Exist other Space Filling Mathematical Structures in the 4th Dimension except the x⁴-Hypercube???...

THANK you...

SPYROU Kosta - Greece--85.74.189.98 (talk) 05:02, 27 April 2014 (UTC)[reply]

Yes. There are three regular space-filling tessellations (or honeycombs) in 4 dimensional Euclidean space. They are the tesseractic honeycomb, the 16-cell honeycomb and the 24-cell honeycomb. There are many other non-regular honeycombs too. Gandalf61 (talk) 12:10, 27 April 2014 (UTC)[reply]

George Olshevsky has listed 143 uniform tilings. Of these, thirteen (including the three cited by G61 above) have a single kind of tile, which may be what the OP seeks. The duals of all 143 are also hyperspace-filling tilings with congruent cells (up to reflection). —Tamfang (talk) 18:41, 27 April 2014 (UTC)[reply]

1. The Cube has 3(-)Edges on a Vertice(.)

2. By the space-filling Cube they are 6(-)Edges on a Vertice(.)

1. The Truncated Octahedron has 3(-)Edges on a Vertice(.)

2. By the space-filling Truncated Octahedron they are 4(-)Edges on a Vertice(.)

I want to know how much (-)EDGES has a Vertice(.) by:

a1. The 8-cell Hypercube has 4(-)Edges on a Vertice(.)

a2. The Hyper-space-filling 8-cell Honeycomb (Tesserakt - Octachoron - Hypercube)

b1. The 16-cell

b2. The Hyper-space-filling 16-cell Honeycomb

c1. The 24-cell

c2. The Hyper-space-filling 24-cell Honeycomb

THANK you VERY MUCH!!!...

Kostas SPYROY - Greece: "Have a nice Day!!!..." — Preceding unsigned comment added by Honeycomp (talk • contribs) 14:54, 27 April 2014 (UTC)[reply]

The number of edges at a vertex is the number of vertices of the vertex figure.

16-cell {3,3,4} (vertex figure is an octahedron): 6.

24-cell {3,4,3} (vertex figure is a cube): 8.

Hypercube tiling {4,3,3,4} (vertex figure is a 16-cell): 8.

16-cell tiling {3,3,4,3} (vertex figure is a 24-cell): 24.

24-cell tiling {3,4,3,3} (vertex figure is a hypercube): 16.

—Tamfang (talk) 18:49, 27 April 2014 (UTC)[reply]

Incidentally, the standard singular form of vertices is actually vertex. Similarly, apex has a plural form apices. Double sharp (talk) 14:55, 28 April 2014 (UTC)[reply]

Let ${\displaystyle \Gamma _{a}(x+a)=x\cdot \Gamma _{a}(x).}$ Then for ${\displaystyle a=1}$ we have ${\displaystyle \Gamma _{_{1}}(x)=\Gamma (x),}$ whose integral expression is ${\displaystyle \int _{0}^{\infty }{\frac {t^{x-1}}{e^{t}}}dt.}$ I was wondering whether such expressions also exist for this generalized version of the ${\displaystyle \Gamma }$ function. — 79.113.194.139 (talk) 04:30, 28 April 2014 (UTC)[reply]

Yes, it's just ${\displaystyle \Gamma _{a}(x)=\int _{0}^{\infty }{\frac {t^{x-a}}{e^{t}}}dt.}$ Widener (talk) 08:27, 28 April 2014 (UTC)[reply]

I think you answered the question that was literally asked, but perhaps not the one that was intended. My guess is that the OP meant to write

${\displaystyle \Gamma _{a}(x+a)=x\cdot \Gamma _{a}(x)}$

How about it, 79.113? Did I guess right? --Trovatore (talk) 09:15, 28 April 2014 (UTC)[reply]

Yes. Sorry. — 79.113.194.139 (talk) 09:44, 28 April 2014 (UTC)[reply]

Then that recursive relationship doesn't uniquely define ${\displaystyle \Gamma _{a}}$. What's the boundary condition? The gamma function was motivated to coincide with the factorial. What is the motivation behind ${\displaystyle \Gamma _{a}}$? 203.45.159.248 (talk) 09:59, 28 April 2014 (UTC)[reply]

Assuming you also want ${\displaystyle \Gamma _{a}(0)=1}$, then ${\displaystyle \Gamma _{a}(x)=a^{x/a}\Gamma (x/a)}$. Sławomir Biały (talk) 11:31, 28 April 2014 (UTC)[reply]

Wow! Unbelievable as always, Slawomir! :-) If you could you also please explain the logic and/or intuition which helped you arrive at this expression ? Thanks! — 86.125.209.133 (talk) 17:56, 28 April 2014 (UTC)[reply]

From the equation ${\displaystyle \Gamma _{a}(x+a)=x\Gamma _{a}(x)}$, I see that the solution should be something like ${\displaystyle \Gamma (x/a)}$. This doesn't quite work, so I try to write ${\displaystyle \Gamma _{a}(x)=f(x)\Gamma (x/a)}$. Imposing the functional equation again gives ${\displaystyle f(x+a)=af(x)}$ which has ${\displaystyle f(x)=a^{x/a}}$ as a solution. (Clearly there will be many functions solving this; I'm not sure what conditions are needed to ensure uniqueness.) Sławomir Biały (talk) 22:02, 28 April 2014 (UTC)[reply]

Resolved

The Cube is a Space-filling Solid...

The Dodecahedron is NOT!!!

Four Dodecahedrons have a empty Corner: 0xyz with 1,5^o

Could it be a SPACE with Space-filling Dodecahedrons???...

Could it be a SPACE where the Icosahedron = 20 Tetrahedrons???...

I can Imagine them BUT could they be Calculated???...

Are these "Mistakes" say something about our Universe???...

THANK you VERY-VERY much!!!...

"Have a nice Day/Night..."

SPYROY Kosta - Greece - Honeycomp (talk) 14:36, 28 April 2014 (UTC)[reply]

In hyperbolic space, you can tile appropriately scaled dodecahedra. Their dihedral angles vary according to their size, so you can make them have dihedral angles of precisely 90° (order-4 dodecahedral honeycomb), 72° (order-5 dodecahedral honeycomb), 60° (order-6 dodecahedral honeycomb), ... , 0° (infinite-order dodecahedral honeycomb). In elliptic space, you can also tile appropriately scaled dodecahedra 3 ({5,3,3}, 120-cell) or 2 ({5,3,2}, dodecahedral dichoron, each dodecahedron takes up a 3-hemisphere) at a corner. Double sharp (talk) 14:47, 28 April 2014 (UTC)[reply]

Though if the order is more than 6 the vertices stick out beyond infinity. And as to {5,3,2}, how do you distinguish it from a great sphere? —Tamfang (talk) 03:13, 29 April 2014 (UTC)[reply]

It's analogous to the pentagonal dihedron {5,2}: the faces of the dodecahedra in {5,3,2} tile a great sphere, just as the sides of the pentagons in {5,2} tile a great circle. Double sharp (talk) 15:24, 29 April 2014 (UTC)[reply]

I don't think this says anything significant about our universe - Euclidean 3-space is only a local approximation of its actual geometry. There's only one space-filling fully regular honeycomb in 3-space - the cube, which has three sets of parallel faces. So if you want to consider the philosophical impact of this, I guess the question is "Why cubes?" AlexTiefling (talk) 15:57, 28 April 2014 (UTC)[reply]

You are probably thinking of the rhombic dodecahedron. This is well-known to fill 3-space, and can be thought of as the 3-space analogy of the regular hexagon (which fills (tessellates) 2-space), though it is not regular. RomanSpa (talk) 19:11, 28 April 2014 (UTC)[reply]

It's sufficiently clear, in my humble opinion, that that's not what he has in mind. —Tamfang (talk) 03:13, 29 April 2014 (UTC)[reply]

Here are two views of a curved space filled with regular dodecahedra [1] [2] — and two views of a space where 20 tetrahedra form an icosahedron [3] [4] —Tamfang (talk) 06:08, 29 April 2014 (UTC)[reply]

Consider some one-way function F(X) that satisfies both F(A + B) = F(A) + F(B) and F(A − B) = F(A) − F(B). Are these properties useful for anything? -Sebastian Garth (talk) 03:44, 29 April 2014 (UTC)[reply]

This is an additive map (also called an additive function, in the first sense given in the lede of this article). A special, but particularly useful case of this is a linear map (also called a linear function). Whenever subtraction is a binary operation (particularly, that it is defined for all values of its inputs, as it would be for integers and real numbers), your first property implies the second, so it would be unnecessary to list it separately. —Quondum 04:39, 29 April 2014 (UTC)[reply]

On second reading, it seems improbable that you'd find a one-way function with this property, if the function's domain and codomain are in the same set, and the addition operation is the same. Of course, there are examples where the domain and codomain are different, for example F(x) = g^x (which also can also be written xg, if the group operation is written as "+"), where g is a generator of some abelian group (with x an integer). This has extensive application in public key cryptography. —Quondum 04:55, 29 April 2014 (UTC)[reply]

Quondum, do you mean a function with this property, but without any other reasonable property (in particular, not bijective) with domain and range being the same is hard to find? YohanN7 (talk) 13:28, 29 April 2014 (UTC)[reply]

I think what he's saying is if the domain of a function and its codomain are both contained in a certain set, and if the '+' and '-' symbols mean the same thing on both sides of the OP's equations, then any linear map is trivial to invert, and thus is not a one-way function. I could be misunderstanding Quondum, and there might be tricks I'm not aware of, but non-esoteric or non-exotic linear maps are usually easy to invert. SemanticMantis (talk) 14:56, 29 April 2014 (UTC)[reply]

While being interesting, the question of whether a function is easy to invert might be besides the topic. Then again, it might not be. The OP asks if they are useful. Functions whose existence relies, for instance, on the axiom of choice are probably not useful in any sense of the word. But their mere existence (and abundance!) is interesting in every sense of the word. YohanN7 (talk) 15:11, 29 April 2014 (UTC)[reply]

Isn't "hard to invert" (for some practical definition of "hard") a necessary feature of a one-way function though? I thought that was the heart of the matter. Surely, if you can use the axiom of choice to define a linear map that is nearly impossible to invert, that would be very interesting, though perhaps of little practical use :) SemanticMantis (talk) 15:23, 29 April 2014 (UTC)[reply]

My interpretation is with SemanticMantis's here. Actually, every additive map is Z-linear, so linearity is maybe not so interesting. Not being bijective is also not necessarily interesting, since this includes every linear map with a nontrivial kernel. In any event, functions with the properties listed by the OP (additivity and one-wayness) are precisely the properties exploited by, elliptic curve cryptography, digital signatures, key agreement protocols etc. So I'd say the answer to the OP's question is an unequivocal "yes". —Quondum 16:08, 29 April 2014 (UTC)[reply]

Oops, I've managed to miss completely that One-way function is part of the OP question. (I blue-linked it in the title so I wont make the same mistake again.) Please just disregard what I have written here if it appears strange. It was written under other premises. YohanN7 (talk) 16:27, 29 April 2014 (UTC)[reply]

Heh-heh – you needn't feel alone. As you may glean from my first post, I initially did the same. —Quondum 16:51, 29 April 2014 (UTC)[reply]

So then how could this could be used to create, say, a public key protocol? -Sebastian Garth (talk) 17:00, 29 April 2014 (UTC)[reply]

As an example, the ECC algorithm is based upon a one-way function that maps integers to "points". There is an addition operator for each, with a homomorphism from the integers to points. This satisfies your description. Given two points and one integer that maps to the one point, you can subtract the points, but you cannot determine what integer to add to the first integer so that the sum maps to the second point. Multiplication of a point by an integer to produce a point is defined by repeated addition. Pretty much every primitive building block in public key cryptography uses essentially this or similar properties. I can't give you a lecture here, but you should be able to learn quite a bit from the articles that I linked, and articles that those link to. —Quondum 18:51, 29 April 2014 (UTC)[reply]

Okay, thank you! -Sebastian Garth (talk) 18:58, 29 April 2014 (UTC)[reply]

You have to do something a bit more than just add the results to get a one way function. If F(1)=b then F(A) will be Ab with the rule given so we just need to divide by b to give the value of A. But see Homomorphic encryption for how people are dealing with this sort of thing. Dmcq (talk) 11:09, 30 April 2014 (UTC)[reply]

Divide by b? When F is a one-way function, this is equivalent to finding a discrete logarithm. —Quondum 00:28, 1 May 2014 (UTC)[reply]

Can a scientific calculator be expected to have a function to allow me to enter 4.6 where the 6 is recurring? I know I could enter 4+(6/9) but I was wondering whether there was an alternative to that. My model is Sharp EL-506V. — Preceding unsigned comment added by 129.215.47.59 (talk) 11:16, 29 April 2014 (UTC)[reply]

I've spent a couple of minutes starting at a Google Images picture of that calculator and I can't see a button for it, I'm afraid. I couldn't say for sure that it's not there, though; hopefully somebody else could confirm that for you. If it helps, I own a Casio fx-83GT PLUS (...Actually, I own six of them...) and there's an icon of a square with a dot above it which can be used for recurring decmials. JaeDyWolf ~ Baka-San (talk) 12:44, 29 April 2014 (UTC)[reply]

Well, any recurring decimal is equivalent to a fraction. At least some scientific calculators allow you to enter mixed numbers, i.e. whole number and fraction. On my HP calculator you can type 4.6.9 to mean 4+6/9 (which, of course, is equivalent to 4.6 recurring). But it's just a shorthand for directly typing 4.66666666667, as the calculator stores only 12 significant digits. If you want a calculator that can store non-integers in fraction (or recurring decimal) form to full precision, that's another matter. --50.100.193.30 (talk) 04:29, 30 April 2014 (UTC)[reply]

My elderly Casio fx-115M can handle rationals with up to three digits in each field and eight digits in all: 999 98/999 or 99 998/999. (If the result of a rational operation exceeds that format, it falls back to floating point, with ten digits.) —Tamfang (talk) 06:28, 30 April 2014 (UTC)[reply]

If you don't already know this, any recurring decimal 0.abcdabcd... may be written (like your 6/9) as ${\displaystyle {\frac {abcd}{10^{n}-1}}}$ (multiplied by another negative power of 10 in cases like ${\displaystyle 0.00142857142857\dots =10^{-2}{\frac {142857}{999999}}}$), which makes your method more useful because more general. Many calculators could be trivially programmed to interpret input like repeating(p,n,o) as ${\displaystyle 10^{-o}{\frac {p}{10^{n}-1}}}$ (so that repeating(8.1,54,3,2)=8.10054054…), but it's clumsy because you have to specify the period (3) and the offset (1) of the repetition. --Tardis (talk) 00:21, 1 May 2014 (UTC)[reply]

Hello Wikipedians.

