Wikipedia:Reference desk/Archives/Mathematics/2014 November 25

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November 25

A Great circle on 2-sphere

Given two points with defined latitude and longitude I want to know if there exist an analytical formula that can describe the great circle drawn through these two points. In particular if such a formula exists I will need to evaluate if a point lying on the 2-sphere and chosen at random is in fact located on the great circle drawn between these two points. Thanks --AboutFace 22 (talk) 01:47, 25 November 2014 (UTC)

The cross product of the two vectors will give you the normal vector of the great circle's plane. The dot product of the candidate vector with that normal vector, which (if I'm not mistaken) is the determinant of all three vectors, will be zero if they lie on the same great circle. —Tamfang (talk) 08:45, 25 November 2014 (UTC)

Also known as the scalar triple product. 129.234.186.11 (talk) 10:04, 25 November 2014 (UTC)

I'll add that if this is is for computational work that doesn't do symbolic manipulation, you won't get exact zeros, and it would likely make sense to accept results below some threshold, e.g. 0<D<\epsilon. SemanticMantis (talk) 15:17, 25 November 2014 (UTC)

Thank you everyone who contributed. Yes, it is for computational work. In particular I am considering a large letter H on a screen outside of 2-sphere, then I will project the letter onto it, the center of projection will be the center of the sphere, and then expand it into a functional basis, do the transforms. I will then need to find all the boundaries inside the sphere. It is not difficult to find projection of key points, corners, etc. I thought I will make my life easier if I can determine the linear boundaries of projection of the letter inside the sphere with a simple rule. I will need it for integration. <Not a home work, unfortunately :-) Many thanks, --AboutFace 22 (talk) 16:50, 25 November 2014 (UTC)

Function for ellipse area inscribed within the two unit square...

Consider the ellipses which are inscribed in a [-1,1]x[-1,1] square. Let A(y) be the function for the area of the ellipse which touches at (1,y) (and thus also at (y,1) and (-1,-y) and (-y,-1)) A(-1)=A(1)=0, A(0)=Pi. Does anyone have a closed form for A(y)?Naraht (talk) 02:45, 25 November 2014 (UTC)

Rotate coordinates 45 degrees and rescale, so your ellipse has equation $x^{2}/a^{2}+y^{2}/b^{2}=1$ and passes through the point (t, 1-t) with slope -1 at that point. So you need to solve the equations $t^{2}/a^{2}+(1-t)^{2}/b^{2}=1$ and $2t/a^{2}-2(1-t)/b^{2}=0$ simultaneously for a, b. Then the area is straightforward. --JBL (talk) 03:02, 25 November 2014 (UTC)

It might make the computations simpler to parameterize the ellipses by (x, y) = (cos t, cos(t + a)), so the ellipses touch the square at (1, cos a), (-1, -cos a), (cos a, 1), (-cos a. -1)) at t=0, π, -a, π, - a resp. Then apply area = 1/2 |∫(ydx-xdy)|, which works out to π|sin a|. --RDBury (talk) 12:06, 25 November 2014 (UTC)

(To be clear, the integral in the comment above is a line integral around the ellipse and it is justified by Green's theorem).--Jasper Deng (talk) 04:54, 26 November 2014 (UTC)

math

10% of 100 — Preceding unsigned comment added by 70.39.186.83 (talk) 2:08, 25 November 2014

Same as 100% of 10. --CiaPan (talk) 13:31, 25 November 2014 (UTC)

This related question, [1] explains how to calculate it, if needed.Phoenixia1177 (talk) 19:58, 25 November 2014 (UTC)

Even better link: [2] (a specific version rather than a diff, plus anchor/section link added). --CiaPan (talk) 21:59, 25 November 2014 (UTC)

Thank you, that was informative to see how it was done too:-)Phoenixia1177 (talk) 11:19, 26 November 2014 (UTC)

What is the significance of quaternions and octonions being normed division algebra

Of course, I understand why they're normed division algebras, but in as much as I've seen (quaternions) used that property doesn't seem to be exploited. I'm aware that it matters if a "strong" derivative is being defined, but it turns out that even with the quaternions defining a strong derivative is far to restrictive. Does the normed division property "help" in some sense?--Leon (talk) 19:11, 25 November 2014 (UTC)

