Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Are the first $n$ digits of $\pi$ equal to the second $n$ digits for some $n\ge1$?

$$\pi \stackrel{?}{=}\underbrace{3.1415926\ldots}_{\text{the first }n\text{ digits}}~\underbrace{31415926\ldots}_{\begin{array}{c}\text{the same }n\text{ digits}\\\text{in the same order}\end{array}}~\underbrace{\ldots\ldots\ldots}_{\text{more digits}}$$

If so, is the smallest such $n$ known?

share|improve this question
    
Just head breaking question! –  Milingona Ana Jun 28 '13 at 9:45
2  
I don't think this is known or even conjectured. –  Chris Eagle Jun 28 '13 at 9:45
4  
Using a crude probabilistic heuristic: if this doesn't happen for $n \le m$, then the chance it happens at all is about $1/9\cdot 10^m$. So if it doesn't happen early it's very unlikely to happen at all. –  Chris Eagle Jun 28 '13 at 9:52

3 Answers 3

up vote 6 down vote accepted

It is very unlikely. If we take the digits of $\pi$ to be "random", the chance of a repeat after $n$ digits is $10^{-n}$. We can exclude however many digits we know do not repeat. Say we know it doesn't repeat by one million digits. Then the chance we have a repeat is less than $$\sum_{i=10^6}^\infty 10^{-i}=\frac {10^{-10^6}}{1-.1}=\frac 1{9\cdot 10^{10^6-1}} $$ which is extremely small.

share|improve this answer
    
Thanks. However, you seem to treat the event of a repeat at $i$ to be independent of, e.g., the event of a repeat at $i+1$. Is that justifiable? –  Glen The Udderboat Jun 28 '13 at 20:47
    
They are not strictly independent, but very close. Given that there is no repeat starting with digit $1,000,000$ it is very slightly more likely that there is a repeat starting at digit $1,000,001$, but it is a tiny effect. Even if it made it ten times more likely, that doesn't matter here at all. –  Ross Millikan Jun 28 '13 at 20:55
    
OK, I'll think a bit about that, because I don't immediately see that. Also, you might actually make it ten times more likely by correcting the $9$ to $.9$. :) –  Glen The Udderboat Jun 28 '13 at 20:58
    
I fixed that in the exponent. Thanks –  Ross Millikan Jun 28 '13 at 21:53

Take a look at Khinchin's constant. The continued fraction coefficients of most real numbers have a have a finite geometric mean that equals Khinchin's constant.

If Pi or one of these other real numbers had a big repeat as described, it would happen after 10 trillion digits (since we know Pi that far), and would introduce a truly huge continued fraction coefficient, enough to skew away from Khinchin. So far, there are only a handful of transcendental numbers that are not Khinchin numbers.

It would be better to look for Khinchin violation elsewhere, since numbers like $log(2)+log(7)+1/e$ can be checked in a split second. You could check sextillions of real numbers with the same amount of effort that it would take to extend Pi another 100 trillion digits.

share|improve this answer
3  
How could any finite repetition effect the value of the Kinchin liimt? –  David Speyer Jun 28 '13 at 15:15
    
I don't see why a single repetition (rather than two or more) would cause a huge CF coefficient. If $q=10^n - 1$, then we'd have a rational approximation $p/q$ with accuracy about $1/10q^2$, only slightly better than a typical convergent. The coincidence seems to me not to be in the goodness of fit, but in the denominator being a repunit. –  Erick Wong Jun 28 '13 at 16:57

Yes. This interesting and useful link shows that the sequence of digits $314$ repeats at the 2,120 decimal digit of Pi.

share|improve this answer
    
Sorry, I wrongly understood the question. –  user64494 Jun 28 '13 at 9:47
    
.. but a nice link... –  draks ... Aug 30 '13 at 21:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.