20171002, 02:39  #1 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{2}·3^{2}·61 Posts 
Repeating Digits of Pi
There is this thread in another forum.
I am interested in members of this board's input. Thanks in advance. https://forum.cosmoquest.org/forumdi...andTechnology ETA: I made correction to my post on the other board: 11 changed to 1010. Last fiddled with by a1call on 20171002 at 03:08 
20171003, 17:13  #2 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{2}·3^{2}·61 Posts 
The conversation has taken an interesting turn.
If an ideal random number generator is guaranteed to output any given number, given infinite trials, can we conclude that any given finite sequence of digits will occur somewhere in the pi's decimal expansion? 
20171003, 20:27  #3  
"Robert Gerbicz"
Oct 2005
Hungary
2·3^{2}·5·17 Posts 
Quote:
The interesting (not solved there) part is that: can you get x if you delete infinitely many digits of a general x number? If x is rational then the answer is yes (Why?). If x is irrational (like pi) then you can still get x: let H={d: 0<=d<10 and d appears in x infinitely many times}. It is clear that there exists k, that after the kth decimal digit of x you see only digits from H. The strategy is: keep the first k digits of x, then skip a digit, you can still find all digits (in order) of x, because these digits are in H, and hence these appears infinitely many times in x. After you have found the (k+1)th digit at the tth position then skip the (t+1)th digit and use induction from here. Note that this is also a (harder) proof for the easier case, when x is rational, and ofcourse H>=2 if x is irrational (though it is not used). Last fiddled with by R. Gerbicz on 20171003 at 20:31 Reason: typo 

20171005, 01:15  #4 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{2}·3^{2}·61 Posts 
As pointed out on the other board, I am having a lot of problems with the normal numbers:
https://en.wikipedia.org/wiki/Normal_number Does anyone else see that the definition is based on the false assumption that a number with infinite digital expansion in base 10 will never have terminating digits in any other base or that it can't have a bias in the density of digits in some bases and not in others? Thanks in advance for any clarifications and/or inputs. 
20171005, 02:17  #5  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
Last fiddled with by science_man_88 on 20171005 at 02:28 

20171005, 02:41  #6  
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{2}×3^{2}×61 Posts 
Quote:
Quote:


20171005, 02:43  #7 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 

20171005, 02:46  #8 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{2}·3^{2}·61 Posts 
The phrase "integer base" does not exist in the article either.

20171005, 04:20  #9  
Jun 2003
12221_{8} Posts 
Quote:


20171005, 05:58  #10 
"Rashid Naimi"
Oct 2015
Remote to Here/There
894_{16} Posts 
Thank you axn.
Will check that out next time I am on a PC. 
20171005, 15:58  #11 
Feb 2017
Nowhere
1010011011110_{2} Posts 
I don't know whether the number pi is normal to base ten (or to any other base). I default to the "assumption of ignorance" that it is normal, because I am ignorant of any reason to think it is not normal. (Also, the decimal digits calculated to date seem to pass the "randomness tests" thrown at them.)
However, the "output of a random number generator" idea for the digits might come to grief, at least in hexadecimal. I might well be wrong here, but if the digits can reasonably be viewed as sequentially put out by a random number generator, then it seems to me there should be no way to calculate the n^{th} digit without calculating all the previous digits first. And at least in hexadecimal, there is a method for finding the n^{th} digit without calculating all the previous digits first. 
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