thanks in advance. -GTBacchus^(talk) 19:54, 5 September 2012 (UTC)

Unless I’m missing something, the first part of the inequality is false.

I will type * for times, and ^ for ”to the power of”.

Because, suppose that z = 2pi*i. Then e^z = e^(2pi*i) = cos(2pi) + i*sin(2pi) = 1 + 0 = 1. So then |e^z - 1| = 0, which is less than 1/2. So z = 2pi*i satisfies the assumption.

But then the first part of the inequality, (1/2)|z| <= |e^z -1|, becomes (1/2)(2pi) <= 0, or pi <= 0, which is just not true. Cardamon (talk) 07:00, 8 September 2012 (UTC)

Yeah, I noticed that too. The inequality applies for z close to 0, not close to other multiples of 2pi*i. It can be fixed by adding the condition that |z|<1 or something.

I seem to have it figured out. The trick is to do the substitution $\zeta =e^{z}-1$, and assume that $|\zeta |\leq {\frac {1}{2}}$; this obviates the problem you mention. Then you really want to show that ${\frac {1}{2}}|\zeta |\leq |\log(\zeta +1)|\leq 2|\zeta |$. This can be accomplished by playing with the series expansion of $\log(\zeta +1)$ and the triangle inequality. -GTBacchus^(talk) 01:59, 11 September 2012 (UTC)

September 6

Implementing an algorithm to generate Proth Primes

Hello all. I'm trying to implement an efficient Proth Prime generator using Proth's theorem, but there are a few concepts that I am a little unclear about. Primarily:

- How many values of 'a' (on a logarithmic scale, perhaps) must be used to make the primality test deterministic?

- How should a given 'a' be chosen? Start at, say, 2 and then simply increment, or what?

- Would it be more efficient to first calculate the Jacobi symbol and test that 'a' is a quadratic nonresidue, or simply select 'a' by some other means?

Apologies for the ignorance, but I'm a computer programmer, not a mathematician! Thanks.

66.87.80.180 (talk) 00:19, 6 September 2012 (UTC)

For the first question, the complexity of finding a such that (a|p) = −1 deterministically is a difficult open problem. Assuming the generalized Riemann hypothesis for quadratic Dirichlet characters, one can always find an a ≤ 2(ln p)² which either has this property or is a proper divisor of p. On the other hand, unconditionally known bounds are essentially exponential, IIRC they are of the form p^ε for some constant ε.

For the second question, the best way is to chose a randomly. If you want to do it deterministically, I’m not aware of any advantage of doing it any other way than testing small a successively as you write, except that in this case, it suffices to test only prime a.

I don’t quite understand the third question, what other means do you have in mind?—Emil J. 11:45, 6 September 2012 (UTC)

Okay, thanks. Taking all of that into consideration, it seems that I might as well just stick with the Miller-Rabin test and abandon Proth Primes altogether then, as it has a much lower known lower bound (and an even lower heuristic lower bound) besides being suitable for primes of any form anyway. Cheers!

66.87.126.188 (talk) 14:50, 6 September 2012 (UTC)

Polynomial equations

hi, i just wonder what could b so wrong in solving math polynomial eqs. by choosing 2 arbitrary values 4 x, x1 n x2 then compute xm=(x1+x2)/2, wrinting P(x)=P(x-xm)+P(xm) then P(x1)=..similary, P(x2)=similary then somthing like P(y-a)=P(y+a)=0 y=xm a=xm-x1

P 4 ur delight, anyway LOL

Florin, Romania — Preceding unsigned comment added by 93.118.212.93 (talk) 13:02, 6 September 2012 (UTC)

Because for a general polynomial P, P(x) does not equal P(x-xm)+P(xm); for instance, $(x-x_{m})^{2}+x_{m}^{2}=x^{2}-2xx_{m}+2x_{m}^{2}$, which is not x^2 unless $x=x_{m}$ which is unlikely since you have effectively taken xm arbitrary. Straightontillmorning (talk) 14:42, 6 September 2012 (UTC)

It seems like the OP might be groping for something like Newton's method. Looie496 (talk) 17:05, 6 September 2012 (UTC)

More on "Project Euler question" from Sept 5

Ok, with "Convergents (p,q) to the continued fraction expansion of sqrt(2)", it is sqrt(2) because the the equation has the form p^2 = 2*q^2 - 1. Had it been p^2 = 99*q^2 -1 it would have been sqrt(99) I presume.

