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May 13

About Mazarae and de:Mazarä

Hello Wikipedia:Reference desk/Mathematics folks,
Article WP:PRODed and then un-PROD-ed
I realio, trulio have only the faintest clue what this article about.
Your thoughts about this?
Pete "Math class is tough - let's go shopping!" AU aka --Shirt58 (talk) 12:24, 13 May 2016 (UTC)

I also have only the faintest clue what the article about. That is not because the topic is difficult; it's the kind of topic I enjoy. It's because the article is very poorly written, probably by some who already knows what a mazarae is and can't appreciate that most readers don't know. Maproom (talk) 22:08, 15 May 2016 (UTC)

I agree with Maproom. It seems to me like Wikipedia_talk:WikiProject_Mathematics might be a better venue, though. --JBL (talk) 01:12, 16 May 2016 (UTC)

I have made the article shorter, and, I hope, more comprehensible. What it was trying to say was really quite simple. Maproom (talk) 14:30, 16 May 2016 (UTC)

One problem is that the English article omits the explanatory diagrams that are found in the German version. -- SGBailey (talk) 15:46, 17 May 2016 (UTC)

I have rewritten the English article: improved the grammar, rearranged material somewhat, cut a few sentences that seemed redundant, and added an example from the German page. --JBL (talk) 21:31, 19 May 2016 (UTC)

predicting grade distributions?

Hey,

I was wondering how accurately one could predict the grade distributions for a class of n students if you know what the lower and upper quartiles and the median are after 60% of the points have been awarded. (The last 40% being a final which is assumed to be of equal difficulty as the other exams).

Thanks — Preceding unsigned comment added by 140.233.174.42 (talk) 18:30, 13 May 2016 (UTC)

See https://www.academia.edu/25108263/Statistical_Induction_and_Prediction Bo Jacoby (talk) 22:02, 13 May 2016 (UTC).

Thanks Bo, but I honestly don't understand most of that. Im sorry — Preceding unsigned comment added by 140.233.174.42 (talk) 12:42, 14 May 2016 (UTC)

You have, say, 100 students. You know about 60%, that is 60 students, that 15 students have grades below the lower quartile, and 15 students have grades between the lower quartile and the median, and 15 students have grades between the median and the upper quartile, and 15 students have grades above the upper quartile. What do you know about the remaining 40 students? The formula gives 10±3.5 students in each of the 4 classes.

   15 15 15 15 predict 40
    10     10     10     10
3.4641 3.4641 3.4641 3.4641

Bo Jacoby (talk) 17:50, 16 May 2016 (UTC).

May 14

Neighborly polytopes

Let P be the Cartesian product of two simplices (not necessarily the same dimension) and let D be the dual of P. When is D neighborly? For example the dual of the product of two triangles is neighborly, but the dual of the product of a line segment and a tetrahedron is not. --RDBury (talk) 05:46, 14 May 2016 (UTC)

When does a "multiple comparisons problems" occur (statistical inference / t-tests)

Hello!

I am wondering whether I am facing a Multiple comparisons problem in my analysis. I set up a "global" (compound/intersection) hypothesis for 2 independent groups, each tested with a t-test. I want to accept the global hypothesis in case both (indepedent) groups are improving on one variable (outcome).

$H_{AB}=H_{A}\cap H_{B}:\mu _{A}>0\quad {\text{and}}\quad \mu _{B}>0$

So the way I see this: both single t-test are highly significant. To accept the global hypothesis I don't have to add anything up. Can I infer that the intersection hypothesis is accepted since each of the two independent tests are significant?

Thanks for advice, --WissensDürster (talk) 08:40, 14 May 2016 (UTC)

I don't think you can infer that. See Simpson paradox. Loraof (talk) 14:21, 14 May 2016 (UTC)

Can you explain this in more detail please? I have two interventions given to two randomized groups. Simpson-Paradox: "in which a trend appears in different groups of data but disappears or reverses when these groups are combined". Combing the data (groups) is not what I want to do (nor test). Combing data of different interventions on different groups isn't my goal. I will rephrase my problem: I just need the correct form of hypothesis to express the following: "There is an intervention on a group and it will increase the performance in the outcome variable and there is another intervention and another group, and their performance on the outcome will also increase" (outcome variable is the same for both groups). kind regards --WissensDürster (talk) 16:02, 17 May 2016 (UTC)

@User:WissensDürster I see. My reading of the multiple comparisons article leads me to believe that you have to be careful about this. You say you want to have a hypothesis that both treatments are effective. I don't think there's a way to do that. You can have a null hypothesis that neither is effective, and the probability of observing what you've observed given that null is (assuming you've used alpha=.05 and they are independent groups) .05×.05=.0025. Loraof (talk) 20:59, 18 May 2016 (UTC)

@User:Loraof. I just need a way of writing this down in mathematical terms. In case it's rather difficult to combine those two:

${\begin{aligned}H_{1}:\mu _{G1|\delta }&>0\quad (H_{0}:\mu _{G1|\delta }\leq 0)\\H_{1}:\mu _{G2|\delta }&>0\quad (H_{0}:\mu _{G2|\delta }\leq 0).\end{aligned}}$

maybe I can separate them? Is it legitimate to call them "SHa" and "SHb". I am relatively sure this won't change the research pattern or results, I just need a precise notation. Subsequent correction for p-value have to be multiplication? I just thought division by 2 is what those Bonferoni methods are trying to do.. I will think more about this. --WissensDürster (talk) 10:14, 19 May 2016 (UTC)

Each null hypothesis has to have an equal sign in it, not a less than or equal to sign. The only possible joint null hypothesis has two equal signs: one effect equals zero AND the other effect equals zero. The alternative hypothesis then would be that they are not both zero. Loraof (talk) 14:33, 19 May 2016 (UTC)

Stupid complex number question

Hi,

I apologize in advance for this probably being a stupid question:

I want to convert (1-i) into polar form. Now, I believe the magnitude of this number is sqrt(1^2 - i^2) = 0??? How is this possible? Also, I get from trig that the angle would be arctan(-i)? What's going on?? — Preceding unsigned comment added by 140.233.174.42 (talk) 14:25, 14 May 2016 (UTC)

Not a stupid question! The point (1-i) in the complex plane is one unit to the right (in the real direction) and one unit down (in the imaginary direction).So its angle is minus 45°, or equivalently 360° - 45° = 315°. Its magnitude is found as the square root of the sum of squares of the coefficients (of 1 and -1, not of 1 and i). So the magnitude is ${\sqrt {1^{2}+(-1)^{2}}}={\sqrt {1+1}}={\sqrt {2}}.$ Loraof (talk) 14:36, 14 May 2016 (UTC)

And the angle can be written as arctan(-1/1) = arctan(-1). Loraof (talk) 14:41, 14 May 2016 (UTC)

oh! thank you! — Preceding unsigned comment added by 140.233.174.42 (talk) 14:45, 14 May 2016 (UTC)

Are complex numbers in polar coords really meaningful ? While it's useful for regular old X and Y coords, to tell us the distance and direction from the starting point, using it for complex numbers seems like saying 1 apple and 1 orange equals 1.41 fruit at a 45 degree angle. StuRat (talk) 14:50, 14 May 2016 (UTC)

They most certainly are, which is most obvious when you multiply them: multiplying a complex number by (1+i) multiplies its magnitude by the square root of 2, and adds 45 degrees to the angle it makes with the x-axis. Double sharp (talk) 14:57, 14 May 2016 (UTC)

There is also no denying that it is useful, as can be seen if you try to work out (1+i)¹⁷ both ways. Double sharp (talk) 15:00, 14 May 2016 (UTC)

So is it just useful as a calculation method, or do complex numbers in polar coords relate to anything in the real world ? If so, what does the angle and magnitude represent ? StuRat (talk) 15:02, 14 May 2016 (UTC)

you're getting philosophical, which will be dismissed by mathematicians..the idea being that all math is just a calculating method, so to speak..and none of it directly relates to the real world (though it can be used for applications in the real world)...68.48.241.158 (talk) 16:19, 14 May 2016 (UTC)

Very much yes! Putting a complex number in its modulus-argument form is frequently very useful, and is no or less meaningful than representing it in its real and complex components. For an example, look at complex number § electromagnetism and electrical engineering, where the argument is often used to represent phase. —  crh ₂₃  (Talk) 16:38, 14 May 2016 (UTC)

In linear difference equations, the characteristic roots may be complex. If so, the magnitude of the largest pair determines whether the real-world quantity being modeled converges to a steady state (magnitude less than 1) or diverges (magnitude greater than 1). The departure of the magnitude from 1 determines how fast the convergence or divergence is. The additive contribution of this pair of complex conjugate roots to the solution for the dynamic variable is $2m^{t}\cdot \cos(\theta t+\delta )$ where m is the magnitude of the roots, t is time, $\theta$ is the angle of one of the complex numbers, and $\delta$ depends on the real and imaginary parts of the complex numbers scaled by their magnitude. Something similar happens with linear differential equations. So the magnitude and angle appear in a solution equation that contains only real elements. Loraof (talk) 16:54, 14 May 2016 (UTC)

Thanks all, for providing examples. StuRat (talk) 22:42, 14 May 2016 (UTC)

A formalism rigorous

When is a formalism rigorous? Is a rigorous formalism just another word for exhaustive or explicit? --Llaanngg (talk) 20:51, 14 May 2016 (UTC)

Presumably this question comes with context. You will get better answers if you provide the context than if you obscure it. --JBL (talk) 22:27, 14 May 2016 (UTC)

It depends with time. I'm sure the Egyptians and Babylonians and Greeks thought what they did was rigorous. However new axioms have had to be added to even Euclids Geometry to fix things missed out in it. Nowadays a formal system would be one that can be checked by a proof checker on a computer. In the future, and I for one welcome our new AI overlords, ;-) artificial intelligences will presumably think our idea of rigor is naive and primitive. Dmcq (talk) 22:52, 14 May 2016 (UTC)

For example: "The objective of this work is to present a rigorous formalism for the solution of engineering problems on vibrations in which the vibrating structure has a discrete distribution of loads." But there are plenty of mathematical or technical texts setting their objective to outline/present/develop a "rigorous formalism" for some problem/field/issue. What would be the difference if their formalism were not rigorous?--Llaanngg (talk) 23:07, 14 May 2016 (UTC)

From formalism (philosophy of mathematics):

Formalism is associated with rigorous method. In common use, a formalism means the out-turn of the effort towards formalisation of a given limited area.

