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May 24

Computability of classical mathematical/geometric theorems

Are classical mathematical/geometric theorems and demonstrations (theorically) computabile in the sense of Turing? It is (theorically) conceivable a computer program able to demonstrate the Pythagorean theorem or other theorems? Thanks in advance (I'm layman, I don't have a mathematical background).--Carnby (talk) 16:15, 24 May 2016 (UTC)[reply]

There are various programs that prove theorems. See e.g. Automated theorem proving. —Kusma (t·c) 16:53, 24 May 2016 (UTC)[reply]

More specifically, see Automated reasoning. — crh ₂₃ (Talk) 17:00, 24 May 2016 (UTC)[reply]

Thanks a lot. Are there simple programs for the layman or do they all require a deep knowledge of logic and maths?--Carnby (talk) 12:34, 5 June 2016 (UTC)[reply]

Definite summation of rational functions

Is there a way to express sums of the form $\sum _{x=0}^{u}{\frac {P(x)}{Q(x)}}$ using digamma functions? For example,

$\sum _{x=0}^{u}{\frac {1}{b+cx}}={\frac {1}{c}}\left(\psi \left({\frac {b}{c}}+u+1\right)-\psi \left({\frac {b}{c}}\right)\right)$

How can this be generalized for $P$ and $Q$ of arbitrary degree? 24.255.17.182 (talk) 20:03, 24 May 2016 (UTC)[reply]

Sure, just use partial fractions. --JBL (talk) 22:42, 24 May 2016 (UTC)[reply]

Unless you're saying that every rational function can be written as a sum of the quotients of degree 0 and degree 1 polynomials that doesn't really help. 24.255.17.182 (talk) 22:56, 24 May 2016 (UTC)[reply]

If Deg(P) < Deg(Q) and Q has no repeated factors then partial fractions does this. If Deg(P) ≥ Deg(Q) then get a quotient and remainder to reduce it to the first case. You'd have to use Faulhaber's formula for the sum of the quotient part. If there are repeated factors then partial fractions also gives you terms of the form d/(bx+c)^k, in which case you'd have to add in the corresponding multigamma function. --RDBury (talk) 05:20, 25 May 2016 (UTC)[reply]

(... which is also essentially just the derivative of the digamma.) --JBL (talk) 13:17, 25 May 2016 (UTC)[reply]

Generalization of circular points at infinity

In the geometry of a complex projective plane, the circular points at infinity seem to be all that is needed to define which curves are circles. Correct me if I've got this wrong, but it seems to me that given a complex projective plane, selecting any two distinct points defines a line (which we could choose to be the line at infinity, and all quadrics that intersect those two points may be defined as circles.

It strikes me that in special relativity, the light cone defines a shape at infinity (two points in the plane, a circle in three dimensions and so on). However, once one has chosen a hypersurface at infinity, only a limited number of points in it need to be chosen to define a quadric in it to act as the light cone. What simple/obvious/standard generalization of the circular points at infinity applies to more than two dimensions? —Quondum 20:45, 24 May 2016 (UTC)[reply]

I'm not quite sure I understand the question, but here's what I got. You want to consider the moduli space of all SO(n)-structures on an n-dimensional real vector space. This is the symmetric space SL(n)/SO(n) (I've used SL rather than GL, because I believe the overall scaling is irrelevant). For example, when n=2, this is the upper half plane (the "circular points at infinity"). Coordinates can be given explicitly in terms of the Iwasawa decomposition on SL(n).

My count of real degrees of freedom is

(n+2)(n-1)/2=-1+n(n+1)/2

. (The

-1

here accounts for the missing scaling.) Sławomir Biały (talk) 21:41, 24 May 2016 (UTC)[reply]