It was said "Alternate convergents (1,1), (7,5), (41,29) etc. satisfy p^2 = 2*q^2 - 1". This is true by inspection, but how did you get alternates in the first place? Had I thought to do such a thing, I would have tried them all and given up when 9 = 3^2 != 2*2^2 - 1 = 7.

I now follow the mechanism that have been used here, but have no feel as to why/how it was decided theat these were good tools / transforms to use. What points in that direction? -- SGBailey (talk) 15:06, 6 September 2012 (UTC)

What points in that direction ? Mostly previous experience, having seen similar problems before and played around with continued fractions. Starting with

$2b^{2}-2b=n^{2}-n$

a natural first step is to complete the square on both sides, giving

$2\left(b-{\frac {1}{2}}\right)^{2}-{\frac {1}{2}}=\left(n-{\frac {1}{2}}\right)^{2}-{\frac {1}{4}}$

Multiplying both side by 4 to get rid of the fractions gives

$2(2b-1)^{2}-1=(2n-1)^{2}$

So if we set

$p=2n-1\quad q=2b-1$

then we want to find integers p and q such that

$p^{2}=2q^{2}-1$

$\Rightarrow {\frac {p}{q}}\approx {\sqrt {2}}$

which suggests we are looking for rational approximations to sqrt(2). The convergents to the continued fraction expansion of sqrt(2) are one such series of approximations, and I happened to know that they satisfy

$p^{2}=2q^{2}\pm 1$

with alternating signs. So it was then just a case of working out the recurrence relations between alternate convergents, and then translating this back into recurrence relations in terms of b and n instead of p and q. Gandalf61 (talk) 15:56, 6 September 2012 (UTC)

Many thanks. That made sense :-) -- SGBailey (talk) 18:54, 6 September 2012 (UTC)

surjective Homomorphism

Hi. Is it true that any two surjective homomorphisms $S_{4}\rightarrow S_{3}$ have the same kernel? By the correspondence theorem there exists a bijection between subgroups of $S_{4}$ containing the kernel of SOME homomorphism and the subgroups of $S_{3}$ but does this imply that this kernel is the same for any homomorphism?--150.203.114.14 (talk) 23:07, 6 September 2012 (UTC)

According to the symmetric group page, S₄ has only one normal subgroup of order 4 (which I think is the one consisting of the 2,2 cycle types). In that case, that's the only possible kernel of size 4 of a group homomorphism from S₄. Rckrone (talk) 02:28, 7 September 2012 (UTC)

This is easily seen from the class equation, where I've put the lengths of the cycles in each conjugacy class in parentheses: 24 = 1 + 3 (2,2) + 6 (2) + 8 (3) + 6 (4). So there's only one way to collect conjugacy classes into a subset with 4 elements, and that just happens to be a subgroup. --COVIZAPIBETEFOKY (talk) 11:36, 7 September 2012 (UTC)

September 7

Hyperbolic Trig?

While it may not be quite the same, I noticed that there are articles for both Trigonometry in a Planar concept as well as spherical geometry. Is there an inverted concept to spherical geometry in Hyberbolic spaces (and this is *not* as far as I can tell related to the Hyperbolic trig functions)Naraht (talk) 12:47, 7 September 2012 (UTC)

If you take a valid formula in spherical geometry and multiply all the lengths by i, you get a valid formula in hyperbolic geometry. A trig function of an imaginary argument becomes a hyperbolic function of a real argument (multiplied by some power of i).

The easiest way for me to understand hyperbolic geometry is with vectors in Minkowski space: (it,x,y,z). Again, the formulae are analogous to those of geometry on the unit hypersphere $w^{2}+x^{2}+y^{2}+z^{2}=1$, but the hyperbolic plane is represented by a unit hyperboloid $-t^{2}+x^{2}+y^{2}+z^{2}=-1$. —Tamfang (talk) 03:37, 8 September 2012 (UTC)

Rocket fin alignment

I have a rocket that has three sets of three fins, but they are not aligned properly. So I traced the circumference of the rocket onto a piece of paper. And I think that if I trisect the circle into 3 60 degree sections, I can use that as my guide to align the fins properly. But the problem is that I cannot find the center of this circle. So how do I trisect that circle into 3 60 degree sections with a pencil, compass, protractor, and ruler at my disposal? Legolover26 (talk) 14:28, 7 September 2012 (UTC)