I am not a mathematical person. I did however wonder at the bus today: For every expression of x^y = z, perhaps y 1,2...n - is there an expression of ax^by = z? Like the way 3^2 = 9 and 3^3 = 27, you could have 1.5*3^2*b = 9 and 1.5*3^3*b = 27. (Or just as well: ax^by = cz.

I don't know if this has any significance. I suspect that if there is an 'a' and 'b' that you can cram in there, they're going to be rather special numbers rather than normal whole numbers. Thank you in advance for any help. 83.108.57.148 (talk) 19:10, 29 April 2014 (UTC)[reply]

OP here, forgot to finish the full subject. I seem to remember you shouldn't adjust those, so there we are I guess. 83.108.57.148 (talk) 19:11, 29 April 2014 (UTC)[reply]

Sure, in fact there are infinitely many a,b that will work. Determining if or how you can get a and b to be integers is much more difficult and subtle (e.g. Diophantine equations). But let's just look at the algebra. x,y are constants.

Then, if we assume ${\displaystyle x^{y}=z}$, we seek to find a, b such that ${\displaystyle ax^{by}=z}$. So, take logarithms of each side, and use properties of logarithms to isolate b. ${\displaystyle ax^{by}=z\rightarrow x^{by}=z/a\rightarrow \log(x^{by})=log({z/a})}$

Then, pull the exponent down, you get ${\displaystyle by\cdot \log(x)=log(z/a)\rightarrow b={\frac {\log(z/a)}{y\cdot \log(x)}}}$.

So, for any (positive real number) a you pick for the right hand side, you can calculate a b (using the known values of x,y,z) that will satisfy the original equation. Make sense? SemanticMantis (talk) 21:15, 29 April 2014 (UTC)[reply]

Ah, a diophantine equation. Brilliant! Thank you for that. 83.108.57.148 (talk) 07:18, 30 April 2014 (UTC)[reply]

You're welcome, glad I could help! Seriously though, Diophantine equations are hard (even for experts in math who don't know much number theory, like me). The best way to start would be to get an introductory book on number theory. I recommend any book published by Dover, they tend to be decent-to-high quality, and rather inexpensive too! Here's one I found searching google books for /Dover number theory/ [5]. SemanticMantis (talk) 16:39, 30 April 2014 (UTC)[reply]

On my phone, one security option is to swipe (without removing the finger from the screen) a pattern on a three by three pattern of dots. There must be at least 4 dots in the pattern. If that was only limitation, then the number of possible swipe passwords would be (for 9 through 4 dots) 9! + 9! + 9!/2! + 9!/3!+ 9!/4! + 9!/5!.

BUT it will not allow you to avoid a center dot when going from one side to another of it. So for example, the following sequences (numbering the dots 1-2-3 across the top, 4-5-6 across the middle, 7-8-9 along the bottom) are not legal 1-3-6-2 since it couldn't get from 1 to 3 without picking up the 2. OTOH, if the center dot of the three has already been used, you can cross, so 2-6-3-1 *is* a legal sequence. (Similarly going from 9 to 3 picks up 6 and from 3 to 7 picks up 5, etc) How many legal sequences are there?Naraht (talk) 20:21, 29 April 2014 (UTC)[reply]

The last bit makes it a little complicated. Without that "skip over an already used number", the answer would be the number of Self-avoiding walks of length >3, on a 3x3 integer lattice. SemanticMantis (talk) 21:20, 29 April 2014 (UTC)[reply]

The fact that the path can go from 2-6 or even 2-9 (though you have to be pretty exact to do a knights move) means that the SAW concept doesn't seem to work.Naraht (talk) 21:28, 29 April 2014 (UTC)[reply]

Right, that's why I said "without" that rule. Still, computing the number for length 9>L>3 SAW will be a good exercise, and give a decent lower bound. SemanticMantis (talk) 22:22, 29 April 2014 (UTC)[reply]

And by computing, you mean having a computer actually run all of the possibilities, right? I can't find any formula...

I think you can calculate such a thing using careful combinatorics and maybe some graph theory. This paper uses analytic methods, but seems to only work with infinite lattices [6]. This paper introduces an algorithm for counting SAW, and I suspect there should be a way to restrict to the simple 3X3 case that you're interested in [7]. In this case, I think you can work it out by hand. Use some arguments from symmetry. Say we just want to count the length 4 walks. First, count all the walks starting at 1. Then that number should be the same for those walks starting at 7,9,3-- the corners. Similarly, the number of walks starting at 2 should be the same as 3,6,8. The walks at 5 only contribute once. You can probably keep track of all of them with a tree_(mathematics) structure, either on paper, or on a computer. SemanticMantis (talk) 19:37, 30 April 2014 (UTC)[reply]

• Actually, someone may have found the right answer here [8], I don't have time to read through it carefully. SemanticMantis (talk) 19:43, 30 April 2014 (UTC)[reply]

I agree that each of the lengths can be split into 4*starts at corner+4*starts at center edge+starts at center with additional symmetry splits.

The way he describes the situation would allow 1-3-2-6, but I don't know how, that wouldn't seem to work on phones like that, but I'll contribute to that page.Naraht (talk) 15:49, 1 May 2014 (UTC)[reply]

I am working on some interesting problem I thought about:

Take an n x n square grid where the squares are colored black and white in an alternating fashion, i.e. like

I want to determine the number of all possible large squares which can be constructed by rearranging the small black and white squares that remain incongruent under rotations and reflections for each n. For example, it is easy to see that for n = 2 there are only 2 possible configurations, namely

and the same configuration where one black and one white square switched places.

For n = 3 I discovered that there are 22 possible configurations. My question is, is there a method for determining the number of possible configuration for arbitrary n? The number appears to be growing quickly for increasing n. For example, for n = 4, I already found over 100 configurations by rearranging the squares 'by hand' (I don't know how many there will be), but this method seems to become practically infeasible for larger n. Is there perhaps a known formula for computing those values? -- Toshio Yamaguchi 10:18, 1 May 2014 (UTC)[reply]

Here is an illustration depicting the 22 possible configurations for n = 3.

-- Toshio Yamaguchi 11:30, 1 May 2014 (UTC)[reply]

Also as a note: I just realized you could get more rotationally and reflectionally incongruent squares by 'inverting' the colors. I wouldn't count a configuration and it's inverse as two separate configurations because they are combinatorially identical. Therefore, the correct number for n = 3 remains 22 for my purposes. -- Toshio Yamaguchi 12:19, 1 May 2014 (UTC)[reply]

There are k black fields. If n is even then k=n²/2. If n is odd then k=(n²+1)/2. The number of ways to choose k fields is the binomial coefficient ${\displaystyle {\tbinom {n^{2}}{k}}}$. There are 8 rotations and reflections, so the number of configurations S_n satisfies the inequality ${\displaystyle {\tbinom {n^{2}}{k}}/8\leq S_{n}\leq {\tbinom {n^{2}}{k}}}$. So 16 ≤ S₃ ≤ 126 and 1609 ≤ S₄ ≤ 12870, in agreement with your result S₃ =22 and S₄ > 100. This is hopefully a step in the right direction. Bo Jacoby (talk) 13:16, 1 May 2014 (UTC).[reply]

Mhm, thanks. I hope to find a way to determine the exact solutions for each n (or a number of them, there are obviously infinitely many). Do you think it might make sense to represent this as a tree or something similar? One can always switch positions of squares, two at a times. If this is repeated, this seems to lead to a tree-like structure where 'branches' always split into two 'sub-branches'. Just some intuition. -- Toshio Yamaguchi 15:33, 1 May 2014 (UTC)[reply]

I've written a program in Mathematica that finds these configurations.

For 3x3 There are 23 configurations, I think the one you're missing is a plus sign.

For 4x4 there are 1674 configurations. Putting these numbers in OEIS gives http://oeis.org/A082963. Apparently there's no known simple formula, but values are known up to 11x11. -- Meni Rosenfeld (talk) 16:39, 1 May 2014 (UTC)[reply]

Well done Meni! Summary: S₀=1, S₁=1, S₂=2, S₃=23, S₄=1674. For n>4 the formula ${\displaystyle S_{n}\approx {\tbinom {n^{2}}{k}}/8}$ is an excellent approximation. Symmetric configurations are rare, so the 8 rotated or reflected pictures are almost certainly different. Bo Jacoby (talk) 21:55, 1 May 2014 (UTC).[reply]

PS Actually, it should be possible to find a fairly simple formula using Burnside's lemma. It will require a summation that might be reducible to a simpler form. I'll try working it out later. -- Meni Rosenfeld (talk) 08:59, 2 May 2014 (UTC)[reply]

One more thing... I didn't give much thought about the additional stipulation that the patterns should also be incongruent under color inversion, because for the given odd example it doesn't matter. My results above are correct if it is disregarded.

However, for even n it does matter. And the result I get with this added is 887 for ${\displaystyle n=4}$. I can't seem to find that sequence in OEIS, but the derivation should be similar. -- Meni Rosenfeld (talk) 15:05, 2 May 2014 (UTC)[reply]

I think it's easier to solve using the Pólya enumeration theorem (this is a more powerful method than Burnside's lemma). You just allow arbitrary numbers of white squares and give them a weight of x. The theorem gives you the generating function taking into account the symmetry, and you then extract coefficient of the power of x^(n^2/2) from the polynomial. Count Iblis (talk) 01:59, 4 May 2014 (UTC)[reply]

Example for n = 3.

1) Identity. All the squares stay where they are, so we get 9 orbits of lenght one. Contribution to the cycle index polynomial is T1^9.

2) 90 degree rotation: Two orbits of length four and one of length 1. The four squares at the corners of the grid from one orbit, and the four suqares inbetween them also form one orbit. The square in the middle stays where it is, so this is an orbit of length one. Contribution to cycle index polynomial is thus T1 T4^2.

3) 180 degree rotation: Four orbits of length 2 and one of length 1. Contribution to cycle index polynomial is T1 T2^4 .

4) Reflection through line through the middle squares: 3 orbits of length 2 and 3 of length 1, so we get a contribution of T1^3 T2^3.

5) Reflection through the diagonal line: 3 orbits of length 2 and 3 of length 1, so we get a contribution of T1^3 T2^3.

We have two rotations over 90 degrees, and for the two types of reflections there are two different ones. The cycle index polynomial is thus given by:

Z = 1/8 [T1^9 + 2 T1 T4^2 + T1 T2^4 + 4 T1^3 T2^3]

Then Polya's theorem says that to obtain the generating function we must substitute:

Ti = 1 + x^i

This gives:

1/8 [(1+x)^9 + 2 (1+x)(1+x^4)^2 + (1+x)(1+x^2)^4 + 4 (1+x)^3 (1+x^2)^3]

Coefficient of x^4 is:

1/8 [Binomial(9,4) + 2*2 + Binomial(4,2) + 4*Binomial(3,2) + 4*Binomial(3,2)*Binomial(3,1)] =

1/8 [126 + 4 + 6 + 12 + 36] = 23

What about general n? It's easy to see that this is just a product of cycle index polynomials of the group acting on just the boundaries of the squares up to and including size n (with step size 2) when you omit the prefactor of 1/(number of elements in the group) = 1/8. Count Iblis (talk) 02:41, 4 May 2014 (UTC)[reply]

Stupid mistake here, you obviously need to multiply the terms from each element of the group only. I'm going to correct what was written below this line. Count Iblis (talk) 13:41, 4 May 2014 (UTC)[reply]

So, for odd n we have:

${\displaystyle Z_{n}={\frac {1}{8}}\left[T_{1}^{n^{2}}+2T_{1}T_{4}^{\frac {n^{2}-1}{4}}+T_{1}T_{2}^{n^{2}-1}+4T_{1}^{n}T_{2}^{\frac {\left(n-1\right)\left(2n-1\right)}{2}}\right]}$

Count Iblis (talk) 13:55, 4 May 2014 (UTC)[reply]

The generating function then becomes:

${\displaystyle {\frac {1}{8}}\left[(1+x)^{n^{2}}+2(1+x)\left(1+x^{4}\right)^{\frac {n^{2}-1}{4}}+(1+x)\left(1+x^{2}\right)^{n^{2}-1}+4(1+x)^{n}\left(1+x^{2}\right)^{\frac {\left(n-1\right)\left(2n-1\right)}{2}}\right]}$

Count Iblis (talk) 14:03, 4 May 2014 (UTC)[reply]

Elementary counting for n=8, the 8x8 chessboard.

