Talk:Affine curvature

What?
This article says that it is a definition for a particular kind of curve - well what kind of curve is it? This article isn't very descriptive... Thanks, zappa 03:18, 10 April 2006 (UTC)


 * you mean "what is an affine plane curve?"? I think a differentiable curve on an affine plane is meant. --MarSch 12:53, 2 December 2006 (UTC)


 * Actually, things have to be slightly stronger than that... my understanding is that the curve has no have non-zero speed, else the curvature isn't defined. Tom pw (talk) (review) 11:49, 30 August 2008 (UTC)


 * In two dimensions, the curve needs to have linearly independent speed and acceleration. More generally, in n dimensions, the first n derivatives of the curve need to be linearly independent.  These are analogous to the requirements for the existence of a Frenet frame (although in that case, you can get by with one order less of differentiability).   siℓℓy rabbit  (  talk  ) 12:31, 30 August 2008 (UTC)
 * Surely if the acceleartion is zero, then the curavture is zero? Tom pw (talk) (review) 13:06, 30 August 2008 (UTC)
 * I suppose that seems reasonable. The approach I am familiar with (in n dimensions) requires you to be able to define the parameterization with respect to special affine arclength.  This fails to exist if the first n derivatives are not independent, since the parameter is chosen so that the determinant of a certain matrix is unity.   siℓℓy rabbit  (  talk  ) 13:13, 30 August 2008 (UTC)
 * Comment: I have emmended my post above. The speed and acceleration need to be independent, not just nonzero.  siℓℓy rabbit  (  talk  ) 13:19, 30 August 2008 (UTC)

I have checked the first formula for the affine curvature given in the article, and it quite clearly needs to have this special choice of parameter. The special affine arclength is one of the two parameterizations of a curve x = x(t) (with suitably generic derivatives up to n-th order), such that the determinant

Consider just the case of two dimensions (m = 2) in which case the affine parameterization can be selected so that the sign is '+' in this formula (in certain other dimensions this is not possible). If x is given the affine parameterization, then
 * $$x''' = k x'$$

for some function k. Indeed, by differentiating
 * $$0 = \det[x'\ x]' = \det[x'\ x']$$

so that x&prime;&prime;&prime; is proportional to x&prime;. The constant k is obtained again by using the affine parameterization condition (1):
 * $$k = k\det[x'\ x]=\det[x'\ x'']$$

which agrees with the formula given in the article apart from a sign (which is probably a matter of convention, at least in dimension 2). siℓℓy rabbit (  talk  ) 13:39, 30 August 2008 (UTC)
 * So zero accerlastion means zero curvarture. What abotu zero speed? Tom pw (talk) (review) 13:48, 30 August 2008 (UTC)
 * No. I think the speed and acceleration must both be nonzero (and linearly independent) in order for the affine curvature to be defined.  It is possible that there are some other definitions of affine curvature floating around in the literature, but the "definition" in the article actually does require this.  siℓℓy rabbit  (  talk  ) 13:51, 30 August 2008 (UTC)
 * We are agreed that $$k =\det[x'\ x]$$... if you put $$x'' =0$$ ionto that, you get $$ k = \begin{array}{|cc|}

x'''_1 & 0\\ x'''_2 & 0\\ \end{array} = 0$$, because if a column of a determinant is all zeros, then the determinant is zero, and hence the curvature is zero, not undefined. Tom pw (talk) (review) 14:01, 30 August 2008 (UTC)
 * k=det[x&prime;&prime;&prime; x&prime;&prime;] for a special choice of parameter (the affine arclength) analogous to the arclength parameter in the differential geometry of curves. Unlike the usual arclength, however, the affine arclength in n=2 dimensions involves the first two derivatives of x.  The bottom line is that the equation:
 * $$k = \det[x'\ x]$$
 * is only good for this choice of parameter. Other choices of parameter will have det[x&prime; x&prime;&prime;] occurring in denominators.   siℓℓy rabbit  (  talk  ) 14:07, 30 August

2008 (UTC)
 * I agree that the particular formulation requires a certain parametisation, but I fail to see why that paramtisation fails if the acceleration is zero.... could you explain that? Tom pw (talk) (review) 16:07, 30 August 2008 (UTC)

(unindent) Certainly. A special affine parameterization is a new parameter s selected so that
 * $$\det[x'(s)\ x''(s)] = 1$$

If x is equipped with an arbitrary parameterization, say x=x(t) where t=t(s) is a regular reparameterization, then
 * $$\begin{align}

1&=\det[x'(s)\ x''(s)]\\ &= \det\begin{bmatrix} \frac{dx}{dt}\frac{dt}{ds} & \left(\frac{d^2x}{dt^2}\left(\frac{dt}{ds}\right)^2+\frac{dx}{dt}\frac{d^2t}{ds^2}\right)\end{bmatrix}\\ &= \left(\frac{dt}{ds}\right)^3\det\begin{bmatrix} \frac{dx}{dt} &\frac{d^2x}{dt^2}. \end{bmatrix} \end{align}$$ so the determinant appearing on the right side cannot be zero if the special affine parameterization is well-defined.