Assuming this is the concept of formalism that is relevant to your sources, it seems to me that this quote implies that "rigorous formalism" is a redundancy meaning "formalism". Loraof (talk) 23:55, 14 May 2016 (UTC)

I think your question is more about the language being used as opposed to what "formalism" is/means...I think the "rigorous" word is just tagged on here to emphasize the formal nature of a formalism...68.48.241.158 (talk) 14:18, 15 May 2016 (UTC)

Couldn't it be that "rigorous" means from basic principles, explicitly describing each step? Llaanngg (talk) 17:52, 15 May 2016 (UTC)

I think that's more along the lines of what "formal" means here...the "rigorous" just meaning here that it's carefully formalized..68.48.241.158 (talk) 17:56, 15 May 2016 (UTC)

May 15

infinity/infinity

If Infinity is divided by infinity, is the answer 1? Or can the answer be anything you want depending upon how the infiniteies were derived?--178.106.99.31 (talk) 17:54, 15 May 2016 (UTC)

Infinity is not a number, so dividing infinity by infinity is undefinable. ←Baseball Bugs ^{What's up, Doc?} carrots→ 17:59, 15 May 2016 (UTC)

No, that's not an answer. Infinity can be a number in some contexts — see for example Riemann sphere, aka the "extended complex numbers". But even in the Riemann sphere, you can't divide infinity by infinity, though you can add it to anything except infinity, and multiply it by anything except zero. --Trovatore (talk) 20:12, 15 May 2016 (UTC)

some infinite sets can be placed in one to one correspondence with other infinite sets, so they can be divided out in a sense along these lines (ie..one to one) but other infinite sets are strictly larger than others (real numbers vs integers) so cannot be divided out one to one)...68.48.241.158 (talk) 18:03, 15 May 2016 (UTC)

I think a case can be made that the Dirac delta function is one special case. I believe also that 0/0 is the same as infinity/infinity and it's apparent that x^2/x is 0 for x=0. These relate to the series expansion above. --DHeyward (talk) 18:25, 15 May 2016 (UTC)

0 divided by 0 is also undefined. ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:28, 15 May 2016 (UTC)

But for continuous functions, it's important to point out that it's not indeterminate just because it evaluates numerically to 0/0. x^2/x is not indeterminant undefined for any value of x even though straight substitution is 0/0. The answer is "0". it's trivial as to why but is the starting point for evaluating series and functions that converge to 0 or infinity at different rates. --DHeyward (talk) 21:37, 15 May 2016 (UTC)

Literally speaking, you cannot evaluate x²/x at x=0. You can take the limit as x approaches 0, and that limit is 0. But that is not the same thing. --Trovatore (talk) 21:45, 15 May 2016 (UTC)

It trivially reduces to f(x)=x^/x=x . It seems odd that f(0)=x^2/x is only 0 in the limit but f(0)=x=0 is not. That seems to defy laws of equivalence/identity. --DHeyward (talk) 22:29, 15 May 2016 (UTC)

It doesn't reduce to x at x=0. The identity ax/bx=a/b assumes x≠0. --Trovatore (talk) 22:38, 15 May 2016 (UTC)

More precisely it can be evaluated in the limit and shown to be 0 and limits are tools for resolving it. That the function f(0)=x/x=1 and f(0)=x^2/x=0 shows that evaluating to 0/0 is indeterminant but not undetermined for any function. They are substitutional identities. The identity ax/bx=a/b does not have any discontinuities and is a/b when x=0. --DHeyward (talk) 22:49, 15 May 2016 (UTC)

No, sorry, that is incorrect. The identity does not hold when x=0. --Trovatore (talk) 22:55, 15 May 2016 (UTC)

I should say, there are things you could correctly mean by what you say. For example, it's true if you mean it to be interpreted in the ring of rational functions over x. In that case, you're not dividing the number ax by the number bx, but rather the function λx.ax by the function λx.bx. (Does anyone know what the html is for $\mapsto$?) But this is not the usual interpretation of the claim; if you want to be correct you have to make that clear.) --Trovatore (talk) 23:11, 15 May 2016 (UTC)

In the cases I provided, they demonstrate "indeterminant" but the different functions are differentiable. It's my understanding that in those cases, the solution is exact and the functions are continuous without exception for x=0. L'Hospital's rule seems to apply. Am I missing something? --DHeyward (talk) 08:08, 16 May 2016 (UTC)

Yes. You're missing what x²/x, evaluated at x=0, actually means. What it means is, you take 0 and square it. Then you take 0. Then you divide the first by the second; that is, 0/0. Not the "form" 0/0, but literally zero divided by zero, which is undefined.

By default, that is what it means. Period, end of discussion. There is no opportunity to apply L'Hospital's rule; it is completely irrelevant. --Trovatore (talk) 08:12, 16 May 2016 (UTC)

But i would not say 0/0 is undefined, rather it is indeterminate. It can have multiple solutions depending on function ans space, including undefined but it can also have very defined solutions such as 1,0,infinty, etc. x²/x doesn't have a discontinuity at x=0. It is 0. When faced with indeterminant substitutions, there are first principle derivations that lead to a solution. Plugging in numbers demonstrates "indeterminant" but it is not the same as "undetermined" or "undefined." Continuous functions are continuous even through values that substitute into an indeterminant form. f(x)=x^2/x is continuous with solutions throughout real values of x including 0. --DHeyward (talk)

You are wrong, Trovatore is right. The formula x^2/x is not defined at x=0. As a function, it has a removable discontinuity. In many contexts one automatically "fills in" values for removable discontinuities, but the "filling in" is a step that has substance, replacing something that isn't defined by something that is. --JBL (talk) 17:03, 16 May 2016 (UTC)

In general, functions are defined by both their domain and the value they give for every member of the domain. So the function $f:\mathbb {R} \to \mathbb {R} ,\ x\mapsto x$ is a different function than $f:\mathbb {R} \backslash \{0\}\to \mathbb {R} ,\ x\mapsto x$. Also in general, given two real functions f and g, $f/g$ is defined as the function whose domain is the set of all elements x in the domain of both f and g for which $g(x)\neq 0$, and which maps every such x to $f(x)/g(x)$. If you apply this definition to the functions $f(x)=x^{2},\ g(x)=x$, you'll see that $x^{2}/x$ does not have 0 in its domain, that is, it is undefined there. So it is a different function from the identity on $\mathbb {R}$. -- Meni Rosenfeld (talk) 20:11, 16 May 2016 (UTC)

Also, "undefined" quite literally means "something which was not defined". The definition of division of real numbers is: "Let $a,\ b\in \mathbb {R}$ where $b\neq 0$. Then $a/b$ is the unique real number c such that $bc=a$." You'll note that this definition doesn't address the case that the denominator is 0, so 0/0 is undefined.

Saying that "0/0" is an indeterminate form is a high-level description of what happens in the limit of functions that approach 0 (namely, that their ratio can be anything, depending on what the functions are). It's not really a statement about 0/0 itself. You can't make statements about it, it's undefined. -- Meni Rosenfeld (talk) 20:16, 16 May 2016 (UTC)

Okay, more book cracking: Removable singularity would be the term and I suppose the proper way to include x=0 in the domain of the function x^2/x is $f(x)={\begin{cases}x^{2}/x&x\neq 0,\\0&x=0.\end{cases}}$ , however even without the explicit inclusion of x=0 in it's domain, it is still continuous. Being a removable singularity (different from a removable discontinuity because it's not discontinuous anywhere as infinitesimal changes still converge - I think that means it can be represented by a conformal map but it's been a while since I did that). It's certainly well defined whether 0 is in the domain or not. (p.s. I hate math.) --DHeyward (talk) 03:20, 17 May 2016 (UTC)

It's both a removable singularity and (as mentioned by JBL) a removable discontinuity. For our purpose it's more relevant to talk about it being a removable discontinuity.

If you don't include 0 in the domain, then sure, the function is continuous in its domain; but it's not continuous on $\mathbb {R}$ (it can't be, without being defined on all of $\mathbb {R}$).

If you hate math, why do you participate in discussions on the math reference desk? -- Meni Rosenfeld (talk) 07:44, 17 May 2016 (UTC)

There's no such thing as a discontinuity at a point not in the domain. --Trovatore (talk) 08:38, 17 May 2016 (UTC)

This seems suspicious. Wouldn't you say that $x^{2}/x$ (aka $f:\mathbb {R} \backslash \{0\}\to \mathbb {R} ,\ x\mapsto x$) has a removable discontinuity at 0? If so, isn't a removable discontinuity a special case of discontinuity? And if not, don't the terms "removable discontinuity" and "removable singularity" lose much of their applicability? -- Meni Rosenfeld (talk) 17:44, 17 May 2016 (UTC)

It's not a discontinuity. "Removable singularity" is OK. --Trovatore (talk) 17:48, 17 May 2016 (UTC)

It seems our article on this mildly supports your version (it considers my notion of removable discontinuity nonstandard but not unheard of). Let's just say that I disagree, this is incompatible with both my experience and my common sense. -- Meni Rosenfeld (talk) 20:56, 17 May 2016 (UTC)

See Indeterminate form. --Kinu ^t/_c 19:34, 15 May 2016 (UTC)

Indeterminate forms are quite common with +-infinity. With only real numbers (i.e. no infinities) there are only 4 indeterminate forms; 0/0, 0 to the 0, the zeroth root of 1, and the logarithm of 1 in base 1. Georgia guy (talk) 20:41, 15 May 2016 (UTC)

Your latter two are not actually traditional "indeterminate forms". There's a temptation to want to come up with an abstract notion of "indeterminate form" and figure out what else it includes besides the traditional ones, but personally I think it's a waste of time. "Indeterminate form" is a historical notion, pretty much superseded now that we have rigorous notions of continuity and so on. It's still useful for teaching calculus, because it serves as a sort of warning system for where common expressions are discontinouous, but I don't believe it's useful to extend it.

So there are exactly seven indeterminate forms: 0/0, 0⁰, ∞−∞, ∞/∞, 0×∞, 1^∞, ∞⁰. Its a closed-ended list and will never be extended, not because there aren't other expressions that are arguably similar, but because there's just not much point. --Trovatore (talk) 20:59, 15 May 2016 (UTC)

Actually, I looked up the dates, and it appears as though the notion of continuous function actually predates the notion of "indeterminate form", so I have to backpedal a little bit on the historical sequence I suggested above. I still think the basic thrust of what I said is correct, though — once you understand continuity, you don't really need "indeterminate forms", but they're still useful as a sort of reminder of things to watch out for, but not useful to extend them. --Trovatore (talk) 21:43, 15 May 2016 (UTC)

Sort of playing the devil's advocate here, but the "indeterminate forms" are precisely those indeterminate operations that can be constructed by the usual concepts of addition, multiplication, and division – with exponentiation thrown in for good measure, although I think there are probably some reasons for regarding that as less fundamental. The often cringeworthy "L'Hopital's rule" tells us that one way of making sense of expressions like $\infty /\infty$ is to change what one means by ∞ (or, for the other forms, 0). It is not the "number" infinity that one is dividing, but rather an order of growth. Thus, I believe, to make sense of limits such as these, one is in effect thinking of the "number" $\infty$ as a point of the Stone-Cech compactification of the real line. (Maybe the "Hewitt compactum"?) Sławomir Biały (talk) 21:49, 15 May 2016 (UTC)

Well, in this context, though, I don't see any good reason to exclude the complex numbers. So if you want to generalize, you probably ought to include e^i∞, or equivalently sin(∞), or i^∞. But my point is that I don't think it's a good idea to generalize. The notion is probably just barely worth learning in calculus, but to take it beyond that is to focus on the wrong things. --Trovatore (talk) 22:07, 15 May 2016 (UTC)

Oh, to be clear, I'm not saying there might not be interesting math to do along these lines. But I would avoid calling it "indeterminate forms", which is a name for a specific pedagogical aid that people need to first understand, then understand the limitations of. --Trovatore (talk) 22:27, 15 May 2016 (UTC)

This is a perennial question, and I think it is worth attempting to give a real answer. I am not trying to generalize it to complex numbers because, in most cases when someone is tempted to write ∞/∞, they are certain that they mean something, and that something usually does not involve complex numbers. For example, when a calculus student writes $\infty /\infty =1$, they need to be told gently that $\infty$ is a concept that does not behave as other numerical quantities do. The "two infinities" are not "the same infinity". Instead, the relevant concept of infinity is really that of an order of growth. For instance, although it is true that the polynomials $n+1$ and $n^{100}+1$ both tend to infinity, they do not go at the same rate, and therefore they occupy "different infinities". This is, of course, merely an intuitive explanation. Part of our task as post-modern mathematicians is to assign meaning to this sort of sensical nonsense. What is the sense in which we really mean that there are different infinities? What is the space of all infinities? Does it have a topology? What algebraic structures does it support? Etc. Also, I disagree about indeterminate forms as merely a pedagogical crutch. And in any case it is not very relevant to the original question whether indeterminate forms as such are a complete list or no.

