Wikipedia:Reference desk/Mathematics

From Wikipedia, the free encyclopedia
Jump to: navigation, search


The Wikipedia Reference Desk covering the topic of mathematics.

Welcome to the mathematics reference desk.
Want a faster answer?

Main page: Help searching Wikipedia

How can I get my question answered?

  • Provide a short header that gives the general topic of the question.
  • Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
  • Post your question to only one desk.
  • Don't post personal contact information – it will be removed. All answers will be provided here.
  • Specific questions, that are likely to produce reliable sources, will tend to get clearer answers,
  • 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 are not a substitute for actually doing any original research required, or as a free source of ideas.



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.
 
Choose a topic:
 
See also:
Help desk
Village pump
Help manual


June 22[edit]

0^0 other values...[edit]

The standard explanation of the fact that 0^0 is undefined is the fact that lim x->0 of x^0 remains 1 and lim y->0 of 0^y remains 1. Considering this as part of the graph z=x^y, is there an easy way to figure out the line of the form y=kx that approaching (0,0), the limit is 1/2 (to pick a number).Naraht (talk) 13:12, 22 June 2017 (UTC)

The limit of x^(kx) is always 1. This can be found in the exercises of any calculus textbook in the chapter on limits. A good rule of thumb when dealing with limits of the form f(x)^g(x) is to begin by taking a logarithm. JBL (talk) 13:24, 22 June 2017 (UTC)


@Naraht: Take
for any real constant k and you'll get a curve approaching (0, 0) which maintains
Example graph at Wolfram Alpha: link. --CiaPan (talk) 13:40, 22 June 2017 (UTC)
You ment to write
for any positive constant k. Bo Jacoby (talk) 06:16, 24 June 2017 (UTC).
@Bo Jacoby: No, I didn't. --CiaPan (talk) 12:02, 26 June 2017 (UTC)
Exponentiation#Zero_to_the_power_of_zero talks about this. In fact entering 0^0 gets there. Dmcq (talk) 14:21, 22 June 2017 (UTC)

The function

is defined for all when is positive, and for all when is a nonnegative integer, and for nonzero when is a negative integer. However is discontinuous when . The limit

is undefined even if

is defined. Bo Jacoby (talk) 06:07, 24 June 2017 (UTC).

Thanx to all. This gave me what I was looking for.Naraht (talk) 04:42, 26 June 2017 (UTC)

@Naraht: You can find some more notes about 00 in the Math Ref.Desk archive from June 30, 2015, the day you asked about the least symmetric triangle :)
Wikipedia:Reference desk/Archives/Mathematics/2015 June 30#Usefulness vs. naturalness
CiaPan (talk) 13:36, 26 June 2017 (UTC)

Symmetric closure of transitive relation[edit]

Is there a transitive relation whose symmetric closure is not transitive? GeoffreyT2000 (talk) 14:50, 22 June 2017 (UTC)