The norm property has a certain intuitive appeal; if you think you should be able to associate quaternions, octonions etc. with geometric objects such as transformations then that they're nicely normed means you can associate a 'size' with them which is preserved by composition of transformations. And this is certainly true of complex numbers and quaternions, used to describe transformations in 2 and 3 dimensions respectively. More abstractly it's simply a way to classify the sequence generated by Cayley–Dickson construction. At each stage you lose something until with octonions you lose being nicely normed (and alternate) as you generate sedenions.--JohnBlackburne^words_deeds 22:57, 25 November 2014 (UTC)

There are useful algebras that have neither a norm nor that are division algebras. In particular, Clifford algebras form a family of associative algebras that are extremely useful, especially in the geometric context. They include the reals, complexes and quaternions, but not the octonions. They define division, but include nonzero zero divisors, so checks for division by zero divisors become a little more complicated. The norm may be replaced by something similar, which can produce a negative result, but as you say, one does not necessary look for a direct analogue. I'm not sure how derivatives tie in, though. —Quondum 23:15, 25 November 2014 (UTC)

Another theorem that follows directly from the nicely normed property on octonions is Degen's eight-square identity. Similar identities exist in four, two and trivially one dimensions but no other.--JohnBlackburne^words_deeds 23:44, 25 November 2014 (UTC)

The result restricting this to the given dimensions depends rather sensitively on the definition of "similar identities". In particular, an additional restriction (bilinearity) must be made (see Pfister's sixteen-square identity). —Quondum 23:15, 26 November 2014 (UTC)

Yes. Similar in the sense of naturally following. You have to do something different to get it to work in more dimensions. The articles have the details.--JohnBlackburne^words_deeds 00:26, 27 November 2014 (UTC)

Another version of the Hurwitz theorem, which makes no explicit mention of norms or sums of squares, is that these are the only dimensions in which a non-trivial triality exists. From that point of view, rational identities like the Pfister one, where not all variables are on equal footing, are automatically ruled out. Sławomir Biały (talk) 00:26, 27 November 2014 (UTC)

Okay, this approach is more interesting/neater, since it gives a minimal (very general) context for the result. In particular, it would suggest (assuming nontrivial triality) that it applies independently of the quadratic form (e.g. that it holds even if some squares are subtracted instead of added), as well as being independent of the underlying field. —Quondum 00:59, 27 November 2014 (UTC)

Yes, that's right. So I suppose this automatically also relaxes what it means to be a "norm", and also what it means to be a "division ring" since it includes various split versions. (Incidentally, it is possible to make sense of a triality in any symmetric monoidal category, and some version of this theorem is probably true even at that level.) But if we naively accept any quadratic form, then I think not all variables are on equal footing because the Dynkin-Satake diagram of SO(Q) no longer has enough symmetries. So I believe that (at least over most fields) this singles out the split cases (although there all of the associated "division rings" are split) and the compact (positive-definite) case. I guess then a related question could be "in what sense are the split octonions a 'normed division algebra'", since they are neither normed nor a division algebra in the usual sense. (Possibly there is some $\mathbb {Z} _{2}$ grading, in the style of the Brauer-Wall group, that sorts this out. Not sure.) Sławomir Biały (talk) 01:19, 27 November 2014 (UTC)

I was thinking in terms of omitting any reference to a quadratic form (and hence also to rotation groups): like vector space duality, the definition of triality should have most value in that it provides (I assume) a more abstract framework of multilinearity without the distraction of the structure of a quadratic form. With duality, despite isomorphism, only once one identifies the vector space with its dual does a quadratic form arise. I would expect that the three vector spaces in triality would be the same: one can refrain from identifying them. The article Triality unfortunately jumps without justification from isomorphism to identification: "... we find that the three vector spaces are all isomorphic to each other, and to their duals. Denoting this common vector space by V ...". It was this idea of a result entirely without needing the concepts of a quadratic form or an algebra that I found intriguing. —Quondum 16:33, 27 November 2014 (UTC)

Trialities need all three spaces to carry a quadratic form, at least in the conventional way of defining them. Whether that structure is "needed" really depends on what one is trying to accomplish, but if you start relaxing conditions, then obviously certain desirable properties will no longer hold. Sławomir Biały (talk) 16:50, 29 November 2014 (UTC)

Normalization of Associated Legendre Polynomials.

