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August 4

Pseudoscalar

Whilst this is supposedly my area of expertise, I feel I need some advice regarding this article. As you can see from the edit history, I removed the Clifford algebra section, not because it is wrong per se, but for the reason that it was inconsistent with the rest of the article. However, despite a Google scholar search, I'm still not too clear on what a pseudoscalar is, or if there really are two distinct definitions.

If we are to take the statement that "The pseudoscalar is the top-grade element of a Clifford algebra" as truth, then it is wrong to state that it need commute with all other elements and change sign under parity inversion. These are both true if the algebra has an odd underlying basis (number of basis vectors), but neither are true if the underlying basis is even.

Unfortunately, some of the same literature that state one definition seem to suggest that the other is equally general. Thus I am unsure how to proceed.

Given that only odd-dimensional space possesses anything like a pseudoscalar as currently defined in the article, I am slightly suggesting scrapping the current content and replacing it with the Clifford algebra definition (the bit I removed, perhaps wrongly). I am also questioning the article's status on the importance scale as "low", given the number of references in physics publications.

This paper seems to introduce the commuting and parity-inversion properties but in passing for a geometry in which this is true. [[1]]

Please may I have some advice. --Leon (talk) 10:21, 4 August 2009 (UTC)

I don't know that much about Clifford algebras, but the definition of a pseudoscalar as a top-grade element makes sense. Top-grade elements form a 1 dimensional space and are invariant under rotations so they "behave like scalars." The Clifford algebra page also corroborates that definition. In even dimensions it's true that the sign of a top-form element doesn't change when you flip everything, but I don't think there is any element that behaves like what you're asking for (besides 0). Can you link a source that makes these claims about the properties of psuedoscalars that isn't specifically talking about 3D? Rckrone (talk) 17:32, 4 August 2009 (UTC)

Sorry, I simply explained badly in the last paragraph. What I mean is that the current definition of a pseudoscalar (element that is scalar-like but flips under parity inversion) only holds for odd-d space, and thus perhaps should be replaced with the top-grade of Clifford algebra definition. The stuff about the pseudoscalar having the sign-change property with odd-d spaces should be mentioned, especially since we're usually considering 3D space.

Please let me know what you think. --Leon (talk) 18:11, 4 August 2009 (UTC)

AHH, regarding your last question: no. But my point is that those properties appear to be given as defining pseudoscalars, in texts that make no mention of Clifford algebras. --Leon (talk) 18:14, 4 August 2009 (UTC)

What I'm asking is can you give an example of somewhere those properties are used as a general definition? For even dimensions, there are no non-trivial elements that satisfy that definition. I think you're right that the general definition should be given as a top-grade element of a Clifford algebra. Although for physics it's not really necessary to get into that stuff. Rckrone (talk) 18:25, 4 August 2009 (UTC)

Showing a function is right continuous

Let $f:[a,b]\to [0,\infty )$ be Lebesgue measurable. Define the function $\omega$ on $[0,\infty )$ by

$\omega (y)=m(\{x:f(x)>y\})$,

where m is Lebesgue measure.

(a) Prove that $\omega$ is right continuous.

I have been working on this for a while, looking at solutions from others even. I used the fact that $\omega$ is right continuous at y if, and only if, for every monotone decreasing sequence $\{y_{i}\}$ which converges to y, we have $\omega (y_{i})\to \omega (y)$. So, we just let $\{y_{i}\}$ be any such sequence and define

$E_{i}=\{x\in [a,b]:f(x)>y_{i}\},\qquad E=\{x\in [a,b]:f(x)>y\}$.

So, what I want to show is $\lim _{i\to \infty }mE_{i}=mE$. The solution I am looking at has

$\lim _{i\to \infty }mE_{i}=m(\lim _{i\to \infty }E_{i})=mE$

and this confuses me. First of all, my textbook never dealt with limits of sets, so although the right equality seems like it should be true, I am not sure. Obviously, the E_i are getting larger and larger and are getting closer and closer to E. But, this is not rigorous. And, I'm not sure why it is okay (if it is) to move the limit inside the measure. Any ideas? Thanks. StatisticsMan (talk) 16:42, 4 August 2009 (UTC)

Measures are indeed continuous from below in the sense required here. The proof is a simple application of countable additivity. Algebraist 16:52, 4 August 2009 (UTC)

Cool thanks. I have seen it with decreasing sequences I don't remember it for increasing. StatisticsMan (talk) 17:15, 4 August 2009 (UTC)

I was wondering. This function $\omega$ seems pretty special. Does it happen to have a name that any one knows of? StatisticsMan (talk) 15:36, 5 August 2009 (UTC)

If f were a probability density function, then ω would be the associated complementary cumulative distribution function. I don't know if anyone uses such terminology more generally. Algebraist 18:44, 5 August 2009 (UTC)

Don't you mean that $\omega (y)=(b-a)(1-F_{X}(y))$ for a random variable $X:=f(U(a,b))\,$? --Tardis (talk) 22:05, 5 August 2009 (UTC)

Solving a system of matrix equations

I have a set of equations of the following form: $X_{i}=ABCCB$, where the product on the right hand side is always a finite sequence of the matrices A B and C. I know the values of all matrices X_i, and I want to solve these for A, B and C. I'm trying to find the right brand of mathematics to deal with this situation, but whatever I google for turns up methods for solving systems of equations with the help of matrices, rather than solving matrix equations like these. I have the following questions. If anybody can point me in the right direction, I'd be much obliged:

1. Assuming all matrices are nxn, and the right hand side always contains a fixed number of matrices, how can I tell whether the problem is over or underdetermined?
2. If there are enough equations for the problem to be overdetermined, is it likely that an analytical least squares solution can be found?
3. What kind of numerical method would be appropriate for this kind of problem? I'm using a kind of genetic algorithm at the moment, but I expect there are less general methods.
4. Is there a way to transform a set of equations like this to the form Ax = b or some other general form?

Thank you. risk (talk) 17:52, 4 August 2009 (UTC)

Well, if somebody has a good clue to this problem, I'm curious to learn it too. In the meanwhile, my obvious remark is that if you have some more information (on the data X_i, or on the form of the equations) things may be different and easier . For instance, if the X_i commute, it could be natural to look for A,B,C commuting too, which changes a lot. In general, a possible approach is maybe to study the map taking (A,B,C) into (X₁, X₂,..) from the point of view of differential geometry and nonlinear analysis, and apply some principle for global invertibility or surjectivity of differentiable mappings (like e.g. Hadamard's theorem: "a C¹ map f:E→E with invertible differential satisfying |Df(x)^-1| ≤ a|x|+b, is a global diffeo"). As to the last point, I would say not: I don't see how one could reduce such a nonlinear problem to linearity. --pma (talk) 21:42, 4 August 2009 (UTC) A further remark: the toy version of your problem: "solving U=AB, V=BA for A and B", already shows that there are nontrivial necessary conditions for the solvability. Assuming that U or V is non-singular (hence both), the equations imply that U and V are similar matrices (U=AVA^-1). This can be checked from their spectral theory, and of course it is also a sufficient condition. Also, some material possibly related with your problem comes back googling "(system) of equations in groups", "equations in matrix groups" and similar combinations.--pma (talk) 08:45, 5 August 2009 (UTC)

Thank you very much. That's already given me a lot to digest. I'll start reading up on your suggestions. I hadn't really realized that the problem was nonlinear. Can you explain how you determined that it wasn't?

I can give you little more information on the equations. The full set contains $n^{k}$ equations, each of the form $X_{i}=Y_{\sigma _{1}}Y_{\sigma _{2}}\cdots Y_{\sigma _{k}}$ where $\{\sigma _{i}\}$ is a sequence with elements in [1, n]. So we have an equation for every such sequence, where $X_{i}$ is always known, and the Y's in the right hand side aren't. The matrices can be assumed to be non-singular. (To make it even more interesting, the end result will be an algorithm where n and d are variables). Do you have any ideas what numerical approaches people would normally use for this kind of problem? I'm using a kind of genetic algorithm at the moment, but I'm sure something more elegant is possible. risk (talk) 13:16, 6 August 2009 (UTC)

I think I should also point out that I'm really looking for more of a least squares solution than an analytical solution. It may be possible to use only the n equations of the form $X_{i}=Y_{j}^{k}$ to derive a unique answer, but if the system doesn't have a unique solution, I would need the error to be distributed evenly over all equations. risk (talk) 13:35, 6 August 2009 (UTC)

FWIW, there is a least squares procedure called seemingly unrelated regression (SUR) that generates the solution to a system of equations when the stochastic term is correlated across equations. Each equation can be represented in matrix form, and so SUR solves the system of matrix equations. Wikiant (talk) 13:46, 6 August 2009 (UTC)

What angle do you need to get a certain length of an arc?

Hi there, fellas!

I had nothing to do today, so in my boredom I decided to develop a cool little graphical app in Java. The thing is, there's a whole lot of arcs and circles in it, and trigonometry wasn't my star subject in high school, and by at this point I'm going insane with all the sines, coses and arc-tangents. So now I have a small little problem which is probably ridiculously easy for anyone with an actual brain, but I would really appriciate some help with it.

The problem is that I want to draw an arc of a circle where I can decide exactly how long the arc will be. So this is the problem: say you want to draw an arc that's P pixels long, on a circle that has a radius R pixels. What is the angle (in radians or degrees) of that arc?

