Wikipedia:Reference desk/Mathematics

Welcome to the mathematics section
of the Wikipedia reference desk.

skip to bottom

Select a section:

• Computing
• Entertainment
• Humanities
• Language
• Mathematics
• Science
• Miscellaneous
• Archives

Want a faster answer?

Main page: Help searching Wikipedia

   

How can I get my question answered?

• Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
• Post your question to only one section, providing a short header that gives the topic of your question.
• Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
• Don't post personal contact information – it will be removed. Any answers will be provided here.
• Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
• Note:
  • We don't answer (and may remove) questions that require medical diagnosis or legal advice.
  • We don't answer requests for opinions, predictions or debate.
  • We don't do your homework for you, though we'll help you past the stuck point.
  • We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

• The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

See also:

• Teahouse
• Help desk
• Village pump
• Help manual

July 10

Cubic of $\sin(1^{\circ })$

$3\sin(1^{\circ })-4\sin ^{3}(1^{\circ })=\sin(3^{\circ })={\frac {2(1-{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+({\sqrt {5}}-1)({\sqrt {2}}+{\sqrt {6}})}{16}}$

$\sin ^{3}(1^{\circ })-{\frac {3}{4}}\sin(1^{\circ })+{\frac {2(1-{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+({\sqrt {5}}-1)({\sqrt {2}}+{\sqrt {6}})}{64}}=0$

$x^{3}-{\frac {3}{4}}x+{\frac {2(1-{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+({\sqrt {5}}-1)({\sqrt {2}}+{\sqrt {6}})}{64}}=0$

After a long travel, I got here

$x={\dfrac {\sqrt[{3}]{{\dfrac {2({\sqrt {3}}-1){\sqrt {5+{\sqrt {5}}}}-({\sqrt {5}}-1)({\sqrt {2}}+{\sqrt {6}})}{2}}+\left(2{\sqrt {8+{\sqrt {3}}+{\sqrt {15}}+{\sqrt {10-2{\sqrt {5}}}}}}\right)i}}{4}}$

$+{\dfrac {\sqrt[{3}]{{\dfrac {2({\sqrt {3}}-1){\sqrt {5+{\sqrt {5}}}}-({\sqrt {5}}-1)({\sqrt {2}}+{\sqrt {6}})}{2}}-\left(2{\sqrt {8+{\sqrt {3}}+{\sqrt {15}}+{\sqrt {10-2{\sqrt {5}}}}}}\right)i}}{4}}$

Now what? יהודה שמחה ולדמן (talk) 09:23, 10 July 2016 (UTC)

There's nothing more that you can do with it. Sin(1°) cannot be written algebraically without using complex numbers, because your first equation is an example of casus irreducibilis. Loraof (talk) 14:44, 10 July 2016 (UTC). (Technically casus irreducibilis refers to cases in which the constant term in the cubic equation is rational, but the same basic idea applies here too.) Loraof (talk) 14:52, 10 July 2016 (UTC)

See also Angle trisection#Angles which can be trisected. Loraof (talk) 14:57, 10 July 2016 (UTC)

OK. Is there a math Generalization for the Angle bisector theorem, lets say an Angle trisector theorem and so on? יהודה שמחה ולדמן (talk) 19:05, 10 July 2016 (UTC)

Not that I know of, but you might be interested in Morley's trisector theorem. Loraof (talk) 20:00, 10 July 2016 (UTC)

Let me solve it for you. 0.141120008 for all practical purposes in building tree houses for kids on earth. 175.45.116.61 (talk) 23:39, 10 July 2016 (UTC)

Tell me when to laugh, genius (sarcasm). יהודה שמחה ולדמן (talk) 09:22, 11 July 2016 (UTC)

I looked at this page here and after calculating I got to $\sin(n^{\circ })=-{\frac {{\sqrt[{3}]{\sin(3n^{\circ })+\cos(3n^{\circ })i}}+{\sqrt[{3}]{\sin(3n^{\circ })-\cos(3n^{\circ })i}}}{2}}$ .