It is easy to construct a diameter: pick two points on the circle. Open the compass to some big enough distance. Draw two arcs centred on the chosen points. The line through the points where the arcs cross is a diameter. Do this two or three times with different points to get more diameters which of course meet at the centre.--JohnBlackburne^words_deeds 15:23, 7 September 2012 (UTC)

Thank you very much! Legolover26 (talk) 16:03, 7 September 2012 (UTC)

Also note that 3 evenly spaced fins on a 360° circle will be 120° apart, not 60°. StuRat (talk) 16:09, 7 September 2012 (UTC)

Yea, I noticed that. I don't know why i said that. Maybe because the three angles of a triangle add up to 180 degrees. — Preceding unsigned comment added by Legolover26 (talk • contribs) 16:52, 7 September 2012 (UTC)

Here is another method you can use if you can't easily trace the rocket (let's say it has an engine sticking out the bottom):

1) Mark a spot on the edge of the rocket. This will be your first fin position.

2) Make a measuring tape by cutting a long strip of paper (if you already have a measuring tape, all the better, with the softer tailor's type being preferred over the stiff metal ones).

3) Wrap the tape around the rocket, starting at the mark, and extending until you return to the mark. Record this position on the tape, as well (or just note the measurement, if the measuring tape is numbered).

4) Then divide that by three, using the ruler to measure, if necessary (a ruler divided into tenths of an inch or mm is better than one divided into eights of an inch, as that makes for some ugly math).

5) Return the measuring tape to the rocket and mark those 2 additional fin locations.

One hint for using any ruler or measuring tape is to start at 1 inch (or cm), not at 0, since the end is often less accurate. Just remember to subtract 1 from your measurement). Some professional rulers don't start right at the end, to avoid this problem. StuRat (talk) 19:36, 7 September 2012 (UTC)

Axiom of empty set

The Zermelo-Fraenkel axioms do not include the axiom of empty set; instead the article contains the line:

The axiom of specification can be used to prove the existence of the empty set ... once the existence of at least one set is established (see above).

The axiom of infinity in the same article uses the fact that the empty set has already been defined. Is it because we implicitly assume that the domain of discourse contains at least 1 set and so by the axiom of specification the empty set?-Shahab (talk) 15:09, 7 September 2012 (UTC)

The axiom of infinity can be rephrased so as not to actually assume that any of the sets named actually exist. For example, to assert that the empty set is in X, simply say that for every y, if y has no elements, then y is an element of X. Then the axiom of infinity a priori only asserts the existence of a set meeting this description, without actually asserting that it contains anything. Then the existence of a set implies the existence of the empty set by the axiom of specification. --COVIZAPIBETEFOKY (talk) 16:25, 7 September 2012 (UTC)

Name of set

The Abel-Ruffini theorem article says "The theorem says that not all solutions of higher-degree equations can be obtained by starting with the equation's coefficients and rational constants, and repeatedly forming sums, differences, products, quotients, and radicals (n-th roots, for some integer n) of previously obtained numbers."

What do you call the set of numbers that are statable by a finite number of sums, differences, products, quotients and radicals of integers? Is there a common name for this? RJFJR (talk) 16:37, 7 September 2012 (UTC)

"Numbers defined by radicals"? — Quondum 18:20, 7 September 2012 (UTC)

Sounds like irrational numbers, to me, unless we include roots of negative numbers, in which case we have complex numbers. StuRat (talk) 18:29, 7 September 2012 (UTC)

They will be irrational, but not all irrational numbers fit that description. The square root of two, for instance, is irrational but can be described in terms of radicals (I just described it that way). --Tango (talk) 22:30, 8 September 2012 (UTC)

Shouldn't your example be an irrational number which can't be described in terms of "repeatedly forming sums, differences, products, quotients, and radicals" to prove your point ? I would think any irrational number could be defined in that way, if we allow for an infinite series. However, when adding the restriction that it be a finite series, then you may well be right. StuRat (talk) 00:40, 9 September 2012 (UTC)

There are more irrationals (aleph1) than there are formulas constructable in a finite number of symbols. I was wondering if they might be related to the computable numbers. RJFJR (talk) 18:44, 7 September 2012 (UTC)

Just have to jump in here — there are $2^{\aleph _{0}}$ irrationals. This may or may not equal $\aleph _{1}$ — the statement that these are equal is the continuum hypothesis, and its truth value is not known.