The number of fields on a chessboard is 64 and the number of black fields is 32. The total number of configurations is ${\displaystyle {\tbinom {64}{32}}}$ = 1832624140942590534.

The number of configurations which are mirror symmetric about the horizontal line (called [-]), is ${\displaystyle {\tbinom {32}{16}}}$ = 601080390. This is the number of configurations on the lower half of the board.

The number of configurations which are mirror symmetric about the vertical line [|] is also 601080390.

The number of configurations which are symmetric with respect to 180 degree rotation is also 601080390.

The number of configurations which are mirror symmetric about both [-] and [|], that is [+], is ${\displaystyle {\tbinom {16}{8}}}$ = 12870.

The number of configurations which are symmetric with respect to 90 degree rotations is also 12870.

The number of configurations which are symmetric with respect to 180 degree rotation, but not 90 degree rotations, is 601080390 - 12870 = 601067520.

The number of configurations which are mirror symmetric about [-] but not [+] is also 601067520.

The number of configurations which are mirror symmetric about [|] but not [+] is also 601067520.

The number of configurations which are mirror symmetric about the rising diagonal [/] is ${\displaystyle \sum _{i=0}^{4}{\tbinom {8}{2i}}{\tbinom {28}{16-i}}}$ = 4965766470. (Note that 8+2⋅28=64 and 2i+2(16-i)=32).

The number of configurations which are mirror symmetric around the falling diagonal, [\] is also 4965766470.

The number of configurations which are mirror symmetric about boths diagonals [X] is ${\displaystyle \sum _{i=0}^{4}{\tbinom {8}{2i}}{\tbinom {12}{8-i}}}$ = 110022.

The number of configurations which are mirror symmetric about [/] but not [X], is 4965766470 − 110022 = 4965656448.

The number of configurations which are mirror symmetric about [\] but not [X], is also 4965656448.

The number of configurations which have 2-fold symmetry but no higher symmetry is 601067520 + 601067520 + 601067520 + 4965656448 + 4965656448 = 11734515456

The number of 8-fold symmetric configurations is ${\displaystyle \sum _{i=0}^{2}{\tbinom {4}{2i}}{\tbinom {6}{4-i}}}$ = 150.

The number of configurations which are mirror symmetric about [+] but not [X] is 12870-150 = 12720.

The number of configurations which are symmetric with respect to 90 degree rotations but not [X] nor [+] is also 12720.

The number of configurations which are mirror symmetric about [X] but not [+] is 11612-150 = 11462.

The number of configurations which have 4-fold symmetry but no higher symmetry is 12720+12720+11462 = 36902.

The number of unsymmetrical configurations is 1832624140942590534 - (11734515456 + 36902 + 150) = 1832624129208037888

So the number of configurations modulo symmetry is 1832624129208037888/8 + 11734515456/4 + 36902/2 + 150 = 229078019084652192.

The Sloan value is 229078019084673798.

My error = 21600. (To be continued). Bo Jacoby (talk) 06:21, 4 May 2014 (UTC). 16:01, 4 May 2014 (UTC). 22:32, 4 May 2014 (UTC). Bo Jacoby (talk) 23:15, 7 May 2014 (UTC).[reply]

If I have an Fokker-Planck Equation for diffusion on a curved manifold I feel like the metric tensor should be appearing somewhere, but I'm not sure where. Is this correct?

Trisecting an arbitrary angle is one of 3 classic problems that the Greek mathemations were never able to solve. As of this date it is considered impossible to trisect an angle using only a compass and unmarked straight edge. Thus this is an unsolved problem for about the past 2000 + years. Any method that claims to effectively trisect an angle must have some foundation in math in order to be considered. It is of course impossible to trisect an arbitrary angle α using only a compass and unmarked straight to an exact degree of accuracy. It is however possible to divide any angle α into n equal segments using a compass and unmarked straight edge to an acceptable desired degree of accuracy with the following procedure. It is essentially a modeling issue, where you need to approximate the length of the arc at radius r. In order to determine the accuracy of the division of an arbitrary angle into n equal segments we need to be able to define what the divided angle would be. Trisecting an angle is part of the larger issue of dividing an angle into n equal angles. This can be Accomplished through the following method: To address this problem consider these preliminary points.

1) π is defined as the ratio of the circumference of a circle divided by the diameter of the circle. Thus c= π . D. Or c = 2 . π . r

2) For a given circle of radius r the circumference is 2π.r and there exists a circle of radius 3r such that the circumference is2π.3.r. The circumference of the circle 3r is 3 times the radius of circle r. Also, any Segment of angle α of the two circles are like wise of the ratio 3:1. (See Figure 1 in Appendix) i.e. a segment of 1/360 would give a circumference of (2.π . r)/360 or (π. r)/180 and for radius 3r, (π. 3r)/180 giving the same ratio of 3:1 and for the general case a ratio of n :1 This relationship holds for any n. See Figure 2 in Appendix

4) The difficulty in taking advantage of this relationship is that the two circles have different curvatures, so it isn’t directly helpful to use this relationship directly. See Figure 3 in Appendix.

5) The following relationship is true for all r and for all α greater than 0 and less than or equal to 360 degrees.

R is all real numbers

N is natural numbers

Let α belong to R greater than 0

Let r belong to R greater than 0

Let ℓ = length of the chord, belonging to R greater than 0

And let n belong to N greater than 1

For r, ℓ1 = 2 sin (α/2).r

For 3r, ℓ3 = 2 sin ((1/3)α/2).3r

Or ℓ3 = 6 sin (α/6).r

For the general case n,

ℓ (n) =2 . n sin ((α/2n) . r Note that the above value for ℓ(n) = 2 . n Sin(α/2n) . r defines the n- section of α for all n. (See Figure 4 in appendix)

6) The next issue is; at what angle α is the ratio of (2.sin (α/2).r) /(6.sin(α/6).r), within the range of error of a compass and unmarked straight edge?

7) The following table addresses this issue. For reference, common lead diameters are as listed below:

.9 mm = .0354″
.7 mm = .0276″
.5 mm = .0197″

α   2sin(α/2)      6sin(α/6)       ratio       error @10 inch

30  .51763809	   .52293416	   .989872     .101
24  .41582338	   .4185389	   .993512     .065
20  .34729636	   .34886894	   .995492     .045
15  .26105638	   .26171632	   .99746313   .025
10  .174314855	   .1745083055	   .9989       .011
6   .1046719125	   .1047144386	   .99959      .0041
5   .087238775	   .0872263351	   .99972      .00284
4   .06979899	   .06980461396	   .9998205    .0018
3   .0523538966	   .052359213	   .9998985    .0013
2   .0349048128	   .034906353	   .99996      .00008
1   .0174530709967 .017453198093   .999993     .00007

The error in degrees is:

30 °	.304
24°	.156
20°	.090
15°	.038
10°	.0111
6°	.0025
5°	.0014
4°	.0007
3°	.0003
2 °	 .00008
1°	.000007

1 minute = .01666667°, 1 second = .000277777°

8) The line segment ℓ can be considered as the face of a polygon of n sides. As the number of sides (n) increases the polygon approaches the form of a circle as n approaches infinity. See Figures 5 to 9 in Appendix.

9) To take advantage of the accuracy of small angles, angles can be bisected to better approximate the circle. Each time you bisect the angle you double the number of segments to divide the angle.

10) To divide the angle α into any other number of segments the same process is followed: i.e. for dividing into 5 equal segments:

ℓ=2 sin (α/2)r

5ℓ= 5(2 sin (α / (5x2))r

Ratio is (2 sin (α/2) r) / (10 sin (α/10) r)

Havent been able to enter the figures from the appendix yet but they are very simple Smithcgrover (talk) 18:29, 1 May 2014 (UTC)[reply]

(Fixed some formatting in an attempt to make the above slightly more readable.) I'm not at all clear what you're asking here. As you say, arbitrary angle trisection by straight edge and compass is not possible in Euclidean geometry. It is not an "unsolved problem". There are various methods by which one can do an approximate trisection, or an exact one if the "straight edge and compass" rules are slightly relaxed, but these are of no great practical interest. AndrewWTaylor (talk) 19:55, 1 May 2014 (UTC)[reply]

Because compass-and-straightedge constructions are of practical interest? --Trovatore (talk) 20:09, 1 May 2014 (UTC) [reply]

You may mock, but in my first (proper) job I impressed a senior person by knowing the compass-and-straightedge construction of a regular pentagon, which he went on to use (for the company logo) in preparing a presentation (which in those days was often done by hand on sheets of flip-chart paper). Rather an over-engineered solution, I grant you. AndrewWTaylor (talk) 20:22, 1 May 2014 (UTC)[reply]

I think what the questioner is asking is not about a general method for perfectly trisecting any angle, but about a method that will provide an approximate trisection that will be indistinguishable from an exact one because of real-world considerations such as the precision of his available drawing implements. This is why he talks about "common lead diameters" - he's using a pencil, and wants to get close enough to perfect trisection that the error is less than the width of his pencil lines. The general answer to this question is that there certainly exist purely compass-and-straightedge constructions that will get you to within any specified level of accuracy of perfect trisection in a finite (though possibly very large) number of steps. With regard to the specific method proposed here, I think I can see what the questioner is getting at, but I haven't followed his reasoning through step by step. RomanSpa (talk) 19:19, 2 May 2014 (UTC)[reply]

Is there a construction, analogous to Taylor series, to build up approximations to a function using derivatives given at two points? Obviously for a given number of derivatives one can find a polynomial, but I'm curious about methods that do not require choosing the degree in advance. —Tamfang (talk) 22:09, 1 May 2014 (UTC)[reply]

Sounds a bit like a spline_(mathematics). Cubic spline, Cubic Hermite spline, and a I think bicubic spline is a term used in some literature. SemanticMantis (talk) 03:27, 2 May 2014 (UTC)[reply]

Obviously for a given number of derivatives one can find a polynomial, but I'm curious about methods that do not require choosing the degree in advance. —Tamfang (talk) 07:21, 2 May 2014 (UTC)[reply]

I'm confused by what you're asking. To clarify: you want an approximation scheme that works on general (continuously) differentiable functions f. It should take as input two points {x1, x2}. It should return an approximating function g, that uses derivatives of f at x1, x2, and also some n, k, that are the highest degrees of differentiation? E.g. such a scheme would produce {ai, bj}, {n,k}, such that ${\displaystyle g=\sum _{0}^{n}a_{i}f^{(i)}(x-x_{1})+\sum _{0}^{k}b_{j}f^{(j)}(x-x_{2})}$, for x in (x1,x2)?

If that's what you mean, then at the least you'll have to specify a tolerance of some sort in advance, e.g. |f-g|(x)<\epsilon, for all x in (x1, x2). Otherwise no algorithm can know in advance how close of an approximation is good enough for your purposes. Well, I suppose you could have the algorithm make a choice based on certain properties of f in the domain, but that's still a choice, you're just pushing it into the algorithm. SemanticMantis (talk) 13:29, 2 May 2014 (UTC)[reply]

The function g is unlikely to be linear in x as you have it ... —Tamfang (talk) 08:58, 3 May 2014 (UTC)[reply]

You can create a polynomial interpolating any number of given derivatives at a pair of points. The interpolating polynomial should approximate the given function. It might not be easy to come up with useful error estimates like you have in Taylor's theorem. Sławomir Biały (talk) 13:46, 2 May 2014 (UTC)[reply]

Yes, obviously you can construct a polynomial of degree ${\displaystyle 2k+1}$ that matches ${\displaystyle f(x_{0})}$ and ${\displaystyle f(x_{1})}$ and the first k derivatives; but if you then decide that k needs to increase, you have to start over, either with a new series of Hermite-like basis functions or with a new family of simultaneous linear equations — whereas with Taylor series you'd add another term without having to recompute anything else. This incremental quality is what I'm asking about. —Tamfang (talk) 08:58, 3 May 2014 (UTC)[reply]

It might be pretty horrible otherwise, but that property would be had by a linear inter/extrapolation between two Taylor series: ${\displaystyle f(x)=(1-\xi )f_{1}(x-x_{1})+\xi f_{2}(x-x_{2})}$ where ${\displaystyle \xi ={\frac {x-x_{1}}{x_{2}-x_{1}}},f_{i}(t)=\sum _{j=0}^{n_{i}}{\frac {t^{j}}{j!}}\left.{\frac {d^{j}f}{dx^{j}}}\right|_{x=x_{i}}}$. --Tardis (talk) 17:26, 3 May 2014 (UTC)[reply]

I haven't done the algebra, but I wouldn't expect the derivatives to match. —Tamfang (talk) 07:17, 4 May 2014 (UTC)[reply]

Oh, sorry; you're right of course. You could use a high-order interpolating function in ${\displaystyle \xi }$ with enough zero derivatives at ${\displaystyle x_{i}}$ to overcome that, but the simplicity (and the "natural" quality) would then suffer. Hm. --Tardis (talk) 04:14, 5 May 2014 (UTC)[reply]

Note that for rational functions it does work this way. The partial fraction expansion is the sum of the principal parts of the Laurent expansions around each pole (including the pole at infinity if the degree of the numerator is equal or larger than that of the numerator). So, perhaps it can be made to work if you recast the problems into one involving expanding in negative powers. Count Iblis (talk) 13:40, 5 May 2014 (UTC)[reply]

This is how you can do it. Suppose we want to expand f(x) at x1, x2, x3,...,xn at orders s1, s2,...,sk respectively such that you can always increase the order at some xk without having to modify the previously computed terms. You can do that by using the functions Q(x;y1,y2,y3,...) = exp[- sum over r 1/(x-yr)^2] which is zero and all its derivatives are zero at the yr that are summed over. You write f(x) as:

f(x) = Q(x;x2,x3,x4,...xn) p1(x-x1) + Q(x;x1,x3,x4,...xn) p2(x-x2) + ....Q(x;x1,x2,x3,x4...xn-1) pn(x-xn)

To compute pr one only needs to expand f(x)/Q(x;x1,x2,...xr-1, xr+1,...xn) to the desired order order. Count Iblis (talk) 21:42, 5 May 2014 (UTC)[reply]

How do I unroll the circumference of a circle as in this picture?

http://upload.wikimedia.org/wikipedia/commons/2/2a/Pi-unrolled-720.gif — Preceding unsigned comment added by 37.238.19.174 (talk) 16:12, 2 May 2014 (UTC)[reply]

You want to know how to make an animated GIF like that one ? Do you have a computer programming background ? First you'd want to create a "primitive" (the wheel shape, skipping the outside edge), then change it's rotation and translate the horizontal coord of the center at each frame. You'd also need some background graphics, like the number line and PI symbol. Finally you'd need to add some logic to "unpeel" the circle, only drawing a portion of it with each frame, with the remainder on the number line. This could all probably be done in a few hours.

