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

December 4

What's the way to know what country is located 180 degrees from some specific city of another country?

What's the way to know what country is located 180 degrees from some specific city of another country? --ThePupil (talk) 00:05, 4 December 2021 (UTC)

@ThePupil: By 180 degrees, I assume you mean "directly opposite" (as in, if you theoretically were able to go through the core of the earth). Those places are called antipodes and there are online tools for finding them, though mathematically you could negate the latitude and subtract/add 180 from the longitude (depending on which hemisphere you're in). For example, Perth is at roughly -32,116, and its antipode is at roughly 32,-64. eviolite (talk) 00:28, 4 December 2021 (UTC)

Linguistic peeve — the alleged word antipode is a barbarism. The word antipodes is both singular and plural.

The singular back-formation antipode has become sanctioned by being so commonly used. If you are etymologically strict, singular antipodes is equally a solecism; Ancient Greek ἀντίποδες (antípodes) is strictly a plural noun, the plural of ἀντίπους (antípous). I have no qualms with pistachio and its pronunciation /pɪsˈtæʃioʊ/ either, although this is spelled pistacchio in Italian and pronounced /piˈstak.kjo/ (pi-STAAH-kyoh).  --Lambiam 11:20, 4 December 2021 (UTC)

Antipodes means "opposite feet". If you're standing at the antipodes from me, your feet (of which you have two) are opposite to mine, so I don't see a problem. I think antipode is clearly substandard; it's on a par with matrice or ephemeride or bicep — note that the last one is also very commonly used, but our biceps article correctly avoids it. The biceps is a single muscle with two heads, which is what the word means. --Trovatore (talk) 20:46, 4 December 2021 (UTC)

But I wouldn't write, "My esteemed opponent's feet (of which they have two, last time I counted them) is opposite mine." My beef is with the use of antipodes as a singular noun. In Ancient Greek, your ἀντίπους is someone having their feet opposite of you. Ancient Greek ἀντίποδες refers to a plurality of such counterfooters.  --Lambiam 23:41, 4 December 2021 (UTC)

Antipodes in English is a singular (and also plural) noun, just as is biceps. I hope you wouldn't write bicep in an encyclopedic context? I agree it could have happened that we had adopted antipus as the singular, but we didn't. I don't agree at all that "antipode has become sanctioned by being so commonly used"; that's just not true. --Trovatore (talk) 18:36, 5 December 2021 (UTC)

In Latin, biceps is singular, a nominalized adjective: "the two-headed one (viz. muscle)". The Latin plural is bicipites. Many muscles have medical names that are nominalized Latin adjectives, such as the longissimus.  --Lambiam 22:01, 5 December 2021 (UTC)

It remains the case that antipodes has long standing as the singular and plural form in English. On the other hand antipode looks terrible, as plural-to-singular back-formations generally do. Maybe a hundred years from now it will be unremarkable (like pea from pease), but for now it still comes across as informal or substandard usage. --Trovatore (talk) 22:05, 5 December 2021 (UTC)

Singular antipode was not invented yesterday either; it's almost as old as pea: [1], [2], [3] .  --Lambiam 22:57, 5 December 2021 (UTC)

It so happens that if you match up the globe with the reflected globe, you'll see that land mostly matches up with water. The antipodes of Los Angeles, for example, is in the Indian Ocean, hundreds of miles from land, and the closest land is some tiny island in the French Southern and Antarctic Lands. You can easily find the exceptions in the map to the right. Some of Argentina and Chile matching up with China/Mongolia, some of the rest of South America with Southeast Asia, New Zealand and Spain, some Pacific islands with Africa. --Trovatore (talk) 01:02, 4 December 2021 (UTC)

Using Eviolite's recipe, we find that the antipode of the Kaʿbah at 21.4225°N, 39.8262°E is at 21.4225°S, 140.1738°W (140.1738° = 180° − 39.8262°) in the southern Pacific Ocean, about 50 km (30 mi) from from the Tematagi atol; see also Tematagi § Antipode of Mecca. I suppose a Muslim performing salah right there could use any direction as the direction of prayer.  --Lambiam 11:39, 4 December 2021 (UTC)

