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

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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? NeonMerlin 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)

t is not time - it is just a bound variable. It could be called u or λ 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)