Also, are you asking about the mechanics of stitching frames together to form an animated GIF ? I suggest using ImageMagick to do that. See some examples of animated GIFs I created using ImageMagick at the bottom of my home page (you need to click to open each one up). StuRat (talk) 17:54, 2 May 2014 (UTC)[reply]

I should also add that you are likely to want to work in polar coords (angle plus radius) for the wheel portion of the pic, and then convert that to rectangular coords (X and Y) for plotting. The polar coords will allow for easier rotation of the spokes in the wheel. StuRat (talk) 23:03, 4 May 2014 (UTC)[reply]

continuing discuission of "http://en.wikipedia.org/wiki/Talk:Naive_set_theory#Definition_of_.22naive_theory.22". So you claim that out of the Limberg-definition and the claim that it's consistent follows that 1=0? Sorry, I can't follow you. I can't imagine how this could have been done. 79.252.242.192 (talk) 22:06, 4 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

So you wanna know, what the word "well-defined" means. It means that any definition of a set M doesn't contain or lead to an inconsistancy about M. How do you wanna show an inconsistancy, when I exclude inconsistancies by definition? (it's night now in Germany, so I probably won't post in the next ours) 79.252.242.192 (talk) 22:01, 4 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

"So you claim that out of the Limberg-definition and the claim that it's consistent follows that 1=0?"

Not really, but close enough. If the claim that it is consistent is provably true from your Limberg-definition, then, yes, 1 = 0 as well as 1 ≠ 0 will follow.

If you want other (and probably better) answers, I'd recommend you to write down your Limberg-definition here and ask a proper question, not continue an argument from an article talk page after having been reverted. Correctness or consistency of your Limberg-definition was not the issue when you were reverted, nor was it the issue on the article talk page. You might be able to raise it as an issue here though, but probably not by just continuing an argument from the talk page (that I probably should have refrained from commenting on, ah well, arguments can be plenty of fun). YohanN7 (talk) 22:51, 4 May 2014 (UTC)[reply]

You want the Limberg-definition of sets. Here we go:

A set is a well-defined collection of objects. A collection C is well-defined means that the definition of C doesn't contain or lead to an inconsistency about C. The objects are called the elements or members of the set. Objects can be anything: either elementary objects or other sets. Elementary objects are numbers, people etc.. They are no sets themselves. There is no requirement that a set be finite. If x is a member of a set A, then it is also said that x belongs to A, or that x is in A. In this case, we write x ∈ A. An empty set (with no elements) is possible, often denoted Ø and sometimes ${\displaystyle \{\}}$.

Hope I didn't forget anything. But I think it is consistent. You wrote "If the claim that it is consistent is provably true from your Limberg-definition, then, yes, 1 = 0 as well as 1 ≠ 0 will follow.". Erm, normally I would say that definitions are not allowed to contain or lead to inconsistencies. So when "the claim that it is consistent is provably true" from my Limberg-definition then 1 = 0 will not follow cause when it is proven to be consistent you can't follow an inconsistency out of it. So I think that you mean that my Limberg-definition contains or leads to an inconsistency. Sorry, I don't see anything wrong. You have to show me, what you think is wrong. I haven't added a definition of a set, yet. And I don't see a contradiction in the definition of the "set"-term (Limberg-definition) itself. And even if I would add set definitions, then inconsistencies were excluded by the Limberg-definition. (Hey, YohanN7, where are you from, cause it's morning now in Germany and you didn't answer yet, (-: ... hm, it's 5 1/2 hours ago now since you posted. Are you busy, asleep or do you not know, what to write, cause my remarks are so convincing (or other reason)?) 79.252.242.192 (talk) 01:53, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

I have told you one thing that is wrong with your theory, and where to read about why (Gödel's incompleteness theorems). I'm not going to spend more time on this because you seem just about knowledgeable enough to not ever believe in your nonsense. You are trolling. YohanN7 (talk) 10:27, 5 May 2014 (UTC)[reply]

You are talking BS! I have told you one thing that is wrong in your entry! And that is: When it is proven that my theory is consistent then you can't say that it leads to 1 = 0. Do I have to explain you what the word consistent means or what?! It means that the theory does not lead to wrong statements!!!!! 93.197.8.254 (talk) 10:52, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

Just to be clear, YohanN7 is referring to Godel's second incompleteness theorem:

For any formal effectively generated theory T including basic arithmetical truths and also certain truths about formal provability, if T includes a statement of its own consistency then T is inconsistent. (quoted from Gödel's incompleteness theorems)

If you believe that your theory includes even a little bit of standard mathematics (e.g. number theory) then it cannot include a proof of its own consistency without being inconsistent. I hope you won't ask for the specific problem with your theory which rules out a consistency proof- this is a typical gambit of people who refuse to believe impossibility proofs when they imply disappointing results. If you don't believe in Godel's incompleteness theorem then nobody here will be enthusiastic about debating it with you. Staecker (talk) 11:57, 5 May 2014 (UTC)[reply]

"When it is proven that my theory is consistent ..."

The thing is, if your theory is a set theory, this proof from your theory will exist if and only if your theory it is inconsistent. Such a proof does lead to 1 = 0 in your theory. Once more, I urge you to read Gödel's incompleteness theorems. (And lower your tone.) YohanN7 (talk) 12:02, 5 May 2014 (UTC)[reply]

"Just to be clear, YohanN7 is referring to Godel's second incompleteness theorem", yeah he said that, but just to be clear, I did not refer to it. I just talked about my definition. "The thing is, if your theory is a set theory, this proof from your theory will exist if and only if your theory it is inconsistent.", so you do assume that my definition is inconsistent. Good that we've cleared that. But I've already told you that definitions are only allowed when they are consistent. So I just would have to discard my definition and come up with a weaker one. You say that a set theory with a little arithmetic and the claim that it is consistent would lead to an inconsistency. Does that mean, the theory is inconsistent or the consistency is just not provable? And then, what about the consistency of Zermelo-Fraenkel set theory? Can we say anything about that? 93.197.8.254 (talk) 12:27, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

Ah, ok, I think the consistency of ZFC is just not provable. But then the 2. Gödel incompleteness theorem only says that my definition (with some arithmetics) cannot be proven consistent. It doesn't mean it was inconsistent. Hm, ok, but then what is your problem with my definition? 93.197.8.254 (talk) 13:01, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

If you include enough sets to get a consistency proof, but make sure never to include any sets which would make it inconsistent, then you will be including very few sets. In particular your theory will not be sophisticated enough to express basic number theory. So likely there is no such thing as "my definition (with some basic arithmetics)". As soon as you allow in number theory (I guess that's what you mean by arithmetic), you will have introduced inconsistencies. This must be true, again, by Godel's second incompleteness theorem. Staecker (talk) 13:10, 5 May 2014 (UTC)[reply]

Häh?! You're talking crap, don't you?! "If you include enough sets to get a consistency proof... ", I didn't know I would get a consistency proof just by adding sets. "but make sure never to include any sets which would make it inconsistent", that what I've said several times already. "then you will be including very few sets. In particular your theory will not be sophisticated enough to express basic number theory.", yes it will since ZFC does too. Or doesn't it? "As soon as you allow in number theory (I guess that's what you mean by arithmetic), you will have introduced inconsistencies.", no, it just says, that consistency can not be proven! 93.197.8.254 (talk) 13:38, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

I'm beginning to feel like you're not listening (no need to say "I told you so", Yohan) so I probably won't respond any more than this. Set theory proofs require sets to be constructed and discussed- if you only allow certain kinds of sets, then this will only allow certain statements to be proved. Conversely if you require that a consistency proof will exist within your system, then this will require certain types of sets to exist in your set theory. If you insist that your theory has enough sets to make a consistency proof, but also no inconsistencies, then this is a major restriction. Godel's theorem says that such a system can be consistent, but will be not be sophisticated enough to express basic number theory. You objection "yes it will since ZFC does too" is false because ZFC does not contain a proof of its own consistency. No mathematically useful version of set theory can, because of Godel's theorem. Staecker (talk) 13:46, 5 May 2014 (UTC)[reply]

No need to say "I'm beginning to feel like you're not listening". I've listened very closely and given you replies to your statements. I'm beginning to feel like you don't wanna listen to what I say. "if you only allow certain kinds of sets, then this will only allow certain statements to be proved.", tell me something I don't know. I've made several comparisons to ZFC and now I do it again, just for you: ZFC only allows certain kinds of sets, either. So in ZFC only certain statements to be proved are allowed, either. So what do you wanna tell me. I don't see a difference to ZFC here. "Conversely if you require that a consistency proof will exist within your system ...", do you mean the word "well-defined" in my definition, or what? "If you insist that your theory has enough sets to make a consistency proof", you mean to formulate a consistency proof? I don't insist that my set definition (the "set"-term definition I mean) has to be proven consistent, since you told me about Gödel. "because ZFC does not contain a proof of its own consistency", neither does my definition. 93.197.8.254 (talk) 14:43, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

Ok- I will reckon it as progress if you've come around and are now agreeing that your system cannot contain a proof of its own consistency. Staecker (talk) 16:45, 5 May 2014 (UTC)[reply]

I cannot remember stating that my system could contain a proof of its own consistency and I'm not going to. But ok... 93.197.8.254 (talk) 17:41, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

I see a breakthrough. Nice.

The problem with your definition is that it was invented yesterday afternoon. It has not been tested for inconsistencies for more than 100 years - as has naive set theory (which really is informal ZF set theory, give or take an axiom or two). Therefore, the latter is more suitable for the article. YohanN7 (talk) 13:09, 5 May 2014 (UTC)[reply]

The solution is simple: Start testing it and give it some time. My definition has a clear advantage. I said that several times already, too. It is a general set definition in comparison to ZFC! You can actually answer (or at least try to) the question, if the statements which are the axioms of ZFC are true or false. For example, is there an empty set? Yes, there is! When A and B are sets, then is there a set that contains the elements from A and those from B together? Yes, there is! And so on. That's why I say that my definition is better than ZFC. 93.197.8.254 (talk) 13:38, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

Yes, you can perhaps start to add axioms to your theory by asking "Does this addition yield inconsistencies?" If no, then add it (as a theorem). The problem is that you will not get very far because your system of axioms (actually theorems in your theory) will soon be large enough that you can't answer the question "Does this ney addition yield inconsistencies?" any longer.

If you could, then I'm afraid Gödel is going to bite you in the ass again. But believe me, this approach has been very very thoroughly investigated in axiomatic set theory over the years. YohanN7 (talk) 13:59, 5 May 2014 (UTC)[reply]

Oh one more thing. I see that new subthreads appear above with sometimes nasty formulations. We have been extremely patient with you, but don't push it too far. I don't personally mind even if you call me an idiot (because I am an idiot), but others aren't as insensitive. Your are balancing on a thin edge. YohanN7 (talk) 14:08, 5 May 2014 (UTC)[reply]

Sorry, what shall I do? I shall add theorems like "this addition yield inconsistencies"? Which addition? Do you mean the statements which are axioms in ZFC or do you mean definitions of sets? I don't get it. 93.197.8.254 (talk) 14:43, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

You might try this: Whenever you don't understand something, acknowledge that as a fact, not as that something being incorrect. Try once again to understand. And again. Sooner or later you will pick up something true and useful – just like you finally actually understood bits and pieces of what Gödel 2 has to say about the consistency of ZF. (I commented on that as a breakthrough above, not for your theory, but for YOU.)