(But it is certainly true that there are at least $\aleph _{1}$ irrationals, and that's a possible reading of your claim, so apologies if that's what you meant.) --Trovatore (talk) 20:32, 7 September 2012 (UTC)

Nope. Pi is a computable number, but not a number defined by radicals (the latter set is presumably a proper subset of the former). — Quondum 19:05, 7 September 2012 (UTC)

You're right. And I now see where you liked Numbers defined by radicals to and that is the answer. Thank you. RJFJR (talk) 20:24, 7 September 2012 (UTC)

"The maximal solvable extension of Q", or "Q^solv" are names. You might have been thinking of constructible numbers which form a subfield, not computable numbers.John Z (talk) 20:55, 7 September 2012 (UTC)

I suspect that "Numbers defined by radicals" is a strictly informal term (perhaps "Numbers contructible from radicals" would be better). John's suggestion would be unambiguous (though I'd have to rely on him that it is in fact the same set; for example, there are solutions to the cubic and quartic equations that I'd assume are in Q^solv but are not constructible from only real radicals of real rumbers and integers). — Quondum 06:33, 8 September 2012 (UTC)

I thought it was called the surd field? Huh, that's red. I know I've seen it somewhere. --Trovatore (talk) 08:40, 8 September 2012 (UTC)

Surd field sounds like a good, snappy, short name, but a quick google seems to show that it used for what we are calling constructible numbers, and should probably redirect there. Quondum, I think you are referring to the fact that the solutions to say a rational cubic with 3 real roots may involve radicals of complex numbers, but by the cubic & quartic formulas they sit in Q^solv. Abel-Ruffini just says that Q^solv is strictly smaller than Q, the algebraic closure of the rationals, the Algebraic numbers, and in particular the roots of most equations with degree bigger than 4 are not in Q^solv. Galois theory identifies extensions with solvable Galois group with ones definable by (iterated) radicals.John Z (talk) 21:00, 8 September 2012 (UTC)

John, yes, that is what I meant. I'm a bit out of my depth here, but it sounds like you're saying that a cubic with 3 real roots is expressible using radicals (by which we mean here real radicals of real numbers, starting with integers), which I was under the impression is impossible in general. I suspect that solvability is critically dependent on the field (R or C). For example, x² + 1 is an irreducible element over R, but is factorizable over C, and related subtleties may affect the interpretation of Galois theory. Or something like that – as I said, this is a bit beyond me. — Quondum 23:20, 8 September 2012 (UTC)

No, you are right, you can't always get all the roots of a cubic with three real roots by only using real radicals of real numbers. You may have to take roots of complex numbers, but that is OK since Q[i] for instance is an extension with a solvable (abelian, even cyclic) galois group; you get it by taking radicals. When we have enough roots of unity in the base, cyclic extensions - ones with cyclic galois group - correspond to taking an nth root of something in the base. Solvable extensions, ones with solvable galois groups correspond to taking iterated radicals. The maximal solvable extension is built up by taking all the iterated radicals, as RJFJR wants. Take all the roots of stuff in Q. Adjoin them to Q & you get some field. Then take all the roots of stuff in there, and go on ad infinitum.

So to get the solution of such a cubic, you can think of taking an extension adjoining i, or the square root of some negative number (given in the cubic formula) & then take some cube root of something in that extension to get the solution of some cubic using the cubic formula. The imaginary parts can cancel out and you land back in R, but to write the solution using radicals, taking them in fields not embeddable in R is necessary.

Another interesting & even more important, probably, subfield of Q^solv is Q^ab, the maximal abelian extension of Q, the biggest extension of Q with abelian galois group. It is a real, non-obvious theorem, the Kronecker–Weber theorem, that this is the same field as that gotten by adjoining all roots of 1, the same field as taking all Cyclotomic fields together, what our Class field theory calls $\mathbb {Q} (\mu _{\infty })$.John Z (talk) 00:05, 9 September 2012 (UTC)

Oops, I see where you are coming from, Quondum. I misread the question, as if RJFJR was asking for the name of the field gotten from taking radicals, then "radicals of previously obtained numbers" etc., as are used in the cubic & quartic formula. But he says "radicals of integers" only, which eliminates many fields generated by solutions of cubics & quartics using radicals. Hmm, not sure of a name for that, though have an idea I need to think about more, with less beer. Googling finds a sloppy statement in a reference work & a book with an incorrect definition that mistakenly says what you (reasonably) thought I was saying, that "solvable by radicals" is the same as "solvable by radicals of rationals/integers", ignoring the dependence on the base field you rightly observed, and thus states an incorrect theorem that implies there is a cubic formula using only radicals of rational numbers! To err is human, but the works of Cardano, Tartaglia, Abel, Galois, Artin etc are divine.John Z (talk) 08:01, 9 September 2012 (UTC)

