User doetoe - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T08:33:59Z http://mathoverflow.net/feeds/user/8216 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76191/does-the-automorphism-group-of-a-cone-determine-the-cone/76195#76195 Answer by doetoe for Does the automorphism group of a cone determine the cone? doetoe 2011-09-23T09:55:29Z 2011-09-23T09:55:29Z <p>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).</p> http://mathoverflow.net/questions/62820/pythagorean-5-tuples/64135#64135 Answer by doetoe for Pythagorean 5-tuples doetoe 2011-05-06T16:44:42Z 2011-05-06T16:44:42Z <p>Another way to generate parametrisations is to repeatedly apply a parametrisation of triples:</p> <p>Say we start with $$(x(s), y(s)) = (\frac{2s}{s^2 + 1}, \frac{s^2 - 1}{s^2 + 1}),$$ satisfying $$x(s)^2 + y(s)^2 = 1.$$ </p> <p>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 $x^2 + y^2 + z^2 = 1$.</p> <p>Repeating this, we get your case:</p> <p>$$(\frac{2u}{u^2 + 1}\cdot\frac{2t}{t^2 + 1}\cdot\frac{2s}{s^2 + 1}, \frac{2u}{u^2 + 1}\cdot\frac{2t}{t^2 + 1}\cdot\frac{s^2 - 1}{s^2 + 1}, \frac{2u}{u^2 + 1}\cdot\frac{t^2 - 1}{t^2 + 1}, \frac{u^2 - 1}{u^2 + 1})$$ 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).</p> <p>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.</p> http://mathoverflow.net/questions/62720/probability-and-math-puzzle-books-references/63190#63190 Answer by doetoe for probability and math puzzle books/references doetoe 2011-04-27T15:45:46Z 2011-04-27T15:45:46Z <p>On the site of the <a href="http://www.imo-official.org" rel="nofollow">International Mathematics Olympiad</a> all past problems can be found. They tend to be quite difficult but no advanced math is required.</p> http://mathoverflow.net/questions/12085/experimental-mathematics/37469#37469 Answer by doetoe for Experimental Mathematics doetoe 2010-09-02T08:22:34Z 2010-09-02T08:22:34Z <p>According to this <a href="http://plus.maths.org/content/butterfly-flap-felt-across-world" rel="nofollow">obituary of Edward Lorenz</a> 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.</p> http://mathoverflow.net/questions/29323/math-puzzles-for-dinner/34846#34846 Answer by doetoe for Math puzzles for dinner doetoe 2010-08-07T16:27:49Z 2010-08-07T16:27:49Z <p>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?</p> http://mathoverflow.net/questions/34699/approaches-to-riemann-hypothesis-using-methods-outside-number-theory/34783#34783 Answer by doetoe for Approaches to Riemann hypothesis using methods outside number theory doetoe 2010-08-06T16:09:44Z 2010-08-07T02:50:45Z <p>I have no idea to what extent the <a href="http://arxiv.org/pdf/math/0211398.pdf" rel="nofollow">idea of Saharon Shelah</a>, about which I read in David Ruelle's popular account <a href="http://books.google.com/books?id=B3A1bjOkOaEC&amp;lpg=PA71&amp;dq=%22riemann%20hypothesis%22%20undecidable%20peano&amp;pg=PP1#v=onepage&amp;q=%22riemann%20hypothesis%22%20undecidable%20peano&amp;f=false" rel="nofollow">the mathematician's brain</a> 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). </p> <p>EDIT: Here is the quote from Shelah's paper:</p> <p>2.3 Dream: Prove that the Riemann Hypothesis is unprovable in PA, but is provable in some higher theory.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/29323/math-puzzles-for-dinner/34846#34846 Comment by doetoe doetoe 2010-08-09T16:29:39Z 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).