How to find 'self locating digits' in irrational numbers

prane

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 :)

SteveL27

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%...louffe_formula

prane

http://upload.wikimedia.org/wikipedi...71ed789415.png

how does that make sense? A forumla for the nth digit that doesn't depend on n?!

mfb

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.