August 8

In a private forum on polytopes, someone whose English is weak (better than my Spanish but not as good as my French) has recently become aware of four-dimensional figures but is unclear on the concept. I'd like to tell him this in Spanish:

You cannot build a polychoron (es:Polícoro) in the real world. But, just as you can make a flat image of a polyhedron on paper, a polychoron can be reduced to three dimensions by the same principles. Some people prefer to consider a series of three-dimensional slices of the four-dimensional object. (I had someone else translate this for me.)

Volunteers? —Tamfang 18:00, 8 August 2007 (UTC)[reply]

Perhaps this would yield more results at the language desk? asyndeton 19:09, 8 August 2007 (UTC)[reply]

I reckon I'll have better luck looking for a math nerd who happens to know Spanish but doesn't read the Language desk than for a language nerd who knows Spanish and geometry but doesn't read the Math desk. I'll give it a bit more time. —Tamfang 04:53, 9 August 2007 (UTC)[reply]

No puedes construir un polícoro en el mundo real. Pero, tan como puedes constuir un imagen plana de un poliedro en papel, puedes reducir un polícoro a tres dimensiónes usando el mismo método. Algunos prefieren considerar un serie de rebanadas tridimensionales del objeto cuadridimensional. This isn't great but it should get the idea across.--Cronholm¹⁴⁴ 05:20, 9 August 2007 (UTC)[reply]

Muchas gracias. I changed puedes to si puede, being uncertain how I stand with this person and what the norms are for tu in Colombia. —Tamfang 16:15, 9 August 2007 (UTC)[reply]

Oops, should be se puede. My Italian (si può) is stronger than my Spanish. —Tamfang 17:18, 9 August 2007 (UTC)[reply]

You are right, it's best to stick with formal unless indicated otherwise. My mistake. If you need any more rough translations you can use my talkpage. Cheers--Cronholm¹⁴⁴ 20:01, 9 August 2007 (UTC)[reply]

Se puede is not formal so much as impersonal ("one can", "it is possible to"). If you wanted to address the listener personally but formally it would just be (Usted) puede. --Trovatore 20:08, 9 August 2007 (UTC)[reply]

You are right Trovatore, I should have said that the tu and vosotros forms aren't appropriate in many places in the Spanish speaking world, except among friends or when speaking to a younger person, etc. At least this is my understanding. --Cronholm¹⁴⁴ 20:23, 9 August 2007 (UTC)[reply]

Like Tamfang, I'm better at Italian than Spanish. In Italian there's a construction called the tu impersonale that directly corresponds to the English usage of "you" that doesn't really mean, you know, you. In Italian that usage is possible (though a little risky) even when talking to someone you would address as Lei, provided it's clear that the tu is being used impersonally rather than referring to the listener/reader. Is there such a thing in Spanish? I know this is a little OT for the math desk, but I'm curious. --Trovatore 22:03, 9 August 2007 (UTC)[reply]

Answering an unasked question: there is in Russian, and it works exactly as you outlined. Tesseran 07:19, 10 August 2007 (UTC)[reply]

Consider the system of differential equations:

${\displaystyle {\begin{bmatrix}\nu (s)\\\eta (s)\\\mu (s)\\x(s)\\y(s)\\\theta (s)\end{bmatrix))'={\begin{bmatrix}a_{11}(s)&a_{12}(s)&a_{13}(s)&a_{14}(s)&a_{15}(s)&a_{16}(s)\\a_{21}(s)&a_{22}(s)&a_{23}(s)&a_{24}(s)&a_{25}(s)&a_{26}(s)\\a_{31}(s)&a_{32}(s)&a_{33}(s)&a_{34}(s)&a_{35}(s)&a_{36}(s)\\a_{41}(s)&a_{42}(s)&a_{43}(s)&a_{44}(s)&a_{45}(s)&a_{46}(s)\\a_{51}(s)&a_{52}(s)&a_{53}(s)&a_{54}(s)&a_{55}(s)&a_{56}(s)\\a_{61}(s)&a_{62}(s)&a_{63}(s)&a_{64}(s)&a_{65}(s)&a_{66}(s)\end{bmatrix))*{\begin{bmatrix}\nu (s)\\\eta (s)\\\mu (s)\\x(s)\\y(s)\\\theta (s)\end{bmatrix))+{\begin{bmatrix}f_{1}(s)\\f_{2}(s)\\f_{3}(s)\\f_{4}(s)\\f_{5}(s)\\f_{6}(s)\end{bmatrix))}$

in other words:

v'(s) = A(s)*v(s) + f(s)

With six boundary conditions. The known functions a(i,j) and f(i) are known as arrays (say, 100 values for s = 0:0.01:1). Without actually numerically solving for the unknowns v(s), how is it possible to know whether this system have (a) a unique solution, (b) infinitely many solutions, (c) no solution. Thanks a ton! deeptrivia (talk) 22:49, 8 August 2007 (UTC)[reply]