Instead of me or anyone else patiently explaining again to you, try to think about what Gödel 2 has to say about your latest claim (your "simple" solution of above). Then think about it again. And again. If you still think you don't understand, them come back and ask here. There is no point in explaining more clearly, because you are not going to accept it unless you discover it almost all by yourself.

Whenever someone tells you that you are wrong, you need to be even more thoughtful before vigilantly responding in a rude way. Given your apparent level of knowledge and experience with the subject, it is close to a certainty that you, in fact, are wrong when anyone here (except me, told you I'm a lunatic) so suggests. YohanN7 (talk) 15:19, 5 May 2014 (UTC)[reply]

You make a fundamental mistake here. Suppose I would say that 2+2=4 and 10 people say I'm wrong, then I still don't give in. But I ask every single one of them for his "solution". Then someone will say 2+2=5, someone will say 2+2=7 and someone will say 2+2=3. Then they will notice that they have to rethink their results. Sorry, that's a pseudo-argument for me. 93.197.8.254 (talk) 15:38, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

I think this Youtube video fits to this situation: "http://www.youtube.com/watch?v=16OrRGwwQ_0". It says: "A seven nation army couldn't hold me back.". 93.197.8.254 (talk) 16:09, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

"Given your apparent level of knowledge and experience with the subject", another fundamental mistake. You've noticed it by yourself already. When I need to know something, I look it up. I only have to know the stuff that is important for this discussion. It's not nesseccary for me to have a deep knowledge of for example Gödels theorems. You said the consistency of my definition cannot be proven and I didn't disagree, although I didn't check Gödels 2. theorem. I said I don't insist on such a proof. Why would I inform myself about Gödels theorem, when I don't insist on such a proof. I said ZFC doesn't have such a proof either. So I don't see a problem here. 93.197.8.254 (talk) 15:58, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

And since you're talking about knowledge and experience, I got a 3. price in the country round of the German Mathematics Olympiad when I was at school. 93.197.8.254 (talk) 16:13, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

Ill respond to one of your items. Then I'll tell you my opinion.

"You said the consistency of my definition cannot be proven and I didn't disagree." Yes, you disagreed several times. Violently.

Similarly for your other items. You simply don't listen.

You should have stayed in school. What do I know? Now you are just a combination of a thinks-he-knows-it-all-crank and a troll. YohanN7 (talk) 16:25, 5 May 2014 (UTC)[reply]

"Yes, you disagreed several times. Violently.", sorry I can't remember a single time. Would you be so kind and give the source to at least one of your assertions? Oh, and lower your tone! 93.197.8.254 (talk) 16:57, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

Sure, this page and Talk:Naive set theory. YohanN7 (talk) 17:01, 5 May 2014 (UTC)[reply]

Oh boy, you're not very familiar with scientific work, heh? I have to explain you everything, heh? Ok, here we go: When you state something, you have to give me the exact place where I can find it, not only a wikipedia page (they can be very long). And you should give a citation (like I did so many times when I answered you). Why is that so hard for you? I thought I would deal with people here who have some skills! 93.197.8.254 (talk) 17:18, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

I haven't seen any scientific work done in these pages. But, since you ask for citations, every post of yours containing 1 = 0, and/or one in response to one of mine that contains 1 = 0.

I am resigning from this discussion. I should have done that a long time ago, but I saw some light in the tunnel somewhere in the middle. It's gone now because you have reverted from mostly being a crackpot to only trolling and being generally unpleasant. You can call me everything and anything, I don't care very much, but I don't enjoy it enough to continue. YohanN7 (talk) 17:45, 5 May 2014 (UTC)[reply]

You wanna resign, fine with me! See I've given you the exact source of my assertions so many times and when I ask you once for it, you're not able to give it to me. You're hallucinating. Stop calling me troll! You're the troll here! I think I'vew found what you mean. I said a single time "I think it is consistent". And nothing more! 93.197.8.254 (talk) 18:05, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

To the curious reader. Here is one quote from Limberg's posts containing 1 = 0 in response to being informed that a set theory proving its own consistency is inconsistent (Gödel's second incompleteness thm):

"You have to do something! YohanN7 is starting to insult me! He writes that when a theory is consistent then he can show that it leads to 1 = 0, though. That's so stupid! Stop him!"

It should illustrate pretty clearly that he either has no clue about what he is talking (is a crank (probable)) or is just in here for the fun of it (is trolling (less probable, but not entirely impossible)). YohanN7 (talk) 23:51, 5 May 2014 (UTC)[reply]

Ok, now I'm sure, you are retarded! It says "when a theory is consistent...", I didn't even refer to my definition. And I've already told you I was not referring to Gödels theorem, either. I think you gave me a misinterpretation of Gödels theorem and I just pointed that out! Oh yeah, and you're the crackpot cause you're hallucinating! 93.197.47.238 (talk) 04:16, 6 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

Limberg wrote what he wrote it in response to this (which is not a misinterpretation of Gödel):

In fact, the very moment you claim you have a definition of set theory, free of paradoxes—as a consequence of the definition itself, your claim (if true) implies that your theory is inconsistent, able of proving 1 = 0, hence containing paradoxes. (YohanN7)

Most posts from Limberg has been of the same nature, not only the ones directed to me. Part insult. Part denial. Part intentional misquotes and part intentional lies so that he can hide his ignorance and continue his journey. In this case, he has later on, sort of, admitted he knew nothing about Gödel for the reason that he didn't need to. See the posts following his sudden insight that ZFC isn't provably consistent. Now, he justifies calling me stupid (and adds retarded) with that he was correcting my misinterpretation of Gödel's second theorem. YohanN7 (talk) 09:42, 6 May 2014 (UTC)[reply]

What you say is wrong, you're the liar and insulter!!! You hide your ignorance and continue your journey. "Limberg wrote what he wrote it in response to this", the word response is ambiguous. It was a reaction, not an answer. I've pointed that out already several times when I wrote, I didn't refer to Gödels theorem. How many times do I have to repeat it? You're slow on the uptake. I said once "I think it's consistent" and didn't repeat it. Why would I, the topic of the discussion changed. You misinterpreted Gödels theorem, didn't change your mind and insulted me! Don't complain about me insulting you back! 93.197.47.238 (talk) 13:16, 6 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

And now, replies aren't supposed to be "responses" or "answers", they are reactions. I never said either that your "reactions" refer to Gödel's theorems. In fact, they couldn't possibly have, because you didn't (and still don't) know about Gödel's theorems, but they were replies reactions to my posts. More insults, now "liar". Where do I misinterpret Gödel's teorems? For reference, I wrote

In fact, the very moment you claim you have a definition of set theory, free of paradoxes—as a consequence of the definition itself, your claim (if true) implies that your theory is inconsistent, able of proving 1 = 0, hence containing paradoxes. (YohanN7)

You did, however, misinterpret it my friend, in your response reaction. I didn't insult you yet, but I might consider doing so.

Slightly off topic: Do you still think we should rename what is today called "naive set theory" in our articles to "Thomas Limberg (Schmogrow) set theory" as you did in your first edits of that article?

It is possible that it would be the right thing to do. Write something about that on the article's talk page please. YohanN7 (talk) 13:49, 6 May 2014 (UTC)[reply]

"Where do I misinterpret Gödel's teorems?", ... oh now I have it ..., "If the claim that it is consistent is provably true from your Limberg-definition, then, yes, 1 = 0 as well as 1 ≠ 0 will follow.", Gödel doesn't say that the consistency claim has to be provably true. He just says the claim has to be made! That's how you misinterpreted it.

Moreover, "I didn't insult you", yes you did (crackpot)! Denial, you're a liar! 93.197.47.238 (talk) 15:53, 6 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

No, Gödel says that if the consistency claim is provably true from the premises, then it is in actuality false (hence not actually provably true). But don't feel bad about this. Many people, even more brilliant than you, as has been mentioned by @Staecker:, feel intimidated. A consistency claim being made isn't enough to prove an inconsistency (it is only entirely pointless). It is enough though, if you can prove the consistency from your premises. Then your theory is provably inconsistent and useless.

I called you a (crank = crackpot) because I believe you are one. It is not an insult, and not a lie. You want to name set theory after yourself and don't know about Gödel. Yet you are intelligent enough to propose an actually sensible tentative definition of what you think set theory ought to be. When directed to fallacies of what you thought was a perfect definition, when these fallacies are pointed out to you, then you hide in a shell and furiously refuse (including insulting people, changing old arguments into something else, etcetera) to acknowledge others actual century-old knowledge. This makes you a crank (or crackpot, which is a synonym) by definition of the term.

You still have the option to contribute to Wikipedia (and even science), but first, you need to dig out of your hole. YohanN7 (talk) 16:55, 6 May 2014 (UTC)[reply]

Stop insulting me!!! I wanted a serious discussion, but you're not able to do this! "because I believe you are one" is not sufficient!!!!! You have to know it!!!!! That's very unscientific! That's religion or something. Stop it!

I may edit my posts afterwards, but I do it directly after I posted them and very quickly (within 2 or 3 minutes). You don't check my posts every minute, do you? So it's not so relevant!

Maybe the spot I've given you was not the one where you misinterpreted Gödels theorem. That was just a sudden idea. (see I immediatly revise myself, when something turns out to be wrong what I said). Gödels 2. theorem demands e.g. "basic arithmetical truths". I told you already, I haven't added any set definitions, yet. I don't think this condition is fulfilled. And even if I would add sets or definitions in number theory, they would have to be well-defined. Thus, inconsistencies are excluded by definition!

You say "If the claim that it is consistent is provably true from your Limberg-definition", what does "from your Limberg-definition" mean? I mean if you (or better: I) wanna show the consistency of my definition then of course you (or I) have to make use of it. Does the "from" mean that other stuff (definitions or something) is excluded? Probably not, I have to define sets and maybe I shall define arithmetic operations so that the demand for number theory is fulfilled. 93.197.47.238 (talk) 18:26, 6 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

You have done a mistake. You told me about Gödel but that was just an answer to my question. You didn't say you wanna refer to it or apply it when you made your assertion. I always suppose, when someone wants to use definitions or views from a source that differ from the ones you have when you're not so much into the topic then this one tells me that. Otherwise it would be unscientific. So now I ask you explicitely: Do I have to have a deeper knowledge of Gödels theorems to understand what you said or is my common knowledge sufficient? Staecker said just later in the discussion that you refer to Gödel, so I suppose the answer is yes. But that's not always so. A source might proof statements which don't contain redefined terms, too. So you have to tell me when that's the case. When you've used terms in your assertion that Gödel has given a different meaning than the commonly known one, you should tell me that (I asked you about the "from" explicitely already). Ok? 93.197.47.238 (talk) 18:26, 6 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

I'm not insulting you my friend. Just being blatant with you.

You are now retreating to that "Maybe the spot I've given you was not the one where you misinterpreted Gödels theorem" where I (YohanN7) supposedly misinterpreted Gödel was a " sudden idea", therefore not an actual reply (or even a reaction?) of yours because I (YohanN7) didn't get Gödel wrong exactly there (but supposedly somewhere else) — hence not even a reflection of yours, just a "sudden idea". Interesting. Wouldn't it be more honest to say you have no clue?

Right, you haven't added any set definitions yet, except for numbers, people, bananas, ...being not sets (see previous post of yours above, not too sure about the bananas). This is quite conflicting with the usual notion, where numbers always are sets (and bananas optionally are, depending on the particular theory).

You still don't get what Gödel has to teach you. This is only natural, since you knew nothing about this two days ago when you started your journey. No Limberg definition in the world will help I'm afraid.

When you exclude inconsistencies by definition, then you are limiting yourself to a non-set theory if you want it provably consistent. Study hard.

