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January 10

Does this continued fraction converge?

I was playing around with [1;2,1,3,1,4,1,5....] at school today, and I'm fairly sure it converges to a value around 1.44. Is there a test that I can run to see if it does converge? Does anyone know if this particular series turns out to equal something interesting? Thanks. 24.18.51.208 (talk) 01:44, 10 January 2009 (UTC)

For a start you may find continued fraction interesting. Every continued fraction with positive integer coefficients will converge; yours converges to about 1.35804743869438. It is irrational (because the continued fraction is infinite), and furthermore is not the root of any quadratic equation (because the continued fraction is not periodic). The continued fraction looks similar to the continued fraction for ${\frac {\tan(1)-1}{2-\tan(1)}}=1.259415845$, which has continued fraction [1;3,1,5,1,7,1,9,...]. Perhaps if you can figure out how to calculate the continued fraction for tan (1) (which has continued fraction [1;1,1,3,1,5,1,7,1,9,...]) then that might give some ideas of how to find the value of [1;2,1,3,1,4,1,5,...]. Eric. 68.18.17.165 (talk) 04:23, 10 January 2009 (UTC)

You may also find this calculator interesting if you just want an evaluation. It gives 641/472 = 1.3580508474576272 = 1, 2, 1, 3, 1, 4, 1, 5 hydnjo talk 04:59, 10 January 2009 (UTC)

Numbers having such continued fractions are a special case of Hurwitz numbers[1]. Using the formula (4) from this article and manipulating it a bit to remove the unwanted preamble, we get

$[1;2,1,3,1,4,1,5,\dots ]={\frac {\alpha -2}{3-\alpha }}\quad {\text{where }}\alpha ={\frac {J_{1}(2)}{J_{0}(2)}}{\text{,}}$

using the bessel functions of the first kind. — Pt _(T) 02:28, 14 January 2009 (UTC)

Pentagons

With only a sheet of paper, a ruler and a pencil, how might I go about drawing a regular pentagon of which each side is 150mm long, without having any construction lines around it afterward? The easiest way seemingly might be to start with one line 150mm long and work out how far away horizontally and vertically from that the end of each other line would be, but how should I do that? 148.197.114.165 (talk) 11:52, 10 January 2009 (UTC)

Can you use a compass? If so, see Pentagon#Construction. You'll need the formula in the previous section to work out what radius to use. Zain Ebrahim (talk) 12:00, 10 January 2009 (UTC)

Actually, this is better because it gives links to details for each step. Zain Ebrahim (talk) 12:13, 10 January 2009 (UTC)

I think I've worked it out. Would this give the right shape?: To draw regular pentagon ABCDE, where s is the length of any side of the pentagon, create a rectangle of size (1+(5^1/2)/2)*s by (((1+(5^1/2)/2)*s)^2 - s/2^2)^1/2. Draw then the line AB, of length s along one of the longer sides of the rectangle, so both lines have their centres at the same point. Draw the line AE from A to the nearest of the shorter sides of the rectange such that its length equals s, then do the same for BC from B to the other side. Find the centre of the long side opposite line AB, mark this as point D and draw lines from this to both C and E.

For a pentagon with sides length 150mm, this rectangle would be 242.7mm by 230.82mm, with A and B 46.35mm from the corners.

Something like that? 148.197.114.165 (talk) 17:48, 10 January 2009 (UTC)

Try it and see. It's easy enough to tell if you have the right shape at the end. --Tango (talk) 23:32, 10 January 2009 (UTC)

max size of a k-component graph

Resolved

 – --Shahab (talk) 14:24, 11 January 2009 (UTC)

This is related to a question I asked a few days back. What is the maximum number of edges possible in a n vertex graph having k connected components where each component has $n_{i}$ vertices. Obviously $\sum n_{i}=n$ and I need the maximum value of ${\frac {1}{2}}\sum n_{i}(n_{i}-1)$. How should I proceed further? I want the final answer to be in terms of n and k. Thanks--Shahab (talk) 13:06, 10 January 2009 (UTC)

As with the k=2 case, the maximal case is when all but one component has 1 vertex. Algebraist 17:00, 10 January 2009 (UTC)

I came to the same conclusion by the following procedure: As $\sum n_{i}^{2}=n^{2}-(k-1)(2n-k)-2\sum _{i<j}(n_{i}-1)(n_{j}-1)$ so if all but component has 1 vertex then the number of edges is ${\frac {1}{2}}(n-k)(n-k+1)$. Since this acts as an upper-bound on the number of edges too (because of the negative sign in $-2\sum _{i<j}(n_{i}-1)(n_{j}-1)$) hence this is the maximal achievable value for the number of edges. Is this proof correct? Secondly, is there a way to maximize the function ${\frac {1}{2}}\sum n_{i}(n_{i}-1)$ subject to the constraints $\sum n_{i}=n,n_{i}\in \mathbb {N}$. Cheers--Shahab (talk) 17:31, 10 January 2009 (UTC)

Shahab asks "is there a way to maximize the function $\textstyle {\frac {1}{2}}\sum n_{i}(n_{i}-1)$ subject to the constraints $\textstyle \sum n_{i}=n,n_{i}\in \mathbb {N}$?" The answer is to take $n_{1}=n$. But perhaps Shahab meant to ask the related question: how to minimize the function $\textstyle \sum n_{i}$ subject to $\textstyle {\frac {1}{2}}\sum n_{i}(n_{i}-1)=N$. Bob Robinson (UGA) and I solved this problem more than 20 years ago but never published it. Consider the greedy approach: choose $n_{1}$ as large as possible so that $\textstyle {\frac {1}{2}}n_{1}(n_{1}-1)\leq N$, then continue in the same fashion with what is left. Our theorem was that this works (i.e. the greedy choice of $n_{1}$ is correct) for all but a finite set of values of N, which we determined. If I recall correctly, there were something like 20 exceptions and the largest was about 86,000. In the exceptional cases, $n_{1}$should be chosen 1 or 2 less than the maximum possible. McKay (talk) 01:59, 11 January 2009 (UTC)

Thanks for the extra info. But my initial question was about maximizing ${\frac {1}{2}}\sum n_{i}(n_{i}-1)$ subject to $\sum n_{i}=n,n_{i}\in \mathbb {N}$ where I meant $\mathbb {N} :=\{1,2\cdots \}$. I cannot take $n_{1}=n$ for then the graph is just $K_{n}$ and I need k connected components in the graph. Cheers--Shahab (talk) 05:15, 11 January 2009 (UTC)

But that question is already answered above: take k-1 components equal to 1 and one component equal to n-k+1. The fact that it is best for k=2 proves that it is best for arbitrary k also. McKay (talk) 11:11, 11 January 2009 (UTC)

I am probably being dense here (& for this I ask you to indulge me) but I cannot understand your last statement. Isn't it possible for the max value to be different then ${\frac {1}{2}}(n-k)(n-k+1)$ and yet be equal to ${\frac {(n-2)(n-1)}{2}}$ for k=2. Can you please clarify?--Shahab (talk) 12:46, 11 January 2009 (UTC)

McKay probably has in mind the following simple proof (also the proof I had in mind above, for what that's worth): suppose we have values n_i subject to $\sum n_{i}=n,n_{i}\in \mathbb {N}$, and suppose there are two values of i for which n_i is not 1. Then by the k=2 case, we can increase ${\frac {1}{2}}\sum n_{i}(n_{i}-1)$ by changing one of these two values to 1 (and the other to their sum minus 1). Hence in the maximal case, all but one n_i must be 1. Algebraist 13:32, 11 January 2009 (UTC)

Exactly, thanks. McKay (talk) 14:28, 13 January 2009 (UTC)

OK. Thanks--Shahab (talk) 14:24, 11 January 2009 (UTC)

Multiplying integers by digit-swapping.

I've been playing around with integers which are multiplied by a factor (≤9) when the RH digit is moved to the LH end. For example, 102564 is quadrupled when the 4 is moved to the start. There is obviously an infinite number of these, seen by considering 102564102564, so I'm interested in the smallest integer for each multiplier. For multiplier k, my analysis gives that the integer comes from an expression with 10k-1 in the denominator, so 102564, for example, must have something to do with thirty-ninths. And 1/39 = 0.0256410256410..., so as a hypothesis the smallest integer for a given multiplier is found by looking at the decimal expansion of 1/(10k-1) and selecting the cycle starting 10. Which brings me to the question - acting on the basis outlined, I've found that the 58-digit integer 1,016,949,152,542,372,881,355,932,203,389,830,508,474,576,271,186,440,677,966 is multiplied by 6 when the final digit is moved to the front, but is there a smaller one?→81.151.247.41 (talk) 19:51, 10 January 2009 (UTC)

There is certainly a name for an integer whose digits get swapped (as you have described) when multiplying with a positive integer less than or equal to 9. Although I can't seem to remember... Can any number theorist help out here? —Preceding unsigned comment added by Point-set topologist (talk • contribs) 20:38, 10 January 2009 (UTC)

According to Sloane's, these are called k-parasitic numbers and the OP's number is indeed the least 6-parasitic number. Algebraist 20:42, 10 January 2009 (UTC)

3x3 Antisymmetric Matrices

Let A be a real 3×3 non-zero antisymmetric matrix. How does one show that there exist real vectors u and v and a real number k such that Au = kv and Av = −ku? What relation do these u,v and k bear to the exponential of the matrix A?

I know the exponential of the matrix A is a rotation matrix, but having not seen an appropriate approach to the first part of the question, I'm unsure as to how to progress with this - thanks.

86.9.125.104 (talk) 23:40, 10 January 2009 (UTC)Godless

If Au = kv and Av = -ku, then A²u = kAv = -k²u, and similarly A²v = -kAu = -k²v. So one approach would be to see what you can show about the eigenvalues and eigenvectors of A², given that A is 3x3 antisymmetric. Gandalf61 (talk) 09:31, 11 January 2009 (UTC)

Three '2's Problem

You have 3 twos, and you must use all of them in any expression of a number. Operations allowed are:

addition (e.g 2+2+2=6)

subtraction (e.g 2-(2-2)=2)

multiplication (e.g 2*(2+2)=8)

division (e.g 2*(2/2)=2)

raising to a power (e.g 2^(2^2)=16)

rooting - i.e. raising to the power of 1/n, where n is an integer (e.g 2^(1/(2+2))=2^(1/4))

logarithm to base 2 or base e - if to base 2, you have to indicate, and that would mean using one of your three twos: otherwise it would be to the base e. (e.g log2(2*2)=2, or log(2*2*2)=3ln(2), although the latter is not an integer)

concatenation - i.e combining say 3 and 4 to get 34 (Using "~" for this: e.g 2~(2/2)=21)

How many distinct positive integers can you form in this way? (note that for example 2+(2+2)=2+(2*2) so this only counts as 1 distinct positive integer).

Thanks,

Mathmos6 (talk) 23:58, 10 January 2009 (UTC)Mathmos6

This problem is fairly trivial: I suggest that you have a go yourself. We won't give you answers here; we can only give you hints at the most. Please first show us your progress (tell us what is the smallest and highest possible positive integer that you can obtain). —Preceding unsigned comment added by Point-set topologist (talk • contribs) 13:01, 11 January 2009 (UTC)

Without concatenation, it's pretty trivial. With concatenation, however, you can get significantly larger integers. --Tango (talk) 16:47, 11 January 2009 (UTC)

See also four fours. -- SGBailey (talk) 19:33, 11 January 2009 (UTC)

Also, RIES is a useful tool for these sorts of problem. **CRGreathouse** (t | c) 04:08, 16 January 2009 (UTC)

January 11

Group of points on the unit sphere

Let G be a finite non-trivial subgroup of SO(3). Let X be the set of points on the unit sphere in R^3 which are fixed by some non-trivial rotation in G. G acts on X - show that the number of orbits is either 2 or 3. What is G if there are only two orbits?

Could really just use a hand getting started on this one if nothing else. If the group is finite, I think all rotations have to be of the form 2pi/k for some integer k, and all axes of rotation have to be at an angle of 2pi/k to all other axes for a (different) integer k, right? That's about as much as I've accomplished so far. Cheers guys,

Spamalert101 (talk) 00:22, 11 January 2009 (UTC)Spamalert101

If the elements of G are not all rotations about the same axis then G must be the rotational symmetry group of one of the Platonic solids. There are three different "types" of points that are fixed by some non-trivial element of G and permuted by the other elements of G - these form the three orbits of G acting on X. One orbit is the vertices of the Platonic solid - I'll let you work out what the other two orbits are (hint: think about dual solids which have the same symmetry group).