I guess this is related to the ultrafilter construction of the hyperreals. Sławomir Biały (talk) 23:17, 15 May 2016 (UTC)

I take issue with Trovatore. Infinity can be added to infinity and the answer is infinity. This video explains how [1]. 80.44.167.65 (talk) 14:34, 16 May 2016 (UTC)

Trovatore's comment is about infinity in the complex numbers. --JBL (talk) 17:06, 16 May 2016 (UTC)

For clarification, this means that Trovatore's infinity is infinity in any direction, rather than positive infinity as defined by Extended real number system, which distinguished the 2 signed infinities. Infinity plus negative infinity is an indeterminate form (a variant of infinity minus infinity that is one of Trovatore's above list of indeterminate forms.) Georgia guy (talk) 17:38, 16 May 2016 (UTC)

More generally, there are many, many kinds of infinity. I don't mean many sizes, I mean many kinds. Some such kinds have a further classification of infinity to different sizes, or directions, or whatever. You can't talk about what "infinity + infinity" is without first specifying what kind of infinity you mean. If we're talking about extended real numbers then yes, $(+\infty )+(+\infty )=(+\infty )$. In the real projective line or Riemann sphere, no, $\infty +\infty$ is undefined. With cardinal numbers, the sum of infinities is their maximum.

https://www.youtube.com/watch?v=23I5GS4JiDg does a very good job of covering the different kinds of infinity that are out there. -- Meni Rosenfeld (talk) 07:51, 17 May 2016 (UTC)

It's true that there are different infinities. But most of these are not relevant in the description of the classical "indeterminate form" ∞/∞, which is a comparison of the order of growth. What (if anything) is the concept of infinity that most closely models this phenomenon? Cardinals and ordinals and general concepts of infinity are not relevant. Clearly the infinities of classical indeterminate forms correspond to certain specific points of the Stone-Cech compactification of R. The question is, which points? Sławomir
Biały 17:54, 17 May 2016 (UTC)

I'd say the Extended real number line (aka two-point compactification of $\mathbb {R}$) and its point $+\infty$ are the relevant concepts here. -- Meni Rosenfeld (talk) 20:56, 17 May 2016 (UTC)

No, that's completely wrong. It fails to clarify the "indeterminate form" $\infty /\infty$. For example, the two functions $n+1$ and $n^{2}+1$ must tend to different points of the compactification. One can detect this easily by the limit $\lim _{n\to \infty }{\frac {n+1}{n^{2}+1}}=0$. If there is only one point at infinity, we cannot (toplogically) distinguish between the limits of these two sequences. They are both "infinity", one is not a "bigger infinity" than the other. Sławomir Biały (talk) 21:31, 17 May 2016 (UTC)

I guess I didn't understand your original question. What I meant is $\infty /\infty$ is indeterminate precisely because we are using only a single size of infinity. We have different rates at which functions can tend to infinity, and only one infinity, so the ratio of functions can tend to anything. Just like $0/0$ is indeterminate - we have different rates of approaching 0, and only one 0, so the ratio can be anything.

You can use an extension of the reals that has different sizes of infinity, like the hyperreal or surreal numbers. But then there's no such thing as "$\infty /\infty$", there are things like ${\frac {\omega }{\omega ^{2}}}={\frac {1}{\omega }}$, which is just a specific size of infinitesimal. The concept of "indeterminate form" becomes moot.

If you mean, what is the correspondence between functions of different growth rates and different points in the extended numbers - if we're talking about Surreals (which is what I'm most familiar with, even though they're not a set), I'd say that the identity function corresponds to $\omega$, and arithmetic operations behave normally (e.g., the function $x^{2}+1$ corresponds to $\omega ^{2}+1$. -- Meni Rosenfeld (talk) 07:27, 18 May 2016 (UTC)

OP here. I thought that x^2/x = x, so why are you talking about the complications of evaluating x^2/x at x=0? Its evident to me that simplifying the function leads to an unambiguous answer. If f(x)=x, then f(x)=0 when x=0. Or am I missing something?--178.106.99.31 (talk) 22:41, 16 May 2016 (UTC)

x^2/x = x is true when x is not equal to 0. If x=0 in that equation, then it says 0/0=0, which is not correct. Gap9551 (talk) 23:02, 16 May 2016 (UTC)

So every time we see x^2/x, we can only simplify it to x if we are sure that x will never be 0? Take the function f(x)=x. This can be evaluated at x=0. Therefore, you are impling that x^2/x != x. Get out of that!--178.106.99.31 (talk) 00:05, 17 May 2016 (UTC)

$x^{2}/x=x$ is only true if $x\not =0$. There are some famous mathematical fallacies that rely on the exceptional case. For example, you can use this to prove that 1=0, which I think we can agree is not true. Sławomir Biały (talk) 00:13, 17 May 2016 (UTC)

However, x=0 is a removable singularity for $x^{2}/x$ however as stated above by someone more knowledgeable than me, x=0 is not within the domain of the function. I don't think that means it's not well-behaved or discontinuous, though. --DHeyward (talk) 03:20, 17 May 2016 (UTC)

Sure, it's well-behaved. It's continuous on its entire domain, which does not include the point 0.

But what you need to get, with no doubts and no half-measures, is that it is not defined at x=0. That's just how terms are evaluated in math. When you evaluate it at x=0, you get a numerator of 0 and a denominator of 0, and at that point you can forget everything else; it's just plain 0/0 and remembers nothing at all about the expression it came from. And 0/0 is not defined. --Trovatore (talk) 04:51, 17 May 2016 (UTC)

I get that, but there is also the observation $f(x)=x^{2}/x$ has a holomorphic function $g(x)=x$ that is a continuous replacement for f(x) over f(x)'s entire domain and removes the singularity at x=0. As I understand it, they are equivalent functions making the singularity meaningless/removeable. --DHeyward (talk) 06:46, 17 May 2016 (UTC)

That is true, but that is in a context with lots more assumptions. The literal interpretation of the notation, in a general context, makes x²/x undefined at x=0.

Here's why I think it's important to emphasize it: I think a lot of times, calculus students learn to do this sort of limit by cancelling stuff out of the numerator and denominator and then plugging in the value where they want to take the limit. Then they get a well-defined value, and lo and behold that is the value they want. It's not a coincidence — but it's not what it means to take the limit. It actually takes a couple of steps to justify that it gives you the correct limit. They aren't hard steps, but they are steps, and more importantly they remind you what's actually going on, which is not just formal algebra on the numerator and denominator. --Trovatore (talk) 07:18, 17 May 2016 (UTC)

I agree with you. My mathematics is mostly focused in engineering rather than theory and it took me a bit to find the terms to to speak to theory. 0/0 is undefined and the functions that return it are indeterminate. My last mathematics course was complex variables many moons ago (post graduate math for optics and electromagnetics) and I like to get refreshed on theory over application so I appreciate the discourse and enlightenment. Engineering would have ended in either a Taylor series or algebra, neither of which address the fundamental singularity that 0/0 is not defined. My first comment was Dirac delta function which, I learned through this discourse is not a function in mathematics theory. Nevertheless, it's practical to see an infinite impulse with zero width has an integral of 1 from -infinity to +infinity. That's more engineering saying that 0/0=1 but it has no mathematical function theory to support it. It's a construct to describe a distribution. I'm not qualified to say where removable singularities exist, only that it's practical to use it. It took me a while to find the theoretical terms mathematicians use to describe it. Engineering, too often, borrows from theory without understanding its limits. --DHeyward (talk) 07:40, 17 May 2016 (UTC)

Technically, one can't speak of "the" function $x^{2}/x$ because specifying the domain is part of defining the function. Having x = 0 in the domain or not amounts to choosing between different functions even if they agree where their domains overlap.--Jasper Deng (talk) 04:17, 17 May 2016 (UTC)

There are conventions for what the domain of a function should be when it is specified by merely a formula. Namely, "the largest domain that makes sense". So, unless stated otherwise (or context suggests otherwise, e.g. we're not talking about real numbers), it's clear that when we say "The function $x^{2}/x$" we mean the function $f:\mathbb {R} \backslash \{0\}\to \mathbb {R} ,\ x\mapsto x$. -- Meni Rosenfeld (talk) 07:57, 17 May 2016 (UTC)

The assumption that computations and limits can be exchanged,

$\lim(f(x))=f(\lim(x))$

is not always valid. It is true that

$\lim _{x\rightarrow 0}{\frac {14+x}{7+x}}={\frac {\lim _{x\rightarrow 0}(14+x)}{\lim _{x\rightarrow 0}(7+x)}}=2$

but it is not true that

$\lim _{x\rightarrow 0}{\frac {14x}{7x}}={\frac {\lim _{x\rightarrow 0}(14x)}{\lim _{x\rightarrow 0}(7x)}}={\frac {0}{0}}$

The function f(x,y)=x/y is not continuous around (0,0), no matter whether 0/0 is defined or not. Nor is x/y continuous in (∞, ∞), no matter whether ∞/∞ is defined or not. Nor is xy continuous in (0, ∞), no matter whether 0∞ is defined or not. Nor is x^y continuous in (0, 0), no matter whether 0⁰ is defined or not. Bo Jacoby (talk) 09:26, 17 May 2016 (UTC).

So, just to be clear, the function f(x)= x is not the same as f(x)= x^2/x??--178.106.99.31 (talk) 20:54, 17 May 2016 (UTC)

I think the interpretation is that $f(x)=x^{2}/x$ can be replaced by $g(x)=x$ over the entire domain of f(x) and since g(x) is defined at the removable singularity f(0), they are equivalent. Equivalence isn't derived from algebra but by properties of functions that are more complicated than substituting $x/x=1$ into f(x) because that identity is undefined for x=0. --DHeyward (talk) 23:16, 17 May 2016 (UTC)

But to answer 178's question directly and literally, no, the two functions are not the same. --Trovatore (talk) 23:32, 17 May 2016 (UTC)

May 16

Rational functions that can't be evaluated anywhere

I was thinking about one of my responses to DHeyward in the "Infinity/infinity" thread above. The point was that, while the identity x²/x=x works only for x≠0, when you're evaluating at real (or complex) numbers, nevertheless it is perfectly valid to write x²/x=x, without restriction, in the field of rational functions over the reals. Here, we aren't evaluating at points; rather, we're dividing one polynomial by another.

So that got me to thinking about what happens in other rational-function fields. For example, if we write Z₅ for the field of integers mod 5, then (someone will correct me if I'm wrong), x⁵-x is a perfectly fine polynomial, distinct from the zero polynomial, even though it equals zero when you evaluate it at any element of Z₅.

Continuing along this line, 1/(x⁵-x) is a perfectly fine rational function over Z₅, even though it does not have a value at any point.

Is this the standard exposition of rational functions over finite fields? Or is it generally useful to identify polynomials that are everywhere equal? Is there a name for the structure where you do make that identification? --Trovatore (talk) 20:35, 16 May 2016 (UTC)

On a related subject, I was once asked to construct the finite field with four elements. I was asked to (as normal) use polynomials over the finite field with two elements, but got stuck when I realized that $x^{2}+x+1=1$ for both 0 and 1. The explanation I was given that this polynomial is not the same as the constant polynomial 1, even though they are everywhere equal.--Jasper Deng (talk) 20:59, 16 May 2016 (UTC)

I guess if you identify polynomials that are equal everywhere, you don't have an integral domain anymore, so you can't form the field of quotients. In my example, x⁵−x factors into x·(x−1)·(x−2)·(x−3)·(x−4). So if you were to identify it with zero, you would have for example that neither x·(x−1) nor (x−2)·(x−3)·(x−4) is zero, but their product is zero. That would be inconvenient.

If you identify polynomials that are everywhere equal, you end up with the set of all functions from the finite field to itself (over finite fields there are more polynomials than functions, while over infinite fields there are more functions than polynomials). —Kusma (t·c) 10:51, 17 May 2016 (UTC)

I'm thinking that the term "field of rational functions" is really a misnomer. You should think of x not as a variable but as a transcendental constant. This seems a bit more natural when the base field is Q; if t is a transcendental number then the field you get by adjoining t to Q is isomorphic to Q(x), where the isomorphism is determined by x↔t. In other words Q(π), Q(e), Q(ln2) are all the same algebraically, so you might as well replace the π, e, ln2 with an indeterminate x, and you shouldn't think of 1/(x⁵-x) as function any more than you think of 1/(e⁵-e) as a function. For Z₅, think of it as being embedded in a field with transcendental elements. Take one of these elements and call it x, the choice is arbitrary but once made x is a constant. The smallest subfield containing x is then isomorphic to the field of rational functions Z₅(x). --RDBury (talk) 11:58, 17 May 2016 (UTC)

This use of "function" (for something that isn't one) pops up in a few other places, notably "generating function." --JBL (talk) 18:04, 17 May 2016 (UTC)

Well, there is a (partial) function naturally associated with it, though. You can reduce the fraction to lowest terms, then evaluate the numerator and denominator at each point of the underlying field and divide. In the case I gave, that turns out to be nowhere defined, but in general it still seems to be a natural interpretation. --Trovatore (talk) 18:15, 17 May 2016 (UTC)

May 17

Integration problem

Hi everyone. I'm studying for my calc 2 final. I was integrating sin^4(x)cos^3(x)dx, and got as an answer 1/5 sin^5(x) - 1/7 sin^7(x) + C.