As I said, it's probably very easy, but my brain is getting very, very tired, and it's a subject I'm not too comfortable with to begin with. 83.250.236.75 (talk) 21:17, 4 August 2009 (UTC)

You know the circumpherence of a circle or that radius...right? You also know that a pie-wedge taken out of that circle has an arc that's the same fraction of the circumpherence as the angle at the center is a fraction of 360 degrees. So for an arc of length 'L' for a circle of radius 'R', you need an angle that is: A = L * 360 / ( 2 * PI * R ) (in degrees of course). In radians, the 2*pi's cancel nicely so you get the much more elegant A = L / R. SteveBaker (talk) 21:56, 4 August 2009 (UTC)

Right, I figured it was something simple like that :) Thanks! 83.250.236.75 (talk) 22:07, 4 August 2009 (UTC)

(ec) Maybe it is safer to redirect this to the science/RD, specifically to Steve Baker. Oh, but he's already here! Anyway, isn't the number of pixels in a segment, or in an arc, depending on its position and not only on its length? Assuming the arc starts from the X-axis and goes counter-clockwise, and that it is within an angle α not larger than π/2 radiants, I imagine that the number of pixels in it is the same as in its vertical projection on the Y-axis. So, if the ratio s:=(no. of pixels in the arc)/(no. of pixels in the horizontal radius) is less than sqrt(2)/2 my guess is α=arcsin(s). Analogously, compare a short arc starting from the Y-axis with its horizontal projection. In general, I guess one reduces to these cases by adding/subtracting arc segments. --84.220.119.55 (talk) 22:19, 4 August 2009 (UTC)

As I said, I'm not terribly good at this, but Steve's solution worked perfectly. The arcs became the right lengths with his simple formula. 83.250.236.75 (talk) 00:59, 5 August 2009 (UTC)

Yeah - it's really dangerous having me on the math desk...I have this annoying tendancy to want to give answers that people can actually understand. Being a computer graphics geek - I understood perfectly that you meant to use the word "pixel" as a unit of measurement (like "inch" or "centimeter") and not in the sense of "a picture element". Indeed if you had wanted to count the number of picture elements along your arc, life would be difficult and painful and we'd CERTAINLY have to send all of the Mathematicians out to buy more pencils then lock the doors after them. It's a tough question to answer because a lot would depend on which of the 30+ known ways of drawing a circle you were actually using. That's because they tend to hit different numbers of pixels. What '84 is trying to tell us (using "math") is that counting the pixels is non-trivial. Consider that a thin straight (diagonal) line drawn from (x1,y1) to (x2,y2) hits the exact same number of pixels as a much shorter horizontal line drawn from (x1,0) to (x2,0) or a vertical line drawn from (0,y1) to (0,y2) - whichever is greatest. Hence, the number of pixels required to draw a circle is something rather complicated. On an infinitely high resolution display, the number of pixels would simply be four time the radius of the circle - but we don't have infinitely high resolution displays - and that makes the answer horribly complex - and (as '84 does not comprehend) critically dependent on which algorithm you use for drawing your circle. Fortunately, I happened to be here asking about 4D cross-products so none of this uglyness proved necessary!  :-P SteveBaker (talk) 13:25, 5 August 2009 (UTC)

Anyway, I was really thinking of you as the most qualified person to answer the OP, and when I saw you there I was glad of my perspicacy. Sure, I suspected that the answer depends on the algorithm and I opted for the most obvious to me  :-P )))) and these pharentheses represents sound waves  ;-) --84.221.69.165 (talk) 22:50, 5 August 2009 (UTC)

4D analog of a 3D cross-product.

I have a function that takes a 2D vector and returns me another 2D vector that's at right angles to it. I have the standard cross-product function that takes two 3D vectors and returns me another 3D vector that's at right angles to both of the other two...what's the equivelant thing in four dimensions? I presume it needs three 4D vectors as input...but I'm having a mental block as to how to mush them together! (Oh - and is there a general name for these functions?) SteveBaker (talk) 21:50, 4 August 2009 (UTC)

The cross product only exists in 3D. The closest thing in higher dimensions is the wedge product. I've never studied it, though, so I'm not sure if it can be used to do what you want. However, you could easily use the dot product to get a system of simultaneous equations to solve. (Just call the desired vector (x,y,z,w) and dot it with each of the vectors you want it to be orthogonal to.) --Tango (talk) 22:20, 4 August 2009 (UTC)

(edit conflict - verbose answer) In 4D (assuming it's just a normal extension of 3D with another right angle 'squeezed in' somewhere unimaginable) then the cross product of 2 vectors is a plane (since there is a whole flat plane that is at right angles to both vectors).

Taking that plane, and then applying the third vector to it gives a single vector (or 2 vectors pointing in opposite directions to be pedantic) - just as when you took a vector in 2d and got another vector at right angles to it.

So yes in 4D 3 (non colinear) vectors define a fourth vector that is at right angles to all three.

To expand - it's easy to count dimensions (aka degrees of freedom) . Curiously it's an extension of this that allows us to show that in 4D one can rotate a vector around a plane 4(dimensions)=2(dimensional contraints given by the plane)+2(degrees of freedom remaining) - thus a vector constrained to rotate about the plane in 4D has 2 degrees of freedom -just the right amount for rotation.

(By the way the maths is dead simple - just linear algebra, and sins/cosines -even easier still if you've got an automatic linear equation solver to do the donkey work)

So to summarise - to derive the formula - first find the equation of the plane P that satisfies the conditions

V1 is perpendicular to the plane

V2 is perpendicular to the plane

Then use your equation for the right angle in 2D and apply it to the plane with the 3rd vector as a constraint - to give the answer (a vector) 83.100.250.79 (talk) 22:24, 4 August 2009 (UTC)

See Levi-Civita symbol.

2D: $\,y_{i}=\epsilon _{ij}a_{j}$

3D: $\,y_{i}=\epsilon _{ijk}a_{j}b_{k}$

4D: $\,y_{i}=\epsilon _{ijkl}a_{j}b_{k}c_{l}$

The 2D formula is shorthand for

$\,y_{1}=a_{2}$

$\,y_{2}=-a_{1}$

The 3D formula is shorthand for

$\,y_{1}=a_{2}b_{3}-a_{3}b_{2}$

$\,y_{2}=a_{3}b_{1}-a_{1}b_{3}$

$\,y_{3}=a_{1}b_{2}-a_{2}b_{1}$

Bo Jacoby (talk) 22:26, 4 August 2009 (UTC).

In my undergraduate multivariate calculus course (which was like Calc 4), our book had a few exercises that gave an analog for any finite dimension, with a subtitle "Cross products in R^n". As you guessed, you need N-1 vectors in N dimensions and you do it exactly like a cross product. That is, you take a matrix and you put e_1, e_2, ..., e_n (instead of i, j, k) in the top row and you put the N-1 vectors in the rest of the rows. Then, you compute the determinant. The exercises are to show it has various properties. Most are obvious from properties of determinants. But, one is that the vector you obtain is orthogonal to all N-1 vectors. My book is Vector Calculus by Susan Jane Colley. My edition is likely not the newest any more but it's a few exercises at the end of Section 1.6 Some n-dimensional Geometry in my edition. StatisticsMan 01:13, 5 August 2009 (UTC)

I think that Tango is correct about the wedge product thing, but incorrect about the number of vectors required (still two). It's important to remember that oriented line segments (vectors) are dual to oriented plane segments (bivectors, which are what you get with a cross product on vectors though you don't realize it!) in R^{3} but not R^{4}. The wedge product is constructed from the Clifford algebra. The lack of duality is manifest in that the wedge of two vectors from R^{4} spans a six dimensional vector space as opposed the three. Might I ask your purpose in knowing about this? I may be able to recommend some resources.--Leon (talk) 13:10, 5 August 2009 (UTC)

OK - the reason is kinda complicated & messy. <sigh> When you multiply a bunch of nice simple rotation matrices together on some real-world device such as a graphics processor, you get roundoff error. That error isn't particularly important in itself because there is typically a degree of interactivity that tends to correct any total error in the angle rotated through. However, after enough times through the multiplication pipeline, the matrices start to become non-orthonormal...which tends to result in all sorts of nasty skews and scalings that are pretty undesirable. This is a common enough problem in computer graphics and we've learned to "repair" 3x3 rotation matrices that seem to be a little the worse for wear by treating the three rows as vectors, normalizing them and then recalculating the third row to make it be orthogonal to the other two. This results in a matrix that may be rotating by slightly incorrect angles - but at least there is no skew or scale going on. You doubtless have "better" ways to do this - but before you leap in with them, I'll point out that I'm doing this a bazillion times a second on graphics hardware that's very efficient at things like normalizing and doing cross products and the like - and sucks badly at doing any of the things you were just about to suggest!

Well, now I need to do the same thing with some 4D rotation matrices...so I need an analogous function that (given three rows of the matrix) allows me to calculate the fourth row to be orthogonal to the other three. SteveBaker (talk) 13:55, 5 August 2009 (UTC)

I Understand. I'm really sorry, but I don't have too many suggestions. I do have one though. Whilst you gave the graphics processor merely as an example of a real world device, if that's what you want to program, I'll ask a follow up question: do you ever use quaternions? They're very common in computer graphics. I won't rant about what I'm suggesting unless (a) you already know about them and (b) are interested in how they can be cleanly extended to 4-D rotations---they are very easy to correct. As it happens, I'm (kind of) studying the field myself, so if this sound helpful... --Leon (talk) 15:14, 5 August 2009 (UTC)

Yes, we use quaternions quite extensively - but (to get technical about it) mostly on the CPU not on the GPU. The main central processing unit (CPU) in the computer...the Pentium chip or whatever...is a totally general purpose machine and it's very easy to program it to toss quaternions around. But the GPU (graphical processing unit) on the graphics pipeline isn't at all general-purpose. It's has a rather unique and specialised architecture that allows it to process 2D, 3D and 4D vectors and matrices with enormous speed...but they don't have any functions specialised for handling quaternions - so we tend to convert everything to matrices just before handing them over to the GPU. That's not to say that it's impossible to handle quaternions in the GPU - just that it's slow and inconvenient. I haven't thought much about using a 4D quaternion (a quinternion?!) - but it would presumably comprise something like a 4D vector plus a rotation around that vector...and that would require 5 numbers. Sadly, the GPU is highly optimised for handling 4 numbers in a group...it would fare badly if forced to toss around 'quinternions'. This is one of those situations where what looks good in abstract mathematics fails miserably when plopped into the middle of a video game that has to do this a million times a second! Also, the use of true 4D geometry is very limited to a few niche applications - it's not something we do a lot of. SteveBaker (talk) 15:35, 5 August 2009 (UTC)

It looks like you may be able to help me more than I can help you! However, I'm not suggesting a quinternion at all, simply ordinary 4-D quaternions interpreted slightly differently. Given your hint that this is for use in a video game, am I at all correct in suggesting you're using this in the context of projective geometry, describing translations as rotations about points infinitely (well, very, anyway) far away? If so, here's a specific link. [[2]]

And now my question for you: can you provide me with links detailing the maths functions native to typical consumer-grade GPUs? Thanks. --Leon (talk) 16:05, 5 August 2009 (UTC)

P.S. Even if my guess at your interest in this is wrong, the link may be helpful nonetheless. The absolute inefficiency of the quaternionic approach is lower when used for 4-D than for 3-D.--Leon (talk) 16:05, 5 August 2009 (UTC)