I can also say why this assumption is necessary in more general terms. In the classical Euclidean geometry of plane curves, the curvature gives a complete description of the structure of the curve in the sense that, if two curves have the same curvature at every point, then one comes from the other by means of a Euclidean transformation; conversely, every specification of a smooth curvature function gives rise to a unique plane curve up to rigid motion. Affine curvature is defined in an analogous fashion: if two curves have the same affine curvature at every point, then they are likewise related by a special affine transformation; conversely, given an affine curvature function there is a unique curve up to an overall special affine transformation. If, however, the rank of the matrix of n-th derivatives drops, then there is an additional discrete invariant (the rank of that matrix) which needs to be considered. In cases of such degenerate points, one can still form a notion of curvature, but it is in some affine subspace of strictly smaller dimension, and the nature of the local geometry becomes rather more complicated.

If we were to agree that the affine curvature vanished when x&prime;&prime;=0, then the affine curvature would fail to classify curves up to special affine motions. Indeed, for any prescribed curvature function k, there exists a unique curve with that curvature whose first two derivatives are linearly independent, up to an overall special affine motion; for a proof, see for instance. If there are two ways for a curve to have vanishing curvature at a point, then the uniqueness assertion falls apart. A more detailed account of these issues in the very general setting of moving frames on a submanifold of a homogeneous space, of which affine curvature is a special case, can be found in and. siℓℓy rabbit (  talk  ) 18:57, 30 August 2008 (UTC)


 * Here is a specific counterexample. The curve
 * $$\beta_1(s) = \left(s, \frac{s^2}{2}\right)$$
 * (a parabola) has zero affine curvature at every point. The curve
 * $$\beta_2(s) = \begin{cases}\left(s, \frac{s^2}{2}\right) & s\ge 0\\ (s,0) & s<0\end{cases}$$
 * has either &beta;&prime;&prime;=0 or vanishing curvature at every point. Nevertheless, these curves are clearly not related by an element of the special affine group.   siℓℓy rabbit  (  talk  ) 13:49, 31 August 2008 (UTC)

I've added a short section on curvature to the (still stubby) article Affine geometry of curves. This article should probably be merged there ultimately, since as I have said the affine curvature pretty much requires the special affine arclength parameterization in order to be defined. siℓℓy rabbit (  talk  ) 14:50, 30 August 2008 (UTC)

Polar
$$ \left|-\frac{ \left(r' \left(2 r+3 r\right)-r r^{(3)}\right)^2}{\left(r^2+2 r'^2-r r\right)^3}+\frac{ \left(r^2+6 r'^2-4 r r+3 r^2-2 r' r^{(3)}\right)}{\left(r^2+2 r'^2-r r\right)^{5/2}}+\frac{3 r^2+2 r' \left(r'+r^{(3)}\right)+r \left(2 r-r^{(4)}\right)}{2\left(r^2+2 r'^2-r r\right)^2}\right| $$


 * Please give a reference for this before adding it to the article. siℓℓy rabbit  (  talk  ) 15:05, 31 August 2008 (UTC)

Incorrect formula
The formula given in the article until now
 * $$\left| \frac{xy-xy}{(x'y-xy')^{5/2}}-\frac{1}{2}\left[\frac{1}{(x'y-xy')}\right]''\right|$$

was incorrect. It was subsequently restored, and sourced to the nonexistent book "Mathematic Encyclopedic Dictionary", Moscow, 1995. p.195. It is obviously wrong for a number of reasons: Finally, I have checked against other references and given the correct formula
 * If this were true, then k would always be positive because of the absolute value, which is clearly wrong.
 * The formula is homogeneous of the wrong degree. If I rescale time, the curvature should remain invariant (or at least have a well-defined transformation law), the above formula is not only not invariant, but the fact that the two terms have different homogeneity implies that there cannot be any nice transformation law.
 * $$k(t)= \frac{1}{3}\frac{4(xy-xy)+(x'y'-x'y')}{(x'y-xy')^{5/3}} -\frac{5}{9}\frac{(x'y-xy')^2}{(x'y-xy')^{8/3}}.$$