My advice is still to jump off that train of yours. It's going to take you nowhere good. YohanN7 (talk) 18:43, 6 May 2014 (UTC)[reply]

Your last paragraph of above is extremely hard to understand. It begins with the mandatory insult: "You have done a mistake..." Then you ask if you need to understand Gödel. You assume this is so (supported by reference to @Staecker:. Then you conclude that you don't need to understand Gödel - or what do you conclude? The paragraph is incomprehensible, sorry. YohanN7 (talk) 18:43, 6 May 2014 (UTC)[reply]

"The paragraph is incomprehensible", no it's not! You're just not able to read my posts (anymore). "I'm not insulting you my friend", yes you are, "crackpot" is an insult! Furthermore, I was actually talking about set definitions like "M := {x|x is even}". So you mean the "set"-term definition. Well, ok, you can interprete it that it contains number theory. Let's suppose for the numbers that are mentioned in the Limberg-definition are arithmetical operations defined. "you hide in a shell and furiously refuse ... to acknowledge others actual century-old knowledge. This makes you a crank", you're hallucinating again. I said several times, you misinterprete Gödels theorem. I didn't say Gödels theorem was wrong!!! "When you exclude inconsistencies by definition, then you are limiting yourself to a non-set theory if you want it provably consistent.", when there is {}, or even {0} and {1}, then there are sets there and I don't see any contradiction. So it is a set theory! 93.197.19.94 (talk) 04:02, 7 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

First a piece of advice. Don't clutter your posts with irrelevancies. It is that that makes them incomprehensible, to the point that I don't want to waste my time deciphering them. Relevant part of your paragraph (about 60-70 characters) : ~

when there is {}, ..., then there are sets... (Limberg)

Yes, there are sets (by your definition), perhaps even a theory of something if you formalize it, but there is no provably consistent set theory in the sense of our discussion. Set theories are powerful enough to include, at the very least, number theory. These can't be proved being consistent using any provably consistent theory (unless they are inconsistent). YohanN7 (talk) 12:13, 7 May 2014 (UTC)[reply]

You just don't get it!!! In ZFC there is the possibility that you get a statement "x∈M" and a statement "x∉M". So you would have a contradiction. With the Limberg-definition this is not possible cause sets have to be well-defined! So what I assume is that Gödels theorem is just not apllicable here. I don't see a way how you could possibly create a contradiction. Would it be about a statement of the form "x∈M" (so the contradiction would be "x∈M and x∉M"), a statement of the form "M is a set" or a "general" statement like "2+2=4"? I don't see any other possibilities. 93.197.19.94 (talk) 04:02, 7 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

Again (see above), there are sets, but no set theory because Gödel says no. YohanN7 (talk) 12:13, 7 May 2014 (UTC)[reply]

I think it's a misunderstanding in the word "from" from your assertion "If the claim that it is consistent is provably true from your Limberg-definition, then, yes, 1 = 0 as well as 1 ≠ 0 will follow.". I thought it suffices to show at my definition that it would be consistent. But I think, you mean, it would actually be made use of the definition to show, it was consistent. Well, in that case, of course, it's possible that one can show an inconsistency. But that would be a beginners mistake. I didn't take that into account, therefore. Of course, one is not allowed to make use of a definition which's consistency he/she wants to prove! That is clear! OMG! See that's the problem, you live in a small world, where you were smart and others were dumb. And then you get a shock when reality proves you wrong! You shouldn't think I was stupid! When I talked about proving the consistency of my definition, I didn't make use of the definition. 93.197.19.94 (talk) 06:46, 7 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

When proving the consistency of your theory, you didn't make use of its definition? What exactly did you use? Magic? Yes, I will get a shock when you prove me wrong. YohanN7 (talk) 12:13, 7 May 2014 (UTC)[reply]

I think what you don't realize is "The second incompleteness theorem only shows that the consistency of certain theories cannot be proved from the axioms of those theories themselves. It does not show that the consistency cannot be proved from other (consistent) axioms. For example, the consistency of the Peano arithmetic can be proved in Zermelo–Fraenkel set theory (ZFC), or in theories of arithmetic augmented with transfinite induction, as in Gentzen's consistency proof." from the article about Gödels theorems. 93.197.19.94 (talk) 06:52, 7 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

Yes, certainly. (I guess you just found that out by the way you present it.) Exactly which consistent theory are you going to use to prove the consistency of yours? ZFC, assuming it is consistent? YohanN7 (talk) 12:13, 7 May 2014 (UTC)[reply]

Fascinating though this conversation is, I don't see a question here that can be readily answered by contributors to the Reference Desk. Have you considered taking this discussion to one of your talk pages? RomanSpa (talk) 15:30, 5 May 2014 (UTC)[reply]

He did have a question. Why isn't it obvious that his version of set theory has none of the flaws of other versions of set theory (and why he can't put it into our articles, naming it after himself). YohanN7 (talk) 16:27, 5 May 2014 (UTC)[reply]

And, in all fairness, he was asked to take the discussion here from the article talk page. YohanN7 (talk) 16:36, 5 May 2014 (UTC)[reply]

"I don't see a question here that can be readily answered by contributors to the Reference Desk", yeah, because I've already answered all questions by myself, so all that's left is to undo the undos of my edits to the article. 93.197.8.254 (talk) 16:52, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

I'll try to answer your questions.

First, note that the article that you have been looking at is Naive set theory. That is, it is informal and intuitive. One "naive" intuition is that any collection of all objects satisfying a particular condition can be turned into a "set". It turns out (e.g. Russell's paradox) that this is not the case, which is why various axiomatic approaches have been developed: they try to avoid these paradoxes. Your definition of a "set" similarly attempts to avoid problems ("A set is a well-defined collection of objects. A collection C is well-defined means that the definition of C doesn't contain or lead to an inconsistency about C."). To the extent that your definition manages to do this, it is no longer "naive". Thus, any edits that you make to the article on Naive Set Theory are inappropriate, because they're in the wrong place.

Second, the approach that you seem to want to follow seems to me to lead naturally to the consideration of Proper classes. As you will see from our article on that subject, the idea that you are proposing has been around for a long time, and has been extensively examined by many brilliant minds. Attempting to append your name to what (although it may seem new to you) is actually a fairly well-established idea looks rather silly.

Third and finally, I think you may have fundamentally misunderstood the nature of Wikipedia. It is a matter of policy that we do not include original research in this project. The reasons for this have already been explained to you. The correct place for new ideas is in the appropriate technical journal. To the extent that your idea is not new, it is already covered in Wikipedia; to the extent that it is original, we do not include it.

RomanSpa (talk) 18:25, 5 May 2014 (UTC)[reply]

" Thus, any edits that you make to the article on Naive Set Theory are inappropriate, because they're in the wrong place.". I've already considered your thoughts. But the thing is, I've just added some things (which are not so relevant I would say) to the definition that already was in the article. The word "well-defined" which avoids the paradoxes you've spoken of already was in the definition. Otherwise I wouldn't have taken the "naive" away. When you wanna show, what a naive set theory is, you should leave off the word "well-defined" so that you can say, that the definition is ambigues (or something) and can lead to paradoxes, justifying it to be called naive.

"seems to me to lead naturally to the consideration of Proper classes", I don't think so. Show it!

"Third", I know it's not new. I want you to recognize that it is not naive. That is no original research. You have stated something wrong (when one can consider the little improvements irrelevant). And I want you to change it. 93.197.8.254 (talk) 18:55, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

You awake the impression that all general set theories are naive and thus to be discarded and only the constructive set theories with its formalized axioms are not naive. But that's not true. 93.197.8.254 (talk) 19:01, 5 May 2014 (UTC) Thomas Limberg (Schmogrow)[reply]

It's night now in Germany, see you tomorrow! — Preceding unsigned comment added by 93.197.8.254 (talk) 20:21, 5 May 2014 (UTC)[reply]

LaTeX probably includes \iiiint for a good reason. 4-dimensional spacetime would be an obvious guess but I'm not sure that it is equivalent to R⁴ in the usual sense of defining a multiple integral. I Googled and found mention of using quadruple integrals for "determining the interference between elements of a compound array of radiating elements" ([9]), but besides some other obvious cases like four-dimensional Euclidean space, I couldn't really find real-life applications of a quadruple Riemann integral. I am aware that quantum mechanics texts often will use "infinite-dimensional" functional integrals (a concept I currently do not grasp).--Jasper Deng (talk) 09:37, 5 May 2014 (UTC)[reply]

The average distance between two randomly chosen points in the unit square is given as a quadruple integral. Generalisations of this e.g. the weighted average between two random points in some arbitrary shaped area, will have useful applications. E.g. if you have particles that can move freely on a surface and you can treat the interaction in perturbation theory, you can compute the average interaction energy to first order in perturbation theory this way. Count Iblis (talk) 13:22, 5 May 2014 (UTC)[reply]

All (AFAIK) relativistic physical theories can be cast into a stationary action formulation. This action is, for a field theory, a spacetime integral. In smooth manifold theory, the suitable objects to integrate (in a coordinate-independent way) are (often top-dimensional) differential forms. In coordinates, these integrals become (are defined as) ordinary multiple integrals in Rⁿ. YohanN7 (talk) 13:39, 5 May 2014 (UTC)[reply]

See Lagrangian density for an integral over 4-dimensional spacetime (though the article uses only a single integral sign with d⁴x) --catslash (talk) 22:41, 6 May 2014 (UTC)[reply]

How do we define ${\displaystyle n^{\infty }}$ and ${\displaystyle n^{(-\infty )}}$ for negative ${\displaystyle n}$??

For values ${\displaystyle n>0}$ we have:

• ${\displaystyle n^{\infty }=\infty }$ for ${\displaystyle n>1}$
• ${\displaystyle 1^{\infty }}$ is undefined
• ${\displaystyle n^{\infty }=0}$ for ${\displaystyle 0<n<1}$
• ${\displaystyle n^{(-\infty )}=0}$ for ${\displaystyle n>1}$
• ${\displaystyle 1^{(-\infty )}}$ is undefined
• ${\displaystyle n^{(-\infty )}=\infty }$ for ${\displaystyle 0<n<1}$

But because of how powers of negative real numbers work, defining their infinity and negative infinity powers is much harder. Any thoughts?? Georgia guy (talk) 18:43, 5 May 2014 (UTC)[reply]

I fixed some of your typesetting.

In general, we don't. It doesn't come up enough to matter. If you're doing some work in a particular area where there's a natural choice, give it that definition for the scope of the work. If parts of that work are interesting or applicable to other areas, maybe it'll catch on and we'll start defining things that way.--80.109.80.78 (talk) 22:54, 5 May 2014 (UTC)[reply]

For integer values of n, ${\displaystyle n^{\infty }}$ diverges as n goes to ${\displaystyle {(-\infty )}}$, so I think it's a fair bet the limit is undefined. AlexTiefling (talk) 23:09, 5 May 2014 (UTC)[reply]

It really depends on whether you regard n as an integer or a real whether 1^∞ is considered as 1 or undefined. Dmcq (talk) 23:39, 5 May 2014 (UTC)[reply]

As suggested by Dmcq, finer points must be brought into play: it is also worth considering whether the exponentiation is considered to be repeated multiplication/division (thus restricting the exponent to integers), or may vary continuously, usually requiring definition through the exponential function. In the latter case, the exponentiation must generally be regarded as undefined for negative bases, and limits even more so. In the former case (restriction to integer exponents), the last bullet should not be considered to hold: the expression diverges rather than converges on ∞, for the reason that the expression alternates in sign as the exponent is increases. Essentially, limiting forms such as these depend upon too unstated material (i.e. context) and it makes no sense to attempt a general definition, but given sufficient constraints, a limit can be determined uncontroversially (including possible divergence). —Quondum 23:47, 5 May 2014 (UTC)[reply]

Yes that's a better way of going about it. Is it repeated multiplication or a use of the exponential function. Dmcq (talk) 11:48, 7 May 2014 (UTC)[reply]

The standard construction of a Pentagon more or less starts with a circle and creates a Pentagon inscribed in the circle (choosing one of the corners with a line through the center). What I would like to find is a construction of a Pentagon from a line and a point not on the line so that the point is one of vertices and the opposite side is on the line. (As a bonus if we can get the equivalent construction for when the point not on the line is one of the "other points", neither opposite or one the line).Naraht (talk) 17:27, 6 May 2014 (UTC)[reply]

Well, you can probably use the standard construction to get a line passing through the Point at an appropriate angle so that it meets the first Line in a vertex; that's a start at least. —Tamfang (talk) 06:18, 7 May 2014 (UTC)[reply]

Does it make sense to ask if the axiom of choice (for example) is physically true or false in our universe in a physical sense? If we ask this, what are the possible answers? (e.g. true/false/neither?) What about other axioms? Do any mathematical axioms have a physical value in our universe? (In much the same way that if you put a finite n number of apples together and keep adding one, you will never run out of apples; but if you put a finite n number of apples together and keep removing one, eventually you reach zero. This means there doesn't exist a k such that positive integer n + k = 0;, and for any positive integer n there is a k such that n-k=0. I presume we can actually prove this using a trivial set of axioms; I further presume that these axioms are in some sense true "physically" in our universe.) Please correct me if I am wrong in any of the above assumptions! 212.96.61.236 (talk) 20:43, 6 May 2014 (UTC)[reply]

It's an interesting question, and probably one you can do entire PhD dissertations on (in philosophy departments, not math departments). However, just to get a quick response in, it's not one that mathematicians usually think much about. Mathematical realists, or at least some of them, hold that (at least some) axioms are either really true or really false, but for the most part they are not taking them to describe physical objects, but rather abstract objects that are nevertheless still real and exist independently of our reasoning about them. --Trovatore (talk) 21:06, 6 May 2014 (UTC)[reply]

are you saying that any conceivable mathematical axiom is totally independent of the actual Universe, and in the actual Universe no basic mathematical axiom is 'true' or 'false' but always 'independent'? That even pi does not have an exact value in our universe, but only under a certain set of axioms mathematicians can entertain, which just so happen to be every set of axioms that are of interest. (No set gives a powerful system with a slightly different version of pi.) So even though every set of axioms that is of interest implies a certain value of pi, and even though our universe sure seems like it shows that, still, we can say with absolute certainty that pi does not have an exact value (regardless of whether we could discover let alone prove it) in our universe, but that every single axiom is totally independent in a physical sense? 212.96.61.236 (talk) 22:05, 6 May 2014 (UTC)[reply]

I'm afraid I don't really understand your question. --Trovatore (talk) 22:10, 6 May 2014 (UTC)[reply]

I think you need to take into account that a correspondence between mathematics and our observation of the physical universe is rather loose. Even the correspondence between a formal system (manipulation of symbols according to a set of rules) and mathematics in general is not clearcut, and with the physical universe the correspondence is even less clear. So, what you might consider to be a physical apple does not have a clear definition: where, in the continuous spectrum of transformation from "an apple" to "something other than an apple" do you draw the line? So already you are operating with some convenient idealization of the universe in which apples are back and white (if you'll forgive the mixed metaphor). As such, it is difficult to imagine how to interpret your question as applying to anything at all. —Quondum 22:23, 6 May 2014 (UTC)[reply]

Well take the Banach–Tarski_paradox. In a literal sense is it physically true in our universe? (I am equally happy to hear that it is false.) I'd just like to know whether either those statements, or any other statements, have a literal truth-value? Or is anything nameable, also independent. 212.96.61.236 (talk) 22:59, 6 May 2014 (UTC)[reply]

Sure, it has a literal truth value. It's true — about the Platonic ideal objects it talks about. In our universe? It's true everywhere, including our universe.