However, putting this into Wolfram Alpha gets me something really weird: http://www.wolframalpha.com/input/?i=integrate+(sinx)%5E4+*+(cosx)%5E3

So my question is, am I correct in my computation, and if so why is WA being strange? Or is that just another way of expressing the same thing? — Preceding unsigned comment added by 140.233.173.11 (talk) 16:32, 17 May 2016 (UTC)

Also, how the heck do I attack #2 on this page? https://math.dartmouth.edu/archive/m11f97/public_html/m12integrals.pdf — Preceding unsigned comment added by 140.233.173.11 (talk) 17:02, 17 May 2016 (UTC)

@140.23.173.11: We will not do your homework for you. The only things I will give you are:

1. Remember that $\cos(2\theta )=\cos ^{2}(\theta )-\sin ^{2}(\theta )$ and that consequently, $\cos(2\theta )=1-2\sin ^{2}(\theta )$. Given that, the proof of the equivalence of the integrals is left as an exercise for you.
2. You want $\int {\frac {dx}{\sqrt {6x-x^{2}}}}$? Have you tried the "completing the square" strategy cited there? What if I rewrote this as $\int {\frac {dx}{\sqrt {9-(x^{2}-6x+9)}}}$?--Jasper Deng (talk) 17:40, 17 May 2016 (UTC)

It's not homework. It's optional. (You can see that yourself on the sheet). I'm just studying. But thanks for your help!

Also, I'm sorry if this is a stupid question, how did you get from $\int {\frac {dx}{\sqrt {6x-x^{2}}}}$ to $\int {\frac {dx}{\sqrt {9-(x^{2}-6x+9)}}}$? — Preceding unsigned comment added by 140.233.173.11 (talk) 18:14, 17 May 2016 (UTC)

@140.233.173.11: How do we complete the square without changing the value of the expression?--Jasper Deng (talk) 18:27, 17 May 2016 (UTC)

um... well I know that x^2+b+c can be written as (x+(b/2))^2 + something ... but this seems different somehow — Preceding unsigned comment added by 140.233.173.11 (talk) 18:36, 17 May 2016 (UTC)

@140.23.173.11: It's no different. You're given an expression of the form $x^{2}+bx$. Do you agree that I can then write it as $x^{2}+bx+(b/2)^{2}-(b/2)^{2}=(x+(b/2))^{2}-(b/2)^{2}$? Do you also agree that the contents of that square root are just this, just with a negative sign in front of the whole expression? What is b equal to in that expression?--Jasper Deng (talk) 18:41, 17 May 2016 (UTC)

Ah. I get it now. Sorry, that was stupid. Appreciate the help — Preceding unsigned comment added by 140.233.173.11 (talk) 18:58, 17 May 2016 (UTC)

Mind if I ask one more? (Maybe for someone other than Jasper :) )

For number 10 on the sheet, I used tan = sin/cos and substituted t= cos x. and eventually got ln(abs(cos x)), but WA laughs at me, and says the answer is 1/2 sec^2(x). What gives? — Preceding unsigned comment added by 140.233.173.11 (talk) 19:04, 17 May 2016 (UTC)

@140.233.173.11: Great substitution! However, you forgot about the denominator, it would seem! What do you get after you make the substitution?--Jasper Deng (talk) 19:12, 17 May 2016 (UTC)

I get $\int {\frac {-tdt}{t^{3}}}$ — Preceding unsigned comment added by 140.233.173.11 (talk) 19:21, 17 May 2016 (UTC)

So that means, $-\int {\frac {dt}{t^{2}}}$ . — Preceding unsigned comment added by 140.233.173.11 (talk) 19:25, 17 May 2016 (UTC)

(edit conflict) @140.233.173.11: Not quite! Let's look at it in more detail. You are using $t=\cos(x)$, so $dt=-\sin(x)dx$. Shouldn't this leave us with just $\int {\frac {-dt}{t^{3}}}$?--Jasper Deng (talk) 19:28, 17 May 2016 (UTC)

doesn't the t in the numerator kill one of the t's in the denominator?— Preceding unsigned comment added by 140.233.173.11 (talk • contribs)

@140.233.173.11: Look closely. Is there a t in the numerator?--Jasper Deng (talk) 19:41, 17 May 2016 (UTC)

Well, it's (sinx/cosx) / (cos^2(x)). In other words, (sinx/cosx) * (1/ cos^2(x)), in other words ... oh. Damn. Sorry. — Preceding unsigned comment added by 140.233.173.11 (talk) 19:45, 17 May 2016 (UTC)

Any rational function of sine and cosine can be integrated explicitly via the tangent half-angle substitution $t=\tan(\theta /2)$, together with the method of residues. Sławomir Biały (talk) 23:01, 17 May 2016 (UTC)

Also, for # 18, I reduced it to $4\int {\frac {dx}{x^{2}+4}}+\int {\frac {dx}{x^{3}+4x}}$ but I am not sure how to continue. — Preceding unsigned comment added by 140.233.173.11 (talk • contribs)

@140.233.173.11: You know how to integrate the first term using trigonometric substitution. For the second, use partial fraction decomposition, viz. ${\frac {1}{x^{3}+4x}}={\frac {1/4}{x}}+{\frac {-x/4}{x^{2}+4}}$. For the last term, what if we let $u=x^{2}+4$?--Jasper Deng (talk) 23:03, 17 May 2016 (UTC)

All those problems are very simple:

$\int {\frac {\tan x}{\cos ^{2}x}}dx=\int \tan x\,d(\tan x)={\frac {1}{2}}\tan ^{2}x+C$

$\int {\frac {dx}{x^{3}+4x}}=-\int {\frac {(1/x)d(1/x)}{1+4(1/x)^{2}}}=-{\frac {1}{8}}\ln |1+{\frac {4}{x^{2}}}|+C$

Ruslik_Zero 13:41, 18 May 2016 (UTC)

At the risk of annoying people, I want to ask one more. How do I attack #27? Is it a partial fractions probleM? — Preceding unsigned comment added by 140.233.173.11 (talk) 19:32, 18 May 2016 (UTC)

If you know enough to ask that question then you know enough to try and answer it yourself. --JBL (talk) 02:15, 19 May 2016 (UTC)

As I said above all those problems are very simple:

$\int {\frac {xdx}{x^{4}-16}}={\frac {1}{2}}\int {\frac {d(x^{2})}{(x^{2})^{2}-4^{2}}}={\frac {1}{16}}\ln \left|{\frac {x^{2}-4}{x^{2}+4}}\right|+C$

Ruslik_Zero 20:28, 19 May 2016 (UTC)

Generalizing curl

I've been told that the generalization of curl to higher dimensions is a non-trivial affair, even in nice Cartesian coordinates. It bothers me that a change from three dimensions to even one dimension higher is really complicated.

The way I am trying to do it is to generalize the circulation-based definition ${\text{curl}}\ \mathbf {F} \cdot {\hat {\mathbf {n} }}=\lim _{A\to 0}{\frac {1}{|A|}}\oint _{\partial A}\mathbf {F} \cdot d\mathbf {x}$ as given in the article. In three dimensions, it is straightforward to show that, when differentiation under the integral sign is permitted, and the curl exists, then it is given in coordinates by the usual formula using partial derivatives. However, in higher dimensions, this runs into problems, namely that if I pick a given coordinate unit vector, there is not a single plane that is orthogonal to it.

I've concluded, and have read, that curl can't be a vector then. Does that invalidate the above definition of the curl?--Jasper Deng (talk) 20:17, 17 May 2016 (UTC)

And yes, I'm aware of differential forms, but the only treatment of them I can understand is that of Rudin, which I've been told is not the best.--Jasper Deng (talk) 20:20, 17 May 2016 (UTC)

There is curl in higher dimensions, just not of "vector fields". The reason that the above formula makes sense is because the dot product $\mathbf {F} \cdot d\mathbf {x}$ is well-defined. The differential $d\mathbf {x}$ is a one-form, which can be integrated against over a curve. In higher dimensions, you need a version of $d\mathbf {x}$ that can be integrated over a higher dimensional set. If A is n-dimensional, and $\partial A$ is $n-1$ dimensional, then we need would need an $n-1$ form to integrate over $\partial A$. In indices, here is such a form:

$dx^{i_{1}}\wedge dx^{i_{2}}\wedge \cdots \wedge dx^{i_{n-1}}.$

There are $n-1$ indices that need to be contracted away, so the field F needs to be an $(n-1)$-vector (that is, a skew-symmetric tensor with $n-1$ indices). The curl in higher dimensions, defined by your formula, turns out to be equivalent to the exterior derivative, which follows from Stokes' theorem. Sławomir Biały (talk) 20:49, 17 May 2016 (UTC)

The connoisseurs will undoubtedly point out that I have played a bit loose with the covariance and contravariance. Perhaps, if they are astute, they may realize that in this description a "vector field" is really not a vector field at all, but is in fact a differential form.

I know, you said you heard of these, but I would like to add some perspective, particularly in relation to your formula. The perspective I wish to advance is that any pair of objects "curl F" and "F" that behave in reasonable ways, if they are to make sense and define a "true formula", then they are actually just differential forms, and curl is actually the exterior derivative. This is what might be called an "abstract" differential form.

The basic idea is that from the essential Stokes formula:

$\int _{A}{\text{curl}}\ \mathbf {F} \cdot d\mathbf {A} =\oint _{\partial A}\mathbf {F} \cdot d\mathbf {x}$

one can infer that F and curl F are dual to currents. Currents encompass the usual thing one may be familiar with from physics: a vector field, for example. But they also include other things like vector fields with delta-function profiles. Included in the menagerie of currents, there are objects like $(A,d\mathbf {A} )$ and $(\partial A,d\mathbf {x} )$ that represent a concept of integration over submanifolds. Currents form a natural homology theory (that is, there is a concept of "boundary" rather than "coboundary"). Anything dual to currents must behave functorially like a differential form rather than a vector field.

Stokes' theorem is a statement about duality between the exterior derivative on the (smooth) differential forms, and the boundary operator in a well-defined homology theory. Sławomir Biały (talk) 00:04, 18 May 2016 (UTC)Sławomir Biały (talk) 21:24, 17 May 2016 (UTC)

Perhaps, to clarify from the "Rudin" perspective, differential forms are things that are dual to chains. So if you ever encounter any kind of "reasonable" object that happens to be dual to chains, then that is actually a differential form. Sławomir Biały (talk) 00:33, 18 May 2016 (UTC)

In any number of dimensions, even three, the curl of a vector field is a bivector field. Bivectors aren't complicated, just unfamiliar.

You can understand the generalized curl via differential forms as Sławomir Biały suggested above, or via geometric algebra (aka Clifford algebra). The geometric algebra of a normed vector space is an algebra of scalars and vectors in which s+s' and v+v' sums, ss' and sv and vs products, and v² squared norm have their usual meanings from vector calculus, and there are no other constraints. Surprisingly, this generates a consistent and nontrivial mathematical structure. In it you can define a total derivative $\partial \!\!\!/$ (see Feynman slash notation), which when applied to a scalar field gives you the gradient, and when applied to a vector field gives you the sum of the divergence and the curl. In two real dimensions, the divergence+curl is a complex number (a value in the even part of the Clifford algebra, which is isomorphic to the complex numbers) with the divergence in the real part and the curl in the imaginary part. In three real dimensions it's a quaternion, and in 3+1 dimensions it's a biquaternion or Dirac spinor. In any number of dimensions it can be understood as an element of the Lie algebra of rotation and scaling, representing the orientation and volume change of a collection of dust particles under the flow defined by the vector field. -- BenRG (talk) 23:58, 17 May 2016 (UTC)

But isn't a bivector just the "ordinary curl"? What if A is not two-dimensional? Sławomir Biały (talk) 00:06, 18 May 2016 (UTC)

You're right, I only generalized to 2D surfaces in arbitrarily many dimensions. The original question was how you generalize the vector normal ${\hat {\mathbf {n} }}$ to more dimensions where there isn't a unique plane perpendicular to a vector. For 2D surfaces in general, the answer is that you use a bivector (a wedge product of vectors that span the tangent plane). The 3D vector normal is the Hodge dual of that bivector. For k-dimensional A you'd use a k-vector. (Or a k-form. I also didn't distinguish between k-vectors and k-forms because Clifford algebra doesn't.)