I have also derived this resuult by hand and then using mathematica from the definition using the special affine parameterization. Here is the code I used in mathematica:

I have also checked this another way in the special case of a parameterization as a graph (x,y(x)), by differentiating the affine curvature along the fourth prolongation of the generators of the action of the special affine group. The curvature is obviously invariant under translations. It suffices to check that it is invariant under the other three generators. Two of these are given by


 * $$X_1^{(4)} = x\partial_y +\partial_{y'}$$
 * $$X_3^{(4)} = x\partial_x - y\partial_y - y'^2\partial_{y'} - 3y'y\partial_{y}-(3y^2+4y'y)\partial_{y}-(10yy'''+5y'y')\partial_{y'}$$

and the remaining generator is the Lie bracket of these two. The formula for the affine curvature originally given in the article isn't killed by either of these vector fields and so is not an invariant under the special affine group. The new one that I supplied, on the other hand, is killed by both of them, and so defines a conserved quantity under the action of the affine group. siℓℓy rabbit (  talk  ) 12:55, 1 September 2008 (UTC)
 * Please do not insert original research here. If you have a source for your variant, please provide it. If you want, I can scan a page from the encyclopedia for your sake. Your derivation (and script for Mathematica) may be correct or incorrect, but without a citation it is pure original research.--Dojarca (talk) 13:29, 1 September 2008 (UTC)
 * As I indicated above, I checked it against the literature first (see the edit, where I think you will find there is a citation). Since there was a discrepancy, to confirm this, I checked the formula in a variety of ways.  None of these would qualify as "original research", since I was only using standard techniques, your own lack of knowledge of the field notwithstanding.  Please do not revert again.   siℓℓy rabbit  (  talk  ) 13:39, 1 September 2008 (UTC)
 * If yiou think your version of analytic geometry is more correct please consider publishing your results. Please stop edit warring.--Dojarca (talk) 13:41, 1 September 2008 (UTC)
 * My version is supported by the source I gave, as well as the references at the bottom of the article. The so-called "reference" supporting your incorrect version of the formula does not appear to exist.  Would you care to provide a proper citation?  I already have (if you would read by edit):  see Guggenheimer, p. 150.  As I have already said, because there was a discrepancy, I personally checked the result by hand and by computer.  I have also given reasons above why your version of the formula is obviously wrong (it has the wrong homogeneity and can never be negative).  You have yet to respond to this in any substantive way, or to provide a reasonable reference.  siℓℓy rabbit  (  talk  ) 13:48, 1 September 2008 (UTC)
 * Well I can scan the page. Also please answer. Does any of the sources you claim support your variant provide the formula exactly in the form you're inserting into the article?--Dojarca (talk) 13:52, 1 September 2008 (UTC)
 * Yes. As I have already said two or three times.  The formula appears in the text of Guggenheimer, published by Dover, and listed in the references section of the article.  In section §7.3.  On page 150.  Near the middle of the page.   siℓℓy rabbit  (  talk  ) 14:02, 1 September 2008 (UTC)
 * Also, please do scan the page. You can email it to me.  If you have email enabled, I can do the same.  siℓℓy rabbit  (  talk  ) 14:06, 1 September 2008 (UTC)

For a plane curve of the form y=y(x), with x=t, the affine curvature is given by k = (-1/2) ((y")-2/3)", according to springer's translation of the Encyclopaedia of Mathematics. siℓℓy rabbit's formula gives this in the case x'=1, while Dojarca's formula does not. R.e.b. (talk) 14:42, 1 September 2008 (UTC)


 * Blaschke (ref given in article) p. 14 gives the same result (after the derivatives have been expanded) also in the special case when the curve is a graph y=y(x), which agrees with my formula.  siℓℓy rabbit  (  talk  ) 14:50, 1 September 2008 (UTC)

There seem to be 2 slightly different things called affine curvature: the equi-affine curvature (often called the affine curvature), and the affine curvature, which is slightly different. This might have something to do with two incompatible formulas for the affine curvature. R.e.b. (talk) 14:59, 1 September 2008 (UTC)

