Wikipedia:Reference desk/Archives/Mathematics/2007 June 4

= June 4 =

Infinity minus one, infinity plus one, double infinity, infinity squared
What, if any, number systems allow the existence of an ∞ greater than any real number such that ∞, ∞-1, ∞+1, 2∞ and ∞2 are all distinct quantities? Neon Merlin  02:09, 4 June 2007 (UTC)
 * I don't think there's any number system in the traditional sense that satisfies these. However, ordinal numbers hit most of these if you replace ∞ with ω. nadav (talk) 02:21, 4 June 2007 (UTC)


 * Ordinal numbers do not extend the real numbers, and so it is not meaningful to say that some ordinal number α is greater than any real number. Surreal numbers, superreal numbers, and hyperreal numbers all extend the real numbers and have the desired property. --Lambiam Talk  06:47, 4 June 2007 (UTC)


 * Or you could start, less ambitiously, by looking at extensions to the integers - see transfinite number, ordinal number and cardinal number. Gandalf61 09:00, 4 June 2007 (UTC)


 * Take a look at the Hilbert's paradox of the Grand Hotel which expalains stuff nicely --h2g2bob (talk) 10:44, 4 June 2007 (UTC)

Intrinsic Coordinates
Good day.

I know how to express a cartesian equation in Intrinsic form using dy/dx as tan phi, and such. But how can one go back, if at all?

I can't find anything on the Intrinsic Coordinates article, or elsewhere on the web that is useful.

I could, say, get to phi = ln s from an equation.

But if I was given phi = ln s, how would I find what the equation was?


 * I'll stick to the notation of the article. In general, let the general point in intrinsic coordinates be given in the form
 * $$(s, \psi(s))\,.$$
 * If instead an "intrinsic equation" of the form s = f(ψ) is given – which is less general, since a curve may in general have the same direction in more than one point − you first need to invert f to get (s, f−1(s)). Now define
 * $$x(s) = x_0 + \int_0^s \cos\psi(t)\,dt\,$$
 * $$y(s) = y_0 + \int_0^s \sin\psi(t)\,dt\,$$
 * This is a parametric representation of the curve, and in fact a natural parametrization. If you succeed in eliminating the variable s from the pair of equations x = x(s), y = y(s), the result is a standard implicit representation of a curve as the locus of solutions of an equation in Cartesian coordinates. (For actually doing this, the integrals need to have solutions in closed form.)
 * Let us see how this works out for the example, where ψ(s) = log s. Let us take, for the sake of simplicity, (x(0), y(0)) = (0, 0). By standard integration methods we find:
 * $$x(s) = \frac{s}{2} (\ \ \cos(\log s) + \sin(\log s))\,$$
 * $$y(s) = \frac{s}{2} (-\cos(\log s) + \sin(\log s))\,$$
 * This can be verified by taking the derivatives with respect to s.
 * Using x = x(s) and y = y(s), we have
 * $$2(x^2+y^2) = s^2\,,$$
 * so we can solve for s (but note that there are two solutions!) and substitute the solution(s) for s in one of the two equations x = x(s) and y = y(s) to obtain a standard implicit representation. However, the result is not pretty. --Lambiam Talk  14:11, 4 June 2007 (UTC)


 * Actually, it isn't too ugly if you are willing to bend the rules a bit:
 * $$\frac{x^2-y^2}{x^2+y^2} = \sin\log(2(x^2+y^2))\,.$$
 * This is not entirely correct; the equation is symmetric in x and y: if (x, y) is a solution, then so are (−x, y) and (x, −y). But the actual curve has no such mirror symmetries, only rotational symmetry . The locus of solutions of this implicit representation consists of two four copies of the curve. --Lambiam Talk  22:57, 4 June 2007 (UTC)
 * By the way, the curve is actually a logarithmic spiral, as can be seen by switching to a different parametrization with parameter t related to s by t = log s. --Lambiam Talk  23:08, 4 June 2007 (UTC)


 * I'm really confused about this
 * $$x(s) = x_0 + \int_0^s \cos\psi(t)\,dt\,$$
 * $$y(s) = y_0 + \int_0^s \sin\psi(t)\,dt\,$$
 * Why do you integrate with dt instead of dS? Why do you integrate in respect to time?
 * 202.168.50.40 23:26, 4 June 2007 (UTC)
 * 202.168.50.40 23:26, 4 June 2007 (UTC)


 * t is not time - it is just a bound variable. It could be called u or &lambda; or anything. But writing


 * $$x(s) = x_0 + \int_0^s \cos\psi(s)\,ds\,$$


 * would be confusing because s then plays two different roles in the same equation, as both a free variable and a bound variable. Gandalf61 04:57, 5 June 2007 (UTC)