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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)

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)

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)

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)