There are no quinternions. You can only have powers of two - 1D, 2D, 4D, 8D or 16D (no higher), and they get progressively less and less well behaved (you may be able to have 32D but they would be so badly behaved as to be useless). See Cayley–Dickson construction. If you started with quaternions are tried to add one more square root of -1 you would find that more would naturally arise as products of existing ones (in the same way as if you add just one root to the complex numbers, you'll find you end up with the quaternions). I'm curious, how do you handle multiplying 4D rotations? I've never done much with them, but I know they aren't as well behaved as their 3D counterparts - you don't necessarily get a rotation if you compose two 4D rotations. --Tango (talk) 18:07, 5 August 2009 (UTC)

Read the paper I linked above! Incidentally, I'm not sure about your final statement, indeed, I think it's wrong, though I'll add that I don't hold too much weight in that opinion as I'm not a geometer. SO(4) is a group, and for your final statement to be true SO(4) mustn't be the group of rotations in R^{4}. ALSO, Mr SteveBaker, whilst I'd advocate at least skimming the aforementioned paper, I know a few people at work who may be able to advise...I'll write here if I have anything else to offer. --Leon (talk) 18:29, 5 August 2009 (UTC)

There is some uncertainty as to which things ought to be called 'rotations' in higher dimensions. Some use the term for everything in SO(4), while Tango seems to be using it to mean only the so-called simple rotations. Algebraist 18:40, 5 August 2009 (UTC)

Sorry, yes, I assumed Steve was talking about simple rotations but I now realise he never actually said that. --Tango (talk) 19:39, 5 August 2009 (UTC)

True. Open suggestion to all who have commented in ways that do not really address the asker's needs: can those section be moved elsewhere? I'm not an experience Wikipedia, so please alert me if this is not a reasonable request.--Leon (talk) 18:47, 5 August 2009 (UTC)

The usual rule on the ref desks is that we don't go off on tangents until the OP has an answer to their original question. Steve has received lots of answers, but only he can say if he's received one that actually helps. So, Steve, do we need to keep working or can we play now? --Tango (talk) 19:39, 5 August 2009 (UTC)

Suggestion for SteveBaker: with an orthogonal matrix, the rows (and columns, obviously) are orthogonal vectors. So, use the Gram-Schmidt Process to "orthogonalize" your matrices. Hope this is helpful.--Leon (talk) 18:43, 5 August 2009 (UTC)

Hi Steve, It seems as if I did not explain my solution properly. I'm sorry. The 2D problem is: You have a vector $a=(a_{1},a_{2})$ We want a vector $y=(y_{1},y_{2})$ such that a and y are at right angles. One solution is $y_{1}=a_{2},\,y_{2}=-a_{1}$. They are at right angle because $y\cdot a=y_{1}a_{1}+y_{2}a_{2}=a_{2}a_{1}-a_{1}a_{2}=0$. The solution is written like this: $y_{i}=\sum _{j=1}^{2}\epsilon _{ij}a_{j}$ where $1=\epsilon _{12}=-\epsilon _{21}$ and $0=\epsilon _{11}=\epsilon _{22}$. The epsilon is antisymmetric: if you swap two indexes the sign is changed. $\epsilon _{ij}=-\epsilon _{ji}$. The 3D problem is: You have two vectors $a=(a_{1},a_{2},a_{3}),\,b=(b_{1},b_{2},b_{3}).$ We want a vector $y=(y_{1},y_{2},y_{3})$ such that a and y are at right angles and such that b and y are at right angles. One solution is: $y_{i}=\sum _{j=1}^{3}\sum _{k=1}^{3}\epsilon _{ijk}a_{j}b_{k}$ where $1=\epsilon _{123}$ and epsilon is antisymmetric: if you swap two indexes the sign is changed. The 4D problem is: You have three vectors $a=(a_{1},a_{2},a_{3},a_{4}),\,b=(b_{1},b_{2},b_{3},b_{4}),\,c=(c_{1},c_{2},c_{3},c_{4}).$ We want a vector $y=(y_{1},y_{2},y_{3},y_{4})$ such that a and y are at right angles and such that b and y are at right angles, and such that c and y are at right angles. One solution is: $y_{i}=\sum _{j=1}^{4}\sum _{k=1}^{4}\sum _{l=1}^{4}\epsilon _{ijkl}a_{j}b_{k}c_{l}$ where $1=\epsilon _{1234}$ and epsilon is antisymmetric. Proof: $y\cdot a=\sum _{i=1}^{4}y_{i}a_{i}=\sum _{i=1}^{4}\sum _{j=1}^{4}\sum _{k=1}^{4}\sum _{l=1}^{4}\epsilon _{ijkl}a_{i}a_{j}b_{k}c_{l}=-\sum _{i=1}^{4}\sum _{j=1}^{4}\sum _{k=1}^{4}\sum _{l=1}^{4}\epsilon _{jikl}a_{j}a_{i}b_{k}c_{l}=-y\cdot a$ so $y\cdot a=0$. Similarily $y\cdot b=0$ and $y\cdot c=0.$ Bo Jacoby (talk) 19:35, 5 August 2009 (UTC).

You know - I can't actually tell whether you answered my question or not! I have to read all the linked documents...which is a slow process because I'm not a mathematician - and Wikipedia math articles are completely reader-hostile to anyone who doesn't already understand what they are saying! All I really wanted was a nice simple set of expressions...like a cross-product...that, given three 4D vectors would give me back a fourth that's at right angles to all of the other three. To pick an obvious example, if I handed it (1,0,0,0), (0,1,0,0) and (0,0,1,0) - it would give me back (0,0,0,1) (or maybe (0,0,0,-1)). I was kinda hoping there would be some nice, simple linear algebra to do that. Bo Jacoby's post that starts: "See Levi-Civita symbol" came close...but left the 4D case unexpanded...since I have yet to learn what the heck a Levi-Civita symbol is and somehow expand the expression for the 4D case, I'm not quite there yet. If someone could expand that one for me - then I think I'd be done and outta your hair! I was aware of the Gram-Schmidt Process - but it's not very nice to do inside a teeny-tiny graphics chip! For a 3x3 matrix, it suffices to say something like (from memory):

• normalize row 1
• normalize row 2
• take the cross-product of row1 and 2 and stuff that into row3
• normalize row 3
• take the cross-product of row1 and 3 and stuff that into row2

Now all three rows are unit vectors and they are all at right angles...so the matrix is now well-behaved - although possibly rotating by a fractionally different amount than it was before...and on a modern graphics chip - you can do that sequence in 5 machine-code instructions! I can easily imagine a 4D analog of that...if there were a 4D analog for a cross product. SteveBaker (talk) 20:31, 5 August 2009 (UTC)

Aha!

There is such a product, yes.

Basically, you take the following determinant: ${\begin{vmatrix}e1&e2&e3&e4\\a&b&c&d\\e&f&g&h\\i&j&k&l\end{vmatrix}}$ where the ei components are the basis vectors.

Does this make sense?--Leon (talk) 20:43, 5 August 2009 (UTC)

$\mathbf {a} \times \mathbf {b} =\det {\begin{bmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\\end{bmatrix}}.$

That's what a cross-product determinant looks like in three-dimensions; I've included this for clarity, such that you can see the similarity. --Leon (talk) 20:53, 5 August 2009 (UTC)

Leon's answer is the one you're looking for, but for what it's worth, you can represent a 4D rotation with a pair of unit quaternions (six degrees of freedom in total), as described in Quaternions and spatial rotation#Pairs of unit quaternions as rotations in 4D space (read the rest of the article too). SO(4) has six dimensions, and in general, SO(n) has n(n−1)/2 dimensions. It's only in three dimensions that the dimension of the rotation group matches the dimension of the space, and only in three dimensions that the notion of a rotational axis makes sense (in general rotations happen in planes, not around axes). Angular momentum in four-dimensional space can't be described by a 4D vector because it has six components, and the same goes for other pseudovectors, like the magnetic field. The wedge product is the correct generalization in those cases, but you don't need to worry about that unless you're doing 4D Newtonian physics. $\sum _{i=1}^{3}\sum _{j=1}^{3}\sum _{k=1}^{3}\epsilon _{ijk}a_{i}b_{j}c_{k}$ is just a fancy way of writing the matrix determinant ${\begin{vmatrix}a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix}}$. Are you sure quaternion arithmetic is inefficient on GPUs? The quaternion product of (s, v) and (t, w) is (st − v·w, sw + tv + v×w), where s and t are scalars, v and w are 3D vectors, and · and × are the 3D dot and cross product. If the GPU has built-in support for the cross product then I'd expect that to be comparable in speed to a matrix-matrix multiply. -- BenRG (talk) 21:46, 5 August 2009 (UTC)

I think every single suggestion given in this section has actually been the same idea just with different notation, so they should all be equally efficient. (The quaternions idea might be slightly different, but simultaneous equations, Levi-Civita symbols and matrix determinants are all exactly the same thing.) --Tango (talk) 23:27, 5 August 2009 (UTC)

Wonderful! I think I have it now. Thanks everyone! SteveBaker (talk) 01:12, 6 August 2009 (UTC)

Good! Ask questions on the talk pages of the reader-hostile math articles. Wikipedia editors try to write reader-friendly articles. The 4D Levi-Civita symbol $\epsilon _{ijkl}$ is the sign of the permutation $ijkl.$ The sign of an even permutation is +1 while the sign of an odd permutation is −1, and the sign is 0 if $ijkl$ is not a permutation of 1234. The 12 even permutations are 1234 1342 1423 2134 2341 2413 3124 3241 3412 4123 4231 4312, and the 12 odd permutations are 1243 1324 1432 2143 2314 2431 3142 3214 3421 4132 4213 4321. So the 4D formula

$\,y_{i}=\epsilon _{ijkl}a_{j}b_{k}c_{l}$

is shorthand for

$\,y_{1}=a_{2}b_{3}c_{4}-a_{2}b_{4}c_{3}+a_{3}b_{4}c_{2}-a_{3}b_{2}c_{4}+a_{4}b_{2}c_{3}-a_{4}b_{3}c_{2}$

$\,y_{2}=a_{3}b_{4}c_{1}-a_{3}b_{1}c_{4}+a_{4}b_{1}c_{3}-a_{4}b_{3}c_{1}+a_{1}b_{3}c_{4}-a_{1}b_{4}c_{3}$

$\,y_{3}=a_{4}b_{1}c_{2}-a_{4}b_{2}c_{1}+a_{1}b_{2}c_{4}-a_{1}b_{4}c_{2}+a_{2}b_{4}c_{1}-a_{3}b_{1}c_{4}$

$\,y_{4}=a_{1}b_{2}c_{3}-a_{1}b_{3}c_{2}+a_{2}b_{3}c_{1}-a_{2}b_{1}c_{3}+a_{3}b_{1}c_{2}-a_{3}b_{2}c_{1}$

This is the nice simple set of expressions that, given three 4D vectors would give me back a fourth that's at right angles to all of the other three. Bo Jacoby (talk) 08:52, 6 August 2009 (UTC).

