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May 7

Pipe joints

Hi.

Probably some textbook questions, but... (Context is in the generation of 3D meshs for a computer graphics application)

(i)(a)For 2 regular prisms A and B in Euclidean, where Prism A and Prisim B's centre line lie in the same plane, but where Prism A and B are at some angle to each other, Is there a generalised algorithm for computing the "vertices of intersection"(?),

(b) Having obtained the 'vertices of intersection', is there an algorithm to clip the prisms.

(ii) There is a transformation that unrolls the clipped prisms to a flat sheet? If so what is it.

Sfan00 IMG (talk) 13:15, 7 May 2014 (UTC)[reply]

Tangents to circles

At the section here, it gives a method for determining the equation of a line that is tangent to two circles, and says that there are "two solutions". However, there are (generally speaking) four lines tangent to two circles, so I don't get it. Is that section wrong, or am I misunderstanding something? 86.169.185.1 (talk) 13:49, 7 May 2014 (UTC)[reply]

They get two solutions first. If you keep reading, the section you linked eventually says "In this way all four solutions are obtained." I didn't check every step of the derivations, but I don't think there are mathematical errors in that section. Basically, this scheme gets all four lines by using two different ways to produce sets of two solutions. There is probably a way to get all four solutions falling out of one master equation, but that would be more complicated, harder to follow, and in my opinion, not worth the effort in the context of this article. SemanticMantis (talk) 14:07, 7 May 2014 (UTC)[reply]

Oh, OK, thanks. I thought the part "Geometrically this corresponds to ..." was the start of an explanation of how to visualise why it works. I wonder if there is some way to present the whole solution at once, explaining which combinations of signs produce which tangent lines. I think this would be easier to follow. 86.169.185.1 (talk) 17:11, 7 May 2014 (UTC)[reply]

There is probably a way to get all four solutions falling out of one master equation, but that would be more complicated, harder to follow, and in my opinion, not worth the effort in the context of this article. ---- Is it not true that all four solutions come from the following equations?

ax₁ + by₁ + c = ±r₁

ax₂ + by₂ + c = ±r₂

I think that explaining it all at once in this way is actually simpler than the current presentation, provided that the issues of duplicate solutions and which solutions correspond to which tangent lines can be handled without too much fuss. 86.179.5.164 (talk) 23:24, 9 May 2014 (UTC)[reply]

are any axioms true in a physical sense in our universe (2)?

First, thanks OP of the previous thread. I have been meaning to ask this for a long time. I create a new thread because I want to specialize the question a bit. The previous thread is now rather long too, risking to go stale for that reason only.

A line of thoughts: Suppose in some physical theory T, particle X exists (in the theory) if and only if mathematical Theorem A (of, say, point set topology) is true. Suppose now Theorem A has implications on the continuum problem. There are certainly such theorems in existence. Modern theoretical physics certainly involves topology.

Admittedly, chances are slim that the premises are fulfilled, but anyway: Suppose now that particle X is discovered experimentally...

You take it from here. YohanN7 (talk) 14:02, 7 May 2014 (UTC)[reply]

Can you clarify: in your example, theorem A has no known proof? If that's what you mean, then I don't think physical existence of a particle can say much about the mathematical necessity of theorem A in a formal system. My reasoning is that, though the particle X in the theory and the particle found in nature are linked by analogy and human models: they are actually quite distinguishable objects. SemanticMantis (talk) 14:11, 7 May 2014 (UTC)[reply]

Or, going the other way, theorem A has a known proof, and you might be suggesting that physical evidence of particle X now somehow entails a "physical truth" of the formal axioms? Sorry, I still don't think so. But philosophy of mathematics is a fun topic here, as long as we don't get too fighty ;) SemanticMantis (talk) 14:18, 7 May 2014 (UTC)[reply]

Ah, yes, I was unclear. Let's say, for definiteness, either of

A is true only if CH is true/false.
If CH is true, then A is true/false.
A is true if and only if CH is true/false.