On the other hand, if the elements of G are all rotations about the same axis then all non-trivial elements of G have the same two fixed points (what are they ?) and hence each of these two points is an orbit, and we have just two orbits. Gandalf61 (talk) 13:44, 11 January 2009 (UTC)

Is your first sentence obvious? Algebraist 14:13, 11 January 2009 (UTC)

Trivial to be precise (I was joking). :) PST —Preceding unsigned comment added by Point-set topologist (talk • contribs) 16:11, 11 January 2009 (UTC)

Having brushed up my geometry with our article Point groups in three dimensions, I can now say that it is not obvious, or even true. Gandalf has forgotten the rotation groups of the regular prisms. In this case, there are 3 orbits. Algebraist 17:22, 11 January 2009 (UTC)

To be more precise, if the elements of G are not all rotations about the same axis then G must be a subgroup of the rotational symmetry group of one of the Platonic solids. Moreover, considering only the icosahedron and octahedron suffice, as the dodecahedron has the same symmetry as the icosahedron, the cube the same symmetry as the octahedron, and tetrahedral symmetry is a subgroup of octahedral symmetry. Eric. 68.18.17.165 (talk) 16:27, 11 January 2009 (UTC)

(Reply to Algebraist) Not entirely obvious, I suppose - there is a proof in Chapter 15 of Neumann, Stoy and Thompson's Groups and Geometry. But maybe there is a simpler (although less geometric) solution to the original problem using Burnside's lemma. Each element of G fixes just 2 points of X, apart from the identity element which fixes all |X| points, so the sum over G of the points of X fixed by each element of G is |X| + 2|G| - 2. But by Burnside's lemma this is a multiple of |G| - in fact it is |G| times the number of orbits. I'll let the OP take it from there. Gandalf61 (talk) 17:25, 11 January 2009 (UTC)

Thanks for the reference. Google books reveals that the correct result is that the finite subgroups of SO(3) are exactly those you mentioned above plus the rotation groups of prisms (including the rectangle as a degenerate prism). The subgroups of these groups Eric alludes to in fact turn out to be on the list already. Algebraist 17:38, 11 January 2009 (UTC)

I agree with all of that - so which part of my original reply do you still think is incorrect ? Gandalf61 (talk) 17:54, 11 January 2009 (UTC)

The first sentence, since it excludes the prism/dihedron case, with dihedral symmetry groups. Algebraist 17:58, 11 January 2009 (UTC)

All of the non-trivial elements of the rotational symmetry group of a prism/dihedron have the same axis. So my first sentence "If the elements of G are not all rotations about the same axis then G must be the rotational symmetry group of one of the Platonic solids" is correct. The prism/dihedron case is covered in my second paragraph which starts "On the other hand, if the elements of G are all rotations about the same axis ...". In short, 3 orbits<=>Platonic sold, 2 orbits<=>prism/dihedron. Gandalf61 (talk) 18:07, 11 January 2009 (UTC)

No, they do not. For the prism associated with the regular n-gon, there are n-1 non-trivial rotations about an axis through the centres of the ends, plus 1 non-trivial order 2 rotation about the centre of each other face and about the midpoint of each edge running between the ends. Thus there are 3 orbits. Algebraist 18:13, 11 January 2009 (UTC)

If one took the prism with vertices (±1,±2,±3), then it has a symmetry group whose intersection with SO(3) has 8 elements, right? It is generated by rotations of 180° about the x, y, and z axes, right? It is not a platonic solid, right? I don't really get the geometric ideas though. I like your counting proof, since it only uses the fact that the elements fix at most 2 points. JackSchmidt (talk) 18:18, 11 January 2009 (UTC)

No, that group has 4 elements. It's the same as the rotation group of the rectangle-in-space, which I included in my list as a degenerate regular prism (the prism on a 'regular 2-gon'). It may gratify you to know that Gandalf's counting argument is used as the basis of the classification of finite rotation groups in the reference he supplied. Algebraist 18:24, 11 January 2009 (UTC)

I agree, 4 elements: the identity and the three 180° degree rotations about the x,y,z axes. I did get that it was the same as the flat rectangle, but I just misremembered you got the dihedral group of order 8. Having no mind for geometry, this seemed reasonable to me, though I was a little concerned that the group appeared to be abelian. I took out my very expensive model of a rectangular prism and applied the rotations carefully to check. Who says a geometry textbook is useless? Glad to know the classification uses ideas I could grasp. Even with 4 elements, this is one of your (A's) counterexamples to G's first sentence, right? Not that it matters that much, since the structure of G's proof is right, and checking the details should always be a part of reading someone's answer. JackSchmidt (talk) 18:37, 11 January 2009 (UTC)

It's not my counterexample, I'm hopeless at geometry. I lifted it from our article. Algebraist 19:58, 11 January 2009 (UTC)

(Reply to Algebraist) Yup, dumb mistake on my part. But it's always good to know that you are waiting to catch any little slips like that from us amateurs. Gandalf61 (talk) 18:44, 11 January 2009 (UTC)

Unidentified trig functions/identities?

Where $\Delta \theta =\theta _{f}-\theta _{s};\quad \Sigma \theta =\theta _{f}+\theta _{s}\,\!$, are

${\frac {\cos(\Sigma \theta )+\cos(\Delta \theta )}{\sin(\Delta \theta )}}={\frac {2\cos(\theta _{s})\cos(\theta _{f})}{\sin(\Delta \theta )}}={\frac {2}{\tan(\theta _{f})-\tan(\theta _{s})}}=?;\,\!$

and

${\frac {\cos(\Sigma \theta )-\cos(\Delta \theta )}{\sin(\Delta \theta )}}={\frac {2\sin(\theta _{s})\sin(\theta _{f})}{\sin(\Delta \theta )}}={\frac {-2}{\cot(\theta _{f})-\cot(\theta _{s})}}=?;\,\!$

(or their reciprocals) recognized as special functions, particularly in regards to spherical trigonometry? ~Kaimbridge~ (talk) 01:13, 11 January 2009 (UTC)

These are elementary consequences of the formulas for sin(A+B) and cos(A+B). What is your question exactly? McKay (talk) 01:26, 11 January 2009 (UTC)

Just if there is some elementary relationship/identity, like the "sine for sides", or in the same way that

${\frac {\lambda _{f}-\lambda _{s}}{{\mbox{arccosh}}(\sec(\phi _{f}))-{\mbox{arccosh}}(\sec(\phi _{s}))}}={\frac {\Delta \lambda }{\int _{\phi _{s}}^{\phi _{f}}\sec(\phi )d\phi }}=\tan(\alpha ){}_{\color {white}.}\,\!$

for loxodromic azimuth, which involves the inverse Gudermannian function. ~Kaimbridge~ (talk) 15:25, 11 January 2009 (UTC)

Conic section slope of directrix

Given a conic section $Ax^{2}+By^{2}+Cxy+Dx+Ey+F$ what is the slope of its directrix? Borisblue (talk) 01:21, 11 January 2009 (UTC)

Sadly, we don't do homework for people. Furthermore, this problem is trivial. Please think about it a bit more and tell us what you have done on the problem so far. PST —Preceding unsigned comment added by Point-set topologist (talk • contribs) 16:10, 11 January 2009 (UTC)

Cubic functions

I'm trying to use cubic (and similar) functions for a little model, but I really don't have much maths. Could a mathematician please explain whether one changes a, b or c to make the curves flatter/steeper, change the y intercept, etc.? The article on quadratic equations does this graphically through a clever image, but a few lines of text would be great. I'd like to add this info to the relevant article as well. --Matt's talk 14:09, 11 January 2009 (UTC) Edited to clarify which article is relevant --Matt's talk 14:17, 11 January 2009 (UTC)

I presume you mean that the cubic is of the form:

$f(x)=ax^{3}+bx^{2}+cx+d\,$

In that case, "flatness" of the curve is measured by its derivative which is (at x):

$f'(x)=3ax^{2}+2bx+c\,$

So only the coefficients a, b and c have an impact on the steepness of the curve (the greater these values are, the greater the steepness; the smaller these values are, the greater the flatness). The y-intercept is given by the image of 0 under f so the value of d equals the y-intercept. If d is 0, the curve passes through the origin. PST

The article elliptic curve might also be of interest to you. PST

And by the way, mathematicians usually use one branch of mathematics in another branch of mathematics. There are numerous examples of this (I might as well let someone else list these examples; there are so many that I can't be bothered!). One interesting example is applying graph theory and the theory of covering maps to prove the well known Nielson-Schreier theorem; i.e every subgroup of a free group is free.

On the same note, there are mathematicians who would prefer not applying mathematics to another field (theoretical mathematicians) and those who would prefer applying mathematics to another field (applied mathematician). From experience, applied mathematicians are generally not so interested in the theoretical parts of mathematics and thus do not choose to learn much theoretical mathematics. But there are special cases. PST —Preceding unsigned comment added by Point-set topologist (talk • contribs) 16:05, 11 January 2009 (UTC)

Elliptic curves aren't really relevant to what the OP is doing, and what does that last paragraph have to do with anything? --Tango (talk) 17:00, 11 January 2009 (UTC)

Have a look at his links to see the relevance of that part. PST

The steepness of the graph for large (either positive or negative) values of x is determined primarily by a. For smaller values of x, the graph will change direction a lot so it's rather more complicated. You may find it helpful to write the cubic as y=a(x-u)(x-v)(x-w), then the steepness for large values of x is, again, given by a, and u, v and w are the x-intercepts. The y-intercept would be -auvw. --Tango (talk) 17:00, 11 January 2009 (UTC)

Re-write equation as:

$y=a\left(x+{\frac {b}{3a}}\right)^{3}+\left(c-{\frac {b^{2}}{3a}}\right)\left(x+{\frac {b}{3a}}\right)+e$

where e is a function of a,b,c and d that I can't be bothered to write out. Then change co-ordinates:

$v=y-e\,;\,u=\left(x+{\frac {b}{3a}}\right)$

$v=au^{3}+\left(c-{\frac {b^{2}}{3a}}\right)u$

so we have placed the cubic's centre of symmetry at the origin. Now we can see that a determines the slope of the cubic far from its centre and $\left(c-{\frac {b^{2}}{3a}}\right)$ determines the slope at its centre, and the number of turning points. Gandalf61 (talk) 17:44, 11 January 2009 (UTC)

Sounds to me like you might be interested Bézier curve and splines in general.Dmcq (talk)

Thank you to all who answered, I've noted this down and plan to add it to the article later. Not as simple as I'd hoped though :-( --Matt's talk 19:47, 18 January 2009 (UTC)

Manifolds

I have a question for those people at the reference desk. This is not homework so you don't need to worry about that. I would like to know why manifolds (and more generally differentiable manifolds) are so important objects of study. I mean, they are in a sense so restrictive in nature (why exclude so many important topological spaces from study?). To me, it seems odd that the figure 8 is excluded from study (why can't you do calculus at the center of the figure 8?). I know that there are more general types of manifolds such as orbifolds and Banach manifolds but is every topological space some type of manifold? If not, what are the minimal conditions required for a space to be some type of manifold? I am pretty sure, for instance, that one can do calculus on the indiscrete space (any function between indiscrete spaces should have derivative 0). But on the other hand, I am also sure that the indiscrete space is not a type of manifold. I certainly understand that manifolds are important but unlike continuity and integration, I don't understand why differentiation cannot be generalized to arbitrary topological spaces (integration to measure spaces). Or it maybe just my lack of knowledge of the subject that I don't know that one can do calculus on arbitrary topological spaces. Thankyou for you help! —Preceding unsigned comment added by 129.143.15.142 (talk) 20:53, 11 January 2009 (UTC)

Note that the indiscrete space is indeed a (0-dimensional) manifold if the space in question has precisely one-point or vacuously a manifold if the space in question is empty. But those are just fine details for you. PST

It's not an area I've ever studied properly, but I can't see how you could define derivatives on a space unless the space was locally metrizable. That rules out some spaces (but not things like a figure of 8). Also relevant is that, depending on your definitions, the long line is not a manifold, but can have a differential structure put on it. --Tango (talk) 22:00, 11 January 2009 (UTC)

Thanks. The definition I follow is the "locally Euclidean" one so I allow the long line to be a manifold. I think one other important thing to consider is that in manifold theory there are important facts about the derivative of a smooth function between smooth manifolds (namely it is a linear isomorphism between tangent spaces). But on one of your notes, the indiscrete space is not locally metrizable (when it has more than one point as someone above mentioned) but I would expect that the derivative of a function between such indiscrete spaes to be 0. You could argue similarly that continuity needs a metric although that is not the case. Isn't there an 'open set formulation' of differentiability? —Preceding unsigned comment added by 129.143.15.142 (talk) 22:16, 11 January 2009 (UTC)

Just to clarify something I said, I think that people don't study calculus on locally metrizable spaces because often in manifold theory, one considers vector spaces also. These are basically my questions (by the way, a real-valued function defined on the figure 8 is differentiable if and only if it extends to a differentiable function on a open set containing the figure 8 but this cannot be generalized to topological spaces that cannot be embedded in Euclidean space; this brings me to another question):

1) Since all Hausdorff, second countable manifolds can be embedded in Euclidean space, why not define differentiability on them as I have just done (i.e differentiable iff extends to a differentiable function on an open superset?)? I know this would exclude all those interesting theories of differentiable structures (like one interesting theorem that there are only 28 non-equivalent differentiable structures on S^7), but for the most part, this definition would suffice.