Geometric shape for four (or more) cylinders around a central cylinder

I am sure it has a specific name that was linked somewhere on discussions of the Energia (rocket) but I can't find it again. JoJo Eumerus mobile (main talk) 20:56, 4 December 2021 (UTC)

Quincunx? Rojomoke (talk) 06:27, 5 December 2021 (UTC)

The term "quincunx pattern" is used for the arrangement of the rocket engines in the S-IC and S-II, two stages of the Saturn V.  --Lambiam 07:07, 5 December 2021 (UTC)

Yes, exactly that! JoJo Eumerus mobile (main talk) 21:31, 5 December 2021 (UTC)

December 5

Improving secrecy

This question is inspired by quantum cryptography, but you don't need to be familiar with that.

Alice and Bob have a method for generating a sequence of random bits $b_{0},\dots ,b_{n-1}$ that they both know, such that Eve only knows 50% of them. More formally, each $b_{i}$ is independently uniformly distributed, and Eve has probability 1/2 of knowing the value of $b_{i}$ (again, independent of anything else). For each bit, Eve knows whether or not she knows it.

If Alice and Bob want Eve to know fewer, they can split the sequence into two and perform bitwise addition mod 2. So $a_{i}=b_{2i}+b_{2i+1}\mod 2$, for $i<n/2$. Then Eve has probability 1/4 of knowing any given $a_{i}$, but the sequence is only half as long.

In general, Alice and Bob can create a sequence of length $n/k$ where Eve has probability $2^{-k}$ of knowing any given bit. Note that if $l$ is the length of the sequence and $p$ is Eve's probability of knowing a bit, this maintains $l(-\log _{2}p)=n$.

Now the question: Can Alice and Bob do better? Is there a process such that $l(-\log _{2}p)>n$? I suspect the answer is no, but I don't know how to show this.--72.80.81.137 (talk) 21:02, 5 December 2021 (UTC)

I should add: it's important that Eve's probability of knowing a given bit remains independent of other bits. Otherwise you could do much better.72.80.81.137 (talk) 21:46, 5 December 2021 (UTC)

I assume that Eve knows the recipe. In the most general formulation, each output bit is produced by a different function, each of which is a function of some subsequence (not necessarily a contiguous segment) from the original sequence, containing the bits that are used in the computation. So taking the output to be $c_{0},...,c_{m-1},$ we could have $c_{4}=f_{4}(S_{4})=f_{4}(b_{3},b_{7},b_{8}),$ in which each input bit is "live", meaning it may influence the outcome. Assume that for some pair $i,j,i\neq j,$ $S_{i}$ and $S_{j}$ share some bit $b_{k},$ so $c_{i}=f_{i}(...,b_{k},...)$ and $c_{j}=f_{j}(...,b_{k},...).$ If we know that Eve knows $c_{i},$ this increases the likelihood she knows $c_{j}$ as well. This will not do for a solid proof, but the requirement of knowledge independence seems to imply that input bits cannot be "reused", but that each can contribute to one output bit only. This would means that you cannot do better than $\textstyle {\sum _{i}-\log _{2}p_{i}=n}.$  --Lambiam 22:38, 5 December 2021 (UTC)

Thinking about how to formalize your argument, I came up with the following: Consider the probability space $\{0,1\}^{n}$, where an element indicates which bits of the original sequence are known to Eve. The measure of a singleton is $2^{-n}$. The probability of Eve knowing every bit of the new sequence is $p^{l}>0$, so there must be at least one point from the space giving this outcome, meaning $p^{l}\geq 2^{-n}$, which gives $p(-\log l)\leq n$, as desired. However, this is assuming that which bits Eve knows of the new sequence depends only on which she knew of the old sequence, and not on the values of the bits in the old sequence. I can't find a way to incorporate the values of the original bits without breaking independence, but I'm not seeing an argument that it's impossible.72.80.81.137 (talk) 05:35, 6 December 2021 (UTC)

