most recent 30 from http://mathoverflow.net 2013-05-21T17:40:08Z http://mathoverflow.net/feeds/user/9498 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44244/what-recent-discoveries-have-amateur-mathematicians-made/44357#44357 Nick S 2010-10-31T17:42:11Z 2010-10-31T17:42:11Z

<p><a href="http://en.wikipedia.org/wiki/Robert_Ammann" rel="nofollow">Robert Ammann</a> had some extremelly important contributions to the study of aperiodic tilings, and to Quasi-crystals.</p>

http://mathoverflow.net/questions/43148/basic-results-with-three-or-more-hypotheses/43212#43212 Nick S 2010-10-22T17:53:17Z 2010-10-22T17:53:17Z

<p>A very simple example (maybe too simple) which seems to be fit this cathegory is the Rolle's Theorem.</p>

http://mathoverflow.net/questions/41736/how-do-you-calculate-prove-the-length-of-n-the-number-of-non-repeating-digits-pr/41743#41743 Nick S 2010-10-11T02:53:40Z 2010-10-11T02:53:40Z

<p>When one writes an irreducible fraction $m/n$ as a periodic digit number all one does is to write</p> <p>$m/n=\frac{a}{999...9000.00}$</p> <p>So the number of digits before the period is the maximum of the power of $2$ and $5$ in $n$, i.e. wirting $n=2^\alpha 5^\beta k$ with $k$ relatively prime to $10$, the number of digits before the period is $\max{\alpha, \beta }$.</p> <p>I think that this will follow for free from the following lemma, whose proof is trivial:</p> <p>Lemma: if gcd$(k,10) =1$ then $k$ has a multiple of the form $999...9$.</p>

http://mathoverflow.net/questions/41473/bounding-the-roots-of-the-sum-of-two-polynomials/41519#41519 Nick S 2010-10-08T16:31:24Z 2010-10-08T16:31:24Z

<p>drvitek's solution actually answers to your question, but the computations migth not be as simple as you hope.</p> <p>All you have to do is find all the roots $x_1,...,x_n$ of the derivative $p'_1$ and the roots $y_1,.., y_m$ of $p_2'$.</p> <p>If the set ${ p_1(x_i)+p_2(y_j) }$ takes both positive and negative values there could be roots (could be false positive).</p> <p>If the set ${ p_1(x_i)+p_2(y_j) }$ takes only positive or only negative values there cannot be any root.</p> <p>But the problem is too vague, you need to provide more details. By the way the following "solution" actually satistfies all your requerements (but it is definitelly not what your are looking for):</p> <blockquote> <p>Blockquote Solution: No matter what $p_1, p_2$ are always report "there migth be roots". This is either true, or false negative</p> </blockquote>

http://mathoverflow.net/questions/41283/is-always-possible-to-slice-a-pizza-in-a-fair-way/41295#41295 Nick S 2010-10-06T16:42:49Z 2010-10-06T16:42:49Z

<p>Intermediate Value Theorem (for two people).</p> <p>We draw a $x$-axis through the centre of the disk. For an $\theta \in R$ we draw an axix which makes an angle of $\theta$-degrees with the x-axis (note that the axix for $\theta+\pi$ gives exactly the opposite direction).</p> <p>Lets look now at the two half disks defined by this $\theta$-axis.</p> <p>Let $f(\theta)$ denote the quantity of peperoni on the left half disk minus he quantity of peperoni on the rigth half disk (since the axis is oriented, left-rigth make sense, one looks in the direction of the axis).</p> <p>$f$ is continuous in $\theta$, and since $f(\theta)+f(\theta+\pi)=0$, either $f \equiv 0$ or $f$ takes both positive and negative values. Now the IVT completes the proof.</p>

http://mathoverflow.net/questions/39853/finding-solutions-to-fx-fx-k/39876#39876 Nick S 2010-09-24T17:33:29Z 2010-09-24T17:33:29Z

<p>DEs are my weak subject so I probably should not comment on this, but if I tried to solve it I would look for a solution of the type:</p> <p>$$A e^{\beta x} \sin(ax+b) + B e^{\beta x} \cos(ax+b) $$</p> <p>My reasoning being: $\sin (ax+b)$ looks like the natural choice but you get an extra constant so you need to introduce an exponential to kill it...</p>

http://mathoverflow.net/questions/44460/positive-function-with-a-zero-riemann-integral/44462#44462 Nick S 2010-11-01T17:30:12Z 2010-11-01T17:30:12Z

Just think about the measures of the sets $$A_n= \{ x \in [0,1] | f(x) \in [ \frac{1}{n}, \frac{1}{n-1} ) \} \,.$$

http://mathoverflow.net/questions/43924/when-are-infinitely-many-points-in-the-orbit-of-a-polynomial-integers Nick S 2010-10-28T03:39:49Z 2010-10-28T03:39:49Z

The open question 1 should be: If the orbit under zero contains infinitely many (different) integers, is it true that some power of $P$ takes integers into integers? The example I posted takes integers into integers, so it is still a counterexample.

http://mathoverflow.net/questions/43012/another-name-for-a-discrete-set Nick S 2010-10-21T06:53:14Z 2010-10-21T06:53:14Z

I think he means permutations...