2) Is the figure 8 some type of manifold?

3) What are minimal conditions for a space to be some sort of manifold?

4) Is the indiscrete space (with more than one point) some sort of manifold (I would expect this naturally although I don't think that it is)?

I have not been getting sleep over this since I first learnt about manifolds (why are they so restrictive in nature?) so I would appreciate any help from the knowledgeable people at WP! —Preceding unsigned comment added by 129.143.15.142 (talk) 22:26, 11 January 2009 (UTC)

1) While such manifolds can be embedded in Euclidean space, they can be embedded in lots of different ways which wouldn't yield the same definition of differentiability (for example, a 2-sphere can be embedded in R³ as a cube, in which case the image of a great circle wouldn't be differentiable since it would have corners in it, but with the standard embedding it clearly is). While you may be able to improve things by requiring the embedding to be smooth, you would end up with a circular definition. 2) A figure of 8 isn't a manifold simply because there is no open neighbourhood of the central point that is homeomorphic to the real line. Mathematicians define things the way they do because those definitions are useful. For example, if a space is locally Euclidean at a point you can define its tangent space at that point (which is, itself, a useful thing to do for all kinds of purposes), you can't define the tangent space to a figure of 8 at that central point (it has two tangents there, so you would end up with the union of two lines, which isn't a vector space). 3) A manifold is a space that is locally Euclidean everywhere (possibly with some extra conditions, depending on who you ask), that is the minimal condition (I'm not really sure what you mean by "some sort" of manifold, there are generalisations of manifolds, but they are really manifolds any more even if the word may appear in their name, you could argue that "topological space" is a generalisation of "manifold", but that doesn't mean much). 4) No, the only neighbourhood of any point in an indiscrete space is the whole space, which can't be homeomorphic to any Euclidean space because no Euclidean space (beyond R⁰, I guess) is indiscrete. --Tango (talk) 00:04, 12 January 2009 (UTC)

Thanks for the response! I guess you didn't understand what my question was. I wanted to know whether the figure 8, despite not being a manifold, is an orbifold, a Banach manifold etc... I understand your answer to question 1 (thanks!). But as I mentioned, I still don't know question 2 but I am pretty sure that the figure 8 is not an orbifold because it is not homogeneous whereas it could be a non-pure Banach manifold (as Banach spaces are always homogeneous)? Basically, I know that the indiscrete space is a Banach manifold (it is itself a Banach space with any vector space structure).

I understand that mathematicians define things that are useful but my question is: why are people still doing calculus?! I mean isn't it too late for that? There are so many important branches of mathematics (that are general such as measure theory and topology) but differential topology just seems to specialized for me. I just somehow think that broader definitions of manifolds (Banach and orbi-) allow a wider class of spaces but it would be great to know what this class is by purely topological methods. I have another question (thanks for answering all these questions for me; I appreciate it):

5) I would expect one to be able to do calculus on any locally path connected, first countable space but why is this not the case?

Thanks again!

This is the first interesting (compared to other questions) question that I have seen on this reference desk. And yet no-one except for Tango has answered it (disappointing!). But anyhow, differenciation involves the notion of 'direction' and arbitrary topological space don't have this notion unless you have a local vector space structure defined on them (for instance locally Euclidean). However, this is just the most likely reason as to why people don't study non-locally Euclidean topological spaces.

P.S What Tango said was correct but one could speak of 'calculus on the circle with respect to particular embeddings'. So this could make matters more interesting.

PST

You know, making differential calculus on a topological space it is more or less like skating on a football pitch. Not smooth enough, no fun. You ask, why the abstract definition of differentiable manifolds with atlases if they are essentially submanifold of R^m? Sometimes Whitney embedding is of help indeed, but in other cases it is not, and would only add an unnecessary additional structure, that makes things more complicated. For the same reason (finite dimensional, real) vector spaces are defined in abstract and nobody would define them as Rⁿ. As to the Banach manifolds, they are just manifolds modelled on Banach spaces, so the difference wrto the classical ones is just the possibly infinite dimension.... what should they have to do with a figure "8"? (a figure 8, by the way, is not a submanifold of R ² but can be seen as an injective immersion of R). Anyway, in general your query seems to address to the general issue: why certain structures are still studied if they are less general of others? The obvious reason is that they have nicer properties, so you can do more with them, even if you can apply them less. What makes successful a mathematical structure is not the maximal generality, but the right generality. That's why (for instance) Banach spaces are a winning specimen and you meet them everywhere, while the more general Topological Linear Spaces are still interesting, but not at all comparable, and usually you meet them only at the Zoo. And Differentiable Manifolds also; a great category, among other things it's an important bridge between Topology and Analysis. --78.13.140.37 (talk) 15:48, 12 January 2009 (UTC)

Thankyou for the response! I now have a somewhat better understanding of why manifolds are so restrictive. But I guess I still have a few questions...

1) The figure 8 is certainly not a submanifold of the plane, I agree. But I am still confused as to why you can't do calculus on it. I mean at the center, you can approach from four directions. If the change is the same in each direction, the function is differentiable. How about that? Similarly, the comb space is not a manifold (in any sense) but you can still do calculus on it. My other question is basically why so important spaces are excluded from study.

2) I like differentiation but continuity and integration are generalized as much as possible. I guess that perhaps this is the best you can generalize differentiation naturally but I feel (as in 1)) that you can generalize a bit further. So, why shouldn't it be possible to do calculus on a locally contractible first countable space?

3) My third question is why people are still doing calculus! I mean mathematics is about theory; not calculations and yet differential geometry seems to be precisely that.

4) What are the most general manifolds? Is every topological space some kind of manifold (can one, given a locally homogeneous space, find a Banach space that looks locally like it and sort of 'patch up this local pieces' to form a Banach space and conclude that every such space is a Banach manifold?)?

Thanks (I would greatly appreciate it if someone could answer each question at a time like Tango)!

I'm not entirely sure about 1) and 2). 3) The main reason for doing pure maths is because it is fun. You can often have more fun by restricting yourself to a specific subclass of objects since more theorems hold about them. For example, if you study right angled triangles, you can find Pythagoras' Theorem, if you study general triangles you can't. Often, once you've found theorems for a specific subclass you can try and generalise them, but often the generalised form isn't as elegant (compare Pythag to the cosine rule). It also helps if it has useful applications, since people are more likely to give you funding to do it. Calculus has enormous numbers of applications throughout the physical sciences and elsewhere, point-set topology and general measure theory don't (they have applications, certainly, but not to the same extent). 4) A manifold is a topological space that is locally Euclidean, nothing else is a manifold. Some generalisations include the word "manifold" in their name, but that doesn't mean much. If there was a type of object that all topological spaces fell into that that type of object would just be called "topological space". --Tango (talk) 18:59, 12 January 2009 (UTC)

As to point 1, I would add that even though the figure 8 is not a submanifold, it is an object that can be efficiently studied by means of differential geometry, and you can apply all differential calculus you want. As you are suggesting, locally at the crossing point it's just union of two nice transverse arcs. Notice that similar sets appear quite naturally in differential geometry: e.g as a level set of a function at a critical non degenerate value. Or it's a local embedding of S¹ etc. Notice also that many more or less classical curves are figure eight (e.g., the Lemniscate of Bernoulli, some Lissajous figures, the three body problem periodic solution by Chenciner & Montgomery [2], etc). Anyway, maybe an answer to your first question is just that you do not need to give a name to an object and to include it in a formal class in order to study it, and that, in particular, a space is not excluded from the methods of differential calculus just because it is not a smooth manifold. As to point 2, notice also that a generalized derivative of functions (always on Rⁿ or on smooth manifolds) is given by the very wide concept of weak differentiation of a distribution, that allows you to differentiate not only L¹_loc functions but even objects that are not even functions, like measures. You may reflect on the way this extension is done by means of functional analysis. You just restrict your attention on a very special and nice setting, that is the space $\scriptstyle {\mathcal {D}}:=C_{c}^{\infty }(\mathbb {R} ^{n})$ of all infinitely differentiable functions with compact support on Rⁿ; here you have the very nice and classical notion of partial derivatives $\scriptstyle \partial _{i}$. For the very reason that $\scriptstyle {\mathcal {D}}$ is so cute and "small", it's dual $\scriptstyle {\mathcal {D}}^{\prime }$ is a very very wide pot, that contains in a natural ways not only $\scriptstyle {\mathcal {D}}$ itself, but also functions of Sobolev classes, locally integrable functions, Radon measures, and all sort of weird and wild objects. Then, each partial derivative operator $\scriptstyle \partial _{i}$ on $\scriptstyle {\mathcal {D}}$ naturally gives rise to a transpose operator on $\scriptstyle {\mathcal {D}}^{\prime }$, in much the same way transposes of matrices are defined for the finite dimensional R^m. These transpose operators (with a change of sign) turn out to be an extension of the partial derivatives to $\scriptstyle {\mathcal {D}}^{\prime }$. What I mean is that there is a kind of moral in this (you are free to choose what moral). Now, this is not exactly the extension of calculus you are asking in 2, for you are thinking also to a more general kind of domain, rather than more general object to differentiate. Still I can imagine various situation and ways in which smooth differential calculus may be applied. This depend on what is your precise problem indeed... I do not agreee completely with your point 3, for theory and calculus are not two opposite issues after all. Abstract theories indeed reduce difficult problems to situations where you can do elementary computations (you can find yourself a lot of examples). I do not completely understand your 4... well, the only relevance of Banach spaces in this context that I can think is that every metric space can be isometrically embedded in a Banach space (this is usually referred to as the Frechét-Kuratowski embedding; a both easy and useful thing: once you put your metric space into a Banach, you can consider the operations or concepts there, like closure , that gives you a completion, convex hull, epsilon-neighbourhoods, etc). As to the question: what's the more general concept of manifolds, it seems a bit too generic. Not Banach manifolds however; you may introduce classes of topological spaces with atlases in order to describe global objects made patching together (in a given way) local objects (picked in a given category). In conclusion, I agree with everything Tango said, except his statement that people do pure math mainly for fun... although I do not have counter-examples to it. ;) --PMajer (talk) 11:17, 13 January 2009 (UTC)

PS: Just to speak, in fact if you put a metric space M into a Banach space X via the FK embedding, then you may define "differentiable functions" on M those functions that are restriction to M of functions on some open nbd of M in the Banach space. The Whitney extension theorem, that can be stated also in the context of Banach spaces, will then give you a characterization of these functions (but do not ask me what to do with this definition) --PMajer (talk) 15:39, 13 January 2009 (UTC)

Thanks! I appreciate your help. I know that calculus has lots of applications everywhere but maths is for the sake of it. First priority should be maths itself. Although non-Hausdorff spaces, for instance, pretty much have no applications, I consider them interesting in their own right: or even orthocompact spaces and such. Anyway, thanks again!

January 12

A Pascal's triangle question I've never seen before

Dear Wikipedians:

I'm used to the following type of questions which applies Pascal's triangle, it asks how many ways to form the word "November":

    N
   O O
  V V V
   E E
    M
   B B
  E E E
 R R R R

as you can see each line is always 1 character more or less than the line above it, however, recently my teacher gave the following variation, which totally stumped me:

    N
   O O
  V V V
  E E E
   M M
 B B B B
  E E E
   R R

as you can see there are now lines with same number of characters as previous line, and also line with 2 characters more than the previous line, how do I handle this?