Based on the view given by $c_{i}=f_{i}(S_{i}),$ Eve knows $c_{i}$ if and only if its value is the same for all assignments to $b_{0},\dots ,b_{n-1}$ that agree with what Eve already knew of the original sequence – which I understood to mean that she knows the actual values of $b_{e}$ for $e\in E,$ where set $E\subseteq \{0\,...,n{-}1\}$ is some set of indices, specific to Eve and known to her, the size of which is about ${\tfrac {1}{2}}n.$ This means that the knowability of $c_{i}$ does not depend on any values of bits of the original sequence that are not among those she is already privy to. If $c_{4}=~{\rm {if}}~b_{3}~{\rm {then}}~b_{7}~{\rm {else}}~b_{8},$ and $3$ and $7$ are in $E$ but $8$ is not, then Eve knows $c_{4}$ iff $b_{3}=1.$ So knowability of an output bit can depend on the actual value of a (known) input bit.  --Lambiam 10:52, 6 December 2021 (UTC)

Density of logical valid statements

Let say I can encode each statement in Propositional calculus, for example by letters and ignore syntax invalid statements (for example a->->b). I ask what is the density of the valid statements (in strings with size not larger than n)? --Exx8 (talk) 23:04, 5 December 2021 (UTC)

I don't think you're going to be able to come up with a satisfying expression in terms of n for a fixed n; it depends too much on the details of the coding.

You might have more luck with the asymptotic behavior — I would be strongly tempted to conjecture that logically valid statements have asymptotic density zero given any reasonable coding. That's basically because a valid statement has to be true in every structure of a given signature, even without imposing any non-logical axioms whatsoever on the structure. That seems like an unusual condition that's going to get more and more unusual as statements get more complex.

However I don't have either a proof or a source; maybe someone else does. --Trovatore (talk) 23:51, 5 December 2021 (UTC)

Oh, I just noticed that you specified the propositional calculus, whereas I was thinking of first-order predicate calculus. My conjecture is still the same, though. A valid statement has to be true no matter what values you assign to the propositional variables. Again that seems like something that's not going to happen very often. --Trovatore (talk) 23:53, 5 December 2021 (UTC)

Among the formulae of length 3, in which case it does not matter whether the notation is infix or Polish, when there are $v$ propositional variables, the density of tautologies is $1$ over $v{+}2.$  --Lambiam 00:34, 6 December 2021 (UTC)

At least, this is the case using only logical constants $\top$ (true), $\bot$ (false), $\neg$, $\wedge$ and $\vee$. The density ${\tfrac {1}{v{+}2}}$ then also applies for formulae of length 1 and 2, making it plausible that this may remain constant for longer formulae. Adding $\rightarrow$ may make a difference, but if the ${\tfrac {1}{v{+}2}}$ conjecture holds for formulae with the original set of five constants, this too will have a non-vanishing asymptotic density.  --Lambiam 04:02, 6 December 2021 (UTC)

With v propositional variables there are only 2^v propositional functions, so sure, if you cap the number of variables, it makes sense that you would have nonzero asymptotic density (maybe equal to 1/2^v?). But that doesn't seem like the natural reading of the question. As the size grows without bound, the number of variables would also be expected to grow without bound. --Trovatore (talk) 06:27, 6 December 2021 (UTC)

Then we need to be precise as to the formal language. In most texts, including our own article, the set of propositional variables is infinite. So then, of the infinitely many formulae of length 1, only one is a (trivial) tautology. For formulae of length greater than 2, we additionally need to specify a distribution over the formulae; otherwise one can enumerate them so as to asymptotically achieve any desired density between 0 and 1.  --Lambiam 07:40, 6 December 2021 (UTC)