But it seems that $\sin(30^{\circ })=-{\frac {{\sqrt[{3}]{\sin(90^{\circ })+\cos(90^{\circ })i}}+{\sqrt[{3}]{\sin(90^{\circ })-\cos(90^{\circ })i}}}{2}}=-1$ if we take the real cube root of 1. The imaginary roots give us ${\frac {1}{2}}$. why? יהודה שמחה ולדמן (talk) 13:26, 12 July 2016 (UTC)

Every angle 3theta has not one but three angles theta. In the case of 3theta = 90°, these solutions for theta are 30°, 30°+120°=150°, and 30°–120°=–90°. The sines of these angles are 1/2, 1/2, and –1 respectively. Loraof (talk) 14:48, 12 July 2016 (UTC)

You mean $3\theta =90^{\circ }+360^{\circ }K$ . But how would these 3 values give different results for $\sin(90^{\circ })$ ? יהודה שמחה ולדמן (talk) 15:19, 12 July 2016 (UTC)

90°+360°k is the same thing as 90°, and is normally written as simply 90°. These three values of theta have three individual values of their sines: 1/2, 1/2, and –1. No matter which of these you substitute into the cubic equation, you get sin 90°=1. That shows that each of them is a solution of the equation in which the value of sin(3theta) is specified as 1. Loraof (talk) 20:35, 12 July 2016 (UTC)

Fast Boolean Matrix Multiplication

Hi, what is the the minimal k for which there exists an algorithm that solves matrix multiplication for boolean matrices in time $n^{k}$? I know that the answer for real-valued matrices is $n^{\omega }$ (where $\omega \approx 2.37$), but I don't know what's the best algorithm for boolean matrices. 31.154.81.27 (talk) 10:36, 10 July 2016 (UTC)

Edit: I also assume the matrices to be upper triangular.31.154.81.27 (talk) 10:50, 10 July 2016 (UTC)

According to our article it can be done in O(n²) expected time. --RDBury (talk) 11:24, 10 July 2016 (UTC)

Thank you! But the algorithm citated in that article has matrices for which it runs in cubic time (it has quadratic time for random pair of matrices). Do you know some algorithm that runs in expected/ deterministic quadratic time for every pair of boolean matrices? 31.154.81.27 (talk) 11:49, 10 July 2016 (UTC)

I notice that the article mentions two distinct types of Boolean matrix: built on the operations AND/OR and AND/XOR respectively, and this might confuse things a bit. I assume that the former is the subject of this question, and the primary topic of the article.

The reference does make the following statement:

At present there is only one algorithm known to multiply Boolean matrices in O(n^β) operations, β < 3. It has been observed [see Fisher and Meyer (1971), Furman (1970), Munro (1971)] that two Boolean matrices may be multiplied in O(n 1og₂7) operations. One uses the method of Strassen (1969) to obtain the real integer product of A and B and normalizes the result by replacing all nonzero entries with 1.

The description appears to suggest that this is a deterministic time. As to an algorithm that would run in exactly (or even arbitrarily close to) quadratic time for arbitrary matrices, such seems unlikely to exist, and even less likely to be known. I guess that the algorithm giving close to O(n²) relies on the sum-of-products being known as soon as one nonzero term is found, thus not having to evaluate the remainder of the sum. —Quondum 15:30, 10 July 2016 (UTC)

Thank you for the citation. I hadn't noticed it when reading this article. It's helpful for me! 31.154.81.65 (talk) 05:51, 12 July 2016 (UTC)

July 11

Dice game

$x$ players each roll a $y$-sided die (containing numbers from 1 to $y$). If they roll a 1 they leave the game. The remaining players roll again, and so on until no players are left. What is the expected number of rolls rounds? (does this problem have a name?) 24.255.17.182 (talk) 01:02, 11 July 2016 (UTC)

The problem is easiest when all the players are independent. You did not specify whether you meant the total number of rolls by everyone (whose expectation is x times the expected value of a geometric distribution with $p={\frac {1}{y}}$, or $xy$), or the total number of rounds before everyone is out. The latter case is a bit difficult to compute, I'll revisit it later today probably.--Jasper Deng (talk) 01:24, 11 July 2016 (UTC)

Yes, all the players are independent, and I meant the total number of rounds before all the players are out. 24.255.17.182 (talk) 01:26, 11 July 2016 (UTC)

Let $p={\frac {1}{y}}$ be the probability to drop out and n be the number of players (that's more natural than x). Denote by $a_{n}$ the result. It is easy to see (by conditioning on the first step) that $a_{n}$ satisfies the recursion

$a_{0}=0$

$a_{n}={\frac {1+\sum _{k=0}^{n-1}a_{k}p^{n-k}(1-p)^{k}{\binom {n}{k}}}{1-(1-p)^{n}}}$

I don't know how to find a closed form, but the first values (for $y=6$) are

$a_{1}=6$

$a_{2}={\frac {96}{11}}$

$a_{3}={\frac {10566}{1001}}$

$a_{4}={\frac {728256}{61061}}$

$a_{5}={\frac {3698650986}{283994711}}$

I couldn't find a match for this on OEIS. -- Meni Rosenfeld (talk) 14:40, 11 July 2016 (UTC)

I found a match for $p={\tfrac {1}{2}}$. https://oeis.org/A158466, https://oeis.org/A158467.