Thanks,

76.65.14.151 (talk) 00:09, 12 January 2009 (UTC)

I assume the goal is to move from top to bottom without "jumping too far". If no rules are specified then I would guess (and state that as an assumption if possible) that if an M is directly above a B then you can only move directly down to that B, so two of the B's cannot be reached. If the question form allows it then you could also make two answers where the other assumes you can move left-down, straigth down, or right-down, so from each M you can move to 3 B's. PrimeHunter (talk) 00:27, 12 January 2009 (UTC)

I'm not sure how to interpret the question in the second case - how do you form the word? Are you allowed to move diagonally when letters are directly on top of each other? If not, then the question is easily reduced to one like the first case. If you are, then it depends on precisely what is and isn't allowed. Incidentally, I don't see how Pascal's triangle comes into it - for the first case the answer is just 2ⁿ where n is the number of lines which are longer than the line before (since, when going to those lines, you have a choice of 2 letters to go two, for all the other lines you just have 1 choice).That misses a few options, sorry. Seeing Yanwen's post below, I see where Pascal's triangle comes into it.--Tango (talk) 00:32, 12 January 2009 (UTC)

The question is understated, can we have some more information please; such as an explicit copy of the said question. In your post you only reference the question you are trying to answer without actually stating what it is that you are trying to do. I too fail to see any relation to pascals triangle. Or the fact that the layers form a word. Or what you mean by 'handle this'. —Preceding unsigned comment added by 92.16.196.156 (talk) 01:41, 12 January 2009 (UTC)

If you take one letter from each row you get the word "NOVEMBER", the question seems to be how many different routes are there from the top row to the bottom. --Tango (talk) 02:52, 12 January 2009 (UTC)

One way to do it is to add up the numbers above it. Taking your first example:

    N
   O O
  V V V
   E E
    M
   B B
  E E E
 R R R R

You start with N. There is only 1 way to get to the two O's. Then, from the two O's, there are 2 ways to get to the center V, and 1 way to get to the V's on the side. And you just keep going until you get this:

    1
   1 1
  1 2 1
   3 3
    6
   6 6
  6 12 6
 6 18 18 6

Then just add up the numbers on the bottom: 6+18+18+6=48

The solution to the variation would then be:

    1
   1 1
  1 2 1
  3 4 3
   7 7
 7 14 14 7
  21 28 21
   49 49

49+49=98

(The two digit numbers throws off the alignment, but hopefully you can still read it alright.)--Yanwen (talk) 03:18, 12 January 2009 (UTC)

Hi, I'm the original question poster. Thank you for all the input! I think Yanwen's solution is the one that I was looking for. I think the question is really understated and badly formed. But I really appreciate the input from everyone. 70.52.148.139 (talk) 00:53, 13 January 2009 (UTC)

Leibniz notation

In calculus, dx, dy, du, etc. seem to be manipulated as if they were variables. This confuses me.

When you're only dealing with first derivatives, this seems pretty straight forward. If you have y = f(x), then ${\frac {dy}{dx}}={\frac {d}{dx}}f(x)=f'(x)$. You can multiply both sides by dx to get dy = f'(x)dx and then integrate $\int dy=\int f'(x)dx$ --> y = f(x) + C

But when it comes to second derivatives, you get ${\frac {d^{2}y}{dx^{2}}}$. Does that mean that $d^{2}x$ and $dx^{2}$ are different things? What's the difference. It seems to me like $d^{2}y=d(dy)$ is an infinitesimal change in the infinitesimal change in y. Isn't that just 0, because dy is already infinitesimally small? And is $dx^{2}{\overset {?}{=}}(dx)^{2}$ in the same way that $\sin ^{n}(x)=(\sin(x))^{n}$? --Yanwen (talk) 03:01, 12 January 2009 (UTC)

Maybe this will help.--Shahab (talk) 09:06, 12 January 2009 (UTC)

Politely speaking, I don't think that link will be of much use to him (most calculus students don't know linear algebra and differential topology). I think the best help would be to note that dy/dx approaches the derivative (if the function is differentiable) as dx approaches 0.

Treating dx, dy etc. as if they were variables is like crossing the road before you see the green man - it's usually okay if you have some experience and take care, but if in doubt, don't do it. For example, "cancelling" the dus appears to work in the chain rule:

${\frac {dy}{dx}}={\frac {dy}{du}}\,{\frac {du}{dx}}$

but if you extend this reasoning to the second derivative you would conclude that

${\frac {d^{2}y}{dx^{2}}}={\frac {d^{2}y}{du^{2}}}\,\left({\frac {du}{dx}}\right)^{2}$

which is incorrect - the correct extension is

${\frac {d^{2}y}{dx^{2}}}={\frac {d^{2}y}{du^{2}}}\,\left({\frac {du}{dx}}\right)^{2}+{\frac {dy}{du}}\,{\frac {d^{2}u}{dx^{2}}}$

A more rigorous approach is to regard d/dx as an operator D which maps each differentiable function y(x) to another function. Then d²y/dx² represents the second iteration of this operator i.e. D(D(y(x))) or D²(y(x)). Gandalf61 (talk) 10:13, 12 January 2009 (UTC)

One of the articles referenced from differential (mathematics), probably differential (infinitesimal), should really treat this but doesn't. A previous discussion related to this is archived at WP:Reference_desk/Archives/Mathematics/2008_September_10#What_are_the_rules_of_treating_differentials.3F. Dmcq (talk) 12:22, 12 January 2009 (UTC)

Yes, d ²x and dx ² are different things. d ²x = ddx = d(dx) and dx ² = (dx) ² = (dx) · (dx) = dx · dx . When x is a variable, then so is the differential dx, and the rules of differentiation a sum d(x + y) = (dx) + (dy) = dx + dy and a product d(x · y) = (dx) · y + x · (dy) = dx · y + x · dy lead to correct formulas. So $\scriptstyle dy={\frac {dy}{du}}\,du$ and $\scriptstyle d^{2}y=d(dy)=d({\frac {dy}{du}}\,du)=d({\frac {dy}{du}})du+{\frac {dy}{du}}d^{2}u.$ You also need to know that if x is a constant, then dx = 0, and if x is an independent variable, then dx is constant and consequently d ²x = 0. But there is a problem in interpreting the differentials as infinitesimals which are zero and nonzero at the same time. Bo Jacoby (talk) 17:04, 12 January 2009 (UTC).

But if we could treat differentials in exactly the same way as other variables then we would have:

${\frac {d^{2}y}{du^{2}}}\,\left({\frac {du}{dx}}\right)^{2}={\frac {d^{2}y}{du^{2}}}\,{\frac {du^{2}}{dx^{2}}}={\frac {d^{2}y}{dx^{2}}}$

which we know is incorrect. So to put differentials on a formal basis, we would need some rules to define when we can treat them just like variables and when we can't. Gandalf61 (talk) 09:52, 13 January 2009 (UTC)

So if treating dy and dx as variables would not yield consistent correct answers, what would be a better way to manipulate differentials? Also, I don't completely understand the page One-form that Shahab referenced. Is it saying that df is a function of x and dx? Wouldn't that just become the difference quotient?--Yanwen (talk) 16:15, 13 January 2009 (UTC)

The "dy and dx as variables" approach can be made rigorous within the framework of non-standard analysis. But my personal opinion is that a rigorous treatment of non-standard analysis adds more complexities than it removes, whereas a non-rigorous treatment leaves us no better off than before. I prefer the differential operator approach - this means abandoning the idea that dy and dx can be treated as variables, but you still reach the same results as before, with less room for error. Gandalf61 (talk) 11:24, 14 January 2009 (UTC)

Apart from nonstandard analysis (which I do not understand), the article on differential algebra explains the formal use of dx. I am a little shaken by Gandalf61's example, though. Bo Jacoby (talk) 12:18, 14 January 2009 (UTC).

Well. The differential is dy = a dx so that dy/dx = a, and da = b dx so that b = da/dx. The second differential is d ²y = b dx ² + a d ²x. Only when x is an independent variable can we set dx = constant and d ²x = 0 and d ²y = b dx ² . Otherwise the second derivative b is a partial derivative, which is not a quotient between differentials. In Gandalf61's example, u is not an independent variable and so d ²y / du ² is a partial derivative. Bo Jacoby (talk) 21:30, 15 January 2009 (UTC).

Lebesgue Integrals

How do you do Lebesgue Integrals? I don't understand it the way our article said it.----The Successor of Physics 11:13, 12 January 2009 (UTC)

I don't know what you mean by 'doing' Lebesgue integrals. But I explain the process to you. Start with a measure space (that is a set together with a sigma algebra (that is, a subcollection of the collection of all subsets of the set). Members of this sigma algebra are called measurable sets. Each measurable set has associated a non-negative real number called its measure (generalizes the notion of 'area' and 'volume'). So in effect, you are defining a function from the sigma algebra to the extended set of non-negative reals. This function is called a measure. A measure space is simply a set, with a sigma algebra and a measure). Call a real valued measurable function on a measure space simple if its range is finite. An indicator function of a measurable set E (as a subset of the measure space) is simply a function that is 1 at E and 0 outside E. Note that every simple function can be expressed as the linear combination of indicator functions. So one can integrate a simple function by factoring out the constants and considering the measure of the sets on which each indicator function is 1. Now the integral of a real-valued non-negative measurable function is simply the supremum of all integrals of non-negative simple functions that are less than or equal to this function.

While reading the article, keep the above in mind. Hope this helps. PST

Try skipping straight to the "Intuitive interpretation" section of Lebesgue integration. If you want to actually calculate the Lebesgue integral of a function f, one way is to find a function g which you can integrate with Riemann integration and which is "sufficiently close" to f; the Lebesgue integral of f is then equal to the Riemann integral of g. And one useful case of "sufficiently close" is if g(x)=f(x) almost everywhere. So, for example, the Dirichlet function is 0 almost everywhere (because the points where it is non-zero, the rationals, are a set of measure 0) so the Lebesgue integral of the Dirichlet function is equal to the Riemann intergal of 0, which is 0. Gandalf61 (talk) 13:06, 12 January 2009 (UTC)

It's useful to remember that all Riemann-integrable functions are Lebesgue-integrable, and the Lebesgue integral evaluates to the value of the (proper) Riemann integral wherever the latter integral exists. So for all the functions one is taught to integrate in elementary calculus, there's no difference. The power of the Lebesgue integral is mostly a matter of the deeper theory. Ray (talk) 14:25, 12 January 2009 (UTC)

To summarize, mathematicians don't really care about what the Lebesgue integral of a particular function is; they are worried about the theory (what functions are Lebesgue integrable, certain properties of the integral, how it behaves for uniformly convergent sequences of functions). PST

It's true that certain kinds of mathematicians don't care about calculating the Lebesgue integral (I for one haven't done this since grad school), but what you say above is certainly false for many pure mathematicians, to say nothing of applied folks who are very interested in calculating things. Staecker (talk) 17:05, 12 January 2009 (UTC)

I could be wrong, but I think the functions those applied folks integrate tend to be at least piecewise continuous, and hence already Riemann/Darboux integrable. Algebraist 17:14, 12 January 2009 (UTC)

I don't think applied mathematicians study Lebesgue integration that is not Riemann integration (or Riemann-Stieltjes integration). This is because I don't consider probabilists to be applied mathematicians. PST

Yes, but if the OP means how does one compute a Lebesgue integral of a function that is only L¹, I'd answer: using all the machinery of the theory, starting with convergence theorems. The power of Lebesgue integration with respect to the Riemann's is not that it allows to treat functions of a wider class, but rather that it allows to do much more general operations on them, both of algebraic and topological nature; and these operations in some cases also allow the computation of integrals. --PMajer (talk) 14:14, 13 January 2009 (UTC)

PS: however, Superwj5: just give these people any L¹ function in bones and flesh, and we are going to compute its integral for you as an example (hopefully) --PMajer (talk) 15:21, 13 January 2009 (UTC)

Thanks a lot!!!----The Successor of Physics 13:35, 16 January 2009 (UTC)

eⁱ

What is eⁱ? It is used in some of Euler's identities and formulae but I can't solve them unless I get eⁱ!----The Successor of Physics 11:31, 12 January 2009 (UTC)

Hint: $i^{i}$ is $e^{-pi/2}$.

It's a transcendental number about which (I think) little can usefully be said except that it is eⁱ. What do you want to know for? Algebraist 12:31, 12 January 2009 (UTC)

You can of course write it as $\cos 1+i\sin 1$, which helps to get an estimate if you have a calculator that can handle trigonometry but not complex numbers. But for any deeper meaning, I don't believe the sine and cosine of 1 radian are very interesting. -- Jao (talk) 12:59, 12 January 2009 (UTC)

Superwj5, if you want to see eⁱ in the complex plane: it is on the unit circle, first quadrant, on one end-point of an arc of lenght 1, the other end-point being e⁰=1. If you understand how works geometrically the multiplication of complex numbers, you can visualize it as the limit of (1+i/n)ⁿ. For a fixed n draw the piecewise linear arc through the points (1+i/n), (1+i/n)²,..(1+i/n)ⁿ; then you can figure out what happens taking the limit of (1+i/n)ⁿ and why the final arc has lenght 1. But what does "Successor of Physics" mean, and why it's cancelled? .) --78.13.140.37 (talk) 14:34, 12 January 2009 (UTC)

Maybe it means "no metaphysics". —Tamfang (talk) 05:17, 14 January 2009 (UTC)

Thanks a lot!!! I want to know it for the equations sin(x)=(e^ix+e^-ix)/2 and cos(x)=(e^ix-e^-ix)/2i. ----The Successor of Physics 13:36, 16 January 2009 (UTC)

what's the problem? these equations just came from e^ix=cos(x)+isin(x) and e^-ix=cos(x)-isin(x), that just tell you what are the real and the imaginary parts of e^ix and e^-ix. --PMajer (talk) 11:28, 17 January 2009 (UTC)

combination and permutation

in a maximum group of 1-50 numbers both inclussive,how many times can a sub group of 5 numbers divide it without repetitions of any group?