Can you clarify whether you’re asking about the density of syntactically valid statements (i.e. well-formed formulae) or logically valid statements (i.e. tautologies)? Regarding Lambiam’s point about variable names, one natural approach would be to consider two strings equivalent if they differ only in variable names. Quick calculation assuming this equivalence: Suppose we only have the conjunction and negation, as well as parentheses and variables (although I suspect the end result will be similar regardless of the details of the encoding). Let $f(n)$ be the number of syntactically valid formulas of length n. Then $f(n)$ is approximately $f(n/2)^{2}+f(n-1)$ (the first term coming from making conjunctions, and the second from negations). This should grow around $2^{n}$. The number of total strings will grow at least like $4^{n}$, since we have four non-variable symbols. So the density of syntactically valid formulas will go to 0 exponentially fast. JoelleJay (talk) 22:29, 6 December 2021 (UTC)

(And logically valid formulas go to 0 even faster.) JoelleJay (talk) 22:46, 6 December 2021 (UTC)

@JoelleJay: As I read the question, it wants to know the frequency of logically valid formulas as a fraction of the number of syntactically valid formulas -- in other words, a quantification of how much faster "even faster" is. --JBL (talk) 16:35, 7 December 2021 (UTC)

Looking at the original question again I agree with you, somehow I skipped over the "ignore" before "syntax invalid statements". I still think it should go to 0 very quickly. JoelleJay (talk) 17:22, 7 December 2021 (UTC)

December 6

Pattern in last 2 digits of polygonal numbers

Legend: X=any digit; E=any even digit; Q=any odd digit (to distinguish from 0 which O can be confused with); X=0 or 5; X=1 or 6; X=2 or 7; X=3 or 8; X=4 or 9

• Triangular numbers end in X0, X1, X3, X5, X6, or X8.
• Square numbers end in 00, E1, E4, 25, Q6, or E9
• Pentagonal numbers end in X0, X1, X2, X5, X6, or X7
• Hexagonal numbers end in X0, X1, X3, X5, X6, or X8.
• Heptagonal numbers can end in any final 2 digits!
• Octagonal numbers end in E0, E1, 33, E5, Q6, or 08.
• Enneagonal numbers end in X0, X1, X4, X5, X6, or X9.
• Decagonal numbers end in X0, X1, X2, X5, X6, or X7.
• Hendecagonal numbers end in X0, X1, X3, X5, X6, or X8.
• Dodecagonal numbers end in E0, E1, Q2, Q3, E4, E5, Q6, Q7, E8, or E9.

Is there any pattern in this sequence?? What final 2-digit endings can a triskaidecagonal number have?? (Please answer using the legend at the top of this section.) Georgia guy (talk) 14:55, 6 December 2021 (UTC)

Considering the properties of modular arithmetic, the final 2-digit endings of the (n+100)-gonal numbers are the same as those of the n-gonal numbers; they only depend on the value of n modulo 100. So in the end it repeats, and there are at most 100 different ending sets. In fact, there are precisely 16. There are some patterns: all (10k+7)-gonal numbers have the same ending set of any final 2 digits, as do the (20k+2)-gonal numbers. The 5k-gonal numbers also share their sets, so the entries above for pentagonal and decagonal are identical. The (10k+9)-gonal numbers and the (20k+14)-gonal numbers also have their final 2-digit sets in common. I see no obvious regularity. Transforming this into your legendary code would be a tedious and unrewarding undertaking that I prefer to skip.  --Lambiam 17:12, 6 December 2021 (UTC)

Grouping the remainders by the resulting sets, I get 3 groups of 15, 3 groups of 5, 5 groups of 6 and 5 groups of 2; I assume this matches up with your results. All sets have either 22, 44, 50 or 100 elements. The fact that the modulus is composite seems to make the results rather messy. It would probably be instructive to first look at the remainder sets mod p where p is an odd prime. In that case the sets have size p or (p+1)/2, and it appears that the groups have size 1 or 2. --RDBury (talk) 18:05, 7 December 2021 (UTC)

Our numbers match up.  --Lambiam 20:39, 7 December 2021 (UTC)

December 7

In Lagrangian and Eulerian specification of the flow field#Material derivative, the specification of field F as F(x, t) is said to be a Lagrangian specification. But in what sense is this specification Lagrangian, and not Eulerian? I could be confused, but it seems like x is just physical space here. Wouldn't a Lagrangian specification be F(x₀, t), where x₀ is a particle label?