Feel free to replace A, CH and the implications as you wish, but please leave theory T's, the particle X's and the experiment E's rôles unchanged. YohanN7 (talk) 14:21, 7 May 2014 (UTC)[reply]

For more definiteness, assume ZF or ZFC. YohanN7 (talk) 14:23, 7 May 2014 (UTC)[reply]

Ok, I think I'll stick with my first response then. SemanticMantis (talk) 14:33, 7 May 2014 (UTC)[reply]

I tend to agree. But, I'd also say that a positive experiment E will have significant weight as a plausibility argument (in the indicated direction, whatever it might be (I numbered the cases above, and allowed for ¬CH)), much stronger than any existing purely philosophical, and perhaps even stronger than some mathematical plausibility arguments (e.g. axiom of symmetry which I don't personally pay much attention to, (and even disagree with)). YohanN7 (talk) 15:08, 7 May 2014 (UTC)[reply]

Using the axiom of choice is extremely ambitious in this context. While real numbers and continuity make intuitive sense in the real world, there could be any number of equivalent mathematical representations of reality that might not require infinite sets, for all we know. This is aside from the obvious point that T is merely a mathematical model of the world from which our observations are drawn, and observation of a particle X may simply invalidate T as a physical theory. I think, too, that mathematics can have various structures depending on choices along the way: it is constructed, and could have many shapes. We might find that when it is built differently (e.g. using a basis other than set theory), something like the axiom of choice might not even fit. —Quondum 16:57, 7 May 2014 (UTC)[reply]

O yes, sure. Mathematics can have many structures.

But is there one that is right in some very very very real (physical reality) Platonic sense? I do not really get your coupling with the axiom of choice. Sure, spacetime may be finite or infinite (modeled as a point set). For purposes of discussion, let us assume that spacetime is a continuum. That makes the discussion easier and more interesting to the original question. YohanN7 (talk) 19:07, 7 May 2014 (UTC)[reply]

There are so many different ways to model reality. Even within existing mathematics, a smooth function on a spacetime continuum occupies a very small space out of all possible spaces. With this and boundary conditions, the total number of possible configuration of the universe could well be finite. So whether spacetime is considered to be a continuum might be irrelevant to the physical reality. I've sometimes wondered whether there isn't a dual universe composed of the same reality, with atom-like structures etc., when looked at in the energy–momentum domain. —Quondum 19:40, 7 May 2014 (UTC)[reply]

With a spacetime continuum and smooth functions representing states, you are unlikely to find a finite number of configurations. You might get (compactness) a countable number of configurations. But this is straying away from the OP anyhow. YohanN7 (talk) 21:32, 7 May 2014 (UTC)[reply]

Yes, I think there is something to be said for plausibility, or "evidence," suggestive power, etc. But then we start mixing up epistemology of math and science quite a bit. I suppose it's not so terrible if we end up someplace like this: claim A may well be true, but unprovable within a given formal system. Because we have some faith in the fidelity of a certain model, an empirical result might give us a high degree of belief that claim A is true in the formal system, and perhaps even that the analogous claim A' is true for the physical concrete real world. SemanticMantis (talk) 17:57, 7 May 2014 (UTC)[reply]

Right. I believe that the deep unanswered questions (at the axiomatic level) will remain unanswered. All we will get are plausibility arguments, wherever they come from. YohanN7 (talk) 19:07, 7 May 2014 (UTC)[reply]

There is nothing wrong with an argument that goes "Supposing that theory T is a faithful model of physical system S ...". Once we have a compelling ToE that is mathematically consistent and is reconciled with observations at all scales, this kind of premise will have ring of plausibility to it. We should not confuse the uncertainty of specific observations with statements about the possible exactness of a model; for example, people (though not all) argue quite strongly for superposition in quantum mechanics being exact. Then this kind of question starts getting quite real. We just have to be very careful that we do not make unwitting assumptions in our math: it seems that we are far more likely to make mathematical mistakes than physical reality is to deviate from a faithful model – at least if we make the assumption that reality is "real", and that we are not brains in vats. —Quondum 19:40, 7 May 2014 (UTC)[reply]

Einstein argued all the way that the true equations of nature are nonlinear. But this is also straying away from the OP. YohanN7 (talk) 21:32, 7 May 2014 (UTC)[reply]

Not that there is anything wrong with straying away from the OP. I just think that these questions are conceptually unrelated to the OP. YohanN7 (talk) 21:32, 7 May 2014 (UTC)[reply]

Solving problem with a matrix inverse

What's a good way to solve a problem of this form?

$\left\Vert (A+cI)^{-1}b\right\Vert ^{2}={\frac {1}{2}}$

A is a matrix, b is a vector, and I'm solving for c, a positive scalar. Am I stuck with iterative methods at this point, or is there a neat trick I'm missing? Thanks. Katie R (talk) 17:47, 7 May 2014 (UTC)[reply]

I should add that

A^{-1}

is already known at this point. Sherman–Morrison formula looks promising, so I guess I'll look into that for now. Katie R (talk) 18:12, 7 May 2014 (UTC)[reply]

Is anything special known about the matrix A? Is it antisymmetric by chance? Sławomir Biały (talk) 19:17, 7 May 2014 (UTC)[reply]