This is what you mean:

Question: Consider the set of numbers from 1-50; call this X. How many different subsets of X exist, each having the property that every member divides 50.

Answer: This problem is trivial and will be removed. Please state it more clearly. If my interpretation is correct, the answer is 6C5 (since there are six factors of 50, and every subset of five elements must be a subset of this set of factors).

PST Preceding unsigned comment by Point-set topologist (talk) 20:49, 13 January 2009 (UTC)

Well, of course the statement about the triviality of your problem reflects at most the personal opinion of PLT, who is however not sure about the interpretation of it. So I do not think it will be removed, and if you would restate it more clearly, even by examples, somebody will certainly give you an answer. --PMajer (talk) 13:48, 13 January 2009 (UTC)

We usually only remove questions if they violate the legal/medical disclaimer, or are cross-postings on other reference desks. PST, if you don't think a question should be answered (e.g., homework, or vague, or trivial), then just don't answer it.

As for the original question, I too cannot understand what you are asking for help with, could you please restate the problem? Eric. 68.18.55.236 (talk) 14:35, 13 January 2009 (UTC)

I agree with the above - we don't remove questions for being trivial and I also don't understand the question. PST's interpretation is possibly correct, but it's far from certain. --Tango (talk) 14:57, 13 January 2009 (UTC)

I feel that sometimes people ask these problems (the question is both grammatically wrong and mathematically imprecise) just to waste our times. If we spend our time and put a lot of effort into answering their questions, can't they be just a little considerate about this when they ask a question (just put a tiny bit of effort into making their questions explicit)? That is what I meant. PST

Sure, it would be nice if they did, but how does removing the question help? --Tango (talk) 22:12, 13 January 2009 (UTC)

Law of excluded middle

Is the law of excluded middle equivalent to the axiom of choice? Stifle (talk) 16:22, 12 January 2009 (UTC)

This is a meaningless question without specifying the base system over which you want the equivalence to hold. The axiom of choice implies the law of excluded middle over a fairly weak fragment of intuitionistic set theory (something like extensionality, separation, and pairing suffices, see Diaconescu's theorem). OTOH, I do not know of any nontrivial (i.e., not already including AC) nonartificial theory where the law of excluded middle implies the axiom of choice. — Emil J. 16:37, 12 January 2009 (UTC)

January 13

the complex plane

Please give a basic description of the complex plane, I'm having trouble understanding the article [3]. I am a student with knowledge through algebra 2 and geometry. Thanks. —Preceding unsigned comment added by 69.136.118.113 (talk) 01:46, 13 January 2009 (UTC)

A complex number has the from a+ib, where i is the square root of -1 and a and b are real numbers. Therefore, you can think of complex numbers of being points on a plane, a+ib would be the point (a,b). We call the set of all complex numbers, thought of as points in that manner, the complex plane. Does that help at all? If not, it would help if you explained in more detail what you are struggling with. --Tango (talk) 01:57, 13 January 2009 (UTC)

Please respond to Tango's comment or this question may be removed. PST (Point-set topologist)

If you remove this question, I will restore it. I do not understand why you are so anxious to delete perfectly valid threads recently. Algebraist 21:08, 13 January 2009 (UTC)

Likewise, why on Earth would we remove a perfectly reasonable question? --Tango (talk) 22:11, 13 January 2009 (UTC)

Well, the person who has asked the question (probably) has not read your response. Unless, in a particular thread, we get a reply from the OP, it is unlikely that he has read it (and hence the thread no longer has any purpose). When I wrote "may", I used italics to indicate that I myself would not remove it. I thought that it was a policy to remove such threads and therefore the use of "may" (and passive voice: I could have written "I will remove it" but I did not). PST

And just to clarify the relevant "policy" that I referred to, I don't see the point in having trivial threads here. PST

(ec)I was misled by your edit summary 'will remove question if OP does not show that he has read Tango's response', which seems to clearly state that you intended to remove the thread. No such policy exists; if you want one, suggest it on the talk page, but I have to say I think it very unlikely you will get any support. Add: your failure to understand a refdesk practice does not constitute a policy. Algebraist 22:20, 13 January 2009 (UTC)

It will be archived in a few days, that's the only removal that happens for good faith questions that don't involve medical or legal advice (removing medical/legal questions is controversial, but it does happen). --Tango (talk) 22:26, 13 January 2009 (UTC)

To Algebraist: "Your failure to understand a refdesk practice does not constitute a policy". That is why I used " " when I referred to "policy" (you probably did not pick this up because of the edit conflict). I don't have such a strong view that such threads should be removed so I won't say that I will remove threads in future (unless, for a different reason, I truly feel that the thread should be removed but in that case I will consult first). But if someone asks what is 5/6, is that a thread that should be removed (as I explained, I don't see the purpose in having trivial threads)?

Another questioner that run away forever... PST's removing policy is so strict, but also so nice, that doesn't matter! :-) --PMajer (talk) 23:46, 13 January 2009 (UTC)

Forever? It's been less than 24 hours... --Tango (talk) 23:47, 13 January 2009 (UTC)

No need to remove, just tell them to use a calculator, or give a link to long division or something. If you remove the thread you risk them not understanding why and putting the question back, which gets us nowhere. --Tango (talk) 23:47, 13 January 2009 (UTC)

PST, maybe they are right... after all "triviality" is a relative concept. Usually a question is not trivial for the questioner, otherwise he or she would answer by himself or herself... and of course it is very likely that it is trivial for you. But let's be patient with everybody ;-) --PMajer (talk) 23:57, 13 January 2009 (UTC)

Yes, I might as well go with their rules since I am not too concerned if my policy is not accepted. But is just seems that 80% of people post a question here but forget that they have done so! This wastes considerable time for us. The laughable thing is that some of their (more or less) trivial questions start a "heated debate" here (by us) when they are never going to see it (this is one such instance). :) PST

Yes consensus is a good idea with polcy. Four tildas ~~~~ to sign a posting is standard policy. Dmcq (talk) 10:33, 14 January 2009 (UTC)

You don't know how many people read the question, are satisfied with the answer, and just don't say anything. While a "thank you" is nice, I see little point in removing questions from people that don't say it. This debate is about policy, it has little to do with the question. When we get into debates about a question it's usually because it's something we find interesting and we enjoy the debate regardless of whether or not it is useful to the OP. --Tango (talk) 16:12, 14 January 2009 (UTC)

Thank you for your help, sorry it took a few days to respond. —Preceding unsigned comment added by 69.136.118.113 (talk) 23:13, 15 January 2009 (UTC)

Don't be. Despite PST's comment, you are in no way obliged to respond here. Algebraist 23:18, 15 January 2009 (UTC)

is a straight line a right angle triangle?

Technically speaking, can I consider a straight line to be a right angle triangle? If one of the sides of a right angle triangle is reduced to zero, then the other side is the same as the hypotenuse. So the Pythagorean theorem would say, $a^{2}+0^{2}=a^{2}$ which seems to hold up.

Duomillia (talk) 02:26, 13 January 2009 (UTC)

I suppose if you really wanted to you could consider a line segment to be a degenerate right triangle with two vertices at the same point. What do you want to do this for, exactly? Algebraist 02:29, 13 January 2009 (UTC)

Well, if you must know I want to use a degenerate triangle for degenerate purposes... no, wait, pretend you didn't hear that...

Following the same logic, could a point be considered an extremely degenerate triangle where all three vertices are the same point? Or for that matter, ditto for a rectangle, hexagon, ect? Not that there is any practical uses for it, but it's interesting to realise. Duomillia (talk) 02:46, 13 January 2009 (UTC)

I don't see why not. --Tango (talk) 03:06, 13 January 2009 (UTC)

Because then the side lines are undefined, and internal angles are undefined (you can't say if it is regular etc.), and diagonals either. Because generally a polygon
'is traditionally a plane figure that is bounded by a closed path or circuit, composed of a finite sequence of straight line segments (i.e., by a closed polygonal chain)',
and a line segment is in turn defined as
'a part of a line that is bounded by two distinct end points'.
CiaPan (talk) 06:58, 13 January 2009 (UTC)

Surely it is better not to allow that a line segment "is" a triangle. What sort of an "is" could that be, after all? Better to say that there are triangles with equal perimeter whose angles are 180°, 0°, and 0° (for the case of 90°, 90°, and 0°, see later), and that they are not all congruent with each other. Suppose that triangle T₁ has sides of length L (non-zero, let us say), 0.6L, and 0.4L, and triangle T₂ has sides of length L, 0.8L, and 0.2L. While T₁ is superimposable on T₂ in the sense that all points constituting T₁ can be "covered" by T₂, the two are not congruent because their vertices are not superimposable. Now, if there is a special relation of "coverage" that holds between T₁ and T₂, surely that same relation holds between any line segment of length L and each of T₁ and T₂. So I propose this instead: for every line segment of non-zero length L there is a class of triangles that it covers and that cover it: all triangles with largest side of length L and with perimeter 2L.

Clearly for every such line segment there are in fact infinitely many classes of polygons, of number of sides 2 (a very special case, only possible in degeneracy) to ∞. Clearly also each of the polygons covered by any such line segment also covers every other such polygon. Clearly also, for every such line segment there are infinitely many classes of polyhedra that it covers, of number of faces 2 (again a very special degenerate case) to ∞ (perhaps excluding 3, or allowing it as a stranger sort of degenerate case; and I assume that coverage is a matter of points on edges only, not all points on surfaces). And clearly for every n-sided polygon and for every natural number m there is an infinite number of classes of (n+m)-sided polygons – and (n+m)-faced polyhedra – that it covers and that cover it and each other. A similar extension may be made associating line segments with "polyhedra" of arbitrarily many dimensions above 3, and with ellipsoids and their higher-dimensional equivalents.

Returning to triangles, there seems to be no special difficulty in allowing triangles with angles 90°, 90°, and 0°. Triangles with sides of length L (non-zero), L, and 0 are all congruent with each other; and they too cover and are covered by a line segment of length L, and so with all of the polygons, polyhedra, and so on that cover and are covered by that line segment.

Excuse the extreme informality of all that. Not my field. And as for L equal to zero, I say nothing at this stage.

Note that not all such "collinear" polygons must have perimeter 2L to be coverable by a line segment of length L. That restriction applies only to triangles. For example, a polygon might have "internal" angles 0°, 360°, 0°, and 0°, and four equal sides of length L. Another polygon might have the same angles but sides of length L, 0.6L, 0.6L, and L. And so on. Every such polygon must have perimeter equal to or greater than 2L, of course.

–_⊥^{¡ɐɔıʇǝo}Noetica!^T– 09:50, 13 January 2009 (UTC)

Usually it wouldn't be considered a triangle, it depends on the circumstances though. For instance if the points are all on a line then there is no interior and one can't say which way the triangle is oriented. Of course it is always possible to assign an orientation as if the figure has an infinitesimal side. It is like asking what is the longitude at the north pole. A degenerate case is as good a description as any. Dmcq (talk) 12:32, 13 January 2009 (UTC)

Lefschetz numbers

I'm trying to understand the Deligne–Lusztig theory for characters of groups of Lie type. It appears that most of the l-adic cohomology stuff can be axiomatized out, but it still leaves a fair number of places where it would be convenient to understand what a Lefschetz number was. I think there are probably some differences from our article Lefschetz number and these, since the field has changed from C to some l-adic field, but since it does not appear to depend on l, I have some hope it is pretty close to the C theory. At any rate, I don't care whether one answers in terms of topology or algebraic geometry (C or Ql), but please specify which, just in case they really are different.

• Must the Lefschetz number always be an integer?

• The Lefschetz number of the identity is the Euler characteristic in the topology case; for other automorphisms is it like the Euler characteristic of the quotient?

• Is it easy to classify all the possible Lefshetz numbers for compact real surfaces in the topology case?