August 5

A question about geographical coordinates and significant digits

There are a number of different ways in which a geographical position can be identified using coordinates. For example the location of Mount Whitney in California is listed in different ways by different sources.

The Geographic Names Information System (GNIS) uses two formats:

36.5782684, -118.2934244 (i.e. 36.5782684°N, 118.2934244°W) — Decimal Degrees (DD)

363442N, 1181736W (i.e. 36°34′02″N, 118°17′73″W) — Degrees, Minutes, Seconds (DMS)

The U.S. National Geodetic Survey (NGS) uses the following format:

36 34 42.89133(N) 118 17 31.18182(W)

Others use:

N 36° 05.885 W 115° 05.583 — Degrees, Minutes (DM)

For example when converting DMS to DD format how may decimal places are significant:

36°36′36″ then (36*3600 + 36*60 + 36)/3600 = 36.61°

How can I determine how many decimal places are significant when converting (e.g. from NGS data) to DD format. Is there some general rule I can follow. What about the inverse function (i.e. converting DD to DMS). Assume 36° implies >= 35.5° and < 36.5° etc. --droll [chat] 00:18, 5 August 2009 (UTC)

Find the precision of the initial value and convert it into the new units. For example 36°34′02″ implies uncertainty on the order of 1". Converting 1" to degrees gives you .000277°. You don't want the precision of your new value to be any higher than that of the initial value, so round up your uncertainty to the nearest power of 10 (.001° in this case) and get rid of any digits beyond that one. In this case the result would be 36.567°. Rckrone (talk) 02:59, 5 August 2009 (UTC)

Sorry if being a bit off-topic: When comparing geodesic data from two different sources, it is also convenient to verify that coordinates are expressed with the same reference datum. Pallida  Mors 05:38, 5 August 2009 (UTC)

True, I use decimal degrees with 4 decimal places in any maps and give a selection of maps a user could use. They're not too badly correlated now but in the past you could get to completely the wrong building with different maps. By the way I try to indicate the entrance rather than the body of a building. Dmcq (talk) 09:33, 5 August 2009 (UTC)

By the way I'd convert the precision up rather than down if comparing things, then subtract the two and see if the difference is smaller than what you're checking for. Who is going to actually read the numbers? Normally they'd just go to another computer. Dmcq (talk) 09:42, 5 August 2009 (UTC)

Mandelbrot set

If one iteration is done on the whole Mandelbrot set, is the result the whole of the set or just a subset? --Masatran (talk) 09:43, 5 August 2009 (UTC)

Clarification: Let the points that constitute the Mandelbrot set be the set S1. Take each point of this set and apply the iteration just once. Let the resultant set of points be S2. From the definition, S2 is a subset of S1. Is S2 a proper subset, or is it the whole of S1? --Masatran (talk) 05:45, 6 August 2009 (UTC)

An iteration of what? (Isn't that you mean the Julia set of a map, instead?) --pma (talk) 10:14, 5 August 2009 (UTC)

Did you mean to take test the iterated condition only once - eg using

zn+1 = zn2 + c

Only to go as far as z₁

That I think would be a subset, or more specifically a single quadratic equation.83.100.250.79 (talk) 13:10, 5 August 2009 (UTC)

And what do you mean? I can't understand neither of you. Can you formulate your question precisely?--pma (talk) 14:38, 5 August 2009 (UTC)

See Mandelbrot set first paragraph - using the formula - one iteration

Mathematically, the Mandelbrot set can be defined as the set of complex values of c for which the orbit of 0 under iteration of the complex quadratic polynomial z_n+1 = z_n² + c remains bounded.

The first iteration (for a general point [x,y]) ie x+iy yields a quadratic formula for the boundary condition, the second iteration a quartic, and so on...

Actually I think it's a subset of a subset - the first subset being a specific value of c, the second subset being the set of iterations forming the first subset. The questioner mentioned "iterations" - one obvious use of iterations in the mandelbrot set is that decribed above.HappyUR (talk) 15:07, 5 August 2009 (UTC)

Who knows what the questioner meant. At the moment I see no question...--pma (talk) 15:31, 5 August 2009 (UTC)

To construct the Mandelbrot set you start with z₀ = 0, then iterate z_n+1 = z_n² + c. If |z_m| is greater than 2 for some m then you can be sure that z_n diverges to infinity for this value of c, so c is not in the Mandelbrot set. So after one iteration you throw out the points with |c| > 2 and you are left with a circle radius 2. After two iterations you throw out points such that |c² + c| > 2 and you are left with a cardioid (I think) that intersects that real axis at c = −2 and c = 1. At each iteration you throw out more and more points - so you are left with a subset of the points you had one iteration before, but this is a superset of the Mandelbrot set itself. Gandalf61 (talk) 16:00, 5 August 2009 (UTC)

Yes that's what I was thinking, and you are right it is a superset (I think) (not subset) since some points will not 'escape' until n is big (for a given c).

Not a cardioid though - a distorted dumbell (peanut) shape with one reflection symmetry - I think the number of inflexions increase by 2 (or maybe doubles) for the boundary with each iteration. maybe/maybe not - a quartic in x and y (from c=x+iy) anyway83.100.250.79 (talk) 16:15, 5 August 2009 (UTC)

You are right - it's not a cardioid. In polar co-ordinates it is the region inside the curve $r^{4}+2r^{3}\cos(\theta )+r^{2}=4$ Gandalf61 (talk) 10:19, 6 August 2009 (UTC)

When the moment comes of interpreting meaningless questions, the RD\M becames like a school of Rabbinic studies in Yiddish jokes. --pma (talk) 22:21, 5 August 2009 (UTC)

Or the Three Stooges as people fall over each other and get into fights.83.100.250.79 (talk) 23:31, 5 August 2009 (UTC)

twice as wide

We have a question in the german wiki reference which can only be answered by a native speaker. How would you understand the sentence: "Now find at least three positions where you can put the light and the card to make a shadow twice as wide as the card." Do they mean a shadow width twice as wide as the edge width of the card or twice as wide as the whole card? --Ian Dury^{Hit me} 15:50, 5 August 2009 (UTC)

I'm not sure what you mean by "edge width". Do you mean "thickness"? In which case, it's not what is meant - width and thickness are completely different words in English. If they meant thickness, they would have said thickness. --Tango (talk) 18:10, 5 August 2009 (UTC)

Clever bloke... In fact I can see now what my unreflectet reproduction from someones expression at our reference desk did. What he meant with "edge width" is the length of the edge... Anyway, thank you very much. --Ian Dury^{Hit me} 18:35, 5 August 2009 (UTC)

I still don't understand... surely the length of the short edge is the same as the width of the card, and the length of the long edge is the same as the length of the card. --Tango (talk) 19:36, 5 August 2009 (UTC)

Looking at the original de: thread, the OP there asked whether the expression can be used to mean a card with doubled area ("Fläche") as well as a card with doubled width. Which, of course, it cannot. —JAO • T • C 19:54, 5 August 2009 (UTC)

Gravity and Mass

Question already posted on the RD/Science, where it fit better indeed. It should maybe be removed from here... --pma (talk) 23:10, 5 August 2009 (UTC)

Indeed. See Wikipedia:Reference desk/Science#Gravity and_Mass. -hydnjo (talk) 00:40, 6 August 2009 (UTC)

August 6

Rational function?

The definition of a rational function is a polynomial divided with a polynomial. So I've been wondering- is $f(x)={\frac {P_{m}(x)}{Q_{n}(x)}}$ a polynomial if $a_{0}=3$ and $b_{0}=2$ (so that $f(x)={\frac {3}{2}}$)? Is $f(x)={\frac {1}{1}}$ a rational function- and if so, is $f(x)=1$ a rational function even thought it isn't a ratio (because if ${\frac {1}{1}}$ is a rational function and since ${\frac {1}{1}}=1$...). And as a final question, is ${\frac {0}{0}}$ a rational function even though it's undefined? 0 is a polynomial (the zero polynomial) so it's technically the ratio between two polynomials- does as function have to be defined? (sorry for asking so many questions, this has been piling up in my head for some time and I'm too tired to stylize my question) --BiT (talk) 01:48, 6 August 2009 (UTC)

Oh never mind the ${\frac {0}{0}}$. A quick read through the "where $P\,$ and, $Q\,$ are polynomial functions in $x\,$ and $Q\,$ is not the zero polynomial. The domain of $f\,$ is the set of all points $x\,$ for which the denominator $Q(x)\,$ is not zero." answered my question --BiT (talk) 01:49, 6 August 2009 (UTC)

Yes, functions which are constantly a rational number (which includes integers) are (constant) rational functions. As are polynomials (just take Q=1). A function does have to be defined everywhere on its domain - that is the definition of a domain, really - so 0/0 would be, at best, a function defined on the empty set, ie. the empty function. --Tango (talk) 02:05, 6 August 2009 (UTC)

Is "The domain of $f\,$ is the set of all points $x\,$ for which the denominator $Q(x)\,$ is not zero." (a quote from Rational function) really true? This implies that f : [0,infty) -> R defined by f(x) = 1/(x² + 1) is not a rational function. Is this common usage? (Depending on the context, it might even be taken to imply that g : R -> R defined by g(x) = 1/(x² + 1) is not a rational function either, because its domain leaves out a whole lot of points (all non-real complex numbers except ±i) where the denominator is not zero.) —JAO • T • C 05:40, 6 August 2009 (UTC)

That section is talking specificially about real functions. (There's another section about the complex version). It should probably say that explicitly. I'll change it. Rckrone (talk) 06:37, 6 August 2009 (UTC)

Actually, after thinking about it, the article is fine as is. It's important to keep in mind the field that you're dealing with when talking about rational functions. What looks like the same function can have different meanings depending on what field you're dealing with. x² + 1 as a polynomial over the reals is nowhere zero, so the domain of 1/(x² + 1) as a rational function over the reals is all of R. x² + 1 as a polynomial over the complex numbers has zeros at i and -i, so the domain of 1/(x² + 1) as a rational function over the complex numbers is C\{i,-i}. —Preceding unsigned comment added by Rckrone (talk • contribs) 06:53, 6 August 2009 (UTC)

To answer the other part of the question, it's generally not ok to restrict the domain and still call the function a rational function without qualifying it in some way. When someone says "rational function" or similarly "polynomial", there are certain properties we expect, which may not hold on some smaller domain. For example the fundamental theorem of algebra doesn't hold if we allow "polynomials" to include functions with restricted domains. Rckrone (talk) 07:03, 6 August 2009 (UTC)

I'd say there is a slight abuse of language due to obvious historical reasons. Strictly speaking a rational function (say with real coefficients) is not a function, its primary meaning being that of an element of the field of fractions R(x) of R[x]. Correspondingly, the term "domain" of a rational function f denotes primarily a certain subset of the coefficient ring, attached to the algebraic object f (and of course it turns out to be the set-theoretic domain of a certain set-theoretic function). Note that "domain" is used with special meanings in other contexts: e.g the "domain of a convex function" f:E →RU{+∞} is the subset of the normed space E where f(x)<+∞, although strictly speaking the domain of f is the whole E.--pma (talk) 07:34, 6 August 2009 (UTC)

I find myself using the term "rational functions" to refer literally to functions more often than anything else. At least this suggests that this usage is fairly common.