I think that $\partial \!\!\!/=\mathrm {d} +\delta$ where $\delta$ is the codifferential, so the Clifford version of the generalized Stokes theorem (for arbitrary-dimensional A in arbitrary-dimensional space) would combine the usual Stokes theorem and its Hodge dual. But I don't know how to write it. -- BenRG (talk) 02:55, 18 May 2016 (UTC)

I think the natural k-vector in this business is the dA in

$\int _{A}\operatorname {curl} F\cdot dA.$

For example, in two dimensions, we parameterize a surface by u and v using a vector valued function r, say. Then

$dA={\frac {\partial \mathbf {r} }{\partial u}}du\wedge {\frac {\partial \mathbf {r} }{\partial u}}dv$

which is a 2-vector. Sławomir Biały (talk) 10:19, 18 May 2016 (UTC)

Well as BenRG says the curl would normally be considered the Hodge dual of the exterior derivative of a 1-form and so be a k-2 form rather than a 2-form. I suppose that is just a bit of historical cruft though and it would really be better to just forget about the Hodge dual business when talking about curl and things like that. That has k(k-1)/2 dimension and in 2 dimensions the curl is a scalar and for 3 dimensions it is 3 dimensional vector. For 4 dimensions it has 6 dimensions and so can't be mapped into 4-space. I think it takes a bit of imagination to think of the space of 2-dimensional plane areas in 3-space as forming a 3 dimensional vector space never mind the six dimensions needed for 4-space areas. However you can get a 4-dimensional vector space of 3-dimensional volume in 4-space,, that would form the basis for a non-standard generalization. Dmcq (talk) 12:59, 18 May 2016 (UTC)

Yes, and I'm saying that it's more natural to regard the curl as a 2-form, if we are taking Jasper Deng's formula as a definition. 2-forms, not 2-vectors, are dual to cycles. Of course, once you start throwing in Hodge duality, metric tensors, etc, there are identifications that can be made. But the "natural duality", as I see it, is between cycles and forms, not cycles and k-vectors. Sławomir
Biały 13:43, 18 May 2016 (UTC)

May 18

Smallest prime factor of 2^1277 - 1

Has anyone yet found the smallest prime factor of 2^1277 - 1?? Georgia guy (talk) 16:36, 18 May 2016 (UTC)

No factors are listed here, so probably not. -- BenRG (talk) 19:28, 18 May 2016 (UTC)

Is there a @PrimeHunter: in the house? (Probably won't give a different answer from Ben's, but might shed more light on this issue). -- Meni Rosenfeld (talk) 23:59, 18 May 2016 (UTC)

No factors are known. It's the smallest Mersenne number without a known factor. Based on http://www.mersenne.org/report_ECM/ the smallest factor probably has more than 60 digits but ECM gives no guarantees, and if a factor is found it isn't known whether it's the smallest unless the number is completely factored. Integer factorization records#Numbers of a special form mentions some of the work to factor smaller Mersenne numbers. One project used around 7500 core years. PrimeHunter (talk) 00:33, 19 May 2016 (UTC)

Seeking References

Can I have some free links to references on solving Laplace equation and legendary polynomials?Sayan19ghosh99 (talk) 17:05, 18 May 2016 (UTC)

You mean Legendre polynomials?--178.106.99.31 (talk) 22:07, 18 May 2016 (UTC)

Also, Laplace's equation ?--178.106.99.31 (talk) 22:41, 18 May 2016 (UTC)

May 19

Banach–Tarski paradox

It's immediately obvious that, as stated, the theorem can't be right. I took it to think overnight, and came up with, what I imagine, is a solid disproval. And because of how easily it is disproved, I came to think if it's some inside math joke, or are there people seriously thinking they are smart for exploiting imperfections to make ridiculous statements instead of fixing them.

Let's just walk over strong form of it, which reads as:

Given any two bounded subsets A and B of an Euclidean space in at least three dimensions, both of which have a nonempty interior, there are partitions of A and B into a finite number of disjoint subsets, A = A₁ ∪ ... ∪ A_k, B = B₁ ∪ ... ∪ B_k, such that for each i between 1 and k, the sets A_i and B_i are congruent.

First of all, it states it can use finite number of elements, that's nice but redundant, see below. Second of all, it states that matching elements are congruent, that puts major restrictions on the statement. And third of all, it says it works with any bounded sets, which is demonstrably false.

In order for two items to be congruent, they must by definition have same shape and same size, therefore same volume. In order for two sets to match, they must have same number of elements. Therefore, any attempts at using infinite sets would be contradictory to the nature of the theorem, that is that sets would precisely match. Additionaly, it means that individual elements will have defined and finite volume. Furthermore, same number of elements of same volume resolves to same total volume. Since subsets are constructed from original sets by partitioning, they must have same total volume as original sets, and it will not be relevant whether volume is well defined for individual elements or not - even if one were to use infinite set.

That establishes that this theorem can only work with finite subsets to begin with, and that both sets A and B have to have same volume.

Now I will further argue that the theorem can only work for pairs of sets of very specific shape. Unfortunately, I lack mathematical education to readily bring up proper terms, so instead I will make some local definitions.

Let "local curvature" be an inverse of a vector from center of a sphere which surface exactly follows local surface of a given set, to the point on that local surface, multiplied by normal vector in this point. So that it's larger in magnitude for curvier surfaces and smaller for flatter, is zero for a plane and is infinity with undefined direction for a point.

Let "net curvature" be a set of "local curvature"-"area" pairs for each distinct module of local curvature, integrated over entire surface area with that curvature, with zero pairs excluded. Objects bound by flat surfaces will have empty net curvature, and so are objects with a bump compensated by a notch of same exact curvature and area. Objects with curved surfaces that are not compensated for with surfaces of same area, opposite sign and equal magnitude will have non-empty net curvature.

Axiom: out of two matching surfaces, one is a fragment of another. Therefore, local curvature in every corresponding point of matching surfaces is identical. If local curvature wasn't identical, surfaces would have been divergent and therefore not matching.

Theorem: sectioning a set into two subsets by arbitrary surface doesn't change net curvature. Sectioning a set this way will produce two subsets that will have matching surfaces of same area where they were sectioned, by definition. Thus net curvature of a set consisting of those two subsets will not include that fragment of subsets surface area because they cancel out due to having same area and module but opposite signs, and the only remaining curvature will be that of the original set. It is therefore impossible to produce a subset with different net curvature than original set by sectioning.

If two original sets didn't had same net curvature, there inevitably will be fragments with surfaces that can not be matched to any remaining surfaces from another subset. Since partitioning doesn't changes net curvature due to producing two identical surfaces with opposite curvature and no net change, the residual area of specific curvatures that can't be matched to any surfaces in another set will remain constant regardless of specific partitioning operations performed. It is thus impossible to partition sets A and B with non-matching net curvature so that resulting subsets could be precisely matched. If you were to attempt sectioning vacant fragment to cancel out a parasitic surface, you would have just created another parasitic surface of same exact curvature.

Summarizing all of above, the theorem only works for objects with same volume and same local shape.

For example, consider sectioning square to match a triangle. They both would have empty net curvature. Or they both could have a bump of same size and curvature, which would resolve to same net curvature between the objects. You simply section off two sides of a square at an angle to produce a trapezoid, put two resulting straight triangles on the top, and section off the bump and put it in new location. Now consider a solid block with spherical void inside it, and a solid ball. No matter how you partition the ball, there will always be convex surfaces that you can't get rid of, that can not be matched anywhere on the block, and vice versa with concave surfaces of the block's void.

The question is, if above text resolves to mathematically correct statements, in which case it disproves most of the theorem. Raidho36 (talk) 00:43, 19 May 2016 (UTC)

Most of this text is not even wrong. Take, for example, the statement that if two sets are congruent, they must be finite. This is false. An example is two unit line segments in the plane. These are infinite sets (they contain infinitely many points; consider Zeno's paradox), but they are nevertheless congruent, they have finite length, etc. Next, the statement "they must by definition have same shape and same size, therefore same volume" presupposes that the sets have volume to begin with. The sets involved in the Banach-Tarski paradox are unimaginably complicated. In fact, it is quite literally impossible to construct them. Their existence is only guaranteed by the axiom of choice. The sets have a well-defined outer measure. And indeed these outer measures are equal to one another. But outer measure is not additive. If you have two disjoint sets, the outer measure of their union need not equal the sum of the outer measures. This fact, that outer measure does not satisfy properties that one normally associates with "volume" is rather surprising. Instead, there is a natural class of sets, the measurable sets, on which the outer measure behaves more or less the way one expects volume ought to. In fact, any set that can be explicitly constructed (without the use of the axiom of choice) is a measurable set. And indeed, the Banach-Tarski paradox is actually false if we confine attention to measurable sets only. So it is hardly surprising that any set you can manage to construct without any formal mathematical training cannot possibly satisfy the Banach-Tarski paradox. There are much more general sets, like fractal sets, that can also be constructed using lots of sophisticated techniques, but they too are measurable, and hence cannot be the mysterious sets referred to in the Banach-Tarski theorem. The decomposing sets of the Banach-Tarski theorem are non-measurable. These are very strange sets indeed, because all though we "know" they exist because of axioms in set theory, it is impossible to exhibit one explicitly. Sławomir Biały (talk) 01:00, 19 May 2016 (UTC)

First, apologies for not noticing this was the refdesk :-/

Second, I haven't read the whole thing, but the main point I think you're missing is that the pieces we're considering don't actually have a well-defined volume. I think Sławomir goes into more detail on that (I haven't read his contribution in detail either), but that's the first thing to realize. --Trovatore (talk) 01:21, 19 May 2016 (UTC)

Really, the best thing you could do, as a first step, would be to understand the Vitali set. It shows a very similar thing, by a very similar (but much simpler) technique. It breaks up the unit interval into a countable infinity of pieces that are just shifts of each other (wrapping around). That shows that the pieces can't have a well-defined "length".

Once you've grokked that, Banach–Tarski won't seem quite so strange anymore. The big difference is that the Vitali-set construction uses infinitely many pieces, whereas you can do Banach–Tarski with five. That is a difference, but ultimately not such a huge one. --Trovatore (talk) 01:58, 19 May 2016 (UTC)

I would first like to point out that I did not made a statement that congruent sets must be finite, I made a statement that they must have same shape and size, by definition of congruence. For some shapes or sizes volume is not well defined but it shall be equal nonetheless. If original sets A and B are equipotent with each pair being congruent, then they have equal total volume, regardless whether volume of individual elements is even defined. Since subset is generated from original set by partitioning, it must retain same total volume. The notion of total volume doesn't comes from sum of volumes of individual fragments, it comes from their congruence and equipotence of their sets, individual fragments may well be non-measureable. As per theorem, they don't have to add up to a complete shape, they just have to be congruent and their respective sets be equipotent. And finally, partitioning preserves area of surfaces that are not matching to any surfaces from another set and therefore belonging to non-congruent elements, such as surface of a ball that can not be matched anywhere on a cube, because partitioning of any set to create opposite matching surface simultaneously generates another such surface, and makes no net change in the area of surfaces that can not be matched.

I see where this coming from with axiom of choice, however it clearly violates some very basic principles that can not be sidestepped by clever definitions. — Preceding unsigned comment added by Raidho36 (talk • contribs) 10:44, 19 May 2016 (UTC)

If they don't have a volume, it doesn't really make sense to say that the volume is the same. It's neither the same nor different. There's just no such thing as the volume. --Trovatore (talk) 10:47, 19 May 2016 (UTC)

1) Subset is created by partitioning of original set, preserving the total volume. 2) Subsets have congruent elements, that is they are identical minus rigid transformations that make no change in shape or size. 3) Subsets are equipotential. To summarize, sets have same number of elements, which are all identical pair-wise, and therefore must have same total volume. Particular volumes of individual elements are not relevant.Raidho36 (talk) 10:56, 19 May 2016 (UTC)

I'm not sure exactly which subsets or elements you're talking about. The subsets I'm talking about are the ones you partition one ball into, and reassemble to form two spheres of the same size as the original. Are those the subsets you mean? Their "elements" are just points, and of course any two points are congruent. The "number" of elements is the cardinality of the continuum, and you can definitely have two sets that both have that number of points, but which do not have the same volume.