See R.e.b. (talk) 15:02, 1 September 2008 (UTC)
 * Here are the scan and photos:User:Dojarca/math. Silly rabbit please scan your sources.--Dojarca (talk) 15:04, 1 September 2008 (UTC)
 * Translation:
 * Differential invariant of affine unimodular group is the quantity $$\kappa = | \left( x y - x y \right) \phi^{-5} - \frac{1}{2} (\phi^{-2}) |, \text{ where } \phi=(x'y-xy')^{1/2}$$, so $$\kappa$$ is the object of affine differential geometry (called affine curvature of the plane curve).''
 * --Dojarca (talk) 15:27, 1 September 2008 (UTC)


 * It appears that this disagreement is the result of a simple typographical error. The exponent in &phi; should be 1/3 rather than 1/2. In other words, $$\phi=(x'y-xy')^{1/3}\,.$$ JRSpriggs (talk) 17:06, 1 September 2008 (UTC)
 * Yes, that does fix it. Thanks for the observation.  The absolute value also needs to go, since negative affine curvature indicates an osculating hyperbola rather than ellipse.   siℓℓy rabbit  (  talk  ) 17:11, 1 September 2008 (UTC)
 * Anyway it is not consistent with what Silly rabbit derived.--Dojarca (talk) 17:13, 1 September 2008 (UTC)
 * Aak. No I just checked it more carefully.  It's still wrong.  siℓℓy rabbit  (  talk  ) 17:21, 1 September 2008 (UTC)

The scan is here: Image:Guggenheimer_Affine_curvature.jpg. It will probably be deleted as invalid fair use soon, so get it while you can. siℓℓy rabbit (  talk  ) 17:32, 1 September 2008 (UTC)
 * Well it gives the formula in vector form. I suggest to keep all three of the above formulas in the form they appear in the sources, give proper citations and note inconsistancy between the formulas (possibly the sources use different terminology).--Dojarca (talk) 17:43, 1 September 2008 (UTC)
 * It really isn't original research to convert from a vector form to a scalar form, by the way. Also, so far three different sources have been given, *all* of which agree with the formula given in Guggenheimer (and which I had added to the article).  *None* of which agree with the obscure Russian tertiary source that you gave.  I think we should eliminate your formula per WP:WEIGHT, unless other primary and secondary sources are forthcoming.  siℓℓy rabbit  (  talk  ) 17:48, 1 September 2008 (UTC)
 * It in not obscure source, this encyclopedia was published by the same state puplishery that published GSE. Also the author of the article is N.V.Yefimov, dean of mechanical-mathematical faculty of the Moscow State University in 1960s (Russian Wikipedia has article about him: ru:Ефимов, Николай Владимирович).--Dojarca (talk) 17:55, 1 September 2008 (UTC)
 * If it isn't obscure, why have you resisted repeated entreaties to give a proper citation for the work?  siℓℓy rabbit  (  talk  ) 17:58, 1 September 2008 (UTC)
 * I gave you the citation, translation, the scan and the photos of the book. Where did I resisted?--Dojarca (talk) 18:07, 1 September 2008 (UTC)
 * This is *not* a citation: "Mathematic Encyclopedic Dictionary", Moscow, 1995. p.195. Citations usually have things like editors, publishers, and whatnot.  Also, a google search turned up no hits at all for "Mathematic Encyclopedic Dictionary".   siℓℓy rabbit  (  talk  ) 18:10, 1 September 2008 (UTC)
 * Chief editor Yu.V.Prokhorov. Publishery: "Great Russian Encyclopedy". Google returns 2240 hits.--Dojarca (talk) 18:28, 1 September 2008 (UTC)

significance
Hi,

Could you please add some material in the lead on the significance of this invariant? Katzmik (talk) 13:54, 1 September 2008 (UTC)


 * Prior to yesterday, it was a stub filled with errors. In fact, in light of the above dispute, I can safely say that not a single formula given in the article was entirely accurate.  Consider it a work in progress.   siℓℓy rabbit  (  talk  ) 13:58, 1 September 2008 (UTC)


 * Thanks very much. I find this concept intriguing.  I have to admit I am not familiar with it.  I have the impression that the key to understanding it may lie in presenting affine arclength in an ample fashion.  Could there be a sentence about this in the introduction? Katzmik (talk) 14:07, 1 September 2008 (UTC)

a naive question
Is the arclength parametrisation of the circle, also its affine arclength parametrisation? Katzmik (talk) 14:48, 1 September 2008 (UTC)