Apparently this is related to skip lists. It also appears that there is no elementary closed form.

There is an expression as a non-recursive sum though:

$a_{n}=\sum _{k=1}^{n}{\frac {(-1)^{k}{\binom {n}{k}}}{(1-p)^{k}-1}}$

-- Meni Rosenfeld (talk) 14:49, 11 July 2016 (UTC)

Countable real analysis?

I was wondering if one could get rid of uncountable sets in real analysis by not taking the step to define real numbers. Instead of defining R first (which involves a limiting process but it yields something that does not exist within the countable domain) and then using that, one could try to calculate whatever one wants to calculate and move the analogue of the continuum limit at the end. Count Iblis (talk) 04:03, 11 July 2016 (UTC)

There are a many workers who, probably because of some grave moral failing, don't like uncountable sets and have examined such questions. For example, you might look into the work of Errett Bishop. In general, these systems have not been widely adopted. --Trovatore (talk) 04:11, 11 July 2016 (UTC)

Sorry, was this a dig against us morally depraved individuals that aren't stricly "mathematical realists"? I will be sure to add this to my already extensive list of personal failings.  ;-D Sławomir Biały (talk) 19:48, 11 July 2016 (UTC)

It was a little piece of ironic hyperbole. I often sense this sort of moralism coming in the opposite direction from advocates of restricted ontologies and methodologies, who somehow can't see the beauty of the true thoughts of God. Turning the tables a little. --Trovatore (talk) 21:20, 11 July 2016 (UTC)

Yes, what I've seen so far from reading about such attempts is that it gets quite complicated. My perspective is that when doing math, say you calculate an integral or you prove some theorem, you are only ever going to manipulate a finite number of symbols using a finite number of rules. So, in that sense all of math can be re-interpreted as discrete math. This means that it should be possible to eliminate all concepts referring to uncountable quantities. Count Iblis (talk) 05:58, 11 July 2016 (UTC)

The number of concepts that we will ever enumerate is finite, but the number of potential concepts is infinite (e.g. the natural numbers). Similarly, the set of possible limiting processes in your description is necessarily uncountable, even if you choose to avoid referring to it as a set and only ever enumerate a finite number of them. —Quondum 06:20, 11 July 2016 (UTC)

Unless one wants to go down the route of rejecting the law of the excluded middle as in Constructivism (mathematics). Dmcq (talk) 08:32, 11 July 2016 (UTC)

Well, that was my first response; Bishop is the earliest constructivist to have come up with a well-developed approach to analysis. Of course it's philosophically misguided, but an interesting system all the same, so I'm told; I've never really gotten around to diving into it. --Trovatore (talk) 08:59, 11 July 2016 (UTC)

Perhaps Skolem's paradox will make them feel happier? Basically just consider uncountable as a word used in the theory rather than anything about real life. Dmcq (talk) 10:49, 11 July 2016 (UTC)

I'm not exactly sure what you're suggesting here. The content of Skolem's paradox is really only that first-order logic is not strong enough to capture the notion of countability (or more relevantly, absolute uncountability). It doesn't imply that it doesn't really exist. --Trovatore (talk) 18:07, 11 July 2016 (UTC)

I do see that you get uncountable quantities as soon as you allow countable infinities. So, I guess one needs to resort to the same tricks as we use in theoretical physics where we impose a cut-off, and only at the end take the limit to infinite cut-off. So, this is like ultrafinitism, but then with the twist that we have a one parameter family of ultrafinitistic models with larger and larger cut-offs. Count Iblis (talk) 19:44, 11 July 2016 (UTC)

Having a limit to the counting numbers strikes me like the xkcd comic Counting sheep. ;-) Dmcq (talk) 23:43, 11 July 2016 (UTC)

See also SMBC. --Trovatore (talk) 23:51, 11 July 2016 (UTC)