Anyway, I agree with Rckrone that a function whose domain is artificially restricted (as opposed to implicitly restricted to our current structure of interest) should not be called rational.

Regarding Tango's comment, any constant function is rational - the constant need not be a rational number. -- Meni Rosenfeld (talk) 16:59, 6 August 2009 (UTC)

Thanks for all the enlightening comments. Especially the fact that "rational function" is often used in English to denote what really is a "rational expression" set me straight from being fooled by my own language, which upholds that distinction. —JAO • T • C 05:01, 7 August 2009 (UTC)

Thank you for your answers and now I just want to verify what Tango said- so you're saying that rational numbers (like 4⁄5), integers (like 5 or 3) and polynomials (like $3x^{2}+4x+16$) are all rational functions? Just to be completely clear. If that's so, then why isn't the introduction to the 'rational function article' formulated like the rational number article? That is to say: "In mathematics, a rational number is any number that can be expressed as the quotient a/b of two integers, with the denominator b not equal to zero. Since b may be equal to 1, every integer corresponds to a rational number." If I understand Tango correctly a better way of putting it would be something like:

“ In mathematics, a rational function is any function that can be expressed as the ratio ${\frac {P_{m}(x)}{Q_{n}(x)}}$ of two polynomial functions, where the denominator $Q_{n}(x)$ is not equal to zero. Since $Q_{n}(x)$ may be equal to 1, every integer corresponds to a rational function. ”

I know I basically copied the introduction from the article about rational numbers- but I think the revision would help other people understand what I was having trouble with. What do you think? --BiT (talk) 23:06, 7 August 2009 (UTC)

As Meni points out, it's not just rational numbers, any real number will do, I made a mistake there. I'm not sure I understand your proposal - the rational function article does say essentially that. --Tango (talk) 00:26, 8 August 2009 (UTC)

1. I actually didn't fully understand Meni- what do you mean by "any real number will do"? (refresh my challenged-at-learning-definitions memory, real numbers include both rational and irrational numbers right?) Does that mean rational functions can be equal to ${\sqrt {x}}$... I feel like I've made some mistake here?
2. Ah it seems I've recalled the article on rational functions incorrectly. I didn't know it said "a rational function is any function which can be written as the ratio of two polynomial functions" and thought it said "a rational function is any function written as the ratio of two polynomial functions", my bad. I do however think that some parts of the introduction I proposed should be included. The "where the denominator $Q_{n}(x)$ is not equal to zero" (I realise it appears in the definition section but I think it belongs in the introduction) and the part about "every integer/real number can be represented as a rational function" --BiT (talk) 00:55, 8 August 2009 (UTC)

As for point one, the numerator and denominator still have to be polynomials in x, which ${\sqrt {x}}$ clearly is not. It's just that the coefficients of these polynomials may be any number (for the meaning of "number" we are currently interested in). So ${\frac {{\sqrt {2}}x^{2}-\pi x+e}{{\sqrt[{7}]{\frac {18}{13}}}x-e^{-\pi ^{2}}}}$, although involving a lot of irrational numbers, is a perfectly rational function (over the real numbers). A constant irrational, like f(x) = e, is just a special case of this. —JAO • T • C 08:42, 8 August 2009 (UTC)

Ah of course, I didn't think of irrational numbers being the coefficients. Back in my head I thought the coefficients had to be integers for some reason. Thanks, I'm learning a lot of new interesting things here. --BiT (talk) 10:17, 8 August 2009 (UTC)

I'm glad that you have clarified this point, but I'm not quite sure that other people will have trouble with the same point as you. Apart the case of truly ambiguous sentences, everybody meets his/her personal point to clarify, and the reason why a certain point is obscure is linked to the reader's personal story (I'm thinking to my personal experience too of course). In most cases this psychological reason is something really doomed to remain obscure, and not so important after all. A rational function is a quotient of two polynomials: why isn't this clear enough? I fear that if you make explicit all possible clarifications the resulting very long explanation will appear going-to-nowhere, and not clear at all. (By the way: what's the need of subscripts m and n ? That's obscure to me). Last remark: if one wants to specify the nature of the coefficients, one says, for instance: "a rational function with coefficients in the field k". Most often, one refers to real or complex coefficients, so in particular, as Meni Rosenfeld pointed out, any constant (rational or irrational) is a rational function; moreover of course P(x)/0 is not a rational function, because it is not even a quotient: it's devil-knows-what). --pma (talk) 00:58, 8 August 2009 (UTC) PS: But maybe the case of polynomials and constants may be added in the subsequent section of examples, that sounds nice.

Fair enough, I just somehow found something wrong with $5$ being a rational function, so I decided to go here for confirmation. "A rational function is a quotient of two polynomials: why isn't this clear enough?"- because the integer $5$ is not a quotient of two polynomials, it's not a quotient at all.

As for the subscripts, my calculus book sometimes used these subscripts to denote the power of the polynomial function. I actually find it handy to use, is it wrong? --BiT (talk) 10:17, 8 August 2009 (UTC)

It's important to distinguish between an object (e.g., a number) and its representation(s). The digit "5" does not represent the number 5 as a ratio. However, the number 5 itself is a ratio - it is 5/1. Similarly, using the symbol "5" to represent the function $f:\mathbb {R} \to \mathbb {R} ,x\mapsto 5$ is not a representation of the function as a quotient of polynomials - but the function is the quotient of the polynomials 5 and 1.

This confusion is also manifested in you thinking that "a rational function is any function written as the ratio of two polynomial functions" could make sense literally. Writing a function differently does not change the function itself. The properties of a function can depend on how it is possible to write it, but not on the specific way you choose to write it.

Regarding subscripts - I wouldn't call it wrong, but it can be unnecessary and confusing when the degree of the polynomial is not of key importance. -- Meni Rosenfeld (talk) 18:29, 8 August 2009 (UTC)

Matrix function

I was reading this article, and began to wonder: is there a type of function in use that describes a matrix based on every entry's position within the matrix itself? For example, consider a matrix in the form

$\mathbf {A} ={\begin{bmatrix}a_{1,1}&\cdots &a_{1,n}\\\vdots &\ddots &\vdots \\a_{m,1}&\cdots &a_{m,n}\end{bmatrix}},$

and a function g of real integers m,n such that

$g:a_{m,n}\mapsto m^{n}\,\!$

and such that we can call say, for example, that for $2\leq m\leq 5$ and $2\leq n\leq 5$,

$g(a_{m,n})={\begin{bmatrix}4&8&16&32\\9&27&81&243\\16&64&256&1,024\\25&125&625&3,125\end{bmatrix}}.$

Do such functions as g have a name, are they used, and which article should I read? —Anonymous Dissident^Talk 08:51, 6 August 2009 (UTC)

$g(a_{m,n})\,$ does not depend on $a\,$ so you would write $g_{m,n}=m^{n}\,\!$ Bo Jacoby (talk) 09:00, 6 August 2009 (UTC)

I had a feeling about that. I was initially going to use g(m,n), but opted with g(a_m,n). Never thought of g_m,n. —Anonymous Dissident^Talk 10:11, 6 August 2009 (UTC)

Let me see if I understand the question properly. You have an m × n matrix $A:=(a_{i,j})\,$ where $1\leq i\leq m$ and $1\leq j\leq n$. You want a function f so that $f(a_{i,j}):=i^{j}?\,$ We could record the behaviour of f by a function $F:{\mbox{Mat}}_{m,n}(\mathbb {R} )\to \mathbb {Z} ^{m\times n}$, where

$F(A)=(1^{1},\ldots ,1^{n},2^{1},\ldots ,2^{n},\ldots ,m^{1},\ldots ,m^{n})\ .$

Well, if this is the case then it sounds like F would be a pretty dull function. The infinite dimensional space of all m × n matrices would get sent to a single point. For example, consider the space of 2 × 2 matrices with real entries (denoted by ${\mbox{Mat}}_{2,2}(\mathbb {R} )$). We have

$F\left({\mbox{Mat}}_{2,2}(\mathbb {R} )\right)=(1,1,2,4)\sim \left({\begin{array}{cc}1&1\\2&4\end{array}}\right)$. ~~ Dr Dec (Talk) ~~ 16:51, 9 August 2009 (UTC)

I think you made a mistake when you said that the space of all m × n matrices is infinite dimensional. It has dimension mn.--Shahab (talk) 08:05, 10 August 2009 (UTC)

Easy contour integral

Hey all, I did this contour integral and a friend did it and we got answers that differ by a sign. I can't find mistakes in either! So, here is my solution and hopefully you can tell me if I made a dumb mistake that I can't see.

Find $\int _{\gamma }z^{1/2}dz$ where $\gamma$ is the lower half of the unit circle from +1 to -1.