Thanks for any help. JackSchmidt (talk) 18:38, 13 January 2009 (UTC)

In the topological case, one can define (Hatcher does) the Lefschetz number in terms of the traces of the induced maps on integral homology mod torsion. In this case, the number is obviously an integer (when it is defined at all), and I think it is obviously the same as the Lefschetz number defined from Q- or C- homology. Algebraist 20:11, 13 January 2009 (UTC)

Cool. I would be very happy to use more of an integer version, both to have it obviously be an integer, and also to have some hope of computing on a computer (though so far I've got no idea how to write down the space I'm taking l-adic cohomology of yet). Would Hatcher be a good place to learn about the Lefschetz number? Is this Hatcher's Algebraic Topology text? Do you happen to think my "all surfaces" question could possibly have an answer (even if it looks like a pain to do it in general; I mean maybe I could do it for spheres and a 1-holed tori)? JackSchmidt (talk) 22:28, 13 January 2009 (UTC)

I doubt Hatcher would be a useful place to look (there's very little there), but you can always see for yourself. I think calculating Lefschetz numbers shouldn't be too hard, at least for sensible spaces and maps (for spheres it's just a matter of computing degree, which is simple enough), but this isn't my field and I'm just going on undergraduate memories here. Algebraist 22:31, 13 January 2009 (UTC)

Yeah Hatcher (p179-189) is pretty brief and focused on fixed points. The worked example was helpful, but I have trouble figuring out "all maps" (even up to homotopy) of a space. I'll see if I can remember what the degree of a self-map of a sphere is (I remember the degree of a circle map, and remember being confused by the Hopf fibration at some point). I'll reread the D-L theory stuff to see if there is any reason I should care about fixed points. I think there sort of is, but at a much more detailed level than just whether or not the Lefschetz number is 0 or not (the actual value matters). Thanks for the help again. JackSchmidt (talk) 22:56, 13 January 2009 (UTC)

Interest calculation (discontinuous)

I want to calculate the end-of-year interest for my savings account, and moreover, I want to understand what I'm actually doing when calculating it. Maths isn't my strongest point, although I am quite interested in it, so I figure asking this here will benefit me in more than one way.

Without further ado, here's the "problem":

• I have a base interest rate of 3.0% per year, calculated based on the daily balance.
• I receive a further 0.6% bonus interest per year based on the lowest balance per quarter. I'm not quite sure how they calculate it, but I think they just take .6% of each quarterly low and let the mean of that be the bonus interest at the end of the year.
• As of today, my balance is €2116.56. My base interest built up until today is €2.02. Assume that the lowest balance this quarter is €2000.00.
• I will deposit €100 each month; I have already deposited the €100 for January.
• Assume I will not make any withdrawals.

Based on the above criteria, (a) what will the cumulative interest be on 1 January 2010 and (b) how do I calculate this?

Feel free to make your own assumptions as to precise dates for my deposits and calculation of the quarterly bonus interest, etcetera.

This kind of maths is quite a bit over my head, especially with the discontinuous aspect (depositing €100 each month) and the quarterly bonus, but I'd nevertheless I'd still like to learn to understand it. Thanks in advance!

P.S. sorry for stretching the page - if someone wants to fix my wiki-markup, be my guest. --Link ^(t•c•m) 20:34, 13 January 2009 (UTC)

I think the easiest way to do with would be to make a spreadsheet (you can do it with pen and paper if you aren't comfortable with computer spreadsheets). An annual interest rate of 3%, compounded daily, corresponds to a monthly rate of $(1+0.03)^{\frac {1}{12}}-1=0.0025$, or 0.25%. To get from one month to the next, do (last month)*1.0025+100. You can then work out the bonus manually, since it's only added on at the end of the year so won't affect the rest of the calculation. --Tango (talk) 20:50, 13 January 2009 (UTC)

If the interest is compounded daily, we probably want 1/365 of the "base interest rate" as the daily rate (0.008219%), and then that gets applied as a multiplier $d\approx 1.00008219$ every day. So, over the course of the year, you get multipliers of $d^{31}\approx 1.002551$, then $d^{28}\approx 1.002304$, $d^{31}$, $d^{30}\approx 1.0024689$, etc. Let us suppose that you deposit before the balance is checked for interest for the first of each month: then at the end of month i (January is 1) you have $b_{i}=(b_{i-1}+100)d^{n_{i}}$ where $n_{i}$ is the number of days in month i and $b_{0}=2116.56-100-2.02=2014.54$. The bonus, supposing it to be added at the end of the year (treating each quarter as precisely 1/4 year) rather than at the end of each quarter, is $0.006{\frac {b_{0}+b_{3}+b_{6}+b_{9}}{4}}$, taking the quarterly low to not include its first month's deposit. (You can use precisely 2000 instead of $b_{0}$ if you wish.) Writing down the full expression for the final balance (before January 1 2010's presumptive deposit) would take up a lot of useless space, but the calculation is trivial: I get $b_{3}=2330.98,b_{6}=2649.98,b_{9}=2971.60$, for a grand total of €3295.67. This isn't guaranteed to be precise, not only because of the question about $b_{0}$, but also because I just used floating-point math and the bank will presumably round (to eurocents, perhaps) as it goes. The interest is of course trivial to calculate since you know what you started with and what you deposited; it's much easier to get it this way than to try to only count interest as you go along. (If, however, the interest is not compounded at all, you just want to average the balance every day and multiply by 3%, so you get $(d-1)\sum _{i=1}^{12}n_{i}(100i+b_{0})$ as the base interest.) --Tardis (talk) 00:23, 14 January 2009 (UTC)

This question is not acceptable (I am just noting which questions I think are appropriate per the recent discussion at the talk page). Too trivial. PST--Point-set topologist (talk) 14:04, 15 January 2009 (UTC)

Convolution

Hi, I take a lecture in Evolutionary Dynamics (I am a biologists, thats why I am asking in the first place) and there was a lecture about Branching processes, specifically the Galton–Watson process. In the lecture script is say this about P(i,j) which is the probability to go from a state i to state j:

${\begin{aligned}P(i,j)=p_{j}^{*i}=\sum \limits _{k_{1}+\cdots +k_{i}=j}p_{k_{1}}\cdots p_{k_{i}}\end{aligned}}$

Then it says:

${\begin{aligned}\{p_{k}^{*i}\}_{k\geq 0}\end{aligned}}$ is the i-fold convolution of ${\begin{aligned}\{p_{k}\}_{k\geq 0}\end{aligned}}$

My problem is with that last sentence since I couldnt find anything in the article Convolution that relates to that statement. I think I understand that ${\begin{aligned}\{p_{k}\}_{k\geq 0}=\{p_{0},p_{1},p_{2}\cdots \}\end{aligned}}$ and thus per analogy that ${\begin{aligned}\{p_{k}^{*i}\}_{k\geq 0}\end{aligned}}$ denotes the set of this expression for all k bigger than zero. But I have trouble to see how they relate or how you can generate the latter from the first by a convolution. Maybe somebody here can help out? Did I just miss this part in the article or is there something strange in that formalism the professor uses? Thanks a lot. Greetings --hroest 22:08, 13 January 2009 (UTC)

This is a discrete convolution, in fact a one-sided discrete convolution, which just means that the $\textstyle p_{k}$ you are considering are zero for all k<0 (or if you prefer, you may think them defined for all k putting $\textstyle p_{k}=0$ for k<0). The one-sided discrete convolution is the same thing as the Cauchy product of power series, in the following sense. If you have two sequences $a:=\{a_{k}\}_{k\geq 0}$ and $b:=\{b_{k}\}_{k\geq 0}$ you can consider their (ordinary) generating functions: then the product of the two generating functions is the generating function of the convolution a*b. In your case you convolute $\{p_{k}\}_{k\geq 0}$ with itself i times, thus:

$\textstyle \sum _{j\geq 0}P(i,j)x^{j}=\left(\sum _{k\geq 0}p_{k}x^{k}\right)^{i}$

I hope this is clear enough --PMajer (talk) 22:58, 13 January 2009 (UTC)

This is the same discrete convolution defined in the article titled convolution, where it says

For complex-valued functions ƒ, g defined on the set of integers, the discrete convolution of ƒ and g is given by:

$(f*g)[n]\ {\stackrel {\mathrm {def} }{=}}\ \sum _{m=-\infty }^{\infty }f[m]\cdot g[n-m]\,$

Simply observe that the sum in the definition above is the same as

$\sum _{k_{1}+k_{2}=n}f[k_{1}]\cdot g[k_{2}].$

Michael Hardy (talk) 03:17, 14 January 2009 (UTC)

.... which is hroest's case, provided f[k] and g[k] vanish for k<0 (as I wrote) --PMajer (talk) 09:34, 14 January 2009 (UTC)

In the Galton–Watson process, Galton was modelling the distribution of surnames, which he assumed to be passed from father to son. Each male in one generation has n sons in the next generation, and we assume $\{p_{0},p_{1},p_{2}\cdots \}$, the probabilities that n=0, 1, 2 etc., are the same for all males and all generations. So the probability that there will be j males in one generation given that there were i males in the previous generation, denoted by $P(i,j)$, is found by summing $p_{k_{1}}p_{k_{2}}\cdots p_{k_{i}}$ across all ordered sets of i non-negative integers $\{k_{1},k_{2}\cdots k_{i}\}$ that sum to j. The set of values $\{P(i,j)\}_{j\geq 0}$ is the "i-fold convolution" of $\{p_{k}\}$. Because of the connection between convolution and multiplying generating functions mentioned above, this way of formulating the problem is particularly useful in cases where the generating function of $\{p_{k}\}$ itself has a simple closed form. Gandalf61 (talk) 10:57, 14 January 2009 (UTC)

Thanks a lot everybody, you really helped me a lot! So I guess I had a problem seeing what the i fold convolution means, but it is just a convolution with itself done i-1 times? As I understand, Gandalf61, if you have for example a Poisson distribution then you could just plug in for $p_{k}$ the value of $POI(k,\lambda )$ for a given parameter and "directly" calculate $P(i,j)$? I understand that you can basically think about this on the level of ${\begin{aligned}\{p_{k}\}_{k\geq 0}\end{aligned}}$ which you convolute i times or on the level of the generating functions which you multiply i times with itself. Thank you all very much --hroest 15:02, 14 January 2009 (UTC)

Yes, I think you have got it. And if $\{p_{k}\}$ follows a Poisson distribution then you get an especially tidy result because the sum of a set of independent random variables, each with a Poisson distribution, also has a Poisson distribution. Gandalf61 (talk) 15:29, 14 January 2009 (UTC)

January 14

Sphere specialists?

I was just wondering whether there are mathematicians specializing only in studying spheres. There are so many important properties of spheres (the well-known Hopf fibration, sphere bundles, parallelizable spheres, how many differentiable structures exist on a given sphere etc...) that I wouldn't be surprised. Any comments are appreciated! Thanks! —Preceding unsigned comment added by 129.143.15.142 (talk) 12:44, 14 January 2009 (UTC)

This sounds like the kind of thing you might see in fiction, like the show Numb3rs. 67.150.254.75 (talk) 13:12, 14 January 2009 (UTC)

The second post should be deleted in my opinion. It seems irrelevant to the question. --Point-set topologist (talk) 13:39, 14 January 2009 (UTC)

Could you not go on about deleting posts please. Personally I get annoyed with administrators saying something and then trying to cover up. I don't think it should be encouraged. Take it up on the talk page if you must. As to the question Hatcher in his book on Algebraic Tpology seems to have made quite a study of the homotopys of the n-sphere. Dmcq (talk) 13:58, 14 January 2009 (UTC)

Who is the administrator here (I am certainly not one) and who is trying to cover something up? I just said that the second post was clearly posted by someone who did not understand the question (i.e a non-mathematician) so the post is irrelevant. Therefore, as promised, I requested someone's opinion on the matter (note that I certainly don't intend on deleting the OP's post). PST (Point-set topologist)

Talk page. Not here. Dmcq (talk) 14:55, 14 January 2009 (UTC)

You might also like Smale's paradox, though why it is called that rather than sphere eversion is beyond me.Dmcq (talk) 14:02, 14 January 2009 (UTC)

Yes, there are mathematicians that specialise in spheres, my tutor is one. I don't really understand what he does, it's differential geometry which I've not studied, but it's something to do with minimal surfaces on higher dimensional spheres (looking at his list of papers it seems he's working his way up the dimensions, I think he's on S⁷ now). --Tango (talk) 16:09, 14 January 2009 (UTC)

${\mathbb {S^{7}}}$ is probably the most interesting. PST

Thanks for your help everyone! —Preceding unsigned comment added by 129.143.15.142 (talk) 11:33, 15 January 2009 (UTC)

• The following is one user's opinion about the question and not a response to the question itself:

This question is acceptable (I am just noting which questions I think are appropriate per the recent discussion at the talk page). PST--Point-set topologist (talk) 14:04, 15 January 2009 (UTC)

Modeling using overlapping probability distributions

First off, this isn't for anything serious - I just got to thinking about it, and am curious as to how one would go about doing it. If you wanted to model, say, rainfall, in a semi-realistic way, one way you could go about doing so is to take the past 20+ years of daily precipitation amounts and create a probability distribution from them. To simulate a period of time, then, you would just sample from that probability distribution to find what each day's rainfall would be. However, that doesn't quite work, as dry days tend to cluster together, as do wet days. So instead of just creating a distribution for just a single day, you create a set of probability distributions for 1,2,3 ... days. How would someone go about choosing a series of daily precipitation amounts such that the series follows all of the different probability distributions for each of the cumulative totals? For the single distribution case, I would integrate my normalized distribution, and use a uniform random number generator and select the precipitation amount which matches the running total of area under the probability distribution. However, I'm not able to wrap my head around how to do it for the multiple simultaneous distribution case. -- 99.154.0.155 (talk) 14:55, 14 January 2009 (UTC)