Ok, SemanticMantis, Meni Rosenfeld, RDBury, John Z, there seems to be consensus that my initial comment can be read in a much sharper way than I intended it. I have struck it out below and replaced it with something that hopefully conveys the point I was trying to make more clearly. --JBL (talk) 15:20, 27 June 2017 (UTC)
@GeoffreyT2000:, you post a lot of substantive questions on the ref desk. Would you please consider including with your questions some indication of what you have attempted before posting the question, to save time and effort for potential answerers? For example, in this instance it is clear that you did not investigate small relations before posting your question, and that would have been a useful piece of information. Thanks. --JBL (talk) 15:20, 27 June 2017 (UTC)
Do you put any thought into your questions before you post them? If so, perhaps you could adapt your posting style to demonstrate it. If not, perhaps you should. --JBL (talk) 17:17, 22 June 2017 (UTC)
Please either help, or do not; being bitey is wasting everyone's time, and reflects poorly on our reference desks. SemanticMantis (talk) 18:10, 22 June 2017 (UTC)
I think JBL's comment is legitimate. IIRC the OP posts a lot of questions. I haven't followed this closely enough to recognize if there's an offending pattern, but if there is, it's justified to call them out on it. The reference desk does have rules. -- Meni Rosenfeld (talk) 18:23, 22 June 2017 (UTC)
Thank you, Meni. I do not think it is unreasonable to expect repeat users to include a sentence of the form "Here is what I have tried before posting this problem: ...." (In this case, such a sentence would not include "looking at relations on a 2 or 3 element set.") --JBL (talk) 18:41, 22 June 2017 (UTC)
It seems to me we don't need such a policy; if you don't like a question then don't answer it. Probably more people will find a question interesting enough to answer if the poster says what they've tried, so maybe it's to the poster's advantage to do that, but it shouldn't be a requirement. --RDBury (talk) 12:25, 23 June 2017 (UTC)
I am not asking for a policy, I am asking for a cultural norm: people shouldn't post thoughtlessly here. --JBL (talk) 12:49, 23 June 2017 (UTC)
While I could not oppose such a norm, one question to ask before responding to posts is whether the response(s) transgress this norm more than the post. To my eyes, the response here that transgresses least is Semantic Mantis's, less than Meni Rosenfeld's (whose transgression is admitted) and rather less than yours. The OP has asked many questions over the years, but they are usually quite intelligent questions, especially considering lack of advanced age. I am often the only one to answer (sometimes only in the archives). In one case I recall, he had obviously put much more thought in the question than my first rather obtuse answer or anyone else's - and I feel I owe him a number of answers to his never answered questions. People have different points of view, things obvious from one may not be obvious from another & the phenomenon of "dumb questions" leading to "interesting discussions" is universal, essential in science and illustrated here. IMHO those expectations of repeat users are not reasonable, and are therefore not "in the rules". "Berating" might be too strong - my candidate adjective is "waspish", but it is a better description than a biased positive self-judgment. So again, the questions in the first response are better addressed to the first responder, than the OP.John Z (talk) 04:24, 24 June 2017 (UTC)
If JBL or anyone else wants to discuss OP's behavior, or suggest any rules on how much thought is required before posting, or rant about "culture" the place to do that is either the OP's talk page or the ref desk talk page. This space is for providing references and helping people get answers, not for berating OP. I remind JBL that his participation here is voluntary, and if he doesn't like OP's questions, he is free to ignore them. The math desk is very slow these days anyway, it's not like we are overloaded with questions. SemanticMantis (talk) 19:03, 23 June 2017 (UTC)
On the same principle, you are not obligated to read or respond to my comments. Moreover, it is clear from your own comments in this thread that you do not believe in a general principle forbidding personal comments on the ref desk. Finally, my comment is not "berating" -- it is critical but polite and to the point, and I am in fact interested in the question of whether GeoffreyT2000 puts any thought at all into the questions that he (I hope this is not presumptive) posts. --JBL (talk) 20:50, 23 June 2017 (UTC)
I have no idea what you're talking about, nor do I care. Asking OP "Do you put any thought into your questions before you post them?" Is rude and unprofessional. Simple as that. I ignore a lot of your comments, but I will continue to speak up when I see you acting like a WP:DICK. SemanticMantis (talk) 16:53, 26 June 2017 (UTC)


I haven't thought about this stuff in a long time but symmetric closure would seem to preserve transitivity. If R is the transitive relation, and S is one symmetric closure, then you know (aRb & bRc) => aRc, and you know that bSa => (aRb or bRa). From here you need only to work out that (aSb & bSc) => aSc. Does that clear it up? SemanticMantis (talk) 18:10, 22 June 2017 (UTC)
I disagree, I think the answer to the OP's question is positive. I can think of at least three very nice examples. -- Meni Rosenfeld (talk) 18:23, 22 June 2017 (UTC)
I think the smallest number of for the underlying set is two. Specifically, if S={0, 1} and R={(0, 0), (0, 1)}. R is transitive but R∪Rt is not. An easy way to check transitivity is to think of R as a Boolean matrix, then R is transitive iff RR⊊R. --RDBury (talk) 00:41, 23 June 2017 (UTC)
Since by now the OP has hopefully had some time to think about it, I'll share what I had in mind (which are less artificial than yours):
  • Real numbers (or any other totally ordered set) and (has to be strict inequality).
  • Positive integers and .
  • Sets and .
As for your example, it can be trimmed down further if you're ok with R being transitive vacuously - . -- Meni Rosenfeld (talk) 08:55, 23 June 2017 (UTC)
Nice one, even more minimal. There are more permutations to think about: Is the transitive closure of symmetric relation still symmetric? Is the symmetric closure of a reflexive relation still reflexive? There are six basic combinations but there are more complex ones too. It looks like the answer is yes for most combinations. --RDBury (talk) 12:25, 23 June 2017 (UTC)
So corrected, thank you. SemanticMantis (talk) 15:27, 23 June 2017 (UTC)