I think you're thinking way too 20th/21st century for this question; the circular points at infinity are kind of a 19th century way of thinking. Not that there's anything wrong with that, the 19th century had some good ways of thinking about things that are sometimes forgotten nowadays. At any rate, to get the circular points at infinity you a take the homogeneous equation for a circle, x² + y² = z², say and intersect that with the line at infinity z=0 to get the homogeneous equation for a pair of points x² + y² = 0. In projective coordinates these are (1, ±i, 0), the two circular points at infinity. But really it's the equation of a conic defined on the projective line. When you follow the same recipe for a sphere you get the equation x² + y² + z² = 0, which is the equation of a conic curve in the plane at infinity, w=0. So the 3-dimensional analog of the points at infinity is the spherical conic at infinity. Any sphere in space will contain this conic just as any circle the plane contains the circular points at infinity. Similarly, in four dimensions you get a conic surface in the hyperplane at infinity. When you start talking about light cones you basically need to change a variable to t and replace its coefficient of i with a real c. In two dimensions (one space and one time) this gives you two real points at infinity (1, ±c, 0) and the conics which pass through these are hyperbolas of the form (x-a)² - c²(t-b)² = k². To get an actual light cone (i.e. a pair of lines) you have to set k=0. In three dimensions (two space and one time) the conic at infinity is x²+y²-c²t² which in some sense is a real circle (only at infinity). The surfaces passing through these are hyperboloids and the light cones are the ones with a singularity. In four dimensions you get x²+y²+z²-c²t²=0 which is in some sense a sphere. The surfaces passing through these are hyper-hyperboloids and again the light cones are the ones with a singularity. In all cases the light cone is formed as the union of all lines connecting the event point and points on the conic at infinity. --RDBury (talk) 04:07, 25 May 2016 (UTC)[reply]

Well, whatever. I think it is interesting regardless of the century. It would be nice to get some clarity about the question, though. Sławomir Biały (talk) 13:23, 25 May 2016 (UTC)[reply]

RDBury's exposition perfectly captures how I look at or understand it. Viewing the circular points as the intersection of a hyperplane and a quadric gives a near-perfect generalization.

Sławomir's response leaves me with some points I'd need to clarify before I can properly interpret it. In my example, the projective plane presumably has a group of motions PGL(C,3), which we restrict by choosing two points of the space on which it acts to stabilize. Since our starting point is a two-dimensional space, I would put n=2, and denote the group PGL(n+1). When regarded as acting on the vector space Cⁿ⁺¹, there are two scalings of interest: that of the vector space (which is irrelevant, since it disappears in action on the quotient space Cⁿ⁺¹/C^×), and another freedom of scaling of the geometry determined by the restriction ("size" is meaningless: any circle still transforms into any other circle; the group is larger than O(n+1,C) by one complex degree of freedom), but we can remove this by a second constraint, which is preservation of oriented n-volume to get SO(n+1,C) – or so my rough reasoning goes.

A little bit of background to where I'm coming from: I'm trying to get a feel for how "fundamental" the different constraints on the PGL group are for the geometry of special relativity. An orthogonal group can be defined as the motions that preserve a quadratic form on the (homogeneous) vector space of dimension n+1, but this does not serve to constrain orientation (the 'S' in 'SO'). Preserving the geometric n-volume as well as the light cone seemed pretty good.

Basically I'm trying to find and aesthetically "clean" (non-redundant, neat) way of defining SO groups. I arrived at this line of questioning by noting that the independent constraints of preserving the light cone and oriented n-volume seem to do so. But it seems that defining a light cone is clumsy; it also does not cover the Euclidean case. —Quondum 16:21, 25 May 2016 (UTC)[reply]

The specific group one is interested in is a very important question (or so it seems to me). It seems like you want to understand the real orthogonal groups, rather than the complex ones? I feel that complex numbers have been brought in a little prematurely and should be left out of the picture until we are sure they are appropriate. The fact that they appear in low dimensions so obviously, it seems to me, is an accidental isomorphism between the groups SO(2) and U(1). The appearance of complex numbers in higher dimensions should be somewhat deeper and involve spinors; for instance, that the spin group of SO(3) is SU(2), so that vectors in three dimensions are associated with two dimensional complex spinors. Also, if we leave the discussion of complex numbers out until later, there are a lot of groups that are "like" the PGL groups. These are the pseudo-unitary groups U(p,q). Over the reals these groups have isomorphic Dynkin diagrams, but distinct root systems characterized by their Satake diagrams. Sławomir Biały (talk) 23:01, 27 May 2016 (UTC)[reply]