I honestly haven't inspected the matrix too closely yet. The elements are generated by:

a_{n,i}=\sum _{j=0}^{N}\sum _{k=0}^{N}\sum _{l=0}^{N}b_{k}b_{l}d_{i-j-k}d_{i-n-l}

Note that d can take negative subscripts. Elements of b and d will range from -1 to 1, but that's about the only constraint on them. Katie R (talk) 19:32, 7 May 2014 (UTC)[reply]

Also b and d will never be all-zero. Katie R (talk) 19:40, 7 May 2014 (UTC)[reply]

Buried in this formula is a (discrete) convolution, which suggests that some Fourier methods might be a means of attack. Are you sure the indices are correct? (It looks like you can pull off the

\sum _{l}b_{l}d_{i-n-l}

term entirely, but I feel like you mean for this to involve j, so that it participates in the other summations.) Sławomir Biały (talk) 19:48, 7 May 2014 (UTC)[reply]

Yes, they are correct, looking at my notes that term was originally seperate, combining it that way was literally the last thing I did to the expression and I have no idea what I was thinking. I have to head home for the day, but thinking of it in the frequency domain is an interesting idea. Hopefully it will get me off to a good start tomorrow. Katie R (talk) 20:08, 7 May 2014 (UTC)[reply]

Looks like I had bunched them together when I pulled a different term out that doesn't show up here, so at least I know why I did it... The matrix can be represented as

a_{n,i}=f_{i-n}g_{i}

, so it has a nice structure.

g_{i}=\sum _{j}f_{i-j}

, but I don't know how much that helps. I'm getting pulled out of algorithm design and into meetings and hardware stuff for a while, but hopefully later today I'll get some time to work out if that structure gets me anything. Katie R (talk) 14:22, 8 May 2014 (UTC)[reply]

1+2+3+4+5+... vs. 1+3+5+...

According to that branch of mathematics which allows summing non-convergent infinite series, we have:

1-1+1-1+...=1/2

1-2+3-4+...=1/4

1+2+3+4+...=-1/12 (note the minus sign)

Let us recasr 1+2+3+4+... as (1+3+5+...)+(2+4+6+...), or (1+3+5+...)+2(1+2+3+...)

First question: Is this recasting valid?

Let us denote 1+2+3+4+... as S, and 1+3+5+... as T.

From above, 1+2+3+4+...=(1+3+5+...)+2(1+2+3+...)=T+2S

Thus S=T+2S, and solving for T, T=-S=+1/12 (the plus sign is for emphasis)

That is,

S=1+2+3+4+5+...=-1/12 7 T=1+ 3+ 5+...=+1/12

Second question: What is going on here? Bh12 (talk) 20:52, 7 May 2014 (UTC)[reply]

It is not true that the sum of the divergent series 1+2+3+... is −1/12. This series diverges properly to +∞. Instead of 1+2+3+..., you actually mean the special value of the Riemann zeta function

\zeta (-1)

, which is equal to −1/12. The "series" 1+3+5+... is understood as the Dirichlet L-function

L(\chi _{2},-1)

where

\chi _{2}

is the unique Dirichlet character mod 2. There are standard functional equations that relate these special functions under very general conditions. But actually, it is possible to do this by hand within the region of convergence of the Dirichlet series and invoke the persistence of functional relations to show that it continues to hold outside. In general, for Re(s)>1, we have

\zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}},\qquad L(\chi _{2},s)=\sum _{n=1}^{\infty }{\frac {1}{(2n-1)^{s}}}.

So, provided

Re(s)>1

, we have

\zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=L(\chi _{2},s)+2^{-s}\zeta (s).

By persistence of functional relations, this holds outside the region of convergence (the Diriclet L-function is known to extend uniquely to a meromorphic entire function). In particular, at

s=-1

, we recover your conclusion. Sławomir Biały (talk) 22:21, 7 May 2014 (UTC)[reply]

Sławomir, diverting the topic slightly, the articles that you link you seem to use a complex exponentiation without defining it specifically. This seems to be very sensitive to the precise definition of exponentiation used. I presume that specifically that the unique real logarithm of the base must be used in the definition of this exponentiation? Would these functions (the ones you linked to) be at all interesting if another branch of this logarithm were to be used? —Quondum 23:08, 7 May 2014 (UTC)[reply]

Changing the branch of the logarithm would multiply all L functions by

e^{-2\pi ims}

for some integer m. This corresponds to a translation by m in the frequency domain, which is surely significant for some reason since the Fourier transform of an L function is basically a theta function (which is a modular form). Sławomir Biały (talk) 10:56, 8 May 2014 (UTC)[reply]

Thanks. This seems to give all sorts of possibilities (even different branches for different terms), but exploring the implications will require a pretty deep knowledge of the area. It's just that I hadn't even thought of the possibility of the ambiguity before. —Quondum 13:33, 8 May 2014 (UTC)[reply]

You seem pretty stupid to me!

