Wikipedia:Reference desk/Archives/Mathematics/2008 November 28

Mathematics desk
< November 27 << Oct | November | Dec >> November 29 >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.


November 28

edit

Capital pi notation

edit

Hi. What means :

 

I do not know how to read Pi notation here.

it is from paper Andrey Morozov : Universal Mandelbrot Set as a Model of Phase Transition Theory--Adam majewski (talk) 13:20, 28 November 2008 (UTC)[reply]

The denominator is the product of the Gm's. See Multiplication#Capital pi notation. Zain Ebrahim (talk) 13:25, 28 November 2008 (UTC)[reply]
You've just inspired me to eat a big piece of pie. :-) StuRat (talk) 15:54, 28 November 2008 (UTC)[reply]

Maybe the whole formule will show the problem better ( it is a product of divisors of m or somthing like that ) :

 

This is not explained on the page about multiplication. --Adam majewski (talk) 16:19, 28 November 2008 (UTC)[reply]

Sorry, I thought you just wanted to know what the capital pi means. I have no idea what those three vertical dots are for - maybe they mean n divides m (as though it was a vertical bar) so the m's in the denominator are all the multiples of n. Just guessing. Zain Ebrahim (talk) 16:42, 28 November 2008 (UTC)[reply]
The first section of the paper you linked to is saying the following. We define:
 
 
so that the set of solutions to Fn(x,c)=0 for a given value of c are the fixed points of fn(x,c). Some of these points will have order n under the iterated map xf(x,c) - but some of them will have an order m that is a factor of n. So we define Gn(x,c) to be the polynomial whose roots are the points with order exactly n. Then
 
because every root of Fn(x,c) is a root of some Gm(x,c) for some m that is a factor of n - and possibly m=n. So
 
You can use a similar process to iteratively find the cyclotomic polynomials. n:m seems to mean m is a proper divisor of n. Gandalf61 (talk) 17:39, 28 November 2008 (UTC)[reply]
You can give an explicit (non-recursive) expression for G in terms of F using the Möbius inversion formula:  , where   is the Möbius function. Algebraist 17:48, 28 November 2008 (UTC)[reply]

Thx for answers. (:-)) See also :

  • [Do1] V Dolotin , A Morozow : On the shapes of elementary domains or why Mandelbrot set is made from almost ideal circles ?
  • [Do2] V Dolotin , A Morozow : Algebraic Geometry of Discrete Dynamics. The case of one variable

--Adam majewski (talk) 08:18, 29 November 2008 (UTC)[reply]

Name of theorem/conjecture

edit

What is the name of the conjecture/theorem that says:

For any positive integer k, there are only a finite number of positive integer solutions (a, b, c, d) such that  .

Dragons flight (talk) 19:19, 28 November 2008 (UTC)[reply]

According to Catalan's conjecture, it is Pillai's conjecture. JackSchmidt (talk) 20:01, 28 November 2008 (UTC)[reply]
Thanks. Dragons flight (talk) 20:26, 28 November 2008 (UTC)[reply]

Google Billboard Riddle

edit

Several years ago, Google ran a billboard ad looking for potential employees:

[1] (work-safe)

The question asked for the "first 10-digit prime found in consecutive digits of e".

How does one solve this?

Acceptable (talk) 22:43, 28 November 2008 (UTC)[reply]

You write a computer program to go looking for it, either equipped with the decimal expansion of e and a list of all 10-digit primes or with algorithms for generating digits of e and testing 10-digit numbers for primality. There's nothing you can do smarter than just searching that I'm aware of. Algebraist 22:48, 28 November 2008 (UTC)[reply]
The prime number theorem says that about 4% of ten-digit numbers can be expected to be prime, so you probably won't have to go very far in e. It's not worth having a list of prime ten-digit numbers. For each ten-digit number a you want to test, divide by each prime p between 2 and 100,000 (the square root of 1010). The first a you find that's not divisible by any of these is your number. 67.150.252.236 (talk) 23:01, 28 November 2008 (UTC)[reply]
There are much better primality tests if you need speed. I'm not sure if that'll matter hear, though, since it's a fairly small problem. Algebraist 23:25, 28 November 2008 (UTC)[reply]
Or the lazy man's way: interpret the question as the "first ten digit prime, that is found in consecutive digits of e" first meaning first in the usual order of natural numbers, then lazily assume e is a Normal number and just give the smallest ten digit prime, whatever it is. This would probably pick out employees who would either save them a lot of money, or embezzle a lot. :-)John Z (talk) 12:17, 29 November 2008 (UTC)[reply]
The number 0000000002 is definitely prime, and the concept of a 10-digit may be considered context-dependent. Bo Jacoby (talk) 23:21, 29 November 2008 (UTC).[reply]
Actually harder than one might think. I found 7427466391 which starts 98 digits in. Dragons flight (talk) 23:35, 29 November 2008 (UTC)[reply]
The easier way to solve it is to look at our article, which says that "a random stream of digits has a 98.4% chance of starting a 10-digit prime sooner." Yes, Wikipedia does, in fact, contain all of the information in the universe. « Aaron Rotenberg « Talk « 03:40, 30 November 2008 (UTC)[reply]
I put "first 10-digit prime found in consecutive digits of e" into the little search box I have for google on my browser and looked at the first answer that came up and it gave the answer. Google contains all the information in the universe and a lot besides :) I'm sure that must be easier. Dmcq (talk) 23:30, 30 November 2008 (UTC)[reply]