 * Actually no, unless the circle has unit radius. In general, the arclength parameterization of the circle of radius R is
 * $$\beta(t)=(R\cos(t/R),R\sin(t/R))$$
 * and so
 * $$\det[\beta' \; \beta''] = 1/R$$
 * which is not equal to one unless R=1. siℓℓy rabbit  (  talk  ) 18:03, 1 September 2008 (UTC)


 * An affine arclength parameterization of the circle of radius R is
 * $$\beta(t)=(R\cos(t/R^{2/3}),R\sin(t/R^{2/3}))$$
 * - siℓℓy rabbit (  talk  ) 18:17, 1 September 2008 (UTC)

What is the affine length of a circle of length R? What is the affine arclength parametrisation of the standard hyperbola? Somehow these elementary questions should be addressed, if not in the lead then in an example or appendix. Actually I now recall learning some of this material in a book called Kreis, Kugel, und Ebene but I can't recall the author. Katzmik (talk) 11:39, 2 September 2008 (UTC)


 * I had intended these issues to be addressed in the section on conics. Actually any conic can, by a suitable special affine parameterization, be put into one of the three normal forms:
 * $$y=x^2/2,\quad kx^2+k^2y^2=1,\quad -|k|x^2+|k|^2y^2=1$$
 * depending on whether k=0,>0,<0, respectively. In each special case, the special affine parameterization is given in the article:
 * $$\beta=(s,s^2/2)\,$$
 * $$\beta(s) = \left(\frac{1}{\sqrt{k}}\sin\sqrt{k}s, -\frac{1}{k}\cos\sqrt{k}s\right)$$
 * $$\beta(s) = \left(\frac{1}{\sqrt{|k|}}\sinh\sqrt{|k|}s,\frac{1}{|k|}\cosh\sqrt{|k|}s\right) $$
 * respectively.  siℓℓy rabbit  (  talk  ) 11:55, 2 September 2008 (UTC)


 * Is there a synthetic description of this invariant? Katzmik (talk) 12:12, 2 September 2008 (UTC)


 * If "a" and "b", the semimajor and semiminor axes, are synthetic enough for you, then (up to agreeing on signs)
 * $$k=1/(ab)^{2/3}.$$
 * This is a special affine invariant because of its relationship to the discriminant. siℓℓy rabbit  (  talk  ) 12:18, 2 September 2008 (UTC)

Thanks for your comments. Katzmik (talk) 12:28, 2 September 2008 (UTC)


 * Not at all. Actually, I would like to add some of this to the article.  Thanks for fleshing out your question from the earlier section a bit.   siℓℓy rabbit  (  talk  ) 12:34, 2 September 2008 (UTC)


 * P.S. The book I was referring to earlier is by

Fejes Tóth, L.: Lagerungen in der Ebene, auf der Kugel und im Raum. (German) Die Grundlehren der Mathematischen Wissenschaften in Einzeldarstellungen mit besonderer Berücksichtigung der Anwendungsgebiete, Band LXV. Springer-Verlag, Berlin-Göttingen-Heidelberg, 1953. It discusses affine length in chapter 2. Katzmik (talk) 14:25, 2 September 2008 (UTC)

Update: JRSpriggs' correction
So, the correction made by JRSpriggs apparently works out to give a constant multiple of the curvature I gave (from the Guggenheimer text) in the case of a graph y=y(x). However, in the general case, the two formulas are seen to differ. In any event, the difference (or actually the sum) of the two formulas works out to give
 * $$\frac{2(xy-xy)}{(x'y-xy')^{5/3}}$$

which clearly vanishes in the case of a curve parameterized as a graph (since x&prime;&prime;=0 in that case). The question, which should be fairly easy to resolve by anyone who knows some differential geometry (or I suspect just calculus), is to verify whether this is actually a differential invariant of the special affine group. siℓℓy rabbit (  talk  ) 17:57, 1 September 2008 (UTC)

Finally I see the problem. I had the wrong sign in my calculation. The two formulas are identical. siℓℓy rabbit (  talk  ) 18:25, 1 September 2008 (UTC)
 * Thank you. That's why we need reliable source even if somewhat appears easily to derive.--Dojarca (talk) 18:32, 1 September 2008 (UTC)
 * Of course, my formula was right all along and yours was wrong. That&prime;s why it is no good copying formulas out of books if you have no idea where the formulas came from.   siℓℓy rabbit  (  talk  ) 19:21, 1 September 2008 (UTC)

Let's try this from scratch
I'm going to amuse myself by attempting to derive the disputed affine curvature formula. I'll assume that all functions are smooth, all integrals exist, etc.; things which should be obvious with the proper hypotheses.