June 25[edit]

Calculating Graham's number[edit]

Hello everyone. As we all know, to calculate Graham's number, we need to start from 3↑↑↑↑3 whose sum is G1 and then continue to calculate 64 layers all the way to G64 which is the final result of Graham's number. What I wanted to know is that is it possible to calculate from beginning to the final result using a standard calculator or any other viable method? It is also mentioned that "Graham's number is so large that the observable universe is far too small to contain an ordinary digital representation of Graham's number, assuming that each digit occupies one Planck volume, possibly the smallest measurable space". So, is it possible to know the actual number of total digits in the final result (G64) of the Graham's number or it can only be guessed? TheGeneralUser (talk) 00:38, 25 June 2017 (UTC)

The total number of digits (i.e. its logarithm) in Graham's number is itself too big to fit in the observable universe. This implicitly means we cannot calculate it using any method, we would literally run out of room before we even get close.--Jasper Deng (talk) 07:11, 25 June 2017 (UTC)
How many logs of logs of logs of logs ..... of Graham's number would we need to take to get to some reasonably representable number? -- Jack of Oz [pleasantries] 22:10, 25 June 2017 (UTC)
Even the number of times you need to take the logarithm to get a writeable number is too large to write out. This really makes Graham's number huge. Georgia guy (talk) 22:30, 25 June 2017 (UTC)

Optimal algorithm for lights out puzzle?[edit]

I would like to solve the following: https://artofproblemsolving.com/community/q1h564755p3301930 . How can I solve the problem via Gaussian elimination? — Preceding unsigned comment added by 37.219.20.248 (talk) 12:53, 25 June 2017 (UTC)

The problem is not always solvable. If n is a multiple of 2m+1 then usually there is no solution. Exclusive or the n/(2m+1) stretches with each other to give a 2m+1 stretch of 1s and 0s. If the result is not all 1s or 0s there is no solution as all 0s is what would result from a solution and all 1s is what one would get from a switch of a 2m+1 stretch - and it would switch back to all 0s. Dmcq (talk) 11:38, 26 June 2017 (UTC)
In fact you get a similar result if there is a number greater than 1 that divides both n and 2m+1. A stretch of 1 is always all 0 or 1 :) Dmcq (talk) 11:42, 26 June 2017 (UTC)
On the business about Gaussian elimination, try working modulo 2. Then multiplying by 1 is equivalent to selecting a vector and adding is equivalent to exclusive orr'ing values. When the matrix has been reduced to upper triangular form you can remove one bit at a time and know the switches required to do that. When n and 2m+1 have a common factor you can still get a solution if one exists this way. Dmcq (talk) 14:45, 26 June 2017 (UTC)
Does Gaussian elimination returns always the optimal solution? --37.219.20.248 (talk) 14:47, 26 June 2017 (UTC)
Just to be clear, although it's not explicitly stated in the problem, the indices on the lights are supposed to be modulo n, so the operation for i=1 switches light 1 through m+1 and n-m through n-1. (I'm going to go with that since it says the light are in a circle.)
In general you'd use Gaussian elimination for something like this, but I think in this case the special form of the basis allows for a faster approach. Instead of looking at the patterns of lights as a vector space over GF(2), think of it as the ring R=GF(2)[x]/(xn+1). You're trying to an element of R as some multiple of f=(xk+1)/(x+1), where k=2m+1. In other words you're trying to find x/f for a given x assuming it exists. If k and n are relatively prime then you can use the Euclidean algorithm to find p and q so p(xn+1)+qf=1, and this makes q=f-1 in R. So x/f is then xq. This should give you O(n log n) time as opposed to O(n3) for Gaussian elimination. Sorry about jumping into ring theory for a problem in linear algebra, but it seemed like the easiest way explain the idea, since trying to talk about it in terms of vectors looked like a lot of reinventing the wheel. --RDBury (talk) 16:27, 26 June 2017 (UTC)
There is just the one solution for every input pattern in terms of which ranges to switch or not. That is assuming you don't go switching the same range more than once which is a bit pointless. Yo can see this as there is the same number of ranges and switches and one can do every pattern. Dmcq (talk) 10:31, 27 June 2017 (UTC)
Maybe I'm confused about the problem, but there are 2n different patterns of lights and 2n possible ways to work the switches. You were saying above that some patterns don't have solutions, which would imply that some patterns have multiple solutions. --RDBury (talk) 21:37, 27 June 2017 (UTC)
Sorry I forgot to put in the bit about n and 2m+1 being coprime. Yes most of the cases where they aren't coprime don't have solutions but when they are coprime there is a single solution for every pattern. Dmcq (talk) 21:57, 27 June 2017 (UTC)

What is the base X where the biggest % of the division of the digits possible at this base works?[edit]

What is the base X where the biggest % of the division of the digits possible at this base works?