In this case, though, there is just no such thing as the volume of the subsets. So your argument cannot proceed. --Trovatore (talk) 11:01, 19 May 2016 (UTC)

Your argument appears to be self-defeating. First of all, you can't assemble anything from non-measureable sets, that would mean that either original object had no size, or it was infinitely large. If sets were measureable, then the whole counter-argumentation you put simply doesn't apply. Second of all, you're stating that they're all "just points", which by definition have zero volume, and the partition subset would therefore be infinite; the theorem clearly states it works with finite sets and individual elements that will therefore have finite size and defined shape. And finally you make contradictory statement, starting with that two sets of points with same cardinality can resolve to different volume, then immediately saying that sets of points (i.e. "this case") have no defined volume. Raidho36 (talk) 11:15, 19 May 2016 (UTC)

" you can't assemble anything from non-measureable sets". Sure you can. Take two non-measurable sets and form their union. There you go: a new set made out of two non-measurable ones. Here "non-measurable" has a precise technical meaning. It does not mean that we cannot estimate the volume of the set (using the outer measure), it means that the set does not behave in reasonable ways under the operations of assembly and decomposition. See Carathéodory-measurable set. Sławomir Biały (talk) 11:26, 19 May 2016 (UTC)

By "can't assemble" I obviously mean "physically", by assembly rather than making unions or whatnot. Raidho36 (talk) 11:30, 19 May 2016 (UTC)

I don't know what you mean, but it seems like you're relying on physical intuition rather than mathematical precision. That's no good. I've already explained in great detail why this theorem is not a "physical" theorem at all. It is about non-constructible sets. Also, please see Carathéodory-measurable set. You might want to read up on measure theory more generally before continuing in this vein. You can have collections of points with the same cardinality as a line segment that have any positive volume you like. You can have collections of points with the same cardinality as a line segment that have zero volume. You can have collections of points with the same cardinality as a line segment whose volume fails to be additive. This latter case is called a "non-measurable set" in this context. Sławomir Biały (talk) 11:40, 19 May 2016 (UTC)

I urge you to re-read my statements more closely. You will then see that I in fact am not confined by intuition, but rather use established mathematical entities as a backbone for my argumentation. Original argument was that you can partition a pea and assemble two copies. However, if non-measureable sets are used, it's impossible. Non-measureable sets can not be used to assemble a measureable set. Raidho36 (talk) 11:45, 19 May 2016 (UTC)

"Non-measureable sets can not be used to assemble a measureable set." False. See Vitali set. "Rather use established mathematical entities as a backbone for my argumentation." You talk about "curvature" and "surfaces". The sets in the Banach-Tarski paradox are not bounded by surfaces. Indeed, most sets have this property, like fractal sets. The tools of smooth analysis are not equipped to make meaningful statements about such kinds of subsets of $R^{3}$. Also, if "physically assemble" means something other than taking a union, you need to be specific. There is no mathematical definition of this term. Sławomir Biały (talk) 11:53, 19 May 2016 (UTC)

If you believe that some issue is being sidestepped by "clever definitions", please correct them. The issue here is that volume is a rather tricky thing to define. It seems like you believe that "volume" counts the number of elements. That is not true. For example, a very large ball and a very small ball have the same number of elements, because there is an explicit bijection between the two sets, but they have different volumes.

So the crux of the matter is this: How do you define the volume of a set? It really does matter how that term is defined here. You can argue that the definition is wrong, but then you need to supply your own definition. This is a lot harder than it might seem. For example, in the above, you are talking about sets as if they have smooth boundaries. What about sets like the set of points with rational coordinates, Cantor dust, or the Alexander horned sphere? How does your definition of volume handle those? Sławomir Biały (talk) 11:05, 19 May 2016 (UTC)

Wait a minute, what? The BTP doesn't say anything about volume, so how could it be "false for some definitions of volume"? The BTP talks about partitioning spherical balls and reassembling them after rigid motions, not about volume. --Trovatore (talk) 11:09, 19 May 2016 (UTC)

Yes, you're right. I've emended my post. Sławomir Biały (talk) 11:19, 19 May 2016 (UTC)

Yes I know bijection doesn't implies that elements are identical, however the theorem states so - it says there would be congruence between elements, meaning they have identical size and shape. I also know that volume may not be defined but it's not relevant because in the "same total volume" rationale, counting volumes of individual elements is completely avoided and only properties of entire set are considered, just like with "same curvature" rationale. This is why argumentation about undefined properties of individual elements doesn't apply. Raidho36 (talk) 11:23, 19 May 2016 (UTC)

"Congruence" does not mean "they have identical size and shape". It means that there is an element of the Euclidean group that maps the one set to another. It may or may not be a "theorem" that this implies the sets have the same "size" and "shape", but you would need to supply definitions of those words first. Sławomir Biały (talk) 11:32, 19 May 2016 (UTC)

Congruence defined as existence of an isometry between the sets. Since isometry preserves distances and is injective, congruence therefore requires that two objects would have identical geometry. Raidho36 (talk) 11:36, 19 May 2016 (UTC)

True, the restriction of the Euclidean metric to each set is preserved by the isometry. So what? You said above that this implies that the sets have the same volume, right? How do you define their volume? The volume is a number, that is not computed from the metric in a direct way. You need to define this number, if you are going to use it in a mathematical proof. Sławomir Biały (talk) 11:43, 19 May 2016 (UTC)

Notion of congruence as identity is important when considering set as a whole, however it's irrelevant when applied to individual elements. Consider a convergent sum function of a divergent function that generates an infinite set. Even if individual values within the set are not defined, its sum converges to a finite value. You can then argue that if another function produces equipotent set with precisely equal elements pair-wise, its sum must converge to same number. Raidho36 (talk) 11:53, 19 May 2016 (UTC)

This is a tissue of mathematical nonsense. Who is talking about "functions" here? We are talking about sets. It is true that a set can be defined by its characteristic function. For a non-measurable set, the characteristic function would also be not a measurable function. It would actually exist at every point (in fact, characteristic functions are equal to 0 or 1 at every point). But non-measurable functions satisfy basically none of the ideas of calculus that you probably want to apply. They are not integrable, not differentiable, etc. "Even if individual values within the set are not defined, its sum converges to a finite value." This is obvious nonsense. A sum of things that are undefined cannot be well-defined. How do you define the sum without having the terms of the sum? Further, I have no idea what the sum of individual values of a set has to do with congruence, volume, or the Banach-Tarski paradox. And I suggest that you probably don't either. "You can then argue that if another function produces equipotent set with precisely equal elements pair-wise, its sum must converge to same number." Again, no idea what you're trying to say here. But equipotence (that is, having the same number of elements) does not imply equal volume, something you have asserted without proof on several occasions in this discussion. I will not be correcting you any more. If you have no wish to learn mathematics, then there really is no purpose to this posting. Sławomir Biały (talk) 12:14, 19 May 2016 (UTC)

I should add, before you get carried away with that challenge, that there already is a very general theorem characterizing the Lebesgue measure (and thus the outer measure). The Lebesgue measure is the unique way of measuring certain "obvious" sets (the Borel sets), which is countably additive on those sets, invariant under the Euclidean group, assigns a value of 1 cubic unit to the unit cube, and transforms under the group of invertible linear transformations by the determinant. So, roughly, any "reasonable" definition of volume on the Borel sets is the Lebesgue measure. Sławomir Biały (talk) 11:19, 19 May 2016 (UTC)

Is it fair to say that this theorem is true in the mathematical realm but not true (or not applicable) in the real world? Or is it somehow true that with proper, very advanced technology it could be implemented in practice? if indeed it's only mathematically true than people would probably be less offended by it if they understood this..??68.48.241.158 (talk) 11:35, 19 May 2016 (UTC)

It has nothing whatsoever to do with the physical world; the concept of an unmeasurable set is completely aphysical. (I do wonder when "argue with a crank" became an accepted reference desk past time, though.) --JBL (talk) 13:26, 19 May 2016 (UTC)

okay, I suspected this but I'm not a mathematician..I've read about this topic on a couple of occasions in the popular press..it might be useful to briefly explain this in the article's intro section for people who arrive there via pop reference..I might put it in the talk page there..68.48.241.158 (talk) 13:56, 19 May 2016 (UTC)

I just had a look at the current intro of Banach-Tarski paradox, and in my opinion the third and fourth paragraphs do a good job of describing the situation. Not that it would be impossible to do better, necessarily, but I don't think anyone could read the intro and come away with the conclusion that they can go home and cut up and orange into two oranges. --JBL (talk) 20:56, 19 May 2016 (UTC)

I think the intro could be improved, but the problem is finding the right RS discussions to cite. Here is a nice bit from this [2] thread on real-life application of B-T: "The Banach-Tarski paradox is an illustration of (one of) the limitations of ℝ3 as a model of the familiar (yet bizarre) ambient space we live in. There are plenty other such incompatibilities. For instance, perfect circles exist in ℝ3, you may want to invite your student to construct one in real life." If you can find any RS discussing the non-physicality of this paradox I can help you incorporate that into the article. SemanticMantis (talk) 21:06, 19 May 2016 (UTC)

I do wonder why you think insulting askers and implicitly criticizing other math desk responders is acceptable - it is not, and it does a disservice to our reference desk. SemanticMantis (talk) 16:00, 19 May 2016 (UTC)

Agree. And anyway the connection between maths and physics is not at all an obvious one, in fact is rather a mystery. Not quite up to mysteries like what is thought but getting there. Dmcq (talk) 16:08, 19 May 2016 (UTC)

Yes, there's not enough evidence to conclude that the OP is a crank. It seems more likely to me that he (I'm just gonna go with "he") is in over his head and unwilling to admit it. See Dunning–Kruger effect. He might be in danger of turning into a crank, though. I'm willing to bet this is how a lot of cranks get their start.

As for "arguing with" — it's my feeling that when you start to sense someone's in over their head, it doesn't do a lot of good to throw outer measure at them; you have to engage them at the level they have the background for (if you engage them at all) while of course not telling lies to children. --Trovatore (talk) 19:34, 19 May 2016 (UTC)

The whole thing was literally sketched on a napkin during lunch break, so it didn't accounted for simple fact that the theorem is not confined to measureable sets. The two described limitations still apply if you want to produce measureable sets by partitioning, and are easy way to test if one (semi)physical object can be re-arranged into another. Raidho36 (talk) 21:51, 19 May 2016 (UTC)

Sławomir Biały mentioned this earlier, but I think it's an important example to show how this kind of reasoning doesn't apply. Consider a solid three dimensional ball A with volume V. Now let B be the set of points in A with three rational coordinates and let C be all the remaining points. Obviously the union of B and C is the original ball A and has volume V. What is the volume of B and C? Both B and C are dense and might superficially appear to have volume V, but in that case volume isn't additive. You can't apply intuition to things like this. CodeTalker (talk) 20:16, 19 May 2016 (UTC)

There are mathematicians who reject the idea of non-measurable sets and who aren't crackpots. See, e.g., Constructivism (mathematics). -- BenRG (talk) 22:24, 19 May 2016 (UTC)

Constructivism means a lot of different things to a lot of different people, so it's hard to generalize, but just at a very initial level, constructivism doesn't play that nice with partitions of sets in the first place. In Errett Bishop's model, for example, there are no discontinuous functions at all, whereas the characteristic function of one of the partitioned subsets must be discontinuous.

There are constructive set theories, or at least ones that some people who call themselves constructivists accept. I don't know an awful lot about them, but my impression is that they tend to be rather weak, and not easily adaptable to this sort of question.

I think a lot of people look at this sort of question and they say, oh, uses the axiom of choice and uses sets that aren't explicitly constructed. Who doesn't like that? Constructivists. But the picture is a lot more complicated than that. --Trovatore (talk) 23:16, 19 May 2016 (UTC)

There is a Solovay model that satisfies ZFC and under which all sets are measurable and axiom of choice fails. I originally thought the theorem may have flaw in its construction, however I came to think that it can't be that simple and the flaw roots deeper than that. After all, paradoxes don't truly exist and if you encountered one it only means that the models you were using are flawed and need amending. If Solivay model is correct, and only measurable sets exist and/or axiom of choice fails, then the theorem immediately breaks down. And it should, because as it stands, it yields absurd results that are obviously wrong. This theorem is meant to be disproved, through improvement of theoretical basis and elimination of open loopholes that allow this kind of behavior to exist at all.

