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September 25

Riemann zeta function

Is there inverse function for zeta function.which it gives value of s. — Preceding unsigned comment added by 151.236.160.69 (talk) 16:00, 25 September 2015 (UTC)[reply]

No, the zeta function is not one-to-one. Sławomir
Biały 17:39, 25 September 2015 (UTC)[reply]

Well, it would exist as a multivalued function, from which you might be able to define a standard branch, much as with log and Log. But if you knew all the values of ζ^-1(0) then you'd have the solution to the Riemann hypothesis. --RDBury (talk) 20:46, 25 September 2015 (UTC)[reply]

Numbers that have digits of 1 and 0 only in every base

This idea came from a science fiction story (I think it was Sagan's Contact, but it was a long time ago so I might have got that wrong). In the story it was discovered that there is a region of digits of pi that consists of only ones and zeroes. Furthermore, the region had this same property when pi was expressed in any base. Is such a thing even possible? I have tried to manually find numbers that are ones and zeroes in multiple bases. Besides the trivial solutions of one and zero, I could only come up with four, which works up to base 4: (100)₂, (11)₃, (10)₄. There does not seem to be any more. Of course "does not seem" is a long way from a proof. I believe I can show that a sequence cannot go on indefinitely. The number must be greater or equal to all the bases in the sequence (except for the trivial solutions). Thus, for any given number, there will be a base that is sufficiently large that it does not work. But could there be any really long sequences? And how to find them? SpinningSpark 21:40, 25 September 2015 (UTC)[reply]

It was Contact (the novel, not the movie), but I don't think you have the details right. See the last paragraph of Contact (novel)#The Galaxy for more info. But you do raise an interesting question. If you don't require the bases to be consecutive then 64 = 1000000₂ = 1000₄ = 100₈ = 10 ₆₄ = 11 ₆₃ and there are similar examples. But even if you're just looking for numbers which are 0's and 1's in base 3 and base 4 you'd pretty much need to search by brute force to find more examples, though you might be able to apply some heuristic arguments to estimate the number of examples. --RDBury (talk) 00:12, 26 September 2015 (UTC)[reply]

82000₁₀ = 10111000₅ = 110001100₄ = 11011111001₃ = 10100000001010000₂. Double sharp (talk) 03:50, 26 September 2015 (UTC)[reply]

(Oh, and 3₁₀ = 10₃ = 11₂.) Double sharp (talk) 03:57, 26 September 2015 (UTC)[reply]

Here is the heuristic argument I was thinking of. The "probability" that a number n has only 0 or 1 digits base 3 is (2/3)^log₃n, which simplifies to n^-.37. Similarly, the The "probability" that n has only 0 or 1 digits base 4 is n^-.5. So the "probability" that n has only 0 or 1 digits in both bases is n^-.87, and the number of such n's is Σn^-.87. But this series diverges so you can expect there will be an infinite number of such n. On the other hand if you try the same thing with base 4 and base 5 you get a convergent series so you can expect only a finite number. The argument assumes the probabilities are independent, so you couldn't apply it to say base 4 and base 8, and it's easy to construct an infinite sequence of numbers that have digits 0 or 1 in those bases. Not a proof and I doubt the theory to formalize this argument exists, at least not yet. --RDBury (talk) 01:01, 26 September 2015 (UTC)[reply]

Another heuristic argument, from a post about a similar topic (the comment by Timothy Gowers): the set of numbers that can be written with only 0 and 1 in base n is a Cantor-like set with dimension log 2 / log n. So find the dimensionality of the intersection of the sets you want! (Just add the dimensionalities of the k sets you're considering, then subtract k − 1.) This also assumes that the probabilities are independent, so including 2 and 4 together would obscure the results (if it's OK in base 4, it has to be OK in base 2); but we'll overlook that flaw and see what happens. If we consider n being {2, 3} or {2, 3, 4}, the answer is positive; for {2, 3, 4, 5} it's negative (around −0.438), and it becomes more and more negative as we consider more bases (e.g. {2, 3, 4, 5, 6} gives −1.05). So it strongly suggests to me that these sets are empty, i.e. there are no really long sequences, and even 82000 is just a singular happy coincidence. Double sharp (talk) 04:28, 26 September 2015 (UTC)[reply]

More heuristic arguments. Note that all these arguments are for starting at base 2 and going up to consecutive bases, which may not be what you want. Double sharp (talk) 04:31, 26 September 2015 (UTC)[reply]

This all strikes me as kind of conflating two rather different things. The original question was talking about Contact, which dealt with the expansions of a real number in various bases, to the right of the radix point. But you seem to be talking about natural numbers, and their expansions to the left of the radix point. I really don't think these things are particularly related. (OTOH Gowers seems to do the same thing in his post, so maybe there's something I'm missing. Or maybe he was having an off day, but no one seems to have called him on it.)