So, I figure I can parametrize this as $\gamma (t)=e^{it}$ as t ranges from 2pi to pi. Then

$\int _{\gamma }z^{1/2}\,dz=\int _{2\pi }^{\pi }e^{it/2}ie^{it}\,dt=-i\int _{\pi }^{2\pi }e^{i3t/2}\,dt={\frac {-2i}{3i}}\left[e^{i3t/2}\right]_{\pi }^{2\pi }=-{\frac {2}{3}}(e^{i3\pi }-e^{i3\pi /2})=-{\frac {2}{3}}(-1+i)={\frac {2}{3}}(1-i)$

My friend used $\gamma (t)=e^{-it}$ as t ranges from 0 to pi. This gives

$\int _{\gamma }z^{1/2}\,dz=\int _{0}^{\pi }e^{-it/2}(-i)e^{-it}\,dt=-i\int _{0}^{\pi }e^{-i3t/2}\,dt={\frac {2i}{3i}}\left[e^{-i3t/2}\right]_{0}^{\pi }={\frac {2}{3}}(e^{-3i\pi /2}-e^{0})={\frac {2}{3}}(i-1)$

Thanks for any help. StatisticsMan (talk) 14:13, 6 August 2009 (UTC)

I started working out, and then realized something: surely $z^{1/2}$ has TWO solutions for non-zero $z$? Couldn't both answers be right? --Leon (talk) 14:41, 6 August 2009 (UTC)

Leon's right. If we account for the two square roots in your parameterization $z=e^{i(t+2\pi n)}$ for $n=0,-1$. Your calculation just picks up a factor of $e^{i\pi n}$ which toggles between your friend's and your own solution. Martlet1215 (talk) 15:22, 6 August 2009 (UTC)

[ec] In other words, the "mistake" is either $\left(e^{it}\right)^{1/2}=e^{it/2}$ or $\left(e^{-it}\right)^{1/2}=e^{-it/2}$. Which one is the real one depends on the specific branch of $z^{1/2}$ intended. -- Meni Rosenfeld (talk) 15:27, 6 August 2009 (UTC)

What does [ec] mean? I have seen it before and didn't know. As to this specific problem, I get what you are saying as far as the two branches and two possible answers but not exactly how to write it up. Are you saying I should use the parametrization Martlet gave and do the integral? Then my "answer" would be two different values, depending on n = 0, -1? And, this is just part (b) of a question. Part (a) is the same function over the upper half of the unit circle from 1 to -1. So, it would also have 2 answers right? Thanks for all your help! I should have titled this "tricky" contour integral. StatisticsMan (talk) 16:04, 6 August 2009 (UTC)

ec=edit conflict --78.13.139.160 (talk) 16:11, 6 August 2009 (UTC)

In this technical sense. — Emil J. 16:32, 6 August 2009 (UTC)

Indeed, and I have mentioned it primarily to clarify that I left my comment as is, without checking its synergy with Martlet's. -- Meni Rosenfeld (talk) 16:46, 6 August 2009 (UTC)

Yep, it does mean that those are the two solutions, and that both should be included - unless the question specifies one and which one is needed. It's pretty usual to have more than one answer to these kind of things though, so don't worry! --Leon (talk) 16:29, 6 August 2009 (UTC)

I disagree. Depending on the branch, there are infinitely many possible solutions to this integral. The question is pretty meaningless without any clarification of the branch (or the assumption of the standard branch). -- Meni Rosenfeld (talk) 16:43, 6 August 2009 (UTC)

For any given z, there are only two values of z^1/2. It follows easily that there are only two choices of the branch which are continuous along the whole integration path, giving the two solutions computed above. — Emil J. 16:51, 6 August 2009 (UTC)

To the best of my knowledge, there is no problem with a (moderately) discontinuous integrand. -- Meni Rosenfeld (talk) 17:01, 6 August 2009 (UTC)

OK, in principle there is no problem even with bestially discontinuous integrands, in this case it is sufficient (and necessary) that the set of points which take a particular one of the continuous branches is measurable. However, in the given context (path integrals in complex analysis) it is tacitly assumed that we are computing integrals of holomorphic functions along continuous paths; there are only two choices how to construe the given example in such way. — Emil J. 17:15, 6 August 2009 (UTC)

You are not bound to use any specific parameterization. You only need to make sure which branch the question is asking about, and find out how this branch applies to the values in your contour. If nothing else is specified, the intended branch is probably $(re^{it})^{1/2}={\sqrt {r}}\ e^{it/2}$ for $t\in (-\pi ,\pi ]$. For this branch, your friend's solution is correct, because $\left(e^{-it}\right)^{1/2}=e^{-it/2}$ for $t\in (0,\pi )$, while $\left(e^{it}\right)^{1/2}=-e^{it/2}$ for $t\in (\pi ,2\pi )$. -- Meni Rosenfeld (talk) 16:39, 6 August 2009 (UTC)

Okay. Thanks. Here no parametrization is given. After all this, my guess is perhaps the whole point of this problem is to recognize there are two branches and get both answers. Otherwise, it's a very easy problem and it is from an old qualifying exam. StatisticsMan (talk) 16:47, 6 August 2009 (UTC)

Again, if you do not require the integrand to be continuous along the path, there are many more than two branches and two answers.

The parameterization is something you choose yourself. What should be given is a branch. -- Meni Rosenfeld (talk) 17:05, 6 August 2009 (UTC)

Maybe a good answer would include the fact that the square root function is multivalued and that an answer including an integer, say n, giving all solutions would be the correct solution. All other attempts are only partial solutions. For example, consider the equation $\sin(x)=0$. What is the solution? Well, we might ask over which interval x lies (this is like the branch problem above), or we could just say

$x\in \{\pi n:n\in \mathbb {Z} \}\ .$

Then there's no doubt: we have every solution! ~~ Dr Dec (Talk) ~~ 17:06, 9 August 2009 (UTC)

Another integral

$f(z)=e^{z+1/z}$. I already calculated the residue at 0 to be $\sum _{n=0}^{\infty }{\frac {1}{n!(n+1)!}}$. Now, I need to calculate the integral around the unit circle with counterclockwise orientation. Of course, I use the residue theorem. It "seems" obvious that 0 is the only singularity but how can I know this for sure? Thanks StatisticsMan (talk) 14:36, 6 August 2009 (UTC)

I did this stuff a long time ago, but if $y$ belongs to the complex numbers, so does $e^{y}$. That's because $e^{y}$ is an Entire function. A simple calculus exercise can demonstrate that everywhere else derivatives exist. --Leon (talk) 14:47, 6 August 2009 (UTC)

Oh yea, good point. For some reason I was thinking of showing the Laurent series has no negative terms. But, it's pretty clear it's differentiable everywhere but 0. Thanks. StatisticsMan (talk) 14:59, 6 August 2009 (UTC)

In other words: the composition of two holomorphic functions is itself holomorphic. Applying this principle in a systematic way is usually enough to find singularities of compound functions like this f. — Emil J. 15:10, 6 August 2009 (UTC)

Thanks! Another thing I knew but didn't think about. Your mentioning it helps my understanding. StatisticsMan (talk) 15:56, 6 August 2009 (UTC)

Assuming you're working over the complex plane, and not the the Riemann sphere, then the function $f(z)=e^{z+1/z}$ has a pole when its value is infinite. Well,

$z+{\frac {1}{z}}={\frac {z^{2}+1}{z}}$

and this is finite for all z ≠ 0. The exponential function is finite for all finite values, so f doesn't have a pole for all z ≠ 0. It's as simple as that! No need to differentiate.

ON A CAUTIONARY NOTE: WE AREN'T HERE TO DO PEOPLE'S HOMEWORK ~~ Dr Dec (Talk) ~~ 17:16, 9 August 2009 (UTC)

This isn't homework. Nor are any of my other recent questions. I am not in any classes. And, both questions of mine that you responded to were questions where I had already done most of the work and I was just asking for a bit of help. Even if these were homework questions, these are types that are allowable here.StatisticsMan (talk) 13:54, 10 August 2009 (UTC)

The math behind general relativity

Hi all:

I have recently begun studying Einstein's general theory of relativity on my own. But man oh man did I not quite expect the math that I was plunging myself into! To me all that math seems to be really bad abuse of all the calculus notations I studied in college. And I ended up not really understanding anything in the book other than the grand conclusions, which I already know from reading relevant entries in Wikipedia.

I studied engineering while in college, so I have only studied the basic higher mathematics, such as single-variate calculus, linear algebra 101 and some discrete math. I did take some system control courses in my junior and senior years that delved into higher order differential equations, and I also studied some multi-variate calculus on my own. As for physics, I have studied freshman's year of mechanics and electromagnetics (didn't get to Maxwell's equations despite prof's promise), and in my sophomore year I picked up some waves, simple harmonic motion, and some quantum theory and special relativity (basically repeated high school physics with calculus thrown in) all rolled up into one second-year course.

My question is, given my background above, what should I do in order to ease the pain with which I read my general relativity textbook? I am thinking of brushing up on my math, I saw many new terms introduced in the general relativity book such as p-forms, manifolds, exterior derivatives, Koszul connection, tensors, Hodge's star operator, wedge operator, etc. I am wondering if there is any one or two math textbook that ties all of these concepts together? Or is there some other better way of "opening up" my mind to general relativity?

On a more curious note, did Einstein himself actually knew all this math and used them when he penned his general theory of relativity back in 1915? (I am asking this because many results in my textbook seems to be derived from papers published in the 1960's, well after Einstein had passed away).

Thank you so much for your help!

L33th4x0r (talk) 22:14, 6 August 2009 (UTC)

There are some GR textbooks out there that don't assume much maths and teach you all the Riemannian geometry and tensor calculus you need. If yours doesn't, find a different one - I suggest one aimed at physicists rather than mathematicians (as a mathematician, I don't have one I can recommend). Einstein would have known quite a lot of maths, but he only came up with the basics of GR (I don't mean to belittle the achievement - getting started is always the hardest bit), a lot of other physicists and mathematicians have added to the theory since then (Einstein added to it after 1915 as well). The more recent results will either be easier ways of doing things that Einstein did the hard way or results used in more recent bits of GR. --Tango (talk) 23:17, 6 August 2009 (UTC)

If you are looking for a physicsy approach you could try Relativity: Special, General, and Cosmological by Wolfgang Rindler. It eases you in through Special Relativity introducing some of the mathematics you'll need for the General stuff later. Also, from a physicist's perspective, most of the terms you mentioned aren't taught in a first, or second course in the subject (at least where I'm from). I think you can safely forget p-forms and wedge operators for a while. Martlet1215 (talk) 00:08, 7 August 2009 (UTC)

General relativity is geometry. I you want to enter the necessary maths, I would suggest you Barret O' Neill's book: Semi-Riemannian Geometry with Applications to Relativity. Of course it is well possible that there exists a textbook on GR, with no or little maths: physicists are able to make it possible such unphysical, paradoxical things... pma (talk) 13:27, 7 August 2009 (UTC)

Thank you all for your help! I feel more confident now. L33th4x0r (talk) 17:16, 7 August 2009 (UTC)

Resolved

L33th4x0r, Although the matter has been resolved, I still wish to give you more advise. I have encountered a similar problem before. I suggest you to read something more understandable such as this [3], especially the long course online.