However, it doesn't mean you can cut up a physical pea and reassemble it into an equally dense pea the size of the Sun. You can't do that, and Banach–Tarski doesn't say you can, because it isn't talking about peas. --Trovatore (talk) 23:03, 6 May 2014 (UTC)[reply]

With reference to the original question: The Banach-Tarski paradox isn't an axiom; it's a consequence of a set of axioms that treat balls as continuous infinitely finely divisible regions of space. Actual balls are collections of atoms; if you divide them too much they lose their ball-ness, and you have an empty bladder, or a pile of sand, or whatever your ball was made from. So in that practical sense, neither the Banach-Tarski paradox nor the underlying axioms are true of actual space and balls. But the principles that led us to choose those axioms in the first place are derived from our intuitive understanding of ordinary space - so although they're not strictly true, they describe everyday experience very effectively. That's why the Banach-Tarski paradox is a paradox: it's obviously false for real objects, but derives from axioms we invented in order to model real objects.

But in general: an axiom cannot, by definition, be falsified. A physical law, on the other hand, may in principle be falsified, even if we are confident it will not be. Accordingly, no mathematical axioms govern real space. They are different categories of statement. AlexTiefling (talk) 23:09, 6 May 2014 (UTC)[reply]

Correction: It's obviously false for physical objects. It's true for real objects. Just not physical real objects. --Trovatore (talk) 23:13, 6 May 2014 (UTC)[reply]

Do you have an example of a non-physical real object handy with which to illustrate the distinction? AlexTiefling (talk) 23:23, 6 May 2014 (UTC)[reply]

Handy? They're everywhere. --Trovatore (talk) 23:30, 6 May 2014 (UTC)[reply]

I feel like I'm missing something. What are non-physical real objects? AlexTiefling (talk) 23:36, 6 May 2014 (UTC)[reply]

The number 2. God. Your soul. --Trovatore (talk) 23:36, 6 May 2014 (UTC)[reply]

I'd hesitate to call '2' an object; it's a mathematical construct used to enumerate things. And the existence of God, and of souls, are theological questions. Answers like this clearly subvert the intention of the original question. (And for what it's worth, I don't think the Banach-Tarski paradox can be applied to any of the purported non-physical real objects you propose.) AlexTiefling (talk) 23:43, 6 May 2014 (UTC)[reply]

So you're not a Platonist. OK, you're allowed to be wrong. --Trovatore (talk) 23:47, 6 May 2014 (UTC)[reply]

I'm not a Platonist, and I'm also not amused by being called wrong. I especially dislike the way you have treated the unfalsifiable assumptions of your own philosophy as so self-evidently shared that you didn't give me a straight answer when I asked for one. I'd appreciate it if you would apologise for this misleading behaviour, and for the insult. AlexTiefling (talk) 00:01, 7 May 2014 (UTC)[reply]

I did give you a straight answer. I do apologize for the belligerent tone. --Trovatore (talk) 00:03, 7 May 2014 (UTC)[reply]

Saying "Handy? They're everywhere" when I asked for an example cannot reasonably be regarded a straight answer. AlexTiefling (talk) 00:05, 7 May 2014 (UTC)[reply]

I give you credit for being able to figure out what I was getting at. --Trovatore (talk) 00:06, 7 May 2014 (UTC)[reply]

I would say that mathematics models the world, but that statements in mathematics aren't true or false the way statements about the world are. That is not to say that mathematical statements describing the world cannot be true or false: but that is an empirical matter, not an ontological one. Rather, it is to say that mathematics has no real essence that exists apart from its own vocabulary and grammar. When we say that a mathematical statement is "true", we don't mean that it is a true statement about the world, but rather that certain grammatical conditions hold that are internal to the language of mathematics. To most mathematicians, this means that a statement is "true" if it is reducible to some collection of axioms using rules for inference. However, many mathematicians also conflate this notion of "truth" with the notion of "truth of a statement about the world", and thus conclude that mathematics must exist in the world in order to be true or false, independently of any vocabulary: that mathematics is the final vocabulary of the world. This is the essentialist viewpoint of mathematics. Sławomir Biały (talk) 23:55, 6 May 2014 (UTC)[reply]

Well, no, the notion that "true" is the same as "reducible to some collection of axioms using rules" was pretty decisively refuted by Goedel. --Trovatore (talk) 00:00, 7 May 2014 (UTC)[reply]

So what does "true" mean then? Sławomir Biały (talk) 00:13, 7 May 2014 (UTC)[reply]

The objects of discourse of foundationally relevant theories, such as arithmetic or set theory, are well-specified up to a canonical isomorphism by their informal descriptions. What "true" means was explained by Tarski. "The apple is red" is true just in case the apple is red. "Every apple is red" is true just in case, for any apple, it's red. And so on. --Trovatore (talk) 00:15, 7 May 2014 (UTC)[reply]

I don't understand your reply. "Every apple is red" is an empirical statement about the world. "Peano arithmetic is consistent" is not. What does it mean to say that "Peano arithmetic is consistent" is true? Sławomir Biały (talk) 00:20, 7 May 2014 (UTC)[reply]

"Peano arithmetic is consistent" is true just in case there does not exist a formal derivation of 0=1 from Peano arithmetic. --Trovatore (talk) 00:21, 7 May 2014 (UTC)[reply]

So if no one is able to find a derivation of 0=1 in Peano arithmetic, it means that "Peano arithmetic is consistent" is true? Sławomir Biały (talk) 00:25, 7 May 2014 (UTC)[reply]

Oh no. It's evidence for it, yes; how strong the evidence is is open for dispute. But there in principle there could certainly be a derivation that we haven't found. In fact, there could be a derivation so long that it won't fit into the observable universe.

Although one has to be a bit careful about the meaning of the word "could" here. When I say there "could" be a derivation, I certainly don't mean there's a possible world in which there is a derivation. I just mean that our failure to find one does not in itself conclusively settle the issue. --Trovatore (talk) 00:33, 7 May 2014 (UTC)[reply]

But inevitably, the assertion "there exists" should raise the question of "where exists?" (which was the purpose of that bit of sophistry on my part, which I was hoping to clarify just as you replied). Someone might come along with such a derivation, in which case we could say (in an in-the-world sense), "there exists". But when a mathematician says "there exists", it is part of the mathematical language-game. Sławomir Biały (talk) 00:43, 7 May 2014 (UTC)[reply]

If it's all a language game, then there is no explanation for the observed failure to find a contradiction, beyond the mere brute fact that none of the derivations that have actually been tried give 0=1. However, if the natural numbers are real objects, then there is an explanation — namely, PA is consistent because it's true, and truth can't be inconsistent. See for example Quine–Putnam indispensability thesis. --Trovatore (talk) 00:53, 7 May 2014 (UTC)[reply]

Sorry to keep doing this, but I think I can clarify my point a little more. Goedel's incompleteness theorem relies fundamentally on the law of excluded middle. This is a rule of inference in the metalanguage used to model Peano arithmetic.

Also, I don't buy the Quine-Putnam thesis. Doesn't it basically justify any kind of essentialism? Like "the apple has redness"? At what point between Aristotle and the present day did "redness" become no longer an existing essence, replaced by the absorption spectra of phytochemicals in the skin of the apple? Sławomir Biały (talk) 01:14, 7 May 2014 (UTC)[reply]

In what way does incompleteness rely on excluded middle? The proof is perfectly intuitionistically valid. Possibly you can find some use of excluded middle in the interpretation; that wouldn't surprise me. But I'd need to know which one you're talking about.

I'm not focusing on the details of the Quine–Putnam argument. The point is that the ontological claims of realism, or at least some of them, have observable consequences that otherwise have no satisfying explanation. Now, you can say that reality doesn't owe you a satisfying explanation, and things just are the way they are, but that way seems to lie instrumentalism about everything. --Trovatore (talk) 01:29, 7 May 2014 (UTC)[reply]

Sorry, I didn't mean to be saying anything deep about Goedel's theorem or any particular interpretation of it, just that the statement "either Peano arithmetic is consistent or it is inconsistent" obviously uses excluded middle. Sławomir Biały (talk) 01:53, 7 May 2014 (UTC)[reply]

Wait-wait-wait. Where someone questions the law of excluded middle "in-world" -- does that mean that possibly there ARE no actual truth statements in the world? (In a binary sense). For example, it is easy to imagine a dream that doesn't "make sense" and basically has no true or false values in it, just impressions images and feelings. is there a chance the Universe is that way as well, and it is meaningless to ask whether any mathematical axiom is true, simply by virtue of the fact that there is no such thing as truth? (It is a social construct only, and has no meaningful physical interpretation)? 212.96.61.236 (talk) 01:55, 7 May 2014 (UTC)[reply]

It's possible, I suppose, but it's not what anyone here has been trying to say, I don't think. The point I'm making is that statements about the observable world are - at least in principle - testable. That is, they have meaningful truth-values, and the way you get from 'undecided' to 'true' or 'false' is by reference to the actual physical world - never to an axiom. If you had an axiom 'all cows are black', and you found a purple cow, you'd invalidate your system. But if you make a prediction 'all cows are black', and you find a purple cow, you can abandon the old prediction in favour of a new one. Axioms define logical systems - that is, what's true or false in the system is dictated by the axioms. But the real world is real - it is not dictated to by any theoretical construct, but is out there for you to go and check for yourself. It doesn't consist of axioms and postulates, lemmas, theorems, sets and proofs. Those things are used to build abstract models of the way some aspects of the world behave. But the world just is. Truth values of statements about the world derive from the world itself. AlexTiefling (talk) 02:57, 7 May 2014 (UTC)[reply]

That's all very well, but completely talks around whether the world itself in fact has truth-values. Further, you can never learn a truth-value by experiment, just as you cannot prove a mathematical theory by numerical experiments, so it really doesn't matter what we find or don't find. It doesn't affect whether there in fact *are* in fact fundamental arithmetic facts about the universe, for example. There might seem to be (via experiment) but this is irrelevant to whether there are. A dream might certainly seem to obey the laws of physics, but since it is just a dream that you are imagining, it does not in fact obey any physical laws precisely. Likewise the universe can certainly seem rational and consistent, without being such. Do we have any way of knowing for sure that even basic arithmetic is "actually" true in our universe? No real answers given so far. 91.120.14.30 (talk) 09:52, 7 May 2014 (UTC)[reply]

I don't personally believe that the world itself has truth values. Truth can only be assigned to statements about the world, and even the truth of those statements depends on their cultural context (think here of the many different kinds of ontological statements made in different religious contexts). It could be that there is some intrinsic truth that is out there in the world, but I don't think it is useful to talk about such essences. Sławomir Biały (talk) 11:15, 7 May 2014 (UTC)[reply]

I'm putting this below another line, so people can continue the thread above it, if it proves an unhelpful diversion. I've often thought a way into the topic is to look at information as an abstract entity. It's rather hard to argue it exists in a pure form in the real world, because of the possibility of error. I know that errors are treated as noise in information theory, but what I mean is this: You find something that you say has 3 "bits" of information. Now all of that depends on the receiver being originally uninformed, but knowing exactly how much noise is in the signal etc. This could only make sense if you could measure physical states perfectly to begin with, so that you could be perfectly uninformed about those 3 bits, and have perfect randomness for each bit. This is physically impossible, or never found in practice. Random number generation that is perfectly random with a perfectly known distribution is probably impossible, even using quantum experiments. If you have a quantum experiment, you still need to set up the apparatus perfectly, eg. so that each outcome has exactly a 50% chance, and I'm sure that doesn't happen. Now flip the argument around, and remember that if you don't believe in information, all of your knowledge of mathematics came from textbooks and information signals buzzing about in your brain, so tell me it isn't real, and I'll be moved to doubt my Platonism. Let me know if I'm getting anywhere, and better yet, let me know if there's a reference that touches on any of this, and links it to the existing philosophy of mathematics. IBE (talk) 11:32, 7 May 2014 (UTC)[reply]

It is certainly the case that there are some axioms the truth or otherwise of which cannot be meaningfully discussed with respect to the physical universe. The first example you mention, the axiom of choice, is one such, because it only has practical application in the consideration of (in a sense that I carefully leave vague!) sets with an infinite number of elements. Since any set in the universe must have a (possibly astronomically large) finite number of elements this axiom has no meaningful applicability or truth value in the real physical universe. A more general answer to your question first requires some careful consideration of your philosophical position. In my case, I am a (somewhat modified) formalist, and adopt the position that your question is in its general sense meaningless, as any answers that may be provided are not in principle testable. RomanSpa (talk) 12:12, 7 May 2014 (UTC)[reply]

We don't know how big the universe is, only the size of the observable universe. The universe may be infinite for all we know. And whether that is irrelevant or not is another unknown, perhaps the best explanations for what we can see will rest upon an assumption that the axiom of choice is untrue in the universe at large. Seems very unlikely but strange things can happen. Dmcq (talk) 14:02, 7 May 2014 (UTC)[reply]

The universe doesn't have to have only finite quantities in it. For example, the universe itself could be infinite in size (why not.) This does not change whether it is meaningful to ask any questions about the universe (even axiomatic ones) that have or may have a truth-value. Either it is meaningless to ask about whether something is "true" in a physical sense (absolutely, mathematically, in a binary, excluded-middle, axiomatic way) or it is not meaningless to ask this question. Are there any questions for which it is meaningful to talk about the existence of a truth-value in the Universe? (Regardless of whether we can learn the value of that truth-value - probably we can't. But that doesn't mean it can't exist.) 91.120.14.30 (talk) 15:06, 7 May 2014 (UTC)[reply]

Hi.