Let &beta;(t) = (x(t), y(t)). First we need to find s(t). s(t) has the formula
 * $$s(t) = \int_a^t \sqrt[3]{\det\begin{bmatrix}\frac{d\beta}{dt}(r)&\frac{d^2\beta}{dt^2}(r)\end{bmatrix}}\,dr$$

(Unlike the article, I'm using r as the dummy variable instead of t because t is also one of the limits.) So, substituting x and y for &beta;, this is
 * $$s(t) = \int_a^t \sqrt[3]{\det\begin{bmatrix}x'(r)&x(r)\\y'(r)&y(r)\end{bmatrix}}\,dr = \int_a^t \sqrt[3]{x'(r)y(r) - y'(r)x(r)}\,dr$$

Next we need &beta;(s) and &beta;'(s). To find those, I'm going to use the chain rule:
 * $$\frac{d\beta}{dt} = \frac{d\beta}{ds}\frac{ds}{dt},$$
 * $$\frac{d^2\beta}{dt^2} = \frac{d^2\beta}{ds^2}\frac{ds}{dt} + \frac{d\beta}{ds}\frac{d^2s}{dt^2},$$
 * $$\frac{d^3\beta}{dt^3} = \frac{d^3\beta}{ds^3}\frac{ds}{dt} + 2\frac{d^2\beta}{ds^2}\frac{d^2s}{dt^2} + \frac{d\beta}{ds}\frac{d^3s}{dt^3},$$

which, after some rearranging, gives
 * $$\frac{d\beta}{ds} = \frac{d\beta}{dt}\Bigg/\frac{ds}{dt},$$
 * $$\frac{d^2\beta}{ds^2} = \frac{\frac{d^2\beta}{dt^2} - \frac{d\beta}{ds}\frac{d^2s}{dt^2}}{\frac{ds}{dt}} = \frac{d^2\beta}{dt^2}\left(\frac{ds}{dt}\right)^{-1} - \frac{d\beta}{dt}\frac{d^2s}{dt^2}\left(\frac{ds}{dt}\right)^{-2}.$$
 * $$\frac{d^3\beta}{ds^3} = \frac{\frac{d^3\beta}{dt^3} - 2\frac{d^2\beta}{ds^2}\frac{d^2s}{dt^2} - \frac{d\beta}{ds}\frac{d^3s}{dt^3}}{\frac{ds}{dt}} = \frac{d^3\beta}{dt^3}\left(\frac{ds}{dt}\right)^{-1} - 2\frac{d^2\beta}{ds^2}\left(\frac{ds}{dt}\right)^{-1}\frac{d^2s}{dt^2} - \frac{d\beta}{dt}\frac{d^3s}{dt^3}\left(\frac{ds}{dt}\right)^{-2}.$$

Well, ds/dt, d2s/dt2, and d3s/dt3 are easy to compute:
 * $$s'(t) = \sqrt[3]{x'y - y'x},$$
 * $$\begin{align}s(t) &= \textstyle\frac{1}{3}(x'y - y'x)^{-2/3}(xy + x'y - yx - y'x)\\&= \textstyle\frac{1}{3}(x'y - y'x'')^{-2/3}(x'y - y'x).\end{align}$$
 * $$\begin{align}s(t) = &-\textstyle\frac{2}{9}(x'y - y'x)^{-5/3}(x'y - y'x)^2 \\&+ \textstyle\frac{1}{3}(x'y - y'x)^{-2/3}(xy + x'y' - yx''' - y'x').\end{align}$$

So now it's a matter of plug-and-chug:
 * $$\frac{d\beta}{ds} = \begin{bmatrix}x'\\y'\end{bmatrix}\cdot\left(x'y - y'x\right)^{-1/3},$$
 * $$\begin{align}\frac{d^2\beta}{ds^2}

