User doetoe - MathOverflow

Answer by doetoe for Does the automorphism group of a cone determine the cone?

2011-09-23T09:55:29Z

Unless there is something I misunderstand, a cone can be isomorphic to a strict subcone (hence no). Take e.g. for D a cone in $\mathbb{R}^2$ generated by two linearly independent vectors, and for C the cone generated by two linearly independent vectors in D that don't generate D (i.e. that are not on the boundary).

Answer by doetoe for Pythagorean 5-tuples

2011-05-06T16:44:42Z

Another way to generate parametrisations is to repeatedly apply a parametrisation of triples:

Say we start with $$(x(s), y(s)) = (\frac{2s}{s^2 + 1}, \frac{s^2 - 1}{s^2 + 1}),$$ satisfying $$x(s)^2 + y(s)^2 = 1.$$

To parametrise solutions to $x^2 + y^2 + z^2 = 1$, write $x^2 + y^2 = 1 - z^2$ which we set equal to $w^2$, so that a parametrised solution $(w(t), z(t))$ to $w^2 + z^2 = 1$ gives rise to a parametrised solution $$(w(t)x(s), w(t)y(s), z(t)) = (\frac{2t}{t^2 + 1}\cdot\frac{2s}{s^2 + 1}, \frac{2t}{t^2 + 1}\cdot\frac{s^2 - 1}{s^2 + 1}, \frac{t^2 - 1}{t^2 + 1})$$ to $x^2 + y^2 + z^2 = 1$.

Repeating this, we get your case:

$$(\frac{2u}{u^2 + 1}\cdot\frac{2t}{t^2 + 1}\cdot\frac{2s}{s^2 + 1}, \frac{2u}{u^2 + 1}\cdot\frac{2t}{t^2 + 1}\cdot\frac{s^2 - 1}{s^2 + 1}, \frac{2u}{u^2 + 1}\cdot\frac{t^2 - 1}{t^2 + 1}, \frac{u^2 - 1}{u^2 + 1})$$ parametrises $$x^2 + y^2 + z^2 + w^2 = 1$$ (to go to your original problem you can throw in another parameter to homogenise, clear denominators and multiply everything by another parameter).

If you apply the same to the parametrisation $(\cos(t), \sin(t))$ of the circle, you get a parametrisation of the 2-sphere by sperical coordinates.

Answer by doetoe for probability and math puzzle books/references

2011-04-27T15:45:46Z

On the site of the International Mathematics Olympiad all past problems can be found. They tend to be quite difficult but no advanced math is required.

Answer by doetoe for Experimental Mathematics

2010-09-02T08:22:34Z

According to this obituary of Edward Lorenz in Plus Magazine, he discovered the seemingly chaotic behaviour of the Lorenz attractor when a small change (due to rounding) in the boundary conditions in a numerical simulation gave rise to hugely different solutions.

Answer by doetoe for Math puzzles for dinner

2010-08-07T16:27:49Z

You and your adversary have a sufficiently large bag of identical coins, and are seated on opposite sides of a rectangular table. You take turns placing coins on the table. The first one that cannot put a coin on the table without overlapping any other coin loses. What is your strategy to always win if you're allowed to start?

Answer by doetoe for Approaches to Riemann hypothesis using methods outside number theory

2010-08-06T16:09:44Z

I have no idea to what extent the idea of Saharon Shelah, about which I read in David Ruelle's popular account the mathematician's brain that uses mathematical logic to prove the RH is promising, but certainly it is different. For as far as I can understand (from Ruelle), it basically comes down to proving that RH is undecidable in Peano arithmetic, in which case the consistency of Peano arithmetic would imply its truth (also in ZFC).

EDIT: Here is the quote from Shelah's paper:

2.3 Dream: Prove that the Riemann Hypothesis is unprovable in PA, but is provable in some higher theory.

What basis does my hope for this dream have? First, the solution of Hilbert’s 10th problem tells us that each problem of the form “is the theory ZFC +φ consistent” can be translated to a (specific) Diophantine equation being unsolvable in the integers, moreover the translation is uniform (this works for any reasonable (defined) theory, where consistent means that no contradiction can be proved from it). Second, we may look at parallel development “higher up”; as the world is quite ordered and reasonable.

Note that there is a significant difference between $\Pi_2$ sentences (which say, e.g., for a given polynomial $f$, the sentence $\varphi_f$ saying that for all natural numbers $x_0 , \ldots , x_{n−1}$ there are natural numbers $y_0 , \ldots , y_m$ such that $f (x_0 , \ldots , y_0 , \ldots ) = 0$) and $\Pi_1$ sentences saying just that, e.g., a certain Diophantine equation is unsolvable. The first ones can be proved not to follow from PA by restricting ourselves to a proper initial “segment” of a nonstandard model of PA. For $\Pi_1$ sentences, in some sense proving their consistency show they are true (as otherwise PA is inconsistent). Naturally, concerning statements in set theory, models of ZFC are more malleable, as the method of forcing shows.

Comment by doetoe

2010-08-09T16:29:39Z

Solution: Chg gur pbva rknpgyl va gur zvqqyr. Gura jurerire lbhe nqirefnel chgf gur pbva, chg vg ng gur fnzr cbfvgvba nsgre ebgngvba ol 180 qrterrf nebhaq gur pragre. (Guvf cbfvgvba vf nyjnlf serr).