The Successor of Physics 13:38, 11 August 2009 (UTC)

August 7

Inferior goods and recessions

Is it normal for the price of inferior goods to rise during a recession? In May of this year, I noticed that since last summer the price of house-brand macaroni-and-cheese dinner had risen from $0.49 to $0.69 per box, and the price of the cheapest beers at the local beer store had gone from $24 to $25.60 per 24-pack. I doubt inflation alone is responsible. (All figures in Canadian dollars.) --NeonMerlin 05:50, 7 August 2009 (UTC)

Food prices have increased in the past few years because of a run-up in the prices of the underlying commodities. See for example 2007–2008 world food price crisis. Among other things the food vs. fuel debate has intensified because of the run-up. The recession is probably a separate issue. 70.90.174.101 (talk) 05:57, 7 August 2009 (UTC)

I agree that it may just be general rising food prices. If it's just the inferior good rising, though, then it could be the recession. As people's income reduces they switch to cheaper alternatives, so the demand for inferior goods increases which will increase the price. It may well be a combination of the two. Of course, if your sample if just two items then it could just be complete coincidence. --Tango (talk) 19:21, 7 August 2009 (UTC)

Giffen good may be relevant, though this seems to be that effect in reverse. AndrewWTaylor (talk) 20:28, 8 August 2009 (UTC)

Line -> Triangle -> Pyramid -> ?

I was thinking today about how the centre of mass is 1/2 the way up a line, 1/3 the way up a triangle, 1/4 the way up a pyramid, and i was trying to describe the corresponding 4-D object it would be 1/5 the way up of. I can visualise it: clearly it would be any 4-dimensional shape with a cross-sectional volume which is uniformly shrinking in the 4th dimension. My friend said a 'pentachoron', but it's my understanding that this is more specific a term than I desire in that it restricts the number of vertices (sort of how tetrahedron is to triangle). What is the object I'm describing called? —Preceding unsigned comment added by 82.35.84.214 (talk) 18:32, 7 August 2009 (UTC)

Pentachoron is the word you want. You need to specify the number of vertices otherwise you haven't specified the shape. A line has 2 vertices, a triangle has 3, a tetrahedron (which is a type of pyramid and the one you want) has 4 and a pentachoron has 5. A general term for all those kind of shapes is a simplex. --Tango (talk) 18:59, 7 August 2009 (UTC)

But the centre of mass of any pyramid (or cone, to use a more common term) is 1/4 of the way from the base to the apex, whatever manner of pyramid/cone it may be. Algebraist 19:02, 7 August 2009 (UTC)

Is it? Ok. Then the desired term is a "4 dimensional cone" or a "cone over a 3 dimensional shape", I don't know of a shorter word (I guess "4-cone" would work). --Tango (talk) 19:17, 7 August 2009 (UTC)

We have a hypercone article, which is concerned with hypercones of spherical base volume (just like Cone (geometry) is mostly concerned with cones of circular base area), but it also hints that the word (just like the word "cone") has a wider meaning too. —JAO • T • C 19:19, 7 August 2009 (UTC)

The "hyper-" prefix is ambiguous - it sometimes means one dimension higher than usual and it sometimes means any number of dimensions higher (and sometimes is used as a general term for any number of dimensions). --Tango (talk) 19:24, 7 August 2009 (UTC)

August 8

Alternative proof that a square matrix is similar to its transpose

A square matrix (over C) is similar to its transpose. This can be shown by a Jordan form. I know it's actually true for any field. For this, I suppose you use Smith normal form. Question: Is there any other way to prove this? (that doesn't use the structure results) Maybe more elementary or conceptually simpler way? More importantly, is there any similar fact or formulation that leads to a generalization of this to a (linear) operator acting on an infinite-dimensional space (e.g., Hilbert space)? -- Taku (talk) 02:03, 8 August 2009 (UTC)

Not quite true in a Hilbert space (without further assumptions). Consider the right-shift operator on l₂(N) taking e_n in e_n+1: its transpose is the left-shift operator, and they are obviously not similar, for the former is injective and the latter is not. --pma (talk) 10:02, 8 August 2009 (UTC)

Definitely not in the fullest generality, because the index of a Fredholm operator (between the same space) doesn't vanish in general. But this actually gives a clue: somehow the result above holds because of the simplicity of geometry, and a Jordan form doesn't capture geometry. I'm actually quite positive that the answer is affirmative. It follows from that A is similar to A^T, that, for example, $\sigma (A)=\sigma (A^{T})$. But the proof of this relies on a Jordan form? which doesn't seem right. (Yes, you can use determinant, but that's basically a weaker result.) -- Taku (talk) 12:06, 8 August 2009 (UTC)

Absolute Values with Factorials

As you know, absolute values are identified conventionally with the formula $|x|={\sqrt {x^{2}}}$. But the problem is that although conventional square roots can only be positive, the true square root can be negative too. Therefore, I set of to find another formula of absolute values. I found this formula $|x|=x^{2}-x\Pi ^{-1}({\frac {\Pi (x)}{x}})$ . Can anyone help me confirm it? Thanks!
The Successor of Physics 04:32, 8 August 2009 (UTC)

The "square root of the square" formula is pretty silly - it only works because the square root symbol is defined to give the positive square root. Why do you want a formula for it? What is wrong with $|x|={\begin{cases}x\,\mathrm {if} \,x\geq 0\\-x\,\mathrm {if} \,x<0\end{cases}}$? --Tango (talk) 04:46, 8 August 2009 (UTC)

The "square root of the square" formula does reference how the Euclidean norm generalize to larger spaces. But I do agree that Tango's formula is more straight forward if we're just dealing with the reals. I also have no idea what Π is in that equation up there. Rckrone (talk) 06:31, 8 August 2009 (UTC)

I'll make it clearer! Thanks!

The Successor of Physics 13:19, 8 August 2009 (UTC)

Oh I see, the Pi function as related to the Gamma function. Pretty sure that formula does not work. First of all Π isn't defined everywhere on the real line, second of all it's not injective on the real line, so Π^-1 is not well defined. Even ignoring those problems the formula would just comes out to x, even for negative x. You can show by Euler's reflection formula that zΓ(z) = Γ(z+1) everywhere that Γ is defined, not just for for z with Re(z) > 0. Rckrone (talk) 07:55, 8 August 2009 (UTC)

But I tested it for a few values and it worked!

The Successor of Physics 13:19, 8 August 2009 (UTC)

1) For x where Π is defined, Π(x)/x = Π(x-1). If we just pretend that Π is injective then Π^-1(Π(x-1)) = x-1 unambiguously. I assume that's the value you want. So the expression on the right is just x² - x(x-1) = x. There could be other values of Π^-1, but you would need Π(x-1) = Π(x+1) for all x < 0 and that is clearly not true in general since Π(x+1)/Π(x-1) = x(x+1).

2) Assume there is a function f on the reals such that |x| = x² - xf^-1(f(x)/x) for all x. Then f^-1(f(x)/x) = x-1 for x > 0 and f^-1(f(x)/x) = x+1 for x < 0. So for all 0 < x < 1, f(x)/x = f(x-1) and f(x-1)/(x-1) = f(x). This implies f(x) = f(x-1) = 0. Then f^-1(f(x)/x) = f^-1(0) = x-1 for all 0 < x < 1 which is a contradiction. No such function f can exist. Rckrone (talk) 17:13, 8 August 2009 (UTC)

Unless you know exactly what you're doing, "I tested it for a few values and it worked" doesn't usually cut it in mathematics. Try applying this approach to the claim "$x^{2}+x+41$ is prime for any integer x" and see what you get.

You should read the introduction of Square root to clear up your confusion on the matter (at least with regards to terminology). Absolute value is obviously also relevant. -- Meni Rosenfeld (talk) 18:12, 8 August 2009 (UTC)

The formula $|x|={\sqrt {x^{2}}}$ (where x is real) is correct, concise, and compatible with some generalizations of real numbers (including complex numbers, where the radicand involves the real and imaginary parts). There is no need to replace it with a different formula, and even if there was, your suggestion is wrong on many levels. -- Meni Rosenfeld (talk) 18:42, 8 August 2009 (UTC)

May I ask what you're doing? In the complex numbers, $|x|={\sqrt {x^{*}x}}$ is what you want. In other systems, maybe something different. Some structures don't really have "absolute values" defined per se. Perhaps an understanding of your needs would help us in gauging how to respond. --Leon (talk) 18:53, 8 August 2009 (UTC)

Set of points with density 1

For a Lebesgue measurable set of real numbers A, let B={X:X is a real number & d(X)=1}, where d(X) is the density of the set A at X. Note that B may not be a subset of A. What property characterizes B? My intuition tells me it should always be an open set, and that B could be any open set of real numbers by choosing the right example of A (eg. B itself); am I right or wrong?

And I have the same question for higher-dimensional Lebesgue measures. --COVIZAPIBETEFOKY (talk) 22:31, 8 August 2009 (UTC)

How about the density at 0 of [-1, 1] with the sets [1/n, 1/n + 1/n³] removed? Rckrone (talk) 23:14, 8 August 2009 (UTC)

That would be a counter example, wouldn't it? Ok, so it doesn't seem like there is a very succinct characterization of B, but now that I think about it, that wasn't really what I needed for my purposes, so it doesn't matter. Just one more question, on which I'm pretty sure I'm correct, but I would like confirmation:

Am I correct in believing that every Lebesgue measurable set is a null set different from exactly one open set of real numbers? A is a null set different from B iff (A-B)U(B-A) is a null set. --COVIZAPIBETEFOKY (talk) 15:25, 9 August 2009 (UTC)

No, you are not. E.g. R itself is a null set different from R\F for any finite F. Algebraist 15:56, 9 August 2009 (UTC)

Right.

Okay, there has to be a name for what I'm looking for, as you have pointed out that merely "open" is not sufficient. I want the set to be open, and no isolated points not in the set.

Let me redefine my B as {X:X is a real number & there is a (sufficiently small) positive real number r such that μ([X-r, X+r] ∩ A) = 2r}.

