Weird reciprocal sums of integers

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Discussion Overview

The discussion centers on the convergence or divergence of reciprocal sums of integers, specifically in the context of unconventional sets defined by binary representations of valid Java programs and other mathematical constructs. Participants explore various sets of integers and their properties, raising questions about the nature of these sets and their implications for convergence.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose examining the sum \(\sum_{x\in S}\frac{1}{x}\) for sets defined by binary representations of valid Java programs, questioning whether such sums converge or diverge.
  • One participant suggests that if the set is symmetric about 0, the sum is usually divergent, but acknowledges the need to exclude 0 from consideration.
  • Another participant notes that if the set of binary numbers is finite, convergence or divergence is not a relevant question, while an infinite set's behavior depends on its specific characteristics.
  • There is a contention regarding the definition of a valid Java program and its implications for the set's finiteness, with some asserting that the set of all integers representing a program is infinite.
  • A participant speculates that the simplest valid Java program would lead to a divergent sum, questioning the impact of adding new bytes to the binary representation on the gaps between valid programs.
  • Another participant introduces the idea that different enumerations of programs could lead to either convergent or divergent sums, depending on the enumeration method used.
  • Concerns are raised about the efficiency of the Java byte-code compiler and its potential impact on the density of valid programs in the set.
  • One participant suggests that to find a convergent set, it may be necessary to identify a set that "grows fast," mentioning Ramsey numbers as a potentially interesting example.
  • Another participant proposes the set of perfect numbers as a candidate for exploration, suggesting that it may lead to a finite sum.

Areas of Agreement / Disagreement

Participants express multiple competing views regarding the convergence or divergence of the sums based on different interpretations of the sets involved. The discussion remains unresolved, with no consensus reached on the nature of the sums or the definitions of the sets.

Contextual Notes

Limitations include the ambiguity surrounding the definition of valid Java programs, the finiteness of the sets being considered, and the specific enumeration methods that could affect convergence. These factors contribute to the complexity of the discussion without clear resolutions.

Dragonfall
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I'd like to see whether weird reciprocal sums of integers in the form \sum_{x\in S}\frac{1}{x}, where S is some unconventional set of integers, converges or diverge. Does anyone know any?

For example, \sum_{x\in S}\frac{1}{x} where S is the set of integers that, when expanded in binary, represents a valid Java program compiled on the x86 architecture. Is it convergent? Divergent?
 
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To answer your second question, would'nt you consider 0 an integer?

edit: Well in that case if S is symmetric about 0 you also see something interesting happening. But with 1/x as a summand the sum is usually divergent.
 
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I suppose we'll have to remove 0 from the sets.
 
The set of binary numbers would have to be finite so convergence or divergence isn't really a question. If the set is infinite, it depends on the set. The set of all squares is convergent, but the set of all even numbers is divergent.
 
"A valid Java program compiled on the x86 architecture" is meaningless.

"A valid program compiled for the Java VM" might not be.
 
uman said:
"A valid Java program compiled on the x86 architecture" is meaningless.

"A valid program compiled for the Java VM" might not be.

Well then let's go with "a valid program compiled for the java VM". Does it converge?

Vid said:
The set of binary numbers would have to be finite so convergence or divergence isn't really a question.

The set of integers is not finite, expressed in binary or otherwise.
 
I meant a set of binary numbers representing a program would have to be finite.
 
But the set of all integers representing a program isn't...
 
Vid said:
I meant a set of binary numbers representing a program would have to be finite.

Like Uman said, the set of all binary numbers which represent a program is not finite.
 
  • #10
Here's a 'top of the head' answer - I haven't actually worked it out. But I suspect the java example converges (edit - diverges). Here's my thinking: the simplest valid java program would just be some declaration of a constant (say a string), along with a bit of overhead. So there would be a sequence of valid programs where the constant value increases by 1 in the coding representation. The question is, what happens when a new byte needs to be added? Can we always get away with just adding one more byte to the binary (I don't know enough about java byte-coding to wnaswer that). Is it ever the case where the consecutive gaps between valid programs are large enough that the sum of their inverses converge? I strongly suspect the answer is no, but I don't have time right now to work out the details. If someone else doesn't beat me to it, maybe I'll work through it this weekend.

edit - oops. I meant to say diverges.
 
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  • #11
It depends on how Java "enumerates" programs. Two different enumerations of Turing machines, for example, might give a convergent and a divergent sum. For example, in one enumeration all valid programs are enumerated at powers of 2, and in the other it's the opposite.

EDIT: In fact, I think the Java example diverges. Any set containing an infinite arithmetic progression leads to a divergent sum, and maybe that Java binaries set contains such a sequence.
 
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  • #12
Dragonfall said:
It depends on how Java "enumerates" programs. Two different enumerations of Turing machines, for example, might give a convergent and a divergent sum. For example, in one enumeration all valid programs are enumerated at powers of 2, and in the other it's the opposite.

Yes. But I'd be very surprised if the java byte-code compiler used any kind of sparse coding, much less exponential. I think it's much more likely that it's going to be linear, so S will look a lot more like {x: x=m + 1} for some minimum m chars of overhead.

If we were talking about Godel-numbering, or some such scheme, then the "weird set" will almost certainly converge. But that's because Godel numbers are constructed so as to have lots of desirable properties when considered "all at once", as a number. There's no need for the Java VM to consider programs in that way - so it will fill the "valid program space" much more efficiently, i.e. much more densely.
 
  • #13
@Dragonfall, I posted while you were editing. Sounds like we're in agreement.

I'm not sure my last post was actually coherent - but I hope my point gets across.
What I'm trying to say is that, in order to find a set you're talking about that converges, you'll have to find some set S that "grows fast". One potentially interesting example that springs to my mind is Ramsey numbers.
 
  • #14
Oh, I thought we were taking the binary from a JAVA program and creating a set of integers from that. Not the set of integers that could represent any JAVA program. My bad.
 
  • #15
Ok what about this, the set of perfect numbers? You don't have to solve the question of whether there are infinitely many, since you can show (somehow) that the sum is finite (or even rational), a la Brun's constant.
 

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