Now that I established that taking jabs at the theorem is pointless, I shall ask a few related questions to shed some light on the whole thing.

1) Is it correct that when you deconstruct set A into non-measurable subsets (which don't have defined Lebesgue metric) to produce derivative set A' that consists of resulting subsets, then set A' wouldn't have defined Lebesgue metric either? If so, then by what mechanism when you use set A' to reconstruct set B, its Lebesgue metric becomes defined?

2) The theorem postulates that [even] non-measurable subsets from sets A and B are congruent pair-wise. Congruence defined as existence of isometry function between two sets. Isometry defined as distance-preserving injective function. Is it correct that if set doesn't have defined Lebesgue metric (particularly geometric size), then distances can't be preserved and therefore mapping function would be simply injective, but not isometric? If so, since there aren't isometric mapping function between paired subests, they are not congruent. More generally, non-measurable sets are not congruent to any other sets, if previous statement is correct. Raidho36 (talk) 23:21, 19 May 2016 (UTC)

I'm not even going to get to your questions; I'm going to address two things in your first paragraph.

• It's a "paradox" in the sense that it violates your intuition. Sometimes, this means the theory is wrong and needs amending. Other times, it means your intuition is wrong and needs amending. This is an example of the second case. There's nothing wrong with using intuition. It's your intuition, meaning you personally, that needs to be changed.
• As for the Solovay model being "correct" — I think you may not be all that familiar with the use of the word "model" as used here. This is not a model in the sense of "a proposed way of accounting for things", as used in physics. It's a model in the sense of model theory. What it shows, in this case, is that ZF by itself, without choice, cannot prove the BTP. But no one (certainly not Bob Solovay!) thinks that the Solovay model is the true universe of sets. --Trovatore (talk) 23:32, 19 May 2016 (UTC)

If you use properties-preserving function on a set then resulting set must have preserved properties of the original set, by definition of the function you were using. If it doesn't, then it's a paradoxical situation of type I as per your description. You should address the questions though, they must be simple enough to answer. Raidho36 (talk) 23:50, 19 May 2016 (UTC)

They are simple enough, I think, the reservation being that I'm not entirely sure I follow the questions. You are not using language in a standard way and I may guess wrong as to what you mean. To be honest, I'm not sure you have the background to follow the answers. You need to learn some real analysis before it's even worth going into it in this kind of detail. It's not something you can learn in fifteen minutes or on the back of a napkin.

Still, I'll take a crack at it: If you partition A into sets A₀, A₁, A₂, A₃, A₄, which are not Lebesgue measurable, and then you look at copies A'₀ etc under rigid motions, then the A'_n copies are also not Lebesgue measurable. Is that what you meant by question 1? It's hard to tell, but it's my best guess.

Second part of question 1: Just because the pieces are non-measurable doesn't mean that their union is non-measurable. I don't know why you think it should mean that, so I don't know how to help you fix your intuition here. But it just doesn't mean that.

Question 2: There is no reason you can't have an isometry between two sets that aren't Lebesgue measurable. This gets back to the "background" question — I don't think you know what Lebesgue measure is, and as I say, you can't understand it in fifteen minutes. It is not the same as a metric; you seem to be mixing up the two things.

Not mixing up, just using wrong term. So the answers are "no" and "no", that's also simple enough. I must assume the Lebesgue measure of union of non-measurable sets can change because reasons. I get that it is not defined for individual subsets so they are not constrained by anything of the sort, and a union function can therefore produce any set. As a rhetorical question, what defines what set union operation will generate? The reason I think non-measurable sets couldn't have isometry comes from my background, where it is a convention that if something is undefined, it's not equal to anything, not even itself. By that definition of identity between undefined entities, it's impossible to construct isometry between non-measurable subsets since there's no injection function between them, they simply can never be matched. That's of course in addition ot previous statement that if geometry is not defined, the mapping function can not preserve it since it doesn't exist to begin with. Raidho36 (talk) 00:25, 20 May 2016 (UTC)

"[I]f something is undefined, it's not equal to anything, not even itself." That's exactly right! That's why you can't make any argument that goes through the volume of the pieces, because they don't have a volume.

Why you think this has anything to do with the existence of an isometry is still obscure. My best guess is still that you are mixing up the notions of measure and metric. An isometry is a bijection that preserves the metric. It must therefore also preserve measure, for sets that have a measure, but there is no obstacle to having an isometry between sets that don't have a measure. --Trovatore (talk) 02:20, 20 May 2016 (UTC)

I have abandoned the idea of preserving measure when I realized the theorem works with non-measurable sets. It is still important that metric is preserved, so the set must have defined metric spaces to begin with.

Anyway, I had too much time to think lately and I eventually came across a thought experiment: if you take an infinite ribbon, and fold it on itsefl infinitely many times but such that it is not allowed to intersect with itself, do you get a solid block? If you fold it uncountably many times? Ribbon is constructed from an infinite plane by sectioning. To put it another way, what is the "thickness" of a set constructed from a folded surface that's not allowed to intersect with itself? More generally, is it correct that if a surface is not allowed to intersect with itself, then folding operation must put the folded part on points immediately adjacent to ones that the surface already occupies? Raidho36 (talk) 22:03, 20 May 2016 (UTC)

The maps can still be measure-preserving; they'll preserve the measure of any measurable set to which they're applied. The A_n don't happen to be measurable, but the map might be (say) from all of R³ to R³. It carries A_n along for the ride. Any measurable subset of R³ will have an image under the map with the same measure as the original. Even if you restrict the map to A_n, there are still measurable subsets of A_n (maybe measure 0, but measurable), and their images will have the same measure as the original.

As for your "ribbon" question, we did that very recently. See Wikipedia:Reference desk/Archives/Mathematics/2016 May 7#how can lines/curves of infinite length have positive area?. --Trovatore (talk) 22:24, 20 May 2016 (UTC)

Thanks, but I will require elaboration. Does this mean that a flat surface has defined finite thickness, even if infinitely small? Raidho36 (talk) 22:32, 20 May 2016 (UTC)

Depends. Is zero finite? I mean, obviously zero is finite, but somehow physicists seem to have trouble figuring that out :-)

The "width" dimension of your ribbon is superfluous; it's easier to think of it as just curve in the plane. And the answer is that, even though the curve itself has zero "thickness", nevertheless it can have positive area. These curves are, of course, very odd, and fail to satisfy some of the attributes you would expect the English word "curve" to carry with it. Similarly, the "pieces" in Banach–Tarski fail to be entirely piece-like in a natural-language sense. However, unlike the B–T "pieces", these "curves" can be defined completely explicitly, and the argument does not require the axiom of choice. -Trovatore (talk) 23:56, 20 May 2016 (UTC)

If a curve have positive area, then by adding a dimenstion, we get that a plane has positive volume. The thought experiment was constructed to establish whether individual points have positive volume. It was important because I derived such definition of measure that implies that individual points have finite sigma-additive volume. Can you please deconstruct it and highlight where what is wrong with it?

Let fractional measure of set A be a known measure of its superset S consisting of all points bounded by it, multiplied by a number that describes what fraction of S set A constitutes.

Any subset is already a fraction of its respective superset. I reason that there must be a number that describes what fraction of points from a superset is in the subset. And indeed, if you bisect a solid unit sphere, you'll get two hemispheres, each of which contains exactly 1/2 of points of original shpere, and each have exactly 1/2 the volume of it. If you can construct a superset with known measure such that you also know what fraction of points are used to build the set you're "measuring", you can derive it's equivalent volume. In an earlier example, consider a solid unit ball that you partition into two balls such that ball A has points with rational coordinates and ball B has the rest. Intuitively, the balls must have smaller volume than original ball because they're both missing some of its points. Using fractional measure, you can establish that they must in fact have smaller volume precisely because they both don't have 100% of original ball's points. Since, as I imagine, between any two infinitely close rational numbers there is infinitely many real numbers, rational ball's fraction must be infinitely small and real ball's fraction must be infinitely close to 1, so rational ball's volume is infinitely close to 0 and real ball's is infinitely close to volume of original ball. Since you can partition a solid ball such that it's every subset contains exactly one point, and they must add up to the volume of original ball since they are all a certain fraction of it, then individual points must have defined, finite and sigma-additive volume. Which sounded absurd to me since that would mean that surface of a set has its own volume and volume of a closure of the same set would be greater by that amount. But I reasoned that if a point didn't had any volume, you must have been able to produce infinitely many disjoint sets containing this point, as per "stacking" thought experiment, which is a contradictory statement and thus the clause automatically resolves to "false". Additionally, since fractional measure is sigma-additive, then there must also exist fractional measure for sets assumed to exist in Banach-Tarski theorem, such that they will add up to the volume of the original set, so you wouldn't be able to partition sets of different volumes such as that they would have produced subsets with identical total fractional volume. This seems logical so I hope you can point out flaws in all of this, most notably resolve the whole "points have defined volume" thing, because I've been assured they do not.Raidho36 (talk) 22:32, 20 May 2016 (UTC)

A ball is not a countable set. Its measure is not the sum of the measures of its points. Although the measure of any countable set is zero. An idealized ribbon of zero thickness is a rectifiable surface, so must have zero volume. You cannot construct a non-rectifiable surface by folding a ribbon. Sławomir Biały (talk) 00:29, 21 May 2016 (UTC)

This is exactly why thought experiment was constructed. If a ribbon can't intersect, its fold can't lie in exactly the same points, it must therefore lie on adjacent points, and the resulting object, instead of remaining flat, is now 2 points thick. It will become twice as thick every time it is folded. Original ribbon may have zero volume, but resulting object would be a solid block consisting of all points in its region. That implies that each individual point must have defined positive measure. If they didn't, the object would have remained flat, which would have violated the definitions by which it was constructed.

Also, consider that this definition of measure makes it sigma-additive for all sets. I would really like to know why it isn't valid. Raidho36 (talk) 00:47, 21 May 2016 (UTC)

Isn't this a case of trying to figure out what 0 times infinity is?Naraht (talk) 00:52, 21 May 2016 (UTC)

There is no such thing as "adjacent" real numbers. And there is no continuous one-to-one function from a ribbon to a cube. Sławomir Biały (talk) 01:11, 21 May 2016 (UTC)

It's important that Raidho36 understand that this is a theorem of ZFC, and it follows inevitably from ZFC's axioms, and intuition doesn't matter.

But don't tell them that their intuition is wrong and needs to be changed. Their intuition is fine, unless they want to work in set theory. One might even call it correct. -- BenRG (talk) 06:38, 20 May 2016 (UTC)

No, that intuition is indeed wrong and needs to be changed. This is a real fact about geometry, not just set theory. It can't be realized physically, as far as we know, but that is a different matter. --Trovatore (talk) 06:48, 20 May 2016 (UTC)

I don't know where that belief comes from, but it isn't from anything we've ever learned about the real world through the evidence of our senses. -- BenRG (talk) 07:02, 20 May 2016 (UTC)

It's the "realer" world than the physical one. See mathematical realism. --Trovatore (talk) 07:26, 20 May 2016 (UTC)

Oh god, here we go again :-D. But seriously, although non-measurable sets are not "constructible", is there any way to have even a picture of one in mind, to guide one's intuition? A fractal, for example, is described by some kind of similarity on different scales. Is there any analogous way to picture the essence of a non-measurable set, in human terms? This was why I thought about Hecke algebras, that there is some essentially arithmetic aspect to groups like R/Q. The book on Borel equivalence relations that you referred me to doesn't seem likely to lead to any useful geometrical intuition. Sławomir Biały (talk) 11:26, 20 May 2016 (UTC)

There's definitely some interesting mathematics involved in trying to understand non-measurable sets. For example, one of the standard constructions of a non-measurable set is to consider the abelian group $\mathbb {R} /\mathbb {Q}$. This is a bit of a weird group to try to understand: it consists of equivalence classes of real numbers x and y, where we identify two reals if they differ by a rational. So this space has an infinitesimal character: every open interval containing 0 has a complete set of representatives of this equivalence relation. In terms of radix expansions, we identify two real numbers if their difference has a terminating base n expansion in some integer base n. A choice of a single representative of each equivalence class defines a subset of the real line, which is non-measurable. In terms of groups, this is a splitting of the epimorphism $\mathbb {R} \to \mathbb {R} /\mathbb {Q}$.