What people try and do is choose as few numbers as possible and as simple a statistical model as possible that gets the distribution as accurately as needed. Beyond that you might as well just use the original data and pick a date at random. You really have to give up trying to do things exactly if you want to model things. The next question is how accurate is good enough, for people wanting funding it is a bit more accurate than the last published model :) There is a nice article about weather forecasting here. Dmcq (talk) 15:17, 14 January 2009 (UTC)

Sounds like you're describing a Markov chain; the basic information of "on how many days did it rain" is redundant given the information about "what patterns of rain/no-rain are seen for every run of three consecutive days", so you just develop a single model with however much memory and go with it. --Tardis (talk) 18:18, 14 January 2009 (UTC)

A Markov chain does seem to be an adequate description. However, I didn't see anything in the article about how one would go about actually calculating the probability of transition for this case. To clarify, I'm not actually interested in weather forcasting, I'm just using it as a conceptual model for this general type of problem. Also, it's not just the binary rain/no-rain patterns, but the cumulative amount of precipitation (though if you imposed a resolution cut off of e.g. 1mm/0.1 in, you'd probably be able to treat the problem as a multi-valued discrete valued one, rather than a continuous valued one). How would one go about calculating the transition probabilities such that a sufficiently long chain would match a given set of probability distributions for 1 day, 2 day, 3 day ... cumulative totals? -- 99.154.0.155 (talk) 02:37, 15 January 2009 (UTC)

Ah — I somehow missed that you wanted a distribution over rainfall amounts. I believe that you can still do much the same thing: calculate (by some sort of interpolation on the data) a joint probability density function for the amounts of rainfall $R_{1},R_{2},\dots ,R_{N}$ on N consecutive days (counting even overlapping runs of N days in the data). Then, at day i, draw $r_{i}$ from the restricted distribution $f(r_{i-N+1},r_{i-N+2},\dots ,r_{i-1},R_{i})$ of one variable. --Tardis (talk) 01:29, 16 January 2009 (UTC)  

• The following is one user's opinion about the question and not a response to the question itself:

This question is acceptable (I am just noting which questions I think are appropriate per the recent discussion at the talk page). PST--Point-set topologist (talk) 14:04, 15 January 2009 (UTC)

definition for geometry situation

If I could draw, it would be very easy.

Imagine a circle, draw a square in that circle so that the circumference is cut or "scribed" into four equal lengths. What is the straight line called that squares off that part of the circumference. In this case, there are four straight lines of equal length, however in general, the lines could be uneven in length and the shape inside is randomKilipaka (talk) 22:11, 14 January 2009 (UTC)

I think you want the word chord. Algebraist 22:14, 14 January 2009 (UTC)

Seems that the question, "In this case, there are four straight lines of equal length...", is asking about the "special" chord: ${\sqrt {r^{2}+r^{2}}}$ but I don't know if it has a "special" name. hydnjo talk 04:00, 15 January 2009 (UTC)

I interpreted that last sentence as meaning the OP was asking about the general case and just giving the special case as an easier to describe example. Either way, I don't know of a name for the special case, either. --Tango (talk) 05:19, 15 January 2009 (UTC)

• The following is one user's opinion about the question and not a response to the question itself:

This question is acceptable (I am just noting which questions I think are appropriate per the recent discussion at the talk page). PST--Point-set topologist (talk) 14:03, 15 January 2009 (UTC)

January 15

Trivial questions and removal of posts

Please have a look at the talk page... PST —Preceding unsigned comment added by Point-set topologist (talk • contribs) 11:28, 15 January 2009

Forgive the trivial question, but would you put a link to the talk page, because I do not know where it is. Then, feel free to remove my post ;) --131.114.72.215 (talk) 12:36, 15 January 2009 (UTC)

After you finish reading this message hold the "alt" key and click the letter "t" on your keyboard (i.e alt-t); release these keys instantly and hit "enter". By magic, you will be directed to the discussion. PST —Preceding unsigned comment added by Point-set topologist (talk • contribs) 12:40, 15 January 2009

Hitting alt-t with Firefox on Linux I get the "Web Search" functionality, not the talk page. I have no clue what browser/platform you're running in which gets you the discussion page. For the rest of us, the talk page for any Wikipedia article can be found by scrolling to the top the page and clicking on the tab which says "discussion"- for this page, that goes to Wikipedia_talk:Reference_desk/Mathematics. PST is probably referring to the "Trivial questions and removal of posts" section (direct link to section: [4]). BTW, Wikipedia:Help_desk is the best place to ask these sort of questions. It's like the Reference desk, but about how to use Wikipedia. -- 99.154.0.155 (talk) 14:20, 15 January 2009 (UTC)

alt-shift-t on most modern browsers. See Wikipedia:Keyboard shortcuts. Algebraist 14:25, 15 January 2009 (UTC)

My keyboard doesn't have an "alt" key. --Carnildo (talk) 23:56, 15 January 2009 (UTC)

Here's that "secret" link again. hydnjo talk 02:13, 16 January 2009 (UTC)

Interesting geometric and probability question

I looked at the previous question (about chords and such) and that got me thinking about the following:

What is the probability that a quadrilateral is convex?

My approach would be to start with the first edge. Now consider the second one; this is uniformly distributed on the interval (0, 360). Now for each value in this interval, consider possible edges from there. The fourth edge is uniquely determined. I tried this tactic out and ended up with massive amounts of calculations (integrating trig. functions) which I have not yet finished. Now I don't want anyone to tell me the answer if they know it. But I just want to know whether there is a better way to solve this problem (with very few calculations). If so, an outline would be helpful (without the answer). Thanks! —Preceding unsigned comment added by 129.143.15.142 (talk) 12:33, 15 January 2009 (UTC)

The problem sounds nice, with a nice generalization to convex n-gons, but first you have to state it more completely. In fact, you have to define what is the probability distribution of the 4 vertices. Or equivalently, what is the random way you adopt to construct the quadrilateral. Notice that there is no uniform probability distribution for random points on the plane, so there is no preferred meaning to "a random quadrilateral". More choices are possible, each one giving a different answer. One possible variant is doing it on a sphere, thus choosing at random uniformly and independently 4 points on S². PMaj --131.114.72.215 (talk) 12:55, 15 January 2009 (UTC)

If I am not mistaken, this problem appeared in the Putnam competition. PST

Bertrand's paradox is relevant here: if you aren't careful to decide exactly what 'random' means, you may end up with a lot of different answers. Algebraist 13:18, 15 January 2009 (UTC)

I believe the OP means that the angles are chosen uniformly, and the the lengths of the sides does not affect the convexity (perhaps he assumes the lengths are chosen to not have self intersections). I think then no heavy calculations are needed, just the sum of the angles in a convex polygon. JackSchmidt (talk) 15:25, 15 January 2009 (UTC)

The angles are 4 positive numbers with sum $2\pi$, and can be seen as the baricentric coordinates of a point of a regular tetrahedron. Does your the angles are chosen uniformly refer to the (normalized) Lebesgue measure on the tetrahedron? If so, the probability of getting a convex quadrilateral is just the probability that all 4 coordinates are less than $\pi$ , that is, they represent a point in the inner regular octahedron: therefore it is just the ratio between the volume of the tetrahedron and the volume of the octahedron: that is 1/2, because the complement of the octahedron is made by 4 tetrahedrons of half (linear) size: 1/2=1-4(1/2)³. --PMajer (talk) 16:20, 15 January 2009 (UTC)

I take him to mean: The first angle A is chosen uniformly in the open interval (0°,360°). The second angle B is chosen uniformly in the open interval (0°,360°-A), and then the convexity is now determined by A+B: if it is 180° or more, then the quadrilateral must be convex, if it is less than 180° then the quadrilateral cannot be convex. This gives a larger probability than 1/2 (but again a nice fraction, now counting sub-triangles instead of sub-tetrahedra), so the OP might be able to decide which he meant based on the "50-50" or "better than 50-50" chance. He did specifically ask not to post the answer. I think both distributions are reasonable, so without reading minds it is hard to tell which he meant. JackSchmidt (talk) 17:13, 15 January 2009 (UTC)

You say 'the convexity is now determined by A+B: if it is 180° or more, then the quadrilateral must be convex, if it is less than 180° then the quadrilateral cannot be convex'. This doesn't seem to be true at all. Whatever do you mean? Algebraist 17:19, 15 January 2009 (UTC)

Oh that is too bad. I think this follows the OP word for word, except that the third edge is not specified, and I guess it matters, more or less wrecking my model. If A+B ≥ 180°, then C+D ≤ 180°, so C,D ≤ 180° so it is convex assuming A,B are ok. If A+B < 180°, then it is possibly convex, possibly not. Less of a problem: I guess one needs to subtract out the probability that A or B itself is bigger than 180°. This now leaves 1/4 where the quadrilateral must be convex, 1/2 where it cannot be, and 1/4 where it depends on the third side. Since one does need to specify the third side, I think it no longer makes sense to model it as a two dimensional choice. PMajer's model may be the most reasonable way, though I am not sure this is how the OP wanted the third side chosen. JackSchmidt (talk) 17:52, 15 January 2009 (UTC)

It may be easier to consider the first angle being chosen to be between 0 and 180, since the orientation of the polygon doesn't affect its convexity. --Tango (talk) 17:49, 15 January 2009 (UTC)

Thanks for you help everyone! I actually did not want the answer (although I am not sure it is correct: no offense intended). Maybe I should have written: "Consider the collection of all four-point subsets of the plane that determine a quadrilateral, uniformly distributed". What is the probability that the quadrilateral determined is convex. Let me explain my method.

Method:

I first consider an arbitrary line segment (of length s). Now my second line segment can be subtended at an angle between 0 and 360; without loss of generality, we may assume that the angle is distributed uniformly between 0 and 180 (by symmetry). Suppose, the line segment is subtended at an angle θ and has length l. Consider the extension of the line infinitely. If the third side of the quadrilateral originates from the start of this extension and has angle greater than θ (with respect to the first line segment), the quadrilateral will certainly not be convex. It turns out that the third side must subtend an angle (with respect to the second side) between 0 and α/2 where α equals the sum of θ and the cosine inverse of (s^2 + x^2 - l^2)/2xs. But the angle can also be between α/2 and α: however, in this case, the length matters. It turns out that the length (where the third side subtends an angle of β with the second side extended) must be:

k = (z+s)^2 + l^2 - 2*(z+s)*l*cos(θ)

where z equals the ( (sine of α - β) into (l^2 + s^2 - 2*l*s*cos(θ)) ) divided by (the sine of ( ( the cosine inverse of (s^2 + x^2 - l^2)/2xs ) - α + β ) ). Induce a measure on [0, 1] using the bijection:

(2/π) * arctan (x) : [0, posinf] -> [0, 1]

When the third side subtends an angle between 0 and α/2 with the second side, it can have any length so the probability is 1 for each such angle. For angles between α/2 and α, the max side length is k (as above) which when transformed to [0, 1] gives (2/π) * arctan (k). Putting this together, the net "mass of favorable region" is (probability = mass of favorable region/total mass):

f = (α/2) + [(integral (β=α/2 to β=α)) ((2/π)*arctan (k)) dβ]

The total mass is π + f + q where again we define a similar integral as above for q (q represents the probability that the third side subtends an angle greater than α with respect to the second side). Now this expression, (f)/(π + f + q) represents the probability that the quadrilateral is convex for particular l and θ. We now integrate this expression for θ between 0 and π and l between 0 and 1 (note that this is a double integral). We have then got the probability that the quadrilateral is convex

Q.E.D

This method seems overly long but I am pretty sure it works. Any comments would be greatly appreciated. Thankyou so much!

But as they were saying "the collection of all four-point subsets of the plane that determine a quadrilateral, uniformly distributed" has not a big meaning, for the same reason that, for instance, there is not a uniform probability on the natural numbers (what should be the probability of {1}, if it has to be equal to the probability of {2}, of {3} etc?). What you could do is : compute the probability P_R for 4 points independently and uniformly choosen in the disk of radius R (given that they make a quadrilateral); then compute the limit of P_R as R goes to infinity (no risk that somebody here steals your pleasure of doing the computation). This is maybe close to what you are thinking, and should be a limit probability in some sense. --84.221.209.239 (talk) 20:46, 15 January 2009 (UTC)

You have attempted to conquer the problem that the edges may have indefinite length by definining a transformation to the unit interval. Life becomes much simpler if you assume that the quadrilateral is bounded.