I didn't notice the "integer" part either. For a while there was an article called radical integer that might have provided an answer to this, but it was AfD'd (disclosure: I filed the AfD) on the grounds that the term was not standard and the claimed result was not reliably sourced. Anyway the claimed theorem, if I have it right, is that if a number is obtainable by radicals starting with the rationals, and is also an algebraic integer, then it's obtainable by radicals starting with the integers. Rich Schroeppel coined the term radical integer for (I think) the latter concept.

It would be very nice to know if this result has been published and confirmed. I would be happy to have the content back, once it's no longer sourced to Mathworld or Eric Weisstein. --Trovatore (talk) 08:22, 9 September 2012 (UTC)

September 8

"excessive" shuffling

One of the regulars in my weekly bridge game complains that I shuffle the deck too many times, leading to perverse distributions. Should I take it seriously? —Tamfang (talk) 05:53, 8 September 2012 (UTC)

Well, I suppose certain shuffling patterns might be self-reversing. Let's say you just cut the deck in half, then put the top half on the bottom. The next time you do that, you should reverse the cut, if you manage to cut it in exactly the same spot. However, the odds of you having such a self-reversing shuffle seem quite low. Otherwise, if your shuffle truly randomizes the cards, then repeating it shouldn't make them any less random, on average. StuRat (talk) 06:03, 8 September 2012 (UTC)

Disclaimer: What I'm about to relate was learned by word of mouth, so may be apocryphal. When bridge software was first being developed, the programmers initially thought they had a bug with their shuffling algorithm, because they kept getting very unusual distributions. So they went back and picked over the code, but there was no problem with it. It turns out that when shuffling by hand, most people do a very poor job of randomizing the deck, and so the distributions seemed odd because the programmers were all used to insufficiently randomized decks. So it's possible that your shuffling is leading to unusual distributions, but the fault lies with everyone else, not with you.--121.73.35.181 (talk) 06:34, 8 September 2012 (UTC)

Well a poorly shuffled deck is more likely to give interesting sequences in the cards and a more entertaining game. --Salix (talk): 09:20, 8 September 2012 (UTC)

Actually I think a poorly shuffled deck tends to give less distributional hands. You probably don't notice it except in the tails, which means you probably don't notice it — maybe you don't get as many nine-card suits, but then, how many do you expect?

(In the extreme case, say you don't shuffle at all. Then any trick in which everyone followed suit, results in one card of that suit to each player.) --Trovatore (talk) 10:26, 8 September 2012 (UTC)

Some expert bridge players can anecdotally do a perfect shuffle, which after a number of such shuffles (8, I think it was) I've heard can return the deck to its original state. Under the assumption of adequately unpredictable shuffling, the claim can be ignored as the distribution will become more random with each shuffle. Under the assumption of chosen shuffles (are you a card sharp?), the number of shuffles hardly matters: the shuffler can in principle predict (and control) the result. — Quondum 06:46, 8 September 2012 (UTC)

Here is a source for what 121.73.. wrote. Seven shuffles should be enough to mix a deck of cards thoroughly. Players may not like it because of odd distributions, or they may have another reason to prefer less shuffling ...

• "... These players had figured out that the cards were not being randomly shuffled, and that they could predict the distributions of cards by knowing what the deck looked like at the end of the previous hand." Ssscienccce (talk) 00:21, 9 September 2012 (UTC)

This is really interesting — led me to the Persi Diaconis article and his "seven shuffles" result. It says that after seven shuffles, the total variation distance falls below 0.5 and then is halved every shuffle. If I've correctly understood what that means (which I'm not certain), then in order to be sure you had about the right probability for a nine-card suit, you'd have to shuffle roughly 18 times.

