How to find 'self locating digits' in irrational numbers

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Discussion Overview

The discussion centers around the concept of 'self locating digits' in irrational numbers, specifically using Pi as a primary example. Participants explore methods to identify these digits efficiently, considering both theoretical and practical aspects of the problem.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes a method to find self locating digits by pairing decimal positions with their corresponding digits in Pi, suggesting an algorithmic approach.
  • Another participant notes the challenge of finding a fast method due to the lack of a formula for the n-th digit of Pi that does not require calculating all preceding digits.
  • A reference is made to the Bailey–Borwein–Plouffe formula, which allows for the calculation of the n-th digit of Pi in base 16 without needing prior digits.
  • There is a question raised about the implications of having a formula that calculates a specific digit without depending on n, prompting further inquiry into its validity.
  • A later reply questions the nature of digits in different bases, specifically addressing the concept of the 11th digit in base 10.

Areas of Agreement / Disagreement

Participants express differing views on the practicality of finding self locating digits and the implications of existing formulas, indicating that multiple competing perspectives remain unresolved.

Contextual Notes

The discussion highlights limitations related to the practicality of proposed methods and the dependence on specific bases for certain formulas, without resolving these issues.

prane
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Let us take the most mainstream irrational out there, (Pi).

Now write (Pi) as:

3.
14159265...

Let us number the decimals of Pi.

0 gets paired with 1
1 gets paired with 4
2 gets paired with 1
.
.
.
6 gets paired with 6

Thus 6 is a self locating digit.

My question is then how do we devise a method to find these self locating digits in a fast way.

This is how I've gone about it.

consider λ=(Pi)-3=0.14159265...

now consider digit number n, that is, the digit that is n places along:

0 gets paired with 1
1 gets paired with 4
2 gets paired with 1
.
.
.
6 gets paired with 6
.
.
.
n gets paired with x

We need an algorithm for finding out what x is without writing the whole of λ out.

Consider a new rational number, ρ.

Let ρ_n be the number which terminates at digit n.

Then ρ_n=0.1415926...x
ρ_(n-1)=0.1415926...w where w is the (n-1)th digit etc

Now consider (ρ_n)*10^(n+1) this is equal to 1415926...x. Let us call this new number β.

We can then find what x is by subtracting ρ_(n-1)*10^(n) from β.

Now if the x = n we have a self locating digit.

This method isn't terribly practical as we still have to basically know what ρ_n is.

Maybe I'll come up with an improvement after some thought but in the mean time I'd love to see what you guys come up with :)
 
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prane said:
My question is then how do we devise a method to find these self locating digits in a fast way.


This method isn't terribly practical as we still have to basically know what ρ_n is.

Right, the problem is that we have no formula for the n-th digit of pi that doesn't involve calculating all the preceding digits.

But it turns out that someone did find such a formula. It only works in base 16, but it's still amazing that such an algorithm exists.

http://en.wikipedia.org/wiki/Bailey–Borwein–Plouffe_formula
 
It is a formula for all digits. But if you want to calculate a specific one, you can calculate it quickly, the wikipedia page explains how.

By the way, what happens to the 11th digit in base 10, for example? A digit cannot be "11" there.
 

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