1-1+1-1+... = (1-1)+(1-1)+... = 0+0+... = 0

1-2+3-4+... = (1-2)+(3-4)+... = (-1)+(-1)+... and that (when seen as infinite series) converges against -∞.

1+2+3+4+... (when seen as infinite series) converges against +∞. Thomas Limberg (Schmogrow) (talk) 07:40, 8 May 2014 (UTC)[reply]

FYI Thomas, the associative law does not necessarily hold when the number of terms is infinite, so your first two results are not correct (first equality sign in each is incorrect). YohanN7 (talk) 09:03, 8 May 2014 (UTC)[reply]

The correct conclusion, when viewed as infinite series, the first two simply diverge. You can also say the are not convergent. YohanN7 (talk) 09:41, 8 May 2014 (UTC)[reply]

Thomas, please don't call other people stupid because they don't understand infinite sums of divergent series; especially as your view isn't the accepted one either. AlexTiefling (talk) 10:38, 8 May 2014 (UTC)[reply]

The OP's interpretation (though poorly documented in his post) is also fully supported by Sławomir Biały, who extremely rarely makes a mistake (in which case it is just temporary). YohanN7 (talk) 12:51, 8 May 2014 (UTC)[reply]

The first two of your series are actually a lot less controversial than the last. These are both Abel summable, respectively, to 1/2 and 1/4. No zeta functional regularization is needed. In fact, these are values of the Dirichlet eta function

\eta (s)=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n^{s}}}

which is actually Abel summable for all complex s (and converges to an entire function). There is no need to analytically continue. Sławomir Biały (talk) 12:58, 8 May 2014 (UTC)[reply]

One obviously has to be pretty careful even about relations between these series (let's say under analytic continuity, which seems to be the most powerful for this purpose). For example, on could try to infer (with a single-term shift)

1+1+1... = (1+2+3...) − (1+2+3...) = 0

1+0+0... = (1+1+1...) − (1+1+1...) = 1

Since all of these can presumably be related to L functions and in none of them do I think we've hit a pole but might be wrong, I would have hoped not to have this kind of thing happening. The 1+1+1... series seems to be the "worst" one here. Is there some condition that one must satisfy to relate a specific series to an L function to avoid this problem? Or is the approach of combining infinite series like this (especially with a one-term shift) the problem? —Quondum 13:33, 8 May 2014 (UTC)[reply]

Yes, one does need to be careful about such things. The series 1+1+1+... actually does correspond to a pole of the Riemann zeta function, so these computations are problematic for that reason. Obviously not any old divergent series is going to correspond to some nice L-function (these have some special "automorphic" properties), so these methods have their limitations. There are two very different approaches to summing divergent series in general. One that is popular in spectral theory is that in order to sum the series

\sum \lambda _{n}

, you first look at the associated zeta function

\sum _{n}\lambda _{n}^{-s}

. With appropriate asymptotics on the

\lambda _{n}

(I believe details are in Shubin, "Pseudodifferential operators and spectral theory"), the zeta function analytically continues to a meromorphic entire function. This is sometimes called "zeta function regularization", and is a popular method for assigning a notion of "determinant" to operators on a Hilbert space. Another procedure, which is popular among physicists, I believe is called "anomaly cancellation". The basic idea is to take a Gaussian regularization of the series, and asymptotically expand in some parameter. However the rules for doing this seem somewhat ad hoc to me, although they may seem more natural to a physicist. It is my understanding (at least based on discussions with Tom Branson, who was the real expert on this stuff) that the two techniques are ultimately equivalent, essentially because at some level both are concerned with analytic continuation, and analytic continuation is unique. Sławomir Biały (talk) 19:41, 8 May 2014 (UTC)[reply]

It seems to me that 1+1+1... corresponds to

ζ(0) = -1/2

, with the only pole at

s = 1

. But never mind, this whole topic has a sense of "deeper magic" about it and clearly has valuable application – had I discovered this approach to handling of apparent infinities through analytic continuation during my early college days, I may well have become addicted and thence studied pure maths. The leap from a series to a meromorphic function is where I perceive a degree of ad hoc-ness, since meromorphic functions are generally well-behaved. I wish I had more time to devote to the pleasure of exploring these topics. —Quondum 04:28, 9 May 2014 (UTC)[reply]

Yes, you're right. I was thinking of the harmonic series, which corresponds to the pole at s=1. Sławomir Biały (talk) 11:44, 9 May 2014 (UTC)[reply]