20111116, 23:44  #67  
Apr 2010
234_{8} Posts 
Quote:
Last fiddled with by ccorn on 20111116 at 23:47 

20111117, 00:03  #68  
Mar 2010
2^{6}×3 Posts 
Quote:
Your geometrical interpretation of a problem was known before. In fact, I saw this problem quoted this geometrical way. 

20111129, 01:27  #69  
Apr 2010
2^{2}×3×13 Posts 
Asymptotic plausibility of Literka's first problem
Quote:
The hypercube has lots of symmetries. In particular, its vertices are spaced evenly across S_{[I]n[/I]1}.* Therefore, the ratio f_{n} of the measures of A_{[I]n[/I]1} and S_{[I]n[/I]1} provides sort of an expected value for the fraction of all hypercube vertices that belong to A_{[I]n[/I]1}. If f_{n} were to drop below 1/2 for some n, this would suggest that there are counterexamples. As it turns out (left as an exercise for the readers), where I(z;a,b) denotes the Regularized (incomplete) Beta function. One finds that , the within1standarddeviation probability of a normal distribution. This does not prove Literka's first problem, but at least it does not indicate a contradiction. *: "Spaced evenly" meaning that the Voronoi regions (on S_{[I]n[/I]1}) of the vertices are all congruent. Last fiddled with by ccorn on 20111129 at 02:01 

20120321, 13:45  #70 
Mar 2010
C0_{16} Posts 
A problem just for fun.
Let f be a concave function (example of concave function is f(x)= log x ) defined on the interval [0,1].
Let f_{1} be the first norm of f. Let f_{2} be the second norm of f. First norm means integral of absolute value of f over [0,1]. Second norm means square root of the integral of f^{2} over [0,1]. With the above assumptions show that sqrt(2)*f_{1} >= f_{2} Have a fun. Hint: solution is very simple. Last fiddled with by literka on 20120321 at 13:56 
20120321, 17:16  #71 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3×29×83 Posts 
You mean f'' < 0 for concave?

20120321, 17:31  #72 
Mar 2010
2^{6}·3 Posts 
Yes, but only in the case function has a second derivative. (My theorem concerns broader class of functions.) I used a definition as it is in http://en.wikipedia.org/wiki/Concave_function.
Sorry, I forgot to add an assumption f>=0, of course. I figured out this problem long time ago, when I was a student. That is why I can hardly remember it. Of course without the assumption f>=0 theorem is not true. Last fiddled with by literka on 20120321 at 17:36 
20120323, 18:42  #73 
Mar 2010
2^{6}·3 Posts 
I realized that there is probably an open problem associated with the above theorem. So I decided to write a webpage with a proof of this theorem and with a problem, which I believe is an open problem. See my page www.literka.addr.com/mathcountry/concave.htm.
Last fiddled with by literka on 20120323 at 18:51 
20120324, 13:01  #74  
Jan 2010
germany
2·13 Posts 
Quote:
Thanks for the interesting link ! I love such puzzles. (that are easy to understand, but not so easy to solve) I also have a problem, that has a relation to mersenne numbers. But maybe the problem is just "a problem of the minute". I do not know, yet. 1.) Let p be a prime number with p>=3 and p is not a Wieferich prime. (Wieferich prime p is prime number with property: 2.) Let k be a natural number with k>=2 Is it possible to chose p and k as descibed in (1) and (2), so that is prime and satisfies the congruence: 

20120324, 13:56  #75  
"Forget I exist"
Jul 2009
Dartmouth NS
2×3×23×61 Posts 
Quote:
Code:
forprime(p=1,1000,for(k=2,100*p,if((2^p)<(2*k*p^2+1),print(p","k);break()))) 

20120324, 14:43  #76  
Jan 2010
germany
32_{8} Posts 
Quote:
I see. From what I understand is your breakcondition. But you can also set a maximum limit for k direct. Maybe this speeds up the calculation a little bit ? <> Therefore: > so k must be smaller as 

20120324, 14:48  #77 
"Forget I exist"
Jul 2009
Dartmouth NS
20E2_{16} Posts 
2*m*p+1  2^p1 so really all your asking is can 2*m*p+1 divide 2 Mersennes but with only one p m=k*p is what you can check for.

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