Wikipedia:Reference desk/Archives/Mathematics/2016 May 23

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May 23

Lindstrom's Theorem and Extensions of FO

Is there an extension of first order logic that doesn't have Lowenheim-Skolem, but is compact (and all other such)? In general, is there any work on logics between FO and SO that have nice enough metalogical properties that they aren't "broken" when it comes to actually working with them, even if they still aren't perfect? I realize this is an absurdly vague and general question, but I've always wondered about why SO seems to be the next step talked about after FO, it feels (if that word makes sense) like there should be a whole world of neat systems between them. I haven't really touched this stuff in a few years -- other projects -- so I apologize if this is poorly laid out, or a bit of nonsense (it's also really late here, so a bit sleepy headed); in any event, thanks for any direction:-) Phoenixia1177 (talk) 09:52, 23 May 2016 (UTC)

Have you seen our articles SO_(complexity) and FO_(complexity)? They talk a bit about restrictions and closures. SemanticMantis (talk) 17:35, 23 May 2016 (UTC)

You might be interested in the Barwise compactness theorem for a certain sort of infinitary logic. Not sure if it's exactly what you're looking for, but it seems at the very least close enough that you might be interested. --Trovatore (talk) 07:20, 25 May 2016 (UTC)

Interchanging between surface area and an integral

Is it true that the n-dimensional integral $\int _{||x||>1}{\frac {1}{||x||^{n+1}}}dx$ is equal to the surface area of the (n-1)-dimensional unit sphere? What formula tackles this?--אדי פ' (talk) 19:19, 23 May 2016 (UTC)

Yes. Integrate in spherical coordinates, using the well-known identity $\int _{1}^{\infty }dr/r^{2}=1$. Sławomir
Biały 19:56, 23 May 2016 (UTC)

(edit conflict)You're talking $\int _{||x||>1}{\frac {1}{||x||^{n+1}}}dx$ right? Note that $\nabla \cdot \sum _{i=1}^{n}{\frac {-x_{i}\mathbf {e} _{i}}{||x||^{n+1}}}=\sum _{i=1}^{n}{\frac {x_{i}^{2}(n+1)}{||x||^{n+3}}}-{\frac {1}{||x||^{n+1}}}={\frac {1}{||x||^{n+1}}}$. Note that a unit "surface normal" to every n-dimensional ball is ${\frac {1}{||x||}}\sum _{i=1}^{n}x_{i}\mathbf {e} _{i}$; taking the dot product of this with the vector field I just constructed yields $-{\frac {1}{||x||^{n}}}$. I am not sure if this is correct, so everything from here on should be taken with caution.

Now choose another ball of radius $r>1$. The n-dimensional divergence theorem gives $\int _{1<||x||<r}{\frac {1}{||x||^{n+1}}}dx=\int _{||x||=1}{\frac {1}{||x||^{n}}}dS-\int _{||x||=r}{\frac {1}{||x||^{n}}}dS=S_{1}-{\frac {S_{r}}{r^{n}}}$ where $S_{1}$ and $S_{r}$ are the (n-1)-dimensional surface areas of the unit ball and ball of radius r respectively. Since the n - 1 dimensional surface area should be proportional to the power $r^{n-1}$, the second term vanishes as $r\to \infty$.

Someone has to check me on this. I do notice that it is correct for n = 2 and n = 3, though.--Jasper Deng (talk) 20:57, 23 May 2016 (UTC)

I did realize, however, that what I just computed shows in one way how the divergence theorem generalizes the fundamental theorem of calculus.--Jasper Deng (talk) 21:17, 23 May 2016 (UTC)

Yep the curl discussion above is a bit about that, the fundamental theorem is the scalar version of Stokes' theorem. Dmcq (talk) 21:44, 23 May 2016 (UTC)

@Dmcq: I was even attempting to apply Stokes' theorem directly (to the region between the two spheres) by noticing that ${\frac {1}{||x||^{n+1}}}dx_{1}\wedge ...\wedge dx_{n}$ is the exterior derivative of the (n-1)-form $\sum _{i=1}^{n}{\frac {(-1)^{i}x_{i}}{||x||^{n+1}}}\wedge _{j=1,j\neq i}^{n}dx_{j}$, but I'm currently having trouble making sense of the integral of the latter over each of the sphere in terms of how it relates to my above calculations.--Jasper Deng (talk) 22:08, 23 May 2016 (UTC)

Well what you're doing there is use Cartesian coordinates instead of spherical ones and all those pluses and minuses are used in calculating the Jacobian like Sławomir Biały mentions below. One doesn't gain in calculation. What makes things simpler is a coordinate system like you had with most of the coordinates tangent to the surface of the sphere and a normal and that makes things straightforward in terms of surface elements which expand slightly as one goes out and slightly sloping sides which don't have any overall contribution, and that's what you did. Dmcq (talk) 15:16, 24 May 2016 (UTC)

The divergence theorem is a nice approach. Another (equivalent) approach is to use the formula $A(r)=V'(r)$ expressing the surface area of a sphere of radius r as the derivative with respect to r of the volume of the ball of radius r. We make the following change of variables in the volume integral:

$\mathbf {y} ={\frac {r^{\alpha }\mathbf {x} }{\|\mathbf {x} \|^{1+\alpha }}}$

where $\alpha >0$ is a free parameter. This transforms the interior of the ball of radius r to the exterior of the ball of radius 1. The Jacobian is ${\frac {\alpha r^{n\alpha }}{(\mathbf {x} \cdot \mathbf {x} )^{n(\alpha +1)/2}}}.$ So, we have the identity

$V(r)=\int _{\|\mathbf {x} \|\geq 1}{\frac {\alpha r^{n\alpha }}{(\mathbf {x} \cdot \mathbf {x} )^{n(\alpha +1)/2}}}d\mathbf {x} .$

The case in the OP has $\alpha =1/n$. (There appears to be an extra factor here, but maybe I've made a mistake in calculating the Jacobian, but I do not have time at the moment to find the error.) Sławomir Biały (talk) 12:24, 24 May 2016 (UTC)