Now, is there a simple name for the property that characterizes B, completely? --COVIZAPIBETEFOKY (talk) 16:14, 9 August 2009 (UTC)

I believe this is the difference between the two:

openness: X is in the set implies that it has a neighborhood in the set.

my property: X is in the set iff it has a neighborhood in the set. --COVIZAPIBETEFOKY (talk) 16:42, 9 August 2009 (UTC)

No, never mind, that doesn't quite work. Would "... iff it has a neighborhood in the set, excluding itself" be sufficient? —Preceding unsigned comment added by COVIZAPIBETEFOKY (talk • contribs) 16:45, 9 August 2009 (UTC)

You can say that every measurable set differs from exactly one set like B by measure zero. In other words, you can set up equivalence classes of sets that have a symmetric difference of measure zero, and then every equivalence class has a unique element B such that B is exactly the set of points that have Lebesgue density 1 with respect to B. It seems like there should be some unifying property of the sets like B, but if there is I have no idea what it is. The property you mention doesn't work, since that would imply that every set like B is open, which is not the case. It's definitely an interesting question. Rckrone (talk) 17:36, 9 August 2009 (UTC)

You seem to forget that I redefined B. It's now defined so that it will always be open. I made the redefinition bold above. --COVIZAPIBETEFOKY (talk) 00:07, 10 August 2009 (UTC)

August 9

Constructing Triangles

Can we construct a triangle whose perimeter is given and two of its angles are given. If yes, please give the steps of construction and its justification too. --$\xi ri\pi i\vartheta \alpha \xi$ 04:15, 9 August 2009 (UTC)

Such a triangle always exists, yes. What do you mean by "construct", though? Do you just want to know the side lengths, or do you have something more like Compass and straightedge constructions in mind? Algebraist 04:50, 9 August 2009 (UTC)

Yes, I have compass and straightedge constructions in mind. --ξΓιΠιγαξ 04:58, 9 August 2009 (UTC)

Not only can one say that it exists (provided of course the sum of the two given angles is not too big and the perimeter is positive) but also that it is unique. But I think if you want a compass-and-straightedge construction, that might depend on what the angles are. Michael Hardy (talk) 05:36, 9 August 2009 (UTC)

Assuming you have a way to construct the angles, you can make a similar triangle to the one you want, and find the perimeter of that. By using the ratio of the perimeter you want and the perimeter of your similar triangle, you can find one of your side lengths from the side length of the similar triangle. Rckrone (talk) 05:43, 9 August 2009 (UTC)

Law of sines is relevent here.83.100.250.79 (talk) 12:42, 9 August 2009 (UTC)

You'll have two big issues with your compass and straightedge constructions. First is that it's a compass and not a protractor. It is exceedingly difficult to construct arbitrary angles. Small, whole number fractions of pi radians are doable, but trying to construct an angle like 0.923876 radians is tedious at best. Secondly, it's a straightedge, not a ruler. Compass and straightedge constructions are usually taken to be scale invariant. That is, the internal scaling is self-consistent, but the absolute size is arbitrary. You'd need to have some sort of external scale bar to reference off of for the size of your perimeter. That said, one way to do the construction is to draw a baseline, construct the two angles facing each other, sharing the common baseline. Extend the sides until they meet at the third point of the triangle. Next, you'll have to to scale the triangle appropriately. How you do that will probably depend on how you're measuring the perimeter, and what you're comparing it to. Note that scaling bit can be avoided if you pre-compute the length of the base, and construct the angles that distance apart. -- 76.201.158.47 (talk) 17:44, 9 August 2009 (UTC)

There are different things that can be meant by saying that a length or angle is "given." It's true that there's no way to establish an overall objective scale with compass and straight edge, so if you "have a length" in the sense that you have a number measurement there's no way to produce that. But you can also "have a length" in the sense there are other elements in the construction that need to have the same length. If you're drawing a triangle in a vacuum, the scale is arbitrary, but if your triangle is part of a larger construction it is meaningful to require for instance that the perimeter be the same as the diameter of some circle you've drawn. A similar story holds for angles. If you "have an angle" in the sense that you have a measurement like π/5 radians, it may be difficult or impossible to construct the angle on paper, but if you "have an angle" in the sense that the angle has already been constructed somewhere, then its easy enough to copy it somewhere else. Presumably if you've already constructed it, then it's an angle that can be constructed, but having knowledge of how to construct it isn't necessary in order to copy it. In that sense, if you are given a paper with two angles and a length drawn on it, whether they be part of another compass and straightedge construction, or just elements floating on the page, you can use compass and straightedge construction to build a triangle with those angles and that perimeter. If you are given a paper with "26°, 75°, 14cm" written on it, the length can't be found without a ruler and the angles may be problematic. Rckrone (talk) 19:32, 9 August 2009 (UTC)

I'd have thought it goes without saying that in a straightedge-and-compass problem, saying that the length is given means that there's a segment of that length and you can set the center at one endpoint and take the other endpoint to be on the circumference, so you can draw that circle. Michael Hardy (talk) 09:10, 10 August 2009 (UTC)

the 24 cells of the 24-cell and their intersections

Hello, I studied 24-cell but I am still not sure about this. As I understand [4], the 24 vertices can be explicitly described in a fourdimensional affine space over the reals. If I'm not mistaken, the 24 cells correspond with the 24 affine three-dimensional subspaces, with equations$+/-x_{i}+/-x_{j}=1$. That seems to make sense: each of the 24 vertices described in that article are one exactly 6 such 3-spaces, and every such 3-space contains exactly 6 of those vertices. But I have a problem with the intersection. It seems that if I fix one of these affine 3-spaces, there are exactly 8 others meeting it in 3 vertices, 6 meeting it in 1 vertex, and nine in no vertices at all. This surprises me as I had expected, based on more combinatorial texts, that there would be 8 meeting it in a plane (3 points), 6 meeting it in a line (2 points), 8 meeting it in a point and finally one meeting it in no points at all.

I must have a mistake somewhere, but I found it hard to visualize such a four-dimensional structure. I would be very happy if someone could tell be a bit mor about this.

Many thanks, Evilbu (talk) 13:17, 9 August 2009 (UTC)

I don't think it would be possible to have 6 other cells meeting a given cell only in a line and still have the whole thing regular. An octahedron has 12 edges, so the number would have to be a multiple of 12. The angle at each edge of a regular octahedron is >90°, which means there has to be fewer than 4 cells around each edge for it to be convex, so there must be exactly 3. That would mean that no two cells share only an edge. 3 cells around each edge requires 6 cells around each vertex: 4 that share a face and 1 that shares only a vertex with a given cell. Since an octahedron has 6 vertices, that makes 6 other cells that share only a vertex. Rckrone (talk) 16:46, 9 August 2009 (UTC)

Thanks for your answer, it seems to confirm correctness of my approach with coordinates. In any case, I still expect that, "from the point of view of one fixed cell", there are five types of cells, of size 1,8,6,8,1. So that means that there would be two "kinds" of disjointness (parallel and not parallel). When I said that I was also thinking of more combinatorial texts, like: [5] where metasymplectic spaces are considered. One axiom is: "(M1) the intersection of distinct symplecta is empty, a point, a line, or a plane." My intuition made me very convinced that this 24-cell is an example of such a metasymplectic space, with the 24 points, 96 lines or edges, 96 planes or faces, and finally 24 "symplectons" or cells. (Note that the name 24-cell does not appear in that pdf, but the term "thin F4 geometry" does, and F4 is precisely the "Coxeter group" given in the article 24-cell). So somehow I must have a wrong intuition or something like that? Many thanks,Evilbu (talk) 17:22, 9 August 2009 (UTC)

Functional square root

What is the functional square root of the exponential function $e^{x}$ and how is it derived? Thanks.--12.48.220.130 (talk) 20:05, 9 August 2009 (UTC)

Interesting stuff here and here, for instance. —JAO • T • C 20:14, 9 August 2009 (UTC)

I typed 'Functional square root' into the search box at the top of the page in wikipedia and guess what? it came up with Functional square root rather than a list of possibilities. Amazing thing the search facility. If wikipedia fails (perish the thought) one can always try google. Dmcq (talk) 11:44, 10 August 2009 (UTC)

By the way, there is no unique functional square root, it can be built up of chunks. Also if you try turning it into a power series and assume it is going to work everywhere then the derivatives diverge. Not pretty. e^x-1 is a little nicer to work with if you want to give it a go yourself but still nothing very clean. Dmcq (talk) 11:52, 10 August 2009 (UTC)

Finding a Permutation Matrix

Suppose we have two nxn Adjacency matrixes A, and B and we want to find a Permutation matrix P such that

$A=PBP^{T}$, we also know that since P's Eigenvectors are Orthonormal

$P^{T}=P^{-1}$

We also are given that at least 1 such P exists but not that P is necessarily unique though it might be. In the article on Permutation matrix, there is a section that describes a method of doing this, but as noted the process is apparently flawed.

So my question is what would be a correct method? Also, what makes P unique? A math-wiki (talk) 22:45, 9 August 2009 (UTC)

This is just the graph isomorphism problem. For practical problems, nauty is a very good solution. This is somewhat of a FAQ. You can see the most recent incarnation at Wikipedia:Reference desk/Archives/Mathematics/2008 November 2#Equivalence_of_0-1_matrices. JackSchmidt (talk) 00:13, 10 August 2009 (UTC)

TeX help

Looking for TeX code for integral sign having a horizontal dash midway (Dixmier trace related symbol). This would be similar to $\oint$ but with a horizontal line instead of a closed line. Tried a stike (∫) but prefer better. Ref: Noncommutative integral. If TeX doesn't have such, then question: can custom symbols be created in TeX? How?, please point. Thank you. Henry Delforn (talk) 07:54, 10 August 2009 (UTC)

The MnSymbol library might have what you need. Check here and look on page 29. Otherwise search that PDF for "integral" and you might find some others. Maelin (Talk | Contribs) 12:04, 10 August 2009 (UTC)

Time Simulation in Matlab

Hi,

I want to produce a plot of two parametrised functions in real time, so i can see the path of the particle being drawn out as time increases; in Matlab. I'm currently studying fluid flows so i want to be able to see the path of a particle given the x and y equations in time. My tutor produces simulations with Maple quite easily but apparently it is possible to produce them with Matlab and since I have a copy of Matlab i'd love to know how.

I've done a bit of searching and have come across Simulink models, but they seem very complicated for what i want to be doing (I have a novice level of understanding on Matlab, I can write scripts with ease but models are beyond me).

Thanks for any help.

Pete 124.184.72.1 (talk) 10:45, 10 August 2009 (UTC)