User mrm - MathOverflowmost recent 30 from http://mathoverflow.net2013-06-18T22:53:10Zhttp://mathoverflow.net/feeds/user/2166http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/8056/what-are-good-non-english-languages-for-mathematicians-to-know/8093#8093Answer by mrm for What are good non-English languages for mathematicians to know?mrm2009-12-07T09:53:10Z2009-12-07T09:53:10Z<p>French, German, Russian. It's a pity English dominates so much.
French reads beautifully. (Also if you read older papers by
Hadamard, Stieltjes or Levy, the first thing you'll notice is their
extreme honesty (they don't use difficult terms, they
define everything, they don't try to make their arguments
<em>appear</em> difficult) also calculations are not condemned nor indulged with (they often just write an equation and say in words how the rest of the computation goes)).</p>
http://mathoverflow.net/questions/7039/typical-value-of-totient-function/7147#7147Answer by mrm for Typical value of totient functionmrm2009-11-29T17:49:24Z2009-11-29T17:49:24Z<p>Let me also mention the following:
You can adapt Schoenberg's result to prove that
1/M * {N <= n <= N + M : phi(n) / n <= t} --> F(t)
uniformly in t, where F is a distribution function. The proof goes
by computing the moments sum((phi(n)/n)^k , N <= n <= N + M).
You can probably get a O(loglog N / log N) rate of convergence
(as was done by Levin ... if I recall correctly). </p>
http://mathoverflow.net/questions/7969/irreducible-polynomials-with-constrained-coefficients/8016#8016Comment by mrmmrm2009-12-06T20:19:55Z2009-12-06T20:19:55Zyes, indeed, i realized that the problem is different only after having posted the message -- my browser doesn't display TeX properly so I made a guess on what the question was... (not a good idea)http://mathoverflow.net/questions/7039/typical-value-of-totient-function/7147#7147Comment by mrmmrm2009-11-29T17:59:39Z2009-11-29T17:59:39Z+ If this is of interest the distribution function F(t) decays doubly exponentially at 0, that is F(1/t) << exp(-C*exp(t)) for some constant C. This was investigated by Erdos and more recently Weingartner (<a href="http://www.mrlonline.org/proc/2007-135-09/S0002-9939-07-08771-0/S0002-9939-07-08771-0.pdf" rel="nofollow">mrlonline.org/proc/2007-135-09/…</a>). Asymptotics for 1 - F(t) when t is close to 1 where studied by Tenenbaum and Toulmonde (a reference is in the paper above). In this case the asymptotic behaviour is more tame. There should be no problem adapting all these results to the case of the interval [N; N + M] with M as you described...