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Given that both pi and e's decimal places are completely random and infinite... I was wondering about how people say that "Every phrase ever uttered can be expressed in the digits of pi" -- which, for all intents and purposes, is true. But can e's decimal places (infinite) be expressed in pi's decimal places? Am I even asking a coherent question (like embedding infinite digits into infinite digits)?

For example, the string "27182818" is first found at position 73154827 of pi's digits; it apparently occurs 3 times in a searchable database of over 200M digits of pi. However, apparently the string "271828182" does not occur at all within the first 200M digits of pi.

Now, how about all the infinite digits of e?

Sure, limits to infinity would tell you, purely based on a probabilistic standpoint, that the chance of infinite digits (each with a probably of 1/10) occurring in a sequence of infinite digits approaches 0... but that doesn't FEEL like a satisfying analysis of the question.

vawd_gandi
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2 Answers2

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Even if $\pi$ is normal, it doesn't necessarily contain infinitely long patterns. However, I think your question is ultimately unknown:

My interpretation of this is: let $\pi(i)$ be the $i$'th digit of $\pi$ $(i=0,1,2,\cdots)$ indexed so that $\pi(0)=3,\pi(1)=1,\pi(2)=4,\ldots$. Similarly, let $e(i)$ by the digits of $e$. You're asking if there exists a $k$ such that $\pi(k+n)=e(n)$ for all $n\geq 0$.

If such a $k$ exists, this means you could write $\pi=A+10^{-k}e$, where $A$ is a rational number (actually a finite digit decimal). Unfortunately, I don't think there's a known way of proving/disproving this. Specifically we don't even know if $\pi+e$ is irrational, let alone whether or not $\pi,e$ are algebraically independent.

Community
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Alex R.
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  • More than a rational number, $A$ should have finite decimal representation, no ? – leonbloy Jun 27 '16 at 23:44
  • @leonbloy: Yea absolutely. – Alex R. Jun 27 '16 at 23:59
  • Wow, we don't know $\pi + e$ is irrational? I see what people mean when they say proving statements about irrationality (and transcendentalness) is hard.. – MT_ Jun 28 '16 at 00:05
  • @Soke On the other hand, we know that (at least) one of $\pi+e$ and $\pi\cdot e$ is irrational (in fact, transcendental). (This makes a nice little exercise, if you haven't seen it before. Hint: what degree-2 polynomial has $\pi$ and $e$ as roots?) – Steven Stadnicki Jun 28 '16 at 00:15
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    @StevenStadnicki: I'm not an expert in this area, but that's a very cute observation. Is it possible, for example through something like Vieta's relations, to form enough equations for questions like this to conclude which element of the pair is indeed irrational? Clearly not for $\pi,e$. – Alex R. Jun 28 '16 at 00:19
  • $(x-\pi)(x-e) = x^2 - (\pi + e)x + e\pi $, but $\pi$ and $e$ are transcendental so this polynomial can't have rational coefficients. Where does transcendentality follow from? – MT_ Jun 28 '16 at 00:23
  • @Soke The quadratic formula - both $\pi$ and $e$ can be expressed by a quadratic in $b=-(\pi+e)$ and $c=e\pi$. If both were algebraic then $\pi$ and $e$ would then be algebraic. – Steven Stadnicki Jun 28 '16 at 00:31
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    @AlexR. As far as I know, there isn't any way - given that the two numbers involved can be 'generic' in a very broad sense, it's unlikely to be able to solve questions like that through relatively lightweight means. – Steven Stadnicki Jun 28 '16 at 00:57
  • +1, but I think this answer could be improved with an explanation/definition of what "$\pi$ is normal" means, or maybe a link to an explanation. – Pedro A Jun 28 '16 at 18:07
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It seems difficult to disprove, but also extremely improbable. Following Alex R.'s answer, if it were true then we'd have $e = 10^k \pi - A $ for some positive integers $k,B$. And if were true also in other base (say, 7) we'd also have $e = 7^j \pi - B $ for some other integers $j,B$. But that would imply that $\pi$ is rational. Hence, it cannot be true in all bases.

leonbloy
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