Related properties of complex and split-complex numbers

Am I correct in saying that there an equivalent of the Cauchy–Riemann equations that applies to split-complex numbers? My guess is that for a function $u+jv=f(x+jy)$ expressible in terms of the algebraic operations, these would be:

${\tfrac {\partial u}{\partial x}}={\tfrac {\partial v}{\partial y}}$ and ${\tfrac {\partial u}{\partial y}}={\tfrac {\partial v}{\partial x}}.$

Through a change of basis

$u'={\tfrac {1}{2}}(u+v),v'={\tfrac {1}{2}}(u-v)$ and $x'={\tfrac {1}{2}}(x+y),y'={\tfrac {1}{2}}(x-y),$

I expect that these equations translate to

${\tfrac {\partial u'}{\partial y'}}=0$ and ${\tfrac {\partial v'}{\partial x'}}=0.$

This reflects the separation of split-complex numbers into the direct sum R ⊕ R.

The split-complex numbers and complex numbers are very closely related; e.g. they are each isomorphic to the Clifford algebra over a one-dimensional vector space, the only difference being the sign of the quadratic form, and (considered as real algebras) complexify to the same algebra. My question is (assuming that I have not confused the above): can we use this insight to reflect this back to somehow split the complex numbers into two noninteracting algebras (as with a direct sum)? —Quondum 04:25, 11 July 2016 (UTC)

The first part is correct, yes, and the second part corresponds to using the null basis of ⁠1 + j/2⁠ and ⁠1 − j/2⁠. And this is where I have to stop due to limited knowledge at present, but the conclusion would be surprising since AFAIK there is no way to split C like you can do for ²R, so something must be subtly different for it to work in one case and not in the other. Double sharp (talk) 12:33, 11 July 2016 (UTC)

It does seem to be subtle, and I'm trying to see how far one can take the correspondence. C and ²R each embed in ²C, and retain the property that ${\tfrac {\partial f}{\partial {\overline {z}}}}=0$ for what I'll call an "expressible function". In the case of ²R, this strengthens to an algebraic statement, true without the need to appeal to continuity: we can split an expressible function's input and the function itself as follows –

f(αz) = αf(z), f(αz) = αf(z), or

f(z) = αf(αz) + αf(αz)

where α is the idempotent element ⁠1 + j/2⁠. The partial differential equations follow directly from these, hence these are stronger statements.

The idempotents algebraically decompose any element in ²R over a real basis, but arbitrary R-linear combinations of a basis are not expressible. Conversely, C does not permit algebraic decomposition over a real basis, but arbitrary elements are expressible as an R-linear combination of a basis. These two observations seem to be flip sides of the same coin: they each prevent an arbitrary R-linear function on the algebra, and must in some sense be related. This somehow seems to suggests to me that there must be an algebraic statement valid for C that is the equivalent of my equalities using idempotents above, that somehow strengthens the Cauchy–Riemmann equations to a purely algebraic statement. —Quondum 14:45, 11 July 2016 (UTC)

July 12

my "bold" edit regarding Cantor's Diagonal Lemma

I made an edit here: Cantor's diagonal argument, oldid=729475472 diff=prev

I think I'm right but not 100% sure...see the article talk section too..(I think this is easier than me explaining it all)..68.48.241.158 (talk) 13:23, 12 July 2016 (UTC)

Yes that's fine thanks. Dmcq (talk) 14:59, 12 July 2016 (UTC)

You agree with the edit or you're okay with how I posed the question?68.48.241.158 (talk) 15:09, 12 July 2016 (UTC)

In case people don't want to go looking, question is basically this: is what is known as Cantor's Diagonal lemma technically used in any way at all in the formal demonstration of what is known as Russell's paradox/antinomy..??(I think it is not)68.48.241.158 (talk) 15:42, 12 July 2016 (UTC)

There's a new thread with discussion on the article's talkpage if people want to just go there instead and explain/weigh-in as it will then answer my question/possibly improve the encyclopedia at the same time... (as no one has been here yet anyway)..68.48.241.158 (talk) 18:05, 12 July 2016 (UTC)

What is the expected value of mega sena loterry?