According to our bridge probabilities article, the probability of a nine-card suit is 0.00037, which means 2700-to-1 against. I think I've had about three nine-card suits in my life, which you'd expect in about 5000 hands. I think I've played more than that but I'm not sure (it's been some time since I've played at all); anyway I can't claim that the number is too high or too low with any good statistical significance. --Trovatore (talk) 08:44, 9 September 2012 (UTC)

It could even be that the effect this player is noticing is that your shuffling is actually more random than other people's in the group. It's possible that by shuffling too few times, the suits get distributed more evenly than random (since they'll tend to be clumped after a hand and then spread out at regular intervals). These distributions might feel more random because there's less variance in hand strength, but aren't. I've heard of people complaining about the shuffling in online games (which should be much closer to random than human shuffling), because they have a false conception of what random should look like. That said, most likely everyone's shuffling is fine and it's just confirmation bias on the part of this other player. Rckrone (talk) 01:15, 9 September 2012 (UTC)

Tesseract

i am looking for information to help me better understand the tesseract and the 4D. — Preceding unsigned comment added by 205.142.178.36 (talk) 19:59, 8 September 2012 (UTC)

Rotation of a tesseract about a plane

See the article tesseract, it's great. You can also read Flatland by Abbott and watch Carl Sagan's Cosmos episode 10, "The Edge of Forever." available on line. μηδείς (talk) 20:09, 8 September 2012 (UTC)

As long as we're referencing works of fiction, check out —And He Built a Crooked House by Heinlein. In addition to being a very entertaining (well, at least entertainingly written) story in its own right, the action takes place in rooms comprising the (hyper)faces of a tesseract, and the description of how they are connected is meticulously accurate (I worked it out). (The opening passage about Americans, Californians, Angelenos, Hollywood residents, Laurel Canyon, Lookout Mountain Avenue, and finally the protagonist, is worth memorizing.) --Trovatore (talk) 21:41, 8 September 2012 (UTC)

Possibly of interest, one of my favorite Youtube movies shows the 120-cell (which is to the tesseract as the dodecahedron is to the cube) as a tiling of spherical 3-space. —Tamfang (talk) 04:37, 9 September 2012 (UTC)

Star Trek Special Effect

At time signature 18:12 on this Star Trek episode http://www.youtube.com/watch?v=hornvnjML88 you see the classic yellow-and black rotating interference band from the vintage Star Trek by Spock's head. I know it is caused by two interfering rotating patterns. What are those underlying patterns? Thanks. μηδείς (talk) 20:06, 8 September 2012 (UTC)

I think it's simply two offset radial patterns, as seen here. -- BenRG (talk) 04:07, 9 September 2012 (UTC) *

Exactly! Wonderful. I had been guessing two spirals, which I think would have been way more checkered. Great! μηδείς (talk) 04:21, 9 September 2012 (UTC)

Resolved

Can you give the url of the page in which that image was embedded? μηδείς (talk) 04:49, 9 September 2012 (UTC)

I updated the original link. It's an interesting web site and I hadn't even noticed since I just found the image in Google Image Search. -- BenRG (talk) 05:20, 9 September 2012 (UTC)

Extrapolating sums of powers

Let's say we define G(N, P) = {1^P + ... + N^P}. Now given that A = G(N, 1), B = G(N, 2), and C = G(N, 3) are easy to calculate, is there any way to extrapolate to get the value of D = (G(N, N) mod (N + 1)) using A, B, and/or C (by inspecting the differences between their respective values, or what have you)? If not, can we, at the very least, infer anything whatsoever about D?

76.25.96.107 (talk) 22:08, 8 September 2012 (UTC)

See Faulhaber's formula Dmcq (talk) 22:55, 8 September 2012 (UTC)

Sorry, I should have been clearer: I'm seeking an efficient method for any given N.

76.25.96.107 (talk) 23:04, 8 September 2012 (UTC)

This only works for a subset of N, but is very efficient for that subset (unless I've bungled it): if N+1 is a prime number, then D = N. This follows from Fermat's little theorem. — Quondum 23:32, 8 September 2012 (UTC)

An efficient method is to calculate the first few values and then look up the Encyclopaedia of Integer Sequences, but that is only for a limited number of values ;-) Sticking 1,2,0,4,3,6,0,6 into that gives A204187. Unfortunately except for the case of N+1 a prime the main reference seems to be 'R. K. Guy, Unsolved Problems in Number Theory, A17' which indicates anything simple may be a bit much to ask. Always worth a good look though if it has come up in some different context. Dmcq (talk) 08:56, 9 September 2012 (UTC)

According to the OEIS page, we can expand the "efficient" cases:

• If N+1 is prime, D = N
• If N+1 is a multiple of 4, D = 0
• Otherwise, there appears to be no known efficient way to calculate D.