As some example at base 3:
0 in base 3 is not divisible by 0 in base 3
1 in base 3 is not divisible by 0 in base 3
2 in base 3 is not divisible by 0 in base 3
0 in base 3 is divisible by 1 in base 3, the result is 0
1 in base 3 is divisible by 1 in base 3, the result is 1
2 in base 3 is divisible by 1 in base 3, the result is 2
0 in base 3 is divisible by 2 in base 3, the result is 0
1 in base 3 is NOT divisible by 2 in base 3, the result is 0.11111111111111111
2 in base 3 is divisible by 2 in base 3, the result is 1

That means 5 out of 9 divisions are possible, so 55.5555...%

Is there any base that would have a % bigger than 55.55555....% — Preceding unsigned comment added by 179.197.131.18 (talk) 13:49, 25 June 2017 (UTC)

First, there is no such thing as numbers being divisible in a particular base. Here are the first 10 rows and columns of a table of what numbers are divisible by each other:
       0 1 2 3 4 5 6 7 8 9  ...
    0  N Y Y Y Y Y Y Y Y Y  ...
    1  N Y N N N N N N N N  ...
    2  N Y Y N N N N N N N  ...
    3  N Y N Y N N N N N N  ...
    4  N Y Y N Y N N N N N  ...
    5  N Y N N N Y N N N N  ...
    6  N Y Y Y N N Y N N N  ...
    7  N Y N N N N N Y N N  ...
    8  N Y Y N Y N N N Y N  ...
    9  N Y N Y N N N N N Y  ...
    ...  ...  ...  ...  ...
The 5 Y's and 4 N's in the top left 3 rows and columns represent the 5/9 of 179.197's example. Each time we go to a higher base, we add one more row and column at the right, so we take a larger and large square from this infinite table. But notice how all each time we add a column it consists mostly of N's with just two Y's. Even if the rows added at the bottom were all Y's, in the limit the table would approach half Y's and half N's. So if there is any solution better than 5/9 it has to be for a small base. It clearly doesn't occur up to base 10, as you see from the table, so I don't think there is better solution, but I'm not going to try to prove it. — Preceding unsigned comment added by 76.71.5.114 (talk) 00:26, 26 June 2017 (UTC)

Laws about numbers[edit]

Some Wikipedia articles taught me that in the real world, numbers that start with 1, 2, or 3 occur the most. Why is this law so natural?? (The first article of this kind I found by looking in the see also section of Significant figures. Georgia guy (talk) 13:55, 25 June 2017 (UTC)

According to the Benford's law article, this is because many natural processes are described by power laws, and a lot of data tends to be distributed approximately uniformly across multiple orders of magnitude. Double sharp (talk) 14:53, 25 June 2017 (UTC)
I would also expect there to be an effect that, if a given quantity only ranges from 1-19, (or 1-199, 1-1999, etc.) then more than half of those numbers start with 1, whereas if it ranges from 1-99 (or 1-999, 1-9999, etc.) the number distributions would be about equal. So, when you average in different ranges, you would tend to get more of the lower numbers at the start. StuRat (talk) 06:03, 26 June 2017 (UTC)
In order for this not to be the case, we would need many quantities to only have the high range, like 80-99 (or 800-999, 8000-9999, etc.), to balance it out. There aren't many quantities I can think of that are like that, although percentage school grades are certainly biased towards the high end. StuRat (talk) 14:13, 26 June 2017 (UTC)

June 26[edit]

But as logical operator[edit]

What logical operator can be associated to the coordinating conjunction but from natural language? (Thanks.)--82.137.14.217 (talk) 22:33, 26 June 2017 (UTC)

