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So I figured of a way to map any infinite sequence of (infinitely many non-zero) integers to $2$ numbers. Let the sequence be,

$$ f(1), f(2),f(3),\dots$$

Now, we construct the following:

$$ S =e^{- f(1) \delta} + e^{- f(2) \delta} + e^{- f(3) \delta} + \dots $$

Let us assume, the below converges to a number $c \neq 0$:

$$ c = \lim_{\delta \to 0} S \delta =\lim_{\delta \to 0} (e^{- f(1) \delta} + e^{- f(2) \delta} + e^{- f(3) \delta} + \dots) \delta $$

(This can be calculated using this with $f = e^{-x}$ and $k = \infty$)*. Then, our numbers are $c$ and the $1$ since we raised the $S$ to the power $1$. Let us, now assume the series converges to $0$ or diverges in that case raise the original series to the power $\lambda$. Such that it converges to $c \neq 0 \neq \infty$. Then we have:

$$ c = \lim_{\delta \to 0} S^\lambda \delta =\lim_{\delta \to 0} (e^{- f(1) \delta} + e^{- f(2) \delta} + e^{- f(3) \delta} + \dots)^\lambda \delta $$

Hence, any integer series can be (one-to-many) mapped to $2$ numbers $(c,\lambda)$ where $c$ tells you the type of function and $\lambda$ tells you the density.

Question

Will this scheme always work? Are there any other mappings between an infinite integer sequence and two real numbers which accomplish the same? Is there a nice way to see the inverse mapping and see it's sensitivity to say $(c,\lambda) \to (c + \epsilon, \lambda)$ or $(c,\lambda) \to (c, \lambda + \epsilon)$?

Arctic Char
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More Anonymous
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  • This is rather unclear. What does $S$, for instance, have to do with the rest of the post (or are there supposed to be negative signs in the exponents)? What is "the original series to the power 1"? – Greg Martin Nov 13 '20 at 06:24
  • @GregMartin fixed some typos and added the reference to $S$. The original series is $S$. – More Anonymous Nov 13 '20 at 06:27
  • So $S$ is a generalized Dirichlet series, and you're creating notation that captures the assertion that $S\sim c\delta^{-\lambda}$ as $\delta\to0$ (really you want a one-sided limit, depending on whether the $f(n)$ are eventually positive or eventually negative). I think the notation $S\sim c\delta^{-\lambda}$ is clearer than this new notation. You ask will this "always work" and whether other methods "will accomplish the same", but neither of these is a well-defined mathematical question at this point. – Greg Martin Nov 13 '20 at 08:12

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