I may be wrong, but I think that if the coefficients are continuous, then there is always a unique solution. -- Meni Rosenfeld (talk) 23:57, 8 August 2007 (UTC)[reply]

Comment: See this. It says: "For linear BVPs, where the ODEs and boundary conditions are both linear, the equation g(s, y(b;s))=0 is a linear system of algebraic equations. Hence, generally there will be none, one or an infinite number of solutions, analogously to the situation with systems of linear algebraic equations." deeptrivia (talk) 18:50, 9 August 2007 (UTC)[reply]

Perhaps I got the wrong impression about the nature of your boundary conditions. Can you describe what form they take? -- Meni Rosenfeld (talk) 19:02, 9 August 2007 (UTC)[reply]

The form of BCs is:

nu(1) = K, eta(1) = -K*theta(1), mu(1) = M, x (0) =0, y(0) = 0, theta(0) = 0.

Thanks, deeptrivia (talk) 19:43, 9 August 2007 (UTC)[reply]

Modified problem

In that case, let me pose this problem. Given:

v'(s) = f(v(s), c(s))

With the required number of boundary conditions. The boundary conditions could be simple, like v4(0) = 0, but could also involve nonlinear coupling, e.g. v2(1) = cos(v3(1)). f is a nonlinear function (involving products and trig functions), and c(s) are known. Again, without actually numerically solving for the unknowns v(s) repeatedly, how is it possible to know whether this system have (a) a unique solution, (b) multiple solutions, (c) no solution? Also, if there is a solution, is it 'stable' in the sense that a small change in parameters will produce only a small change in the solution v(s). Thanks, deeptrivia (talk) 01:14, 9 August 2007 (UTC)[reply]

For your original problem. If your conditions are given at one point and are of the form ${\displaystyle \nu (0)=a,\eta (0)=b,\mu (0)=c,x(0)=d,y(0)=e,\theta (0)=f}$, so you actually have an initial value problem then if the coefficients of the matrix are continuous you will have a solution which is unique, by the Picard-Lindelöf theorem. If you actually have a proper boundary value problem then you are in trouble. Consider ${\displaystyle v'(s)=u(s),u'(s)=-v(s)}$ so the equation for ${\displaystyle v}$ is ${\displaystyle v''(s)=-v(s)}$ with general solution ${\displaystyle v(s)=A\cos(s)+B\sin(s)}$. If your boundary conditions are ${\displaystyle v(0)=0,v(\pi )=0}$ then the former of these gives you that ${\displaystyle A=0}$ and the latter doesn't give you any more information, so there are infinitely many solutions, all functions of the form ${\displaystyle v(s)=B\sin(s)}$ are solutions. On the other hand, if you change the latter condition to ${\displaystyle v(\pi )=\epsilon }$ for any non-zero ${\displaystyle \epsilon }$ then you will have no solutions. Stefán 19:09, 9 August 2007 (UTC)[reply]

Yes, it is a proper BVP with three conditions on both ends. The question is, under what general conditions will it have infinite solutions? deeptrivia (talk) 19:47, 9 August 2007 (UTC)[reply]

Well, if you only know the matrix numerically, you are not going to be able to tell ${\displaystyle v'(s)=u(s),u'(s)=-v(s),v(0)=0,v(\pi )=0}$ apart from ${\displaystyle v'(s)=(1+\epsilon )u(s),u'(s)=-v(s),v(0)=0,v(\pi )=0}$ for ${\displaystyle \epsilon >0}$ very small. The former of these has infinitely many solutions but the latter none. Stefán 20:21, 9 August 2007 (UTC)[reply]

Assuming this doesn't happen (i.e., assume the numbers are known exactly, so that it's as good as knowing the function), then what conditions should A,f and the boundary conditions satisfy for unique, no, or infinite solutions? deeptrivia (talk) 20:58, 9 August 2007 (UTC)[reply]

What happens when you differentiate the Lambert W function? The article talks about integration but ignores differentiation. 202.168.50.40 22:55, 8 August 2007 (UTC)[reply]

IIRC, there is a rule that describes the derivative of inverse functions. Furthermore, you can probably sketch the graph of the derivative yourself by studying the graph of Lambert W.

I get ${\displaystyle W'(z)={\frac {W(z)}{z(1+W(z))))}$, but I'm welcome to be corrected. This is undefined at z = 0, but since you can find the derivative via the inverse function, it actually tells you a little about the limit of W(z)/z. Confusing Manifestation 23:11, 8 August 2007 (UTC)[reply]

Yes, this is correct, and it even appears already in the article, in the section "Differential equation"; The OP may have expected something more explicit, as an elementary function of z only, but alas, AFAIK this is impossible. -- Meni Rosenfeld (talk) 23:43, 8 August 2007 (UTC)[reply]