June 4
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? NeonMerlin 02:09, 4 June 2007 (UTC)[reply]
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)[reply]
- 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. --LambiamTalk 06:47, 4 June 2007 (UTC)[reply]
- 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)[reply]
- Take a look at the Hilbert's paradox of the Grand Hotel which expalains stuff nicely --h2g2bob (talk) 10:44, 4 June 2007 (UTC)[reply]
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
![{\displaystyle (s,\psi (s))\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50efc88d8dd06c9da2799b73f81ea4babe2438d3)
- 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
![{\displaystyle x(s)=x_{0}+\int _{0}^{s}\cos \psi (t)\,dt\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ebc2e6f5d0d6f2d1cc0637f657a82bf5881715d)
![{\displaystyle y(s)=y_{0}+\int _{0}^{s}\sin \psi (t)\,dt\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/34210de5e80efad82bcd0da728c5a7dfa00be198)
- 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:
![{\displaystyle x(s)={\frac {s}{2))(\ \ \cos(\log s)+\sin(\log s))\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0c08a95510ac05b20bc7a594e5d6d3a49b8d01b6)
![{\displaystyle y(s)={\frac {s}{2))(-\cos(\log s)+\sin(\log s))\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aad1a6b126165b3f324820ba15448c04c9b27e54)
- This can be verified by taking the derivatives with respect to s.
- Using x = x(s) and y = y(s), we have
![{\displaystyle 2(x^{2}+y^{2})=s^{2}\,,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84f0e2fc1e87b4eaf474833032a326fa42b16b43)
- 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. --LambiamTalk 14:11, 4 June 2007 (UTC)[reply]
- Actually, it isn't too ugly if you are willing to bend the rules a bit:
![{\displaystyle {\frac {x^{2}-y^{2)){x^{2}+y^{2))}=\sin \log(2(x^{2}+y^{2}))\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba08cad0ad897a78e1bb3f8dc130c7b7c31c56cc)
- 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. --LambiamTalk 22:57, 4 June 2007 (UTC)[reply]
- 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. --LambiamTalk 23:08, 4 June 2007 (UTC)[reply]
- I'm really confused about this
![{\displaystyle x(s)=x_{0}+\int _{0}^{s}\cos \psi (t)\,dt\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ebc2e6f5d0d6f2d1cc0637f657a82bf5881715d)
![{\displaystyle y(s)=y_{0}+\int _{0}^{s}\sin \psi (t)\,dt\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/34210de5e80efad82bcd0da728c5a7dfa00be198)
- 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)[reply]
- t is not time - it is just a bound variable. It could be called u or λ or anything. But writing
![{\displaystyle x(s)=x_{0}+\int _{0}^{s}\cos \psi (s)\,ds\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7bb5d4fe521c54d6935fa97c601c5f5ad1c5b075)
- 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)[reply]