Someone posted this on Facebook. Is it true?

[Jamin2112]

I figured this would be most relevant in the Analysis subforum. Someone posted the following on my Facebook page.

I'm not convinced. First of all, how can you prove that the decimal representation of pi has no repetition?

 

[micromass]

The proof that the decimal representation of $\pi$ is nonrepeating is not very difficult. Basically, you need to prove that $\pi$ is irrational. Proofs of this fact can be found in many undergraduate analysis books.

What the picture claims, however, is something stronger. They claim that any finite string of numbers appears somewhere in the decimal expansion of $\pi$. I don't think this has ever been proven, and I hope others may give a reference for this fact.

 

[Jamin2112]

So I'm starting to come up with a formalization of this claim.

If an infinite string of numbers 1-9, meaning an infinite ordered set S={S(1), S(2), S(3), ...}, where each S(k) has k in {0, 1, ..., 9}, does not that have repetition, that means there does not exists an N such that {S(1), S(2), ..., S(N)} = {S(N+1), S(N+2),..., S(2N)} = {S(2N+1), S(2N+2),..., S(3N)} = ... According to this picture, any finite string of numbers 1-9 is contained in the set of all ordered subsets of S.

 

[micromass]

So I'm starting to come up with a formalization of this claim.

If an infinite string of numbers 1-9, meaning an infinite ordered set S={S(1), S(2), S(3), ...}, where each S(k) has k in {0, 1, ..., 9}, does not that have repetition, that means there does not exists an N such that {S(1), S(2), ..., S(N)} = {S(N+1), S(N+2),..., S(2N)} = {S(2N+1), S(2N+2),..., S(3N)} = ... According to this picture, any finite string of numbers 1-9 is contained in the set of all ordered subsets of S.

The picture is wrong. For example

$0.1101001000100001000001...$

So I leave $n$ zeroes between the $1$'s. This has no repetition. But clearly, not every finite string is contained in this decimal expansion. For example $111$ or even $8$ are not present.

 

[Jamin2112]

The picture is wrong. For example

$0.1101001000100001000001...$

So I leave $n$ zeroes between the $1$'s. This has no repetition. But clearly, not every finite string is contained in this decimal expansion. For example $111$ or even $8$ are not present.

Now find a string not contained in the decimal expansion of pi.

 

[MathematicalPhysicist]

Well, this claim seems to be like the claim of the "Bible code".

As in you can't predict anything but you can see it after the event has happened, which is meaningless.

 

[jbunniii]

In order to prove this claim, I believe a sufficient condition is that $\pi$ is a normal number. Unfortunately, it is not known whether $\pi$ is normal. Moreover, it is not even known whether every decimal digit $0,1,\ldots,9$ appears infinitely often in the decimal representation of $\pi$. Therefore, the validity of your claim is unknown.

 

[AnTiFreeze3]

Now find a string not contained in the decimal expansion of pi.

His example had a random, irrational number, but there was still a system which could be referred to, that could also prove that numbers like 8 can't show up. Having n number of zeroes in between 1's doesn't allow for a number like 8 to be present; we don't know of a system like that for pi, thus we can't rule out any strings in it.

 

[pwsnafu]

Now find a string not contained in the decimal expansion of pi.

He doesn't need to. The poster is still wrong. Read the first line again.

1. Pi is irrational
2. The decimal expansion of pi contains every finite decimal number.

The poster uses the words "meaning that", ie 1 is the same as 2. Micromass gave an example of an irrational number where you can't find every decimal number in its expansion.

It doesn't matter if 2 is correct or not. The implication that the poster makes is invalid.

 

[DrClaude]

For those who read French and have access to a good old library, there was an interesting article in Pour la science: Jean-Paul Delahaye, « Les nombres universe », juillet 1996.

In order to prove this claim, I believe a sufficient condition is that $\pi$ is a normal number. Unfortunately, it is not known whether $\pi$ is normal. Moreover, it is not even known whether every decimal digit $0,1,\ldots,9$ appears infinitely often in the decimal representation of $\pi$. Therefore, the validity of your claim is unknown.

There is in French the term "universe number", for which I can't find an equivalent in English, which has less constraints than a normal number. In a "universe number", any sequence of digits will appear at least once, whereas in a normal number any sequence will appear an infinite number of times. So indeed, if π is a normal number, then it is also a "universe number" and would contain every sequence, but there is no proof of π being a "universe number".

 

[Jamin2112]

He doesn't need to. The poster is still wrong. Read the first line again.

1. Pi is irrational
2. The decimal expansion of pi contains every finite decimal number.

The poster uses the words "meaning that", ie 1 is the same as 2. Micromass gave an example of an irrational number where you can't find every decimal number in its expansion.

It doesn't matter if 2 is correct or not. The implication that the poster makes is invalid.

I understand, but I'm still interested in this specific case. Might there be a formula for the nth digit of pi?

 

[micromass]

I understand, but I'm still interested in this specific case. Might there be a formula for the nth digit of pi?

Yes, there is such a formula, but it isn't helpful. See http://en.wikipedia.org/wiki/Bailey–Borwein–Plouffe_formula (the infinite sum can actually be reduced to a finite sum, see later in the article).

Like others has said, it is currently unkown whether all finite sequences occur in the decimal expansion of $\pi$ or not. So your original question can not be answered.

 

[jbunniii]

There is in French the term "universe number", for which I can't find an equivalent in English, which has less constraints than a normal number. In a "universe number", any sequence of digits will appear at least once, whereas in a normal number any sequence will appear an infinite number of times. So indeed, if π is a normal number, then it is also a "universe number" and would contain every sequence, but there is no proof of π being a "universe number".

Interesting, I hadn't heard this term before.

Just one more remark for the OP: it's pretty easy to prove that almost every real number is a normal number, where "almost every" is used in the technical sense that the set of real numbers which are NOT normal has measure zero.

Of course, you can still fit plenty of numbers (uncountably many, even) into a set of measure zero, so this in no way guarantees that $\pi$ is normal.

But this does mean that the property described on the Facebook poster is a very common one shared by almost all real numbers.