&= \begin{bmatrix}x\\y\end{bmatrix}\cdot\left(x'y - y'x\right)^{-1/3} - \begin{bmatrix}x'\\y'\end{bmatrix}\cdot\textstyle\frac{1}{3}(x'y - y'x)^{-2/3}(x'y - y'x)\cdot\left(x'y - y'x\right)^{-2/3} \\ &= (x'y - y'x)^{-1/3}\begin{bmatrix}x - \frac{1}{3}x'(x'y-y'x)/(x'y - y'x)\\y - \frac{1}{3}y'(x'y-y'x)/(x'y - y'x)\end{bmatrix}\end{align}$$
 * $$\begin{align}\frac{d^3\beta}{ds^3} &= \begin{bmatrix}x\\y\end{bmatrix}\left(x'y - y'x\right)^{-1/3} \\&\quad - 2\left(x'y - y'x\right)^{-1/3}\begin{bmatrix}x - \frac{1}{3}x'(x'y-y'x)/(x'y - y'x)\\y - \frac{1}{3}y'(x'y-y'x)/(x'y - y'x)\end{bmatrix} \\&\quad \cdot\left(x'y - y'x\right)^{-1/3}\cdot\textstyle\frac{1}{3}(x'y - y'x)^{-2/3}(x'y - y'x) \\&\quad -\begin{bmatrix}x' \\ y'\end{bmatrix}(-\textstyle\frac{2}{9}(x'y - y'x)^{-5/3}(x'y - y'x)^2 \\&\quad\qquad+\textstyle\frac{1}{3}(x'y - y'x)^{-2/3}(xy + x'y' - yx - y'x')) \\&\quad\cdot(x'y - y'x)^{-2/3} \\

&= (x'y - y'x)^{-1/3}\begin{bmatrix}x - \frac{2}{3}x\frac{x'y-y'x}{x'y - y'x} + x'\left(\frac{4}{9}\left(\frac{x'y-y'x}{x'y - y'x}\right)^2 - \frac{1}{3}\frac{xy + x'y' - yx - y'x'}{x'y - y'x}\right) \\ y - \frac{2}{3}y\frac{x'y-y'x}{x'y - y'x} + y'\left(\frac{4}{9}\left(\frac{x'y-y'x}{x'y - y'x}\right)^2 - \frac{1}{3}\frac{xy + x'y' - yx - y'x'}{x'y - y'x}\right)\end{bmatrix}. \end{align}$$ Wow, that's really hideous. But I'm sure that when we compute k(t), it'll turn out much nicer:
 * $$\begin{align}k

&= (x'y - y'x)^{-2/3}\det\begin{bmatrix}x - \frac{1}{3}x'\frac{x'y-y'x}{x'y - y'x''} & x - \frac{2}{3}x\frac{x'y-y'x}{x'y - y'x} + x'\left(\frac{4}{9}\left(\frac{x'y-y'x}{x'y - y'x}\right)^2 - \frac{1}{3}\frac{xy + x'y' - yx' - y'x'}{x'y - y'x}\right) \\ y - \frac{1}{3}y'\frac{x'y -y'x}{x'y - y'x''} & y - \frac{2}{3}y\frac{x'y-y'x}{x'y - y'x} + y'\left(\frac{4}{9}\left(\frac{x'y-y'x}{x'y - y'x}\right)^2 - \frac{1}{3}\frac{xy + x'y' - yx' - y'x'}{x'y - y'x}\right)\end{bmatrix} \\ &= (x'y - y'x)^{-2/3}\Bigg((xy - xy) - \frac{2}{3}(xy - xy)\frac{x'y-y'x}{x'y - y'x} \\&\qquad + (xy' - x'y)\left(\frac{4}{9}\left(\frac{x'y-y'x}{x'y - y'x}\right)^2 - \frac{1}{3}\frac{xy + x'y' - yx - y'x'}{x'y - y'x}\right) \\&\qquad -\frac{1}{3}(x'y - xy')\frac{x'y-y'x}{x'y - y'x} + \frac{2}{9}(x'y - xy')\left(\frac{x'y-y'x}{x'y - y'x}\right)^2 \\&\qquad- \frac{1}{3}(x'y' - x'y')\left(\frac{x'y-y'x}{x'y - y'x}\right)\left(\frac{4}{9}\left(\frac{x'y-y'x}{x'y - y'x}\right)^2 - \frac{1}{3}\frac{xy + x'y' - yx - y'x'}{x'y - y'x}\right)\Bigg) \\ &= (x'y - y'x)^{-2/3}\left((xy - xy) + (xy' - x'y)\left(\frac{4}{9}\left(\frac{x'y-y'x}{x'y - y'x}\right)^2 - \frac{1}{3}\frac{xy + x'y' - yx - y'x'}{x'y - y'x}\right) -\frac{1}{9}\frac{(x'y-y'x)^2}{x'y - y'x}\right) \\ &= (x'y - y'x)^{-2/3}\left(xy - xy - \frac{1}{3}(xy + x'y - yx - y'x) + \frac{4}{9}\frac{(x'y - y'x)^2}{x'y - y'x} - \frac{1}{9}\frac{(x'y - y'x)^2}{x'y - y'x}\right) \\ &= (x'y - y'x)^{-2/3}\left(\frac{2}{3}(xy - xy) - \frac{1}{3}(x'y - y'x) + \frac{1}{3}\frac{(x'y - y'x)^2}{x'y - y'x}\right).\end{align}$$ OK, that's not so amusing. This doesn't seem to agree with either of the formulas proposed above. What am I doing wrong? Ozob (talk) 21:18, 1 September 2008 (UTC)


