The digits of e in pi
Well, it's late, I'm tired, my mind is wandering, and maybe that's why I thought of this.
Can it be said that somewhere in the decimals of pi=3.14159..., we can find any arbitrary number of digits of e=2.718281828459045...
For example, of course we can find 7 in pi.
But can we find (or even claim) that the first million digits of e are in pi?
It may exist, but I doubt you'd find it.
If each of the digits 0-9 in are equally likely, then the probability that the first digits of e occur somewhere in the first n digits of should be around the following:
The above is only approximate.
Under the assumption that is normal, I believe your probability is wrong as
Originally Posted by jamix
edit ignore this post, its wrong
I know this isn't your question, but on a related note, the first 200 million digits of pi can be searched here.
Thanks for noticing that. Hopefully I'll do better this time.
The question of whether there exists particular objects in sequence, within another RANDOM sequence of n objects can be solved using recurrent sequences.
Letting denote the above probability, and conditioning on \whether or not the first object in the sequence, is also the first object in , we see we have the following relation:
where is the probability of any particular object in occurring in a given position in the sequence.
In this particular question, if we assume random/normal conditions, then we would have and for .
Unfortunately even now we cannot efficiently solve the above for since we have to calculate for each first. This is doable since calculating is easy for any , but I don't suppose anyone sees any shortcuts?
BTW, i never studied this and i am basically guessing.
I'm not sure whether the questioner is really interested in the specific case of "is a part of e inside pi", or the general question of "will an irrational number contain an arbitrary finite sequence if you go through enough digits".
If the interest is in the second question, read on...
The answer to the second question is no, it is not necessary. The simplest way to prove it is to show that an irrational number can be constructed without containing a sequence, x.
x = the finite sequence you are looking for
y = all but the last digit in that sequence
Then the following number is irrational because it never repeats:
(if the last digit of x was zero, youd need to use a different number as a spacer, obviously!)