— Quondum 09:38, 9 September 2012 (UTC)

Cube: 2 Hypercube

I do not want to be a bother to Mathmatics but i was wondering how accurate is the movie Cube 2 Hypercube, i know its all about tesseracts and 4D? Can any one explain how it works to me? — Preceding unsigned comment added by 205.142.178.36 (talk) 23:34, 8 September 2012 (UTC)

Did you not find the above answers to your question helpful? μηδείς (talk) 00:29, 9 September 2012 (UTC)

No this is about the movie in general — Preceding unsigned comment added by 76.16.47.115 (talk) 02:46, 9 September 2012 (UTC)

It's a horror movie. You might do well to ask a specific mathematical question here, or a plot question on the entertainment desk. μηδείς (talk) 04:22, 9 September 2012 (UTC)

Explain how it would work in math — Preceding unsigned comment added by 76.16.47.115 (talk) 05:00, 9 September 2012 (UTC)

The surface of a cube consists of six squares, each of which is connnected via its four sides to four of the other five square that make up the surface of the cube. In the same way, the "hypersurface" of a tessaract consists of eight cubes, each of which is connected via its six faces to six of the other seven cubes that make up the "hypersurface" of the tessaract.

If you take a tour of the squares around a cube, always going straight on i.e. exiting each square on the opposite side that you entered it, then you go through four different squares before returning to where you started. In the same way, if you take a tour of the cubes around a tesseract, always going straight on i.e. exiting each cube on the opposite face that you entered it, then you go through four different cubes before returning to where you started. Gandalf61 (talk) 08:15, 9 September 2012 (UTC)

September 9

Isomorphism between G/N x N ~= G

Is there always an isomorphism between the quotient of a group G and a normal subgroup N with the direct product of N to G?

G/N × N ≈ G

Is there a theorem for that? --helohe (talk) 12:39, 9 September 2012 (UTC)

No, because it's not true. The simplest counterexample is $G=S_{3}$, which contains normal a subgroup generated by the 3-cycles, $N=\langle (123)\rangle$. Then $G/N\cong \mathbb {Z} _{2}$, but $G$ is not abelian, so $G\ncong N\times \mathbb {Z} _{2}$. — Preceding unsigned comment added by COVIZAPIBETEFOKY (talk • contribs) 13:16, 9 September 2012 (UTC)

Finding the constituent numbers in a list that add up to partial totals

I was wondering of there was a methodology I could use for future reference for a problem I had the other day. I had 28 undated invoices, each with amounts and I had three checks that I knew were payments, each of more than one invoice on that list but not which ones and they were not in any order and I had to match the payments to which invoices they covered. Complicating matters, I also knew that the checks were only partial payment of the total invoices--they would match exactly some combination of the invoices but not all. I took me a few hours to actually figure it out by trial and error, playing with the numbers in excel, and I know there has got to be a better way. Let me provide the actual numbers and the solution so you'll know what I mean (I also think there might be a computer program that might solve this but I had no idea how to look for such a program). Here goes:

Invoices

Template:MultiCol $18.43
$23.91
$7.32
$234.65
$10.13
$11.55
$34.60

| class="col-break " | $67.37
42.81
$75.44
$233.45
$98.43
$91.36
$31.04

| class="col-break " | $72.94
$58.38
$26.48
$95.02
$78.45
$66.57
$48.30

| class="col-break " | $57.67
$49.30
$23.86
$87.88
$89.47
$35.33
88.40 Template:EndMultiCol

Checks

check 1: $327.79
check 2: $437.66
check 3: $381.18

As you can see, I've color coded the solution. The red invoice amounts equal the red check exactly, and the same for green and blue and the uncolored were those invoices that were not paid. But I started with no idea which invoices equaled which checks. I never want to have to do this again blindly.--108.54.25.10 (talk) 13:35, 9 September 2012 (UTC)

The most obvious answer is to keep better track of things in the first place (date your invoices), so you don't need to try to figure it out after. However, a trial-and-error program is the obvious solution. If you want, I could even write you one and send it to you. However, there is the possibility of multiple solutions. StuRat (talk) 14:53, 9 September 2012 (UTC)

This is the Subset sum problem. Taemyr (talk) 15:02, 9 September 2012 (UTC)