What is the expected value of mega sena loterry?
Mega sena with a prize of 20000000 brazillian reals would work like this:
People select numbers between 1 and 60, they must pick 6 different numbers and they must be different, their order is irrelevant.To "play" they must pay 3.5 brazillian dollars.
Then at some point of the month, mega sena pick 6 different numbers between 1 and 60, again order doenst matter
If you guessed right the 6 numbers you get 7000000 brazillian reals.
If you guessed right 5 numbers but didnt guessed right 6, you get 3800000 dollars.
If you guessed right 4 numbers but didnt guessed right 5 or 6, you get 3800000 dollars.
Under those specific conditions what is the expected value of mega sena 201.79.78.164 (talk) 20:23, 12 July 2016 (UTC)

The number of ways picking 6 numbers out of 60 is ${\binom {60}{6}}=50063860$

The number of ways picking 6 correct numbers is ${\binom {6}{6}}=1$

The number of ways picking 5 correct numbers and 1 incorrect number is ${\binom {6}{5}}{\binom {54}{1}}=324$

The number of ways picking 4 correct numbers and 2 incorrect numbers is ${\binom {6}{4}}{\binom {54}{2}}=21465$

The expected value is $(A+324B+21465C)/50063860$ where A, B and C are the price money. (Some prizes are in Brazillian Reals and some are in Dollars. Check the numbers and look the rate of change up yourself). Bo Jacoby (talk) 22:50, 12 July 2016 (UTC).

According to our article, the drawing is held twice per week, not once a month. Also, the total prize is based on ticket sales, so it changes from drawing to drawing. Furthermore, R$ 7000000 jackpot and the R$ 3800000 (it should read Reals) for 4 and 5 correct is split among the respective winners, so there's no way to get an expected value without knowing how many tickets were sold. But the total payout for the various types of drawings is fixed at 46% of ticket sales, so on average the expected payout for a R$3.5 ticket is R$ 3.6 × .46 = R$ 1.656. Income tax takes out some of the winnings so the actual expected payout is somewhat less. --RDBury (talk) 00:17, 13 July 2016 (UTC)

July 13

name for equation?

Is there a name for partial differential equations of the form ${\frac {\partial ^{2}}{{\partial x}^{2}}}V=K{\frac {\partial }{\partial t}}V$ ?

I believe this is the heat equation in one dimension. --RDBury (talk) 00:24, 13 July 2016 (UTC)

Thank you! That takes me to Parabolic partial differential equation, which is what I'm looking for. RJFJR (talk) 01:16, 13 July 2016 (UTC)

July 14

Recovering a group from a quotient

If I have a group G and a normal subgroup H of G, is there a natural, general, way to reconstruct G from knowledge of H and G/H? Taking the direct product seems to not work in cases such as $G=\mathbb {Z} /4\mathbb {Z} ,H=G/\{0,{\bar {2}}\}$. I'm also interested in seeing if there's an easy-to-apply sufficient condition for a direct product to work, for example when G has order p² for some prime p and is acyclic, with H the cyclic group of order p.--Jasper Deng (talk) 04:20, 14 July 2016 (UTC)

One way is the semidirect product, which covers more cases than the direct product, but still by no means all. An instructive example is to find the groups of order 8 having a cyclic normal subgroup of order 4. All but one of the five groups of order 8 meet this criterion, so knowing H and G/H actually tells you very little in this case. You can get a bit further in the case that H is contained in the center of G, and there is a whole theory of these central extensions. The general case is covered by the theory of group extensions, but it's probably best not to try to tackle it without a good understanding of of semidirect products and central extensions. --RDBury (talk) 06:35, 14 July 2016 (UTC)

Warren Buffett: luck or ability?

Could Warren Buffet's Berkshire Hathaway's success be explained by luck alone? Or does beating the S&P for four decades imply ability?Llaanngg (talk) 00:25, 14 July 2016 (UTC)

• The above was posted on the Humanities Desk. I'm moving it here in the hope that someone can answer it sensibly from a point of view of regression analysis or some other appropriate statistical test. --69.159.60.163 (talk) 04:33, 14 July 2016 (UTC)

According to this Business Insider graph, between 1964 and 2014, the S&P has gone up about 2300%, while Berkshire Hathaway has risen 1,800,000%, even with a few years where it didn't outperform the S&P. I'd say that that's conclusive evidence for staggering expertise or a really good deal with the Devil. Clarityfiend (talk) 08:38, 14 July 2016 (UTC)

Buffett is an exponent of value investing, and "buy and hold", which is to say he purchases undervalued shares and waits for them to come good. He finances his investments with a large liquidity pool, for example he owns many insurance companies, which have large cash assets and deferred liability. 86.168.123.201 (talk) 10:19, 14 July 2016 (UTC)