Hmm, yeah, the OP first started on Contact and then gave examples of integers. I wonder which the OP had in mind. Double sharp (talk) 04:55, 26 September 2015 (UTC)[reply]

The irrational sequence inspired the question but calculating pi to the necessary number of digits in different bases is way beyond my capabilities. Besides, I already knew that no really long sequences of 1s and 0s had ever been found in pi despite it being calcualted to millions of digits. So yes, I then switched to natural numbers. I looked at the binary digits up to 1023 and calculated what number they would represent in bases up to 10. That did not go high enough to find 82000 in bases 2 and 3 unfortunately. Even if it did I probably would not have found it as I did not search my data comprehensively. Since then I hit on the idea of colour coding the results from different bases and then sorting the entire data set numerically. This makes it very easy to spot sequences. The data starts off with lots of rapid change of colour showing the existence of matching sequences, but rapidly degenerates into long blocks of solid colour, which kind of confirms Double sharp's result that the probability of really long sequences is vanishingly small. SpinningSpark 09:35, 26 September 2015 (UTC)[reply]

Did you try my version below? --Trovatore (talk) 18:24, 26 September 2015 (UTC)[reply]

Well, it's not my result, but thank you! (^_^) Double sharp (talk) 11:09, 26 September 2015 (UTC)[reply]

Did you try my challenge below? --Trovatore (talk) 04:44, 26 September 2015 (UTC)[reply]

Here's an interesting variant you might like: Does there exist an irrational number between 0 and 1 such that, for every natural number base b at least 2, there's a number N greater than 100 (depending on b) such that the digits from position N+1 to 2N are all zero? (Position 1 being the first one past the radix point.)

Variant: The digits from position N+1 to 2N are an exact repeat of the digits from position 1 to N?

Variant: For each b, there are infinitely many N such that the first statement is true?

Variant: For each b, there are infinitely many N such that the second statement is true?

Turns out that each of the variants has an affirmative answer ~~(though not all four for the same irrational, of course)~~. Need a hint? --Trovatore (talk) 01:38, 26 September 2015 (UTC) Oops -- yes all four for the same irrational. --Trovatore (talk) 01:39, 26 September 2015 (UTC) [reply]

PS. see OEIS: A258981 (integers with 0's and 1's base 3 and 4) and OEIS: A146025 (integers with 0's and 1's base 3, 4 and 5). Note that the second one only has three numbers, 0, 1, and the 82000 listed above. --RDBury (talk) 06:46, 26 September 2015 (UTC)[reply]

And as the heuristic arguments above suggest, the first sequence has lots of terms (its series diverges/its dimensionality is positive), while the second one only has three (its series converges/its dimensionality is negative). Double sharp (talk) 11:02, 26 September 2015 (UTC)[reply]

Did you try my challenge above? --Trovatore (talk) 08:11, 26 September 2015 (UTC)[reply]

Seems logical that the first one is possible: choose a number that complies for N=101 and b=2, nothing but 0's from position N+1, call it A, change position 2N+1 to a 1 and call that number B. Convert both to base 3, it's always possible to find a terminating decimal > A and < B, choose a value N larger than the number of digits (and > 100) add 0's and do the same as before, repeat for all b. For the infinitely many variant, you do the same but in order b= 2, 3, 2, 3, 4, 2, 3, 4, 5, 2, 3...

The other variants can be done the same way, I would think. Ssscienccce (talk) 19:23, 26 September 2015 (UTC)[reply]

Yeah, I think that works. You'll have a few details to check in making sure that manipulating the number in one base doesn't change earlier digits you're already relying on in an earlier base, but that should all go through.

You can skip some of those annoyances (and answer this sort of question faster) if you know the Baire Category Theorem. --Trovatore (talk) 19:31, 26 September 2015 (UTC)[reply]