In English, "but" logically means the same as "and", with a non-logical implication that the second word or phrase somehow contradicts the first word or phrase. CodeTalker (talk) 01:22, 27 June 2017 (UTC)
If you mean "not X but Y", you would just use two statements: "~X", and "Y". However, unless "~X" has logical import for the deduction, you'd usually just say "Y". If they are related, as in "~X therefore Y", that would be a single logical proposition. In other words, the conjunction "but" can have different meanings, and you'd need to pick the one that best matches your assumptions.OldTimeNESter (talk) 13:40, 27 June 2017 (UTC)

June 27[edit]

Notation for a differential[edit]

In mathematics, letter-based symbols that have the same, standard meaning in many different contexts (e.g. sin, log) are usually typeset in "roman" style, while variables whose meanings are defined in the context are typically typeset in italics. From a philosophical standpoint, would it be more "correct" to typeset the "d" in a differential (in calculus) in roman style or italic style? Put differently, is it more correct to typeset a derivative as or ? --134.242.92.97 (talk) 23:39, 27 June 2017 (UTC)

By the reasoning in your first sentence, "d" should not be in italics. Loraof (talk) 02:06, 28 June 2017 (UTC)
It should be italicized. --Deacon Vorbis (talk) 05:51, 28 June 2017 (UTC)
My preference is for both the d and the variable (x, y etc.) to be written with the same typeface. Dolphin (t) 06:05, 28 June 2017 (UTC)
I agree, preferentially use italics, ISO notwithstanding. There has been a sort of movement to use the upright d, but it has not really caught on, at least in the mathematics community. I think some physicists or engineers may use it more.
As for the "philosophical" aspect, I'm not too convinced by the distinction of using roman for symbols with a "fixed meaning", and even if I were, I'm not sure that it applies to the d, which doesn't have a clear and uniform "meaning" in and of itself. I suppose it's true that it's not a variable, and it might appeal to the tidy-minded to make such a typographical distinction, but it is not really standard. --Trovatore (talk) 06:15, 28 June 2017 (UTC)
My preference is upright 'd', but Wolfram Mathworld uses italics: http://mathworld.wolfram.com/Derivative.html --CiaPan (talk) 07:49, 28 June 2017 (UTC)
I have a low opinion of MathWorld, but I agree with them in this case. Italics is more standard in mathematics. --Trovatore (talk) 09:25, 28 June 2017 (UTC)
Italics are more standard, although I sympathise somewhat with the arguments for the upright d (and similarly for the upright e and i when they denote constants). Double sharp (talk) 10:53, 28 June 2017 (UTC)
It should be typeset in italics. And you must always use emacs when typesetting it. Anyone who typesets it upright is an unwashed heathen who is not fit to have an opinion on the matter; the same goes for anyone that prefers vi to emacs. Sławomir Biały (talk) 11:45, 28 June 2017 (UTC)
Go revert me then. --CiaPan (talk) 13:22, 28 June 2017 (UTC)
I thought differentials as in differentiable manifolds normally used roman, but looking at Exterior derivative I see it was changed from roman to italic with this edit [1] which just says 'fmt', and a quick look on the web shows that a number of people do use italic nowadays for the exterior deriviative. I guess it would really have needed a \d operator in LaTeX rather that \mathrm{d} or what really should be used which is \operatorname{d} to take off and mathematicians are lazy ;-) Dmcq (talk) 12:40, 28 June 2017 (UTC)
Almost everyone uses italics in mathematics for the exterior derivative. Sławomir Biały (talk) 12:52, 28 June 2017 (UTC)
I had a quick look and the italic d is certainly more common, but six of the first sixteen I looked at via Google Books used upright d. It seems more common when bolded with vectors in bold. And others are definitely for the sloping form even when they have lots of upright Greek letters. Dmcq (talk) 14:03, 28 June 2017 (UTC)

June 28[edit]

Citation for Pooled_variance#Sample-based_statistics[edit]

I have read many relevant sections from textbooks on statistics, but none of them mentioned the formula in the linked section. I hope someone will provide a reference for that. Thanks.

--117.136.2.134 (talk) 00:26, 28 June 2017 (UTC)

Hilbert[edit]

I encountered the concept of a Hilbert space a couple of years ago as the result of a categorization and Wikidata-related problem. Today, I encountered the concept of a Hilbert transform. Apparently they're both related to waves, and they're named for the same guy, but beyond that I'm not clear: are they at all related? My mathematical education stopped with an unsuccessful attempt at introductory calculus. Nyttend backup (talk) 13:14, 28 June 2017 (UTC)