PST —Preceding unsigned comment added by Point-set topologist (talk • contribs) 20:16, 15 January 2009

The question doesn't make sense in that form, there is no uniform distribution for 4 points on the plane. (Basically, the probability of any of the points being in any given bounded region would have to be 0 in order for the total for the whole plane no to be more than 1, but obviously that doesn't work.) --Tango (talk) 20:27, 15 January 2009 (UTC)

The OP's solution works assuming there is a uniform distribution for four points in the interior of a ball. So that is certainly not a flaw with his/her method. PST —Preceding unsigned comment added by Point-set topologist (talk • contribs) 20:36, 15 January 2009

Another possible model is to transport the uniform measure of the Riemann sphere via the stereographic projection onto the plane and choosing independently the 4 points. You get this way the same situation of 131's variant. Note that, thanks to the conformality, the angles are preserved. On the sphere variant it seems to me that, choosing the fourth point X, the quadrilateral is (geodesically) convex if and only if switching X with -X it is not. So this choice also gives 1/2 (maybe it's the same?) --PMajer (talk) 21:03, 15 January 2009 (UTC)

That just gives you one of many non-uniform distributions on the plane, no-one said they didn't exist. --Tango (talk) 22:26, 15 January 2009 (UTC)

are you talking to me? In this case, yes, but I didn't say no-one said they didn't exist ;) Here we are just suggesting some possible natural distributions, just to speak. --PMajer (talk) 22:32, 15 January 2009 (UTC)

"What is the probability that a quadrilateral is convex"? The obvious and stupid answer is that it depends on which quadrilateral you are talking about: if it is a convex quadrilateral, then the probability is one, otherwise it is zero. Want to ask another question such as "What is the probability that a random quadrilateral is convex"? Well, there is no such thing as a random quadrilateral. Every quadrilateral is unique. "What is the probability that a randomly chosen quadrilateral is convex"? That depends on from where it is chosen. You may have a box of quadrilaterals and choose one randomly. The probability depends on the contents of the box. If all the quadrilaterals in the box are squares, then the probability that a randomly chosen quadrilateral is convex is one. The OP's question does not make sense, and so the answer is perfectly undefined. Bo Jacoby (talk) 12:04, 16 January 2009 (UTC).

Well I just don't agree that one cannot speak of a "random quadrilateral", why? According to the customary use, the meaning is: a random variable with codomain in the set of all quadrilaterals (which is a space with its natural topology). But as everybody was saying, the base probability space (the domain) has to be given too, to make the question meaningful. --131.114.72.215 (talk) 15:53, 16 January 2009 (UTC)

Sure, you can have a random quadrilateral, you just need to specify the probability distribution. The OP tried to specify a uniform distribution, which doesn't exist, but plenty of other distributions do. --Tango (talk) 17:09, 16 January 2009 (UTC)

Hi guys,

Assume that the quadrilaterals are bounded within the unit circle so the uniform distribution works. Then my method (have a look please) would work: wouldn't it? Thanks!

January 16

Probability and the universe

If space is infinite (or ∞ ) then what is the probability of the being intelligent life in the remaining cosmos, more specificly how would one express this in mathematical/statistical format? —Preceding unsigned comment added by 166.189.133.93 (talk) 03:45, 16 January 2009 (UTC)

We don't know. It's not even clear that it's a meaningful question. Algebraist 03:49, 16 January 2009 (UTC)

You might be looking for Drake equation. JackSchmidt (talk) 03:53, 16 January 2009 (UTC)

If any given segment of space (say, a cube a billion light-years to a side) has a positive probability of containing intelligent life, then an infinite space almost surely contains intelligent life. But space isn't believed to be infinite in that sense. **CRGreathouse** (t | c) 04:19, 16 January 2009 (UTC)

Assuming such a probability is meaningful, and disjoint volumes of space have independent chances of containing intelligence. Algebraist 04:24, 16 January 2009 (UTC)

Why wouldn't it be meaningful? --Tango (talk) 04:30, 16 January 2009 (UTC)

For the same reason 'given that Alpha Centauri has a planet made of blue cheese, what is the probability that God exists?' might not be meaningful: it asks for a probability for a one-off statement that is either true or false, and it conditions it on another statement which is quite possibly false. Some approaches to the philosophy of probability allow sense to be made of this, but others do not. Algebraist 12:41, 16 January 2009 (UTC)

Well sure, for something like that you need Bayesian probability, but people rarely seem to have a problem with that. You don't even strictly need it for this problem since you have lots of large regions of space to consider and can talk about the frequency at which they contain life (obviously we can't actually observe a large enough number of them to make a useful estimation of the probability that way, but mathematically it is possible). --Tango (talk) 15:36, 16 January 2009 (UTC)

Sure, Bayesianism works here, but the frequency interpretation doesn't. We aren't talking about the probability of life in large regions of space, but the probability of intelligence in the universe at all. To be a frequentist about that, we'd need a lot of spare (infinite) universes. Algebraist 15:40, 16 January 2009 (UTC)

In my answer, I was positing a constant probability, which would mean independence; the meaning would be definitional. But how that applies to reality is unclear, since defining "intelligent life" is difficult, positing constant probability seems cray, even as a heuristic, and assuming infinite (not merely unbounded) extent seems wrong (though see below). **CRGreathouse** (t | c) 04:51, 16 January 2009 (UTC)

Who says space isn't believed to be infinite in that sense? I thought it was still very much an open question. --Tango (talk) 04:30, 16 January 2009 (UTC)

I withdraw the statement, then; that's outside my field of expertise. I had thought that Olbers' paradox ruled out an essentially constant distribution of matter, and quantum effects don't allow a fractal distribution that would escape infinite gravity. But I really don't know what the present state of knowledge is. **CRGreathouse** (t | c) 04:51, 16 January 2009 (UTC)

Olbers' paradox is avoided because the observable universe is finite (which is a consequence of the age of the universe and the speed of light both being finite), but it is unknown whether the universe extending beyond what is currently observable is finite or not. Dragons flight (talk) 06:04, 16 January 2009 (UTC)

Once again, I'm inexpert here, but doesn't gravity still cause a version of Olber's paradox? (Infinite mass causing infinite gravity causing all matter to recede to a point at nearly c?) **CRGreathouse** (t | c) 06:13, 16 January 2009 (UTC)

Gravity propagates at c, just as light does, so the same explanation applies. Algebraist 12:41, 16 January 2009 (UTC)

I don't know about space, but there is supposed to be a large but finite number of galaxies, stars, and planets in the observable universe, which means a finite chance that one of them will host intelligent life (or two of them, if we consider intelligent life to exist on Earth). StuRat (talk) 05:57, 16 January 2009 (UTC)

The observable universe is definitely finite (by current understanding), but the whole universe is almost certainly larger than the observable universe. --Tango (talk) 15:36, 16 January 2009 (UTC)

As an obvious remark I would recall that such a probability is (as it is always a probability) just a subjective measure of (un)certainity, expecially because we are not talking of a reproducible fact. That is, it's just the degree of likelyhood we assign to the existence of intelligent life somewhere, given the information we have about the universe and the formation and evolution of life. Since our information today on these matters is still quite poor, we have to base the extimation on assumptions largely due to the indifference principle, and it's well possible that different people have different probabilities. --PMajer (talk) 12:58, 16 January 2009 (UTC)

If a coefficient is significantly different from 1.00

I have to run the following regression on a set of data and find out whether 'beta1' coefficient is significantly different from 1.00.

y = alpha + beta1*x + beta2*z + noise

Statistical packages (I use SPSS) by default test whether it is different from 0. How can I set up a program in SPSS so that I test whether it is statistically different from 1.00. I will appreciate your help.--24.214.202.118 (talk) 04:28, 16 January 2009 (UTC)

You could just subtract the x-value from each y-value and then test if the coefficient is significantly different from 0... **CRGreathouse** (t | c) 04:52, 16 January 2009 (UTC)

Arclength

Given a curve, perhaps in the plane, it's easy to get a sensible lower bound on its length by using the principle that a line segment is the shortest curve connecting its endpoints. Simply divide the original curve into many small pieces, replace them with line segments, and compute the sum of their lengths. Is there a similarly straightforward way of getting an upper bound on its length? Black Carrot (talk) 04:58, 16 January 2009 (UTC)

Not such a direct way. If your curve u:[a,b]->R² is absolutely continuous you may work directly on $\scriptstyle l(u)=\int _{a}^{b}|u'|dt$. Alternatively: if u_n is a sequence of curves of length less or equal than L and converging uniformly to u then by the lower semicontinuity of length, $\scriptstyle l(u)\leq L$ as well. Additional information would depend on how your curve has been given. For instance if it is given as a fixed point ( i.e. u=T(u) ) of some transformation T, which is a contraction wrto the uniform distance, and for some L you prove that $\scriptstyle l(v)\leq L$ implies $\scriptstyle T(v)\leq L$ for all $v$, then you can conclude $\scriptstyle l(u)\leq L$. --PMajer (talk) 08:27, 16 January 2009 (UTC)

I doubt there's anything quite so simple. Here's a suggestion in the case of a parametric plane curve (parametrized by a nice enough function f(t)), but I'm not 100% sure it's correct yet. Break your curve into small pieces with constant concavity (either to the right or to the left - this amounts to saying that the determinant det(f′(t), f″(t)) has constant sign on the relevant interval, or that the angular coordinate of f′(t) varies monotonically there), and such that the tangent vector changes direction by less than 180° in the relevant interval (this only makes sense in an obvious way if f′ ≠ 0, but it should be possible to adapt it to some cases in which f′ is allowed to be 0 at the endpoints). Say the endpoints of one of these pieces are A and B. Draw the tangents to the curve at A and B and let them meet at C. The length of that piece of the curve is probably less than AC + BC - maybe someone can prove this. This method may not work if the concavity changes direction infinitely many times. 67.150.252.232 (talk) 08:44, 16 January 2009 (UTC)

Sorry, some of my primes and double primes didn't come through above, at least on my browser. Bear in mind that they may be missing. 67.150.252.232 (talk) 08:47, 16 January 2009 (UTC)

A simple answer is: there's no simple method to do that. The Peano curve and various fractal figures demonstrate that you can fit an arc of any length in an arbitrarily small region (say an open ball) of the space — so, testing a curve in any finite set of points is not enough to get an upper bound of its length. You will have to use some global information, as others said above. --CiaPan (talk) 10:57, 16 January 2009 (UTC)

If it's a continuously differentiable parametric curve, then the arc length is finite. If it's twice differentiable, and the modulus of the second derivative is bounded by M, then you can obtain an upper bound in terms of M for the difference between the arc length of the curve and that of a polygonal line approximating it. In fact, the length of the curve obtained between times t = a and t = b differs from the length of the single line segment approximating it by at most M(b - a)² / 2. If the same interval is broken into N equal subintervals, the difference between the arc length of the curve and that of the corresponding polygonal line will be bounded by M(b - a)² / 2N.67.150.254.73 (talk) 15:25, 16 January 2009 (UTC)

Matrix

What is a matrix??? For example a scalar is a magnitude and a vector is a magnitude with a direction but what is a Matrix???----The Successor of Physics 13:46, 16 January 2009 (UTC) —Preceding unsigned comment added by Superwj5 (talk • contribs)

Well there's the article Matrix (mathematics). if the start of an article like that isn't good enough for someone who knows about vectors then it should be improved. Dmcq (talk) 14:05, 16 January 2009 (UTC)

Maybe you are wondering about how to visualize a matrix geometrically. One way is to try to visualize the linear map corresponding to the matrix (e.g. considering how the basis vectors are moved around). Or in some cases, an $N\times M$ matrix may be considered to be a vector in $N\cdot M$-dimensional space. Aenar (talk) 15:35, 16 January 2009 (UTC)

A matrix is by definition and ordered chain of vectors. So if we have an n x m matrix, it can be thought of as m column vectors in Rⁿ (order matters). Note however, that we may not be dealing with Rⁿ. We might just be dealing with matrices over any field K. This means that entries in the matrix are simply elements of K. Since we can add, subtract, multiply and divide by non-zero elements in a field, we can multiply matrices over a particular field. This is why matrices are generalized in this manner. Note also that matrices may be thought of as linear transformations: every non-singular matrix induces a linear isomorphism. Basically, linear transformations of the plane, for example, simply preserve addition of vectors in the plane (as well as other properties). You may find the article on affine transformation of interest. Smooth maps between smooth manifolds induce linear isomorphisms of tangent spaces. These linear isomorphisms can be represented as a matrix; when the dimensions of the manifolds in question equal, this matrix is called the Jacobian matrix. In general, matrices are the most useful when you deal with vector spaces (this article is really good; take a look). You may also find Hermitian matrix interesting. I can go on all day but I guess I should stop here. --PST 16:30, 16 January 2009 (UTC)

In fact, you can have matrices over any ring, not just a field. Although, if it isn't a field, a non-zero determinant is no longer sufficient for a matrix to be invertible. --Tango (talk) 17:06, 16 January 2009 (UTC)