I have a few questions concerning orthonormality of Associated Legendre Polynomials (ALP). I want to stress the word orthonormality as opposed to simply orthogonality. The reason for that is computational. It is a well known fact that when ALP with large indices $l$ & $m$ are computed the functional values grow in magnitude to the point that the exponents overflow. Double precision is required and in some cases even quadruple precision is needed. The normalization diminishes the absolute values of the functions considerably but not universally. I want to make sure that I understand normalization correctly. Wikipedia article on ALP gives two formulas.

$\int _{-1}^{1}P_{k}^{m}(x)P_{l}^{m}(x)dx$ = ${\frac {2(l+m)!}{(2l+1)(l-m)!}}\delta _{k,l}$

Thus the normalization factor here will be:

$N_{1}={\sqrt {\frac {(2l+1)(l-m)!}{2(l+m)!}}}$

I call it normalization in respect to $l$

For my task it is more important to normalize in respect to $m$. It is given by this formula:

$\int _{-1}^{1}{\frac {P_{l}^{m}(x)P_{l}^{n}(x)}{(1-x^{2})}}dx={\frac {(l+m)!}{m(l-m)!}}\,\,\,(m=n\neq 0)$

The normalization factor for each subspace with a given $l$ but differing $m$ should be this:

$N_{2}={\sqrt {\frac {m(l-m)!}{(l+m)!}}}$

I call it normalization in respect to index $m$. I am uncertain about what to do with the weight factor (function) $(1-x^{2})^{-1}$ under the integral, however. Does it have to be included in the normalization formula, perhaps as ${\sqrt {(1-x^{2})^{-1}}}$ ?

I would appreciate if both normalization formulas ($N_{1}$ and $N_{2}$) will be confirmed. Thanks. --AboutFace 22 (talk) 22:49, 25 November 2014 (UTC)

A little bit of context might be helpful here. ALP are often used in situations (e.g. quantum mechanics, seismology) where the interesting expression isn't $P_{k}^{m}(x)$ but rather the spherical harmonics $Y_{k}^{m}(\theta ,\phi )\propto P_{k}^{m}(\cos {\theta })e^{im\phi }$. In particular, the orthogonality conditions that you mention is really an extension of the orthonormality of spherical harmonics, i.e.:

$\int _{\theta =0}^{\pi }\int _{\varphi =0}^{2\pi }Y_{\ell }^{m}\,Y_{\ell '}^{m'}\,d\Omega =\delta _{\ell \ell '}\,\delta _{mm'},$

Where the $\theta$ part of the integral gives rise to the $\int _{-1}^{1}{\frac {P_{l}^{m}(x)P_{l}^{n}(x)}{(1-x^{2})}}dx$ expression when the substitution $x=\cos {\theta }$ is used.

Depending on your application, you may actually get better numerical results computing things (such as integrals) in terms of θ. The normalization conventions for spherical harmonics are shown here and vary a little bit by field, but generally are the same as your expression $N_{1}$ except for factors of 2 π and a choice of plus/minus sign.

Returning to your original question. In general, you can use any constant scaling you want in computations, e.g. $S_{l}^{m}(x)\equiv N\,P_{l}^{m}(x)$, simply by defining a scaling factor N and making sure you account for it at the end of the computation. If you only care about numerical overflow/underflow, then choosing N to ensure normality isn't essential. Either expression $N_{1}$ or $N_{2}$ could be reasonable choices. If you are actually computing orthogonality integrals, then choosing one of those forms might be helpful, but it isn't essential. Pulling out an extra term like $(1-x^{2})^{-1}$ is almost certainly a bad idea though as you would have to account for it's impact in all of your analysis and since it varies with x, the impact would be non-trivial. You also might interested in other scaling choices. For examples, this gives some inequalities for ALP, such as:

$\max {\left\vert P_{n}^{m}(x)\right\vert }\leq {1 \over {\sqrt {2}}}{\sqrt {\frac {(n+m)!}{(n-m)!}}}$

as well as more sophisticated x dependent forms. Dragons flight (talk) 01:21, 27 November 2014 (UTC)

Thank you very much. It is encouraging and helpful. The Spherical Harmonics is the next step and I definitely need to compute integrals (transforms) eventually and very soon. It is good that I should forget about $(1-x^{2})$ weight factor. The inequality is also very helpful because I can use it as an additional means of computation control. --AboutFace 22 (talk) 02:40, 27 November 2014 (UTC)