 * In your computation of the second derivative of &beta;,
 * $$\frac{d^2\beta}{ds^2} = \frac{\frac{d^2\beta}{dt^2}\frac{ds}{dt} - \frac{d\beta}{dt}\frac{d^2s}{dt^2}}{\left(\frac{ds}{dt}\right)^2}$$
 * Btw: there is a mathematica script above that will do the calculation for you. - siℓℓy rabbit  (  talk  ) 21:41, 1 September 2008 (UTC)


 * Here:
 * $$\begin{align}

\frac{d\beta}{ds} &= \frac{d\beta}{dt}\left|\frac{d\beta}{dt} \; \frac{d^2\beta}{dt^2}\right|^{-1/3}\\ \frac{d^2\beta}{ds^2} &= \frac{d^2\beta}{dt^2}\left|\frac{d\beta}{dt} \; \frac{d^2\beta}{dt^2}\right|^{-2/3} - \frac{1}{3}\frac{d\beta}{dt}\left|\frac{d\beta}{dt} \; \frac{d^2\beta}{dt^2}\right|^{-5/3}\left|\frac{d\beta}{dt} \; \frac{d^3\beta}{dt^3}\right|\\ \frac{d^3\beta}{ds^3} &= \frac{d^3\beta}{dt^3}\left|\frac{d\beta}{dt} \; \frac{d^2\beta}{dt^2}\right|^{-1} - \frac{d^2\beta}{dt^2}\left|\frac{d\beta}{dt} \; \frac{d^2\beta}{dt^2}\right|^{-2}\left|\frac{d\beta}{dt} \; \frac{d^3\beta}{dt^3}\right|+\frac{5}{9}\frac{d\beta}{dt}\left|\frac{d\beta}{dt} \; \frac{d^2\beta}{dt^2}\right|^{-3}\left|\frac{d\beta}{dt} \; \frac{d^3\beta}{dt^3}\right|^2\\ &\qquad\quad -\frac{1}{3}\frac{d\beta}{dt}\left|\frac{d\beta}{dt} \; \frac{d^2\beta}{dt^2}\right|^{-2}\left|\frac{d^2\beta}{dt^2} \; \frac{d^3\beta}{dt^3}\right|-\frac{1}{3}\frac{d\beta}{dt}\left|\frac{d\beta}{dt} \; \frac{d^2\beta}{dt^2}\right|^{-2}\left|\frac{d\beta}{dt} \; \frac{d^4\beta}{dt^4}\right| \end{align} $$
 * so

k = -\frac{5}{9}\left|\frac{d\beta}{dt}\; \frac{d^2\beta}{dt^2}\right|^{-8/3}\left|\frac{d\beta}{dt} \; \frac{d^3\beta}{dt^3}\right|^2 +\frac{4}{3}\left|\frac{d\beta}{dt}\; \frac{d^2\beta}{dt^2}\right|^{-5/3}\left|\frac{d^2\beta}{dt^2} \; \frac{d^3\beta}{dt^3}\right| +\frac{1}{3}\left|\frac{d\beta}{dt}\; \frac{d^2\beta}{dt^2}\right|^{-5/3}\left|\frac{d\beta}{dt} \; \frac{d^4\beta}{dt^4}\right| $$

hypersosculating conic
There seems something wrong with the intro where it claims there is a unique conic through 4 points; there is usually a 1-parameter family. Should there be 5 points, or is there a missing condition on the conic? R.e.b. (talk) 19:56, 2 September 2008 (UTC)


 * The conic must also pass through P, so there are five points on it. I had omitted this important condition from the original statement, however.  Thanks for spotting the error. The main point is that it should have fourth order contact with the curve at the point P; so four points, in addition to P, are needed.  siℓℓy rabbit  (  talk  ) 21:07, 2 September 2008 (UTC)