Probably some textbook questions, but... (Context is in the generation of 3D meshs for a computer graphics application)

(i)(a)For 2 regular prisms A and B in Euclidean, where Prism A and Prisim B's centre line lie in the same plane, but where Prism A and B are at some angle to each other, Is there a generalised algorithm for computing the "vertices of intersection"(?),

(b) Having obtained the 'vertices of intersection', is there an algorithm to clip the prisms.

(ii) There is a transformation that unrolls the clipped prisms to a flat sheet? If so what is it.

Sfan00 IMG (talk) 13:15, 7 May 2014 (UTC)[reply]

At the section here, it gives a method for determining the equation of a line that is tangent to two circles, and says that there are "two solutions". However, there are (generally speaking) four lines tangent to two circles, so I don't get it. Is that section wrong, or am I misunderstanding something? 86.169.185.1 (talk) 13:49, 7 May 2014 (UTC)[reply]

They get two solutions first. If you keep reading, the section you linked eventually says "In this way all four solutions are obtained." I didn't check every step of the derivations, but I don't think there are mathematical errors in that section. Basically, this scheme gets all four lines by using two different ways to produce sets of two solutions. There is probably a way to get all four solutions falling out of one master equation, but that would be more complicated, harder to follow, and in my opinion, not worth the effort in the context of this article. SemanticMantis (talk) 14:07, 7 May 2014 (UTC)[reply]

Oh, OK, thanks. I thought the part "Geometrically this corresponds to ..." was the start of an explanation of how to visualise why it works. I wonder if there is some way to present the whole solution at once, explaining which combinations of signs produce which tangent lines. I think this would be easier to follow. 86.169.185.1 (talk) 17:11, 7 May 2014 (UTC)[reply]

First, thanks OP of the previous thread. I have been meaning to ask this for a long time. I create a new thread because I want to specialize the question a bit. The previous thread is now rather long too, risking to go stale for that reason only.

A line of thoughts: Suppose in some physical theory T, particle X exists (in the theory) if and only if mathematical Theorem A (of, say, point set topology) is true. Suppose now Theorem A has implications on the continuum problem. There are certainly such theorems in existence. Modern theoretical physics certainly involves topology.

Admittedly, chances are slim that the premises are fulfilled, but anyway: Suppose now that particle X is discovered experimentally...

You take it from here. YohanN7 (talk) 14:02, 7 May 2014 (UTC)[reply]

Can you clarify: in your example, theorem A has no known proof? If that's what you mean, then I don't think physical existence of a particle can say much about the mathematical necessity of theorem A in a formal system. My reasoning is that, though the particle X in the theory and the particle found in nature are linked by analogy and human models: they are actually quite distinguishable objects. SemanticMantis (talk) 14:11, 7 May 2014 (UTC)[reply]

Or, going the other way, theorem A has a known proof, and you might be suggesting that physical evidence of particle X now somehow entails a "physical truth" of the formal axioms? Sorry, I still don't think so. But philosophy of mathematics is a fun topic here, as long as we don't get too fighty ;) SemanticMantis (talk) 14:18, 7 May 2014 (UTC)[reply]

Ah, yes, I was unclear. Let's say, for definiteness, either of

1. A is true only if CH is true/false.
2. If CH is true, then A is true/false.
3. A is true if and only if CH is true/false.

Feel free to replace A, CH and the implications as you wish, but please leave theory T's, the particle X's and the experiment E's rôles unchanged. YohanN7 (talk) 14:21, 7 May 2014 (UTC)[reply]

For more definiteness, assume ZF or ZFC. YohanN7 (talk) 14:23, 7 May 2014 (UTC)[reply]

Ok, I think I'll stick with my first response then. SemanticMantis (talk) 14:33, 7 May 2014 (UTC)[reply]

I tend to agree. But, I'd also say that a positive experiment E will have significant weight as a plausibility argument (in the indicated direction, whatever it might be (I numbered the cases above, and allowed for ¬CH)), much stronger than any existing purely philosophical, and perhaps even stronger than some mathematical plausibility arguments (e.g. axiom of symmetry which I don't personally pay much attention to, (and even disagree with)). YohanN7 (talk) 15:08, 7 May 2014 (UTC)[reply]

Using the axiom of choice is extremely ambitious in this context. While real numbers and continuity make intuitive sense in the real world, there could be any number of equivalent mathematical representations of reality that might not require infinite sets, for all we know. This is aside from the obvious point that T is merely a mathematical model of the world from which our observations are drawn, and observation of a particle X may simply invalidate T as a physical theory. I think, too, that mathematics can have various structures depending on choices along the way: it is constructed, and could have many shapes. We might find that when it is built differently (e.g. using a basis other than set theory), something like the axiom of choice might not even fit. —Quondum 16:57, 7 May 2014 (UTC)[reply]

O yes, sure. Mathematics can have many structures.

But is there one that is right in some very very very real (physical reality) Platonic sense? I do not really get your coupling with the axiom of choice. Sure, spacetime may be finite or infinite (modeled as a point set). For purposes of discussion, let us assume that spacetime is a continuum. That makes the discussion easier and more interesting to the original question. YohanN7 (talk) 19:07, 7 May 2014 (UTC)[reply]

There are so many different ways to model reality. Even within existing mathematics, a smooth function on a spacetime continuum occupies a very small space out of all possible spaces. With this and boundary conditions, the total number of possible configuration of the universe could well be finite. So whether spacetime is considered to be a continuum might be irrelevant to the physical reality. I've sometimes wondered whether there isn't a dual universe composed of the same reality, with atom-like structures etc., when looked at in the energy–momentum domain. —Quondum 19:40, 7 May 2014 (UTC)[reply]

With a spacetime continuum and smooth functions representing states, you are unlikely to find a finite number of configurations. You might get (compactness) a countable number of configurations. But this is straying away from the OP anyhow. YohanN7 (talk) 21:32, 7 May 2014 (UTC)[reply]

Yes, I think there is something to be said for plausibility, or "evidence," suggestive power, etc. But then we start mixing up epistemology of math and science quite a bit. I suppose it's not so terrible if we end up someplace like this: claim A may well be true, but unprovable within a given formal system. Because we have some faith in the fidelity of a certain model, an empirical result might give us a high degree of belief that claim A is true in the formal system, and perhaps even that the analogous claim A' is true for the physical concrete real world. SemanticMantis (talk) 17:57, 7 May 2014 (UTC)[reply]

Right. I believe that the deep unanswered questions (at the axiomatic level) will remain unanswered. All we will get are plausibility arguments, wherever they come from. YohanN7 (talk) 19:07, 7 May 2014 (UTC)[reply]

There is nothing wrong with an argument that goes "Supposing that theory T is a faithful model of physical system S ...". Once we have a compelling ToE that is mathematically consistent and is reconciled with observations at all scales, this kind of premise will have ring of plausibility to it. We should not confuse the uncertainty of specific observations with statements about the possible exactness of a model; for example, people (though not all) argue quite strongly for superposition in quantum mechanics being exact. Then this kind of question starts getting quite real. We just have to be very careful that we do not make unwitting assumptions in our math: it seems that we are far more likely to make mathematical mistakes than physical reality is to deviate from a faithful model – at least if we make the assumption that reality is "real", and that we are not brains in vats. —Quondum 19:40, 7 May 2014 (UTC)[reply]

Einstein argued all the way that the true equations of nature are nonlinear. But this is also straying away from the OP. YohanN7 (talk) 21:32, 7 May 2014 (UTC)[reply]

Not that there is anything wrong with straying away from the OP. I just think that these questions are conceptually unrelated to the OP. YohanN7 (talk) 21:32, 7 May 2014 (UTC)[reply]

What's a good way to solve a problem of this form?

${\displaystyle \left\Vert (A+cI)^{-1}b\right\Vert ^{2}={\frac {1}{2}}}$

A is a matrix, b is a vector, and I'm solving for c, a positive scalar. Am I stuck with iterative methods at this point, or is there a neat trick I'm missing? Thanks. Katie R (talk) 17:47, 7 May 2014 (UTC)[reply]

I should add that ${\displaystyle A^{-1}}$ is already known at this point. Sherman–Morrison formula looks promising, so I guess I'll look into that for now. Katie R (talk) 18:12, 7 May 2014 (UTC)[reply]

Is anything special known about the matrix A? Is it antisymmetric by chance? Sławomir Biały (talk) 19:17, 7 May 2014 (UTC)[reply]

I honestly haven't inspected the matrix too closely yet. The elements are generated by:

${\displaystyle a_{n,i}=\sum _{j=0}^{N}\sum _{k=0}^{N}\sum _{l=0}^{N}b_{k}b_{l}d_{i-j-k}d_{i-n-l}}$

Note that d can take negative subscripts. Elements of b and d will range from -1 to 1, but that's about the only constraint on them. Katie R (talk) 19:32, 7 May 2014 (UTC)[reply]

Also b and d will never be all-zero. Katie R (talk) 19:40, 7 May 2014 (UTC)[reply]

Buried in this formula is a (discrete) convolution, which suggests that some Fourier methods might be a means of attack. Are you sure the indices are correct? (It looks like you can pull off the ${\displaystyle \sum _{l}b_{l}d_{i-n-l}}$ term entirely, but I feel like you mean for this to involve j, so that it participates in the other summations.) Sławomir Biały (talk) 19:48, 7 May 2014 (UTC)[reply]

Yes, they are correct, looking at my notes that term was originally seperate, combining it that way was literally the last thing I did to the expression and I have no idea what I was thinking. I have to head home for the day, but thinking of it in the frequency domain is an interesting idea. Hopefully it will get me off to a good start tomorrow. Katie R (talk) 20:08, 7 May 2014 (UTC)[reply]

According to that branch of mathematics which allows summing non-convergent infinite series, we have:

1-1+1-1+...=1/2

1-2+3-4+...=1/4

1+2+3+4+...=-1/12 (note the minus sign)

Let us recasr 1+2+3+4+... as (1+3+5+...)+(2+4+6+...), or (1+3+5+...)+2(1+2+3+...)

First question: Is this recasting valid?

Let us denote 1+2+3+4+... as S, and 1+3+5+... as T.

From above, 1+2+3+4+...=(1+3+5+...)+2(1+2+3+...)=T+2S

Thus S=T+2S, and solving for T, T=-S=+1/12 (the plus sign is for emphasis)

That is,

S=1+2+3+4+5+...=-1/12

T=1+ 3+ 5+...=+1/12

Second question: What is going on here? Bh12 (talk) 20:52, 7 May 2014 (UTC)[reply]

It is not true that the sum of the divergent series 1+2+3+... is −1/12. This series diverges properly to +∞. Instead of 1+2+3+..., you actually mean the special value of the Riemann zeta function ${\displaystyle \zeta (-1)}$, which is equal to −1/12. The "series" 1+3+5+... is understood as the Dirichlet L-function ${\displaystyle L(\chi _{2},-1)}$ where ${\displaystyle \chi _{2}}$ is the unique Dirichlet character mod 2. There are standard functional equations that relate these special functions under very general conditions. But actually, it is possible to do this by hand within the region of convergence of the Dirichlet series and invoke the persistence of functional relations to show that it continues to hold outside. In general, for Re(s)>1, we have

${\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}},\qquad L(\chi _{2},s)=\sum _{n=1}^{\infty }{\frac {1}{(2n-1)^{s}}}.}$

So, provided ${\displaystyle Re(s)>1}$, we have

${\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=L(\chi _{2},s)+2^{-s}\zeta (s).}$

By persistence of functional relations, this holds outside the region of convergence (the Diriclet L-function is known to extend uniquely to a meromorphic entire function). In particular, at ${\displaystyle s=-1}$, we recover your conclusion. Sławomir Biały (talk) 22:21, 7 May 2014 (UTC)[reply]

Sławomir, diverting the topic slightly, the articles that you link you seem to use a complex exponentiation without defining it specifically. This seems to be very sensitive to the precise definition of exponentiation used. I presume that specifically that the unique real logarithm of the base must be used in the definition of this exponentiation? Would these functions (the ones you linked to) be at all interesting if another branch of this logarithm were to be used? —Quondum 23:08, 7 May 2014 (UTC)[reply]

You seem pretty stupid to me!

1-1+1-1+... = (1-1)+(1-1)+... = 0+0+... = 0

1-2+3-4+... = (1-2)+(3-4)+... = (-1)+(-1)+... and that (when seen as infinite series) converges against -∞.

1+2+3+4+... (when seen as infinite series) converges against +∞. Thomas Limberg (Schmogrow) (talk) 07:40, 8 May 2014 (UTC)[reply]