A basic trend in mathematics is that instead of trying to study a space directly, it is often better to look at functions on the space. Measurable functions, for example. Because Q acts ergodically, any measurable function on R invariant under Q is constant. I believe this implies that the algebra of measurable random operators on $\mathbb {R} /\mathbb {Q}$ is a type II von Neumann algebra. I believe there is some Hecke algebra construction as well. Is there an account of this anywhere? Sławomir Biały (talk) 23:48, 19 May 2016 (UTC)

This is a very good starting point for the study of Borel equivalence relations. Excellent references are Vladimir Kanovei; Borel equivalence relations. Structure and classification. University Lecture Series, 44. American Mathematical Society, Providence, RI, 2008. x+240 pp. ISBN 978-0-8218-4453-3 and Su Gao; Invariant Descriptive Set Theory. Chapman & Hall/CRC Pure and Applied Mathematics. ISBN-13: 978-1584887935. --Trovatore (talk) 09:11, 20 May 2016 (UTC)

Thanks for that. I'm afraid that treatment is a bit orthogonal to my own interests, but it looks like the paper "Ergodic equivalence relations, cohomology, and von Neumann algebras" by Moore and Feldman, referenced there, looks readable to me. Sławomir Biały (talk) 10:17, 20 May 2016 (UTC)

Actually, this example is discussed in part I of Murray and von Neumann from the point of view of von Neumann algebras. I haven't been able to find any treatment from the Hecke algebra perspective though. Sławomir Biały (talk) 10:42, 20 May 2016 (UTC)

Reality

One comment that I've heard on the Banach-Tarski paradox is that Reality is simply a crude approximation to some set of Mathematical Axioms. To me, the paradox is simply a close cousin to the fact that the even integers and the odd integers both have the same cardinality as the integers (which is similar in that the set is cut into two pieces which can both be mapped back to the original). What I think throws people off is that the "volume" of the sphere is bounded, *but* the mapping in the paradox uses the fact that there are an infinite number of zero-measure points, and comes up with a cool way to both cut them into 4 groups (actually a few more, in the exact proof) such that there is always a point in the old sphere that can map to anyone you need in each of the new spheres. Since in all cases, a point that maps into sphere A has a point that maps into sphere B less than epsilon away, looking at the volume of the subsets before they get mapped really doesn't lead to anything useful.Naraht (talk) 00:29, 21 May 2016 (UTC)

May 20

Power tower

Let $x=(5/10)^{(7/10)^{(9/10)^{(11/10)^{...}}}}$. Does $x$ converge to a finite value? 24.255.17.182 (talk) 05:49, 20 May 2016 (UTC)

0. The first 3 terms equal 0.646176... and then the powers exceed 1 so every step gets smaller. -- SGBailey (talk) 07:17, 20 May 2016 (UTC)

Well, but how do you know they get smaller without bound (above 0)? They could still approach a nonzero limit. They don't, I think, if I've understood the rule for the successive exponents, but "every step gets smaller" is not a sufficient argument. --Trovatore (talk) 07:30, 20 May 2016 (UTC)

Oh, wait, I was associating the exponents backwards. Oops. That makes it harder. Not sure of the answer; will think about it more. --Trovatore (talk) 07:32, 20 May 2016 (UTC)

I think SGBaily is right that, after the first three steps, the part past 9/10 increases (without bound? I think so). In that case, the part starting with 9/10 goes to 0. So in the limit, the expression approaches $(5/10)^{(7/10)^{0}}=(5/10)^{1}$. Still some details to check.... --Trovatore (talk) 07:37, 20 May 2016 (UTC)

You are correct. SGBailey forgot that exponentiation is right-associative.

If you have a power tower $a^{a^{a^{a\ldots }}}$, it goes to infinity for every $a>e^{1/e}=1.44467...$. So in the OP's power tower, the part that starts with 15/10 is greater than the power tower of all 15/10, so it is infinite. This means the parts starting with 13/10, and with 11/10, are infinite.

Then that part starting with 9/10 goes to 0, 7/10 goes to 1, and the whole tower becomes $5/10=1/2$. -- Meni Rosenfeld (talk) 08:33, 20 May 2016 (UTC)

Knot theory

Can knot theory help people untie complex knots in the real world? I know knot theoretic knots are closed but do they have applications to real knots? 2001:630:12:2428:7139:A0F9:1688:C7C2 (talk) 15:30, 20 May 2016 (UTC)

Somewhat. Here's an interesting paper [3] using knot theory to understand real-world knots arising from random agitation. Turns out prime knots are highly over represented compared to compound knots! My WP:OR is that understanding basic mathematics of basic knots helps me deploy knots better in the real world. Knot sum and satellite knot (and analogous operations) can be used as a sort of syntax that can be useful doing something like securing a load or tying up a bear bag. Real-world knots are often examples of tangle_(knot theory), and things analogous to Reidemeister moves help in untying real knots. Basically, untying real knots is a practiced skill. Like any practiced skill, knowing some theory can help, but the extent to which it helps is debatable and a somewhat a matter of opinion. If you know a little knot theory, then I suggest you start learning more about real knots and then you can decide for yourself :)

Now, you didn't ask this, but there are some cool applications of knot theory to real-world scientific problems where things are knot-like. DNA is a prime example. Knot theory can be used to understand structural/informational properties of DNA supercoiling, how circular DNA moves in gel electrophoresis, and lots of other things. If you want more along those lines let me know and I'll be happy to provide refs. SemanticMantis (talk) 16:04, 20 May 2016 (UTC)

Equal distribution (infinitely many sides on a die)

(moved from science desk) If I roll infinite sided die with faces labelled from zero to one are the odds of it coming up higher than, say, 0.8 equal to it coming up lower than 0.8? I mean there as many numbers above 0.8 as there are below so it seems like it they should be. 88.164.119.137 (talk) 14:52, 20 May 2016 (UTC)

probably move to math...do you mean the die will have every real number between 0 and 1 (an infinite sum, I do believe)??..I'm curious how this will be answered..it's possible the question will be considered nonsensical..68.48.241.158 (talk) 15:16, 20 May 2016 (UTC)

The article almost surely almost surely provides some background to the problems when dealing with the intersection between infinite sets and finite sets. --Jayron32 15:19, 20 May 2016 (UTC)

It depends a bit on how you conceptualize this die. But the reasonable and naive thing to do is to identify it with the uniform distribution on [0,1]. That is analogous to the fair die in the case of finitely many sides. In that case rolling the die is the same as throwing a dart (whose tip is a point) at the unit interval. The probability of the dart hitting any given number is zero, and the probability of it hitting an interval is the size of that interval. So, for your example, the probability of getting a roll in (0.8,1] is 0.2, while P(x in [0,0.8))=0.8. The number of points in the interval is not as important as the size or measure of the interval. This is the distinction between cardinality of a set versus the measure of a set. So the measure of the set is the size of the event_(probability), and we use that to compute probabilities of given events. For intervals, the measure of the interval is just its length (l((a,b))=|b-a|), but for more complicated sets/events, you'd use tools form measure theory to compute the probabilities. SemanticMantis (talk) 15:43, 20 May 2016 (UTC)

op here. Thank-you for that clear and well sourced response. Does the same reasoning hold for a countably infinite sided die (faces labelled with, say, all the positive integers)? 92.90.17.111 (talk) 15:59, 20 May 2016 (UTC)

yes the cardinality couldn't matter...if infinite the chance of landing on a particular number would still have to be zero...and the discussion as far as measure would have to be same too...(it's possible there are fine points involved here that haven't been mentioned)..68.48.241.158 (talk) 16:10, 20 May 2016 (UTC)

or maybe I'm wrong as far as the measure as there's no defined interval like between 0 and 1...?? But I'm not sure this question is technically sensical..68.48.241.158 (talk) 16:17, 20 May 2016 (UTC)

I think the move from [0,1] to $\mathbb {N}$ is importantly different - the former accepts a uniform distribution while the latter does not. Never underestimate the weird things that can happen when switching back and forth between sets of countable and uncountable cardinality ;) SemanticMantis (talk) 16:39, 20 May 2016 (UTC)

You're welcome! For the follow up - No, I don't think so. What's the uniform distribution on the natural numbers? We can define it on an interval, but not on the whole line, and not on the positive integers. To put a probability distribution on a set we have to have an integrable function on it. For a discrete uniform distribution, we just set the probability at 1/n for n many choices. But there's no obvious way to extend that to all $\mathbb {N}$, because the infinite sum of a finite value will always be infinite. See here [4] [5] for some discussion of why we can't have uniform distributions on the whole line. It comes down to the axioms of probability. Improper priors are a way to look at things that are almost like probability distributions, but are not. Here [6] is a paper that examines things that are in some sense "almost uniform" distributions on the natural numbers. I haven't read that but it does look promising. Another thing to do would just be to have the die not be fair. Then you could put e.g. a Geometric_distribution on it. If you want more on this sub-question, I do recommend making a new post on the math desk, and including a link to this question if you do so. SemanticMantis (talk) 16:39, 20 May 2016 (UTC)

Thank-you for another excellent and interesting response. If I knew how to link to this question, I would certainly do as you suggest for my followup and repose the same question on the math desk. 2A01:E34:EF5E:4640:CDC3:3824:8EB5:8B7C (talk) 17:22, 20 May 2016 (UTC)

You know, now I'm curious what anyone else might come up with for the countably many sides case, i.e. analogs of a uniform distribution on N. So I'm moving it over presently. SemanticMantis (talk) 17:49, 20 May 2016 (UTC)

further discussion

Anyone have any other ideas for how a "fair" $\mathbb {N}$-sided die might behave or be modeled? Thanks, SemanticMantis (talk) 17:49, 20 May 2016 (UTC)

See natural density. --JBL (talk) 18:18, 20 May 2016 (UTC)

So that is roughly "fair", but fails to be countably additive, right? I think it would be finitely additive though, and if so, then it fits in well with the Schirokauer and Kadane piece I linked above, though I'm not sure if the natural density would satisfy the various uniformity and shift invariance properties they discuss. The idea that all finite sets have density zero and arithmetic progressions ax+b have density 1/a matches pretty well with what we'd intuitively want though, and this is a conceptually easier notion than the things they discuss. Thanks! SemanticMantis (talk) 19:39, 20 May 2016 (UTC)

Yes, that's right: finite sets have density 0, but of course countably infinite sets can have positive density, so no countable additivity. In the Schirokauer--Kadane piece, I think you want to see Example 2.5 to find the (different) name by which they are calling natural density. --JBL (talk) 22:25, 20 May 2016 (UTC)

My comprehension falters already on reading "infinite sided die with faces labelled" where I don't distinguish between "sides" and "faces" and can imagine only a spherical ball with no place on which to write a label. (A so-called spherical dice that has internal trickery that loads it to settle in a finite number of positions seems ruled out.) Unless the die is constrained to roll in only one direction, its settling state after rolling needs to be defined in two, not one, dimensions. If we are interested in the Euclidean norm of the complex-number settling state of a sphere of unit circumference, it lies in the range 0 to 1.4142... and the value 0.8 is somewhat displaced from the center of the range at 56.57...%. AllBestFaith (talk) 22:42, 20 May 2016 (UTC)

I'm not a mathematician but...what!? Is this for real???68.48.241.158 (talk) 01:07, 21 May 2016 (UTC)

No, there's nothing of value. --JBL (talk) 01:29, 21 May 2016 (UTC)

Polynomial optimization problem

Description of the problem: Given $f:\mathbb {C} \rightarrow \mathbb {R}$ and $a\in \mathbb {R}$, find a polynomial $p$ with integer coefficients such that $p$ has a root $r$ with $f(r)=a$ (and constrain the degree and range of coefficients). What kind of algorithms exist to find a set of coefficients within the range such that $|f(r)-a|$ is minimized, beyond bruteforce? 24.255.17.182 (talk) 01:35, 21 May 2016 (UTC)