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$\newcommand{\range}{\operatorname{Range}}$I have a function with an infinite number of variables which can be produced as the limit of a set of finite functions with increasingly many variables.

$x_i \in \mathbb{N}$ (does not include $0$)

$T_1(x_1) = \frac{2^{x_1}}{3^1}-\frac{3^0}{3^1}$

$T_2(x_1,x_2) = \frac{2^{x_1+x_2}}{3^2}-\frac{3^0\times 2^{x_1}+3^1}{3^2}$

$T_3(x_1,x_2,x_3) = \frac{2^{x_1+x_2+x_3}}{3^3}-\frac{3^0\times 2^{x_1+x_2}+3^1\times2^{x_1}+3^2}{3^3}$

$$T_d(x_1, \dots, x_d)= \frac{2^{x_1+\dots+x_d}}{3^d}-\frac{1}{3^d}\sum_{a=1}^{d-1} 3^{d-a-1}\times2^{x_0+\dots+x_a}$$

$$T=\lim_{d\to\infty} T_d(x_1,\dots,x_d)$$

I'd like to know the range of T. I suspect it is:

$$\range(T)=\left\{\frac{2k-1}{3^{n-1}}:k,n \in \mathbb{N}\right\}$$ But I have no idea how to prove this. As I have no experience dealing with functions formulated in this manner. Any help / advice is greatly appreciated.

  • It may be worth noting that if all $x_i = 2$ with $i>d$ then $T(x_1, \dots, x_d) = T(x_1,\dots,x_d,2,2,2,2,2,\cdots)$
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  • $\begingroup$ You need to be precise about what you mean by "$T=\lim_{d\to\infty} T(x_1,\dots,x_d)$", as this statement makes no sense for the usual meanings of "limits". $\endgroup$ – Eric Wofsey Oct 25 '16 at 6:17
  • $\begingroup$ @EricWofsey Any idea on some notation I could use? To me that notation makes perfect sense for what it is I am trying to do. I appreciate that this doesn't automatically make it the correct notation. I guess what i'm after is the limit of the ranges? Since $Range(T(x_1) \subset Range(T(x_1,x_2)) \subset Range(T(x_1,x_2,x_3)) \subset \cdots Range(T(x_1, \cdots , x_d))$ $\endgroup$ – Ben Crossley Oct 25 '16 at 6:33
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    $\begingroup$ You can say you want to take the union of the ranges. (By the way, you should really give different names to those functions, like let $T_d$ be the $d$-variable version. Also, you should say what their domains are!) $\endgroup$ – Eric Wofsey Oct 25 '16 at 6:36
  • $\begingroup$ Can't believe I didn't pop the domains in. School boy error. I'll fix that now. If I changed the notation to $T_1, T_2, \dots$ would I leave out the arguments? The reason I have been using only T is because they are essentially all the same formula, but with all arguments not in use automatically set to 2. $\endgroup$ – Ben Crossley Oct 25 '16 at 6:41
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First calculate the range of $T_d$ for all $d$.

Then, just show that for all $d$, $\range(T_d) \subset \range(T)$ -- reading the comments it seems like you have already done that.

Then you have that $\bigcup_{d=1}^{\infty} \range(T_d) \subset \range(T) $. I assume that $\bigcup_{d=1}^{\infty} \range(T_d) =\left\{ \frac{2k-1}{3^{n-1}}:k,n\in \mathbb{N} \right\}$ although admittedly I haven't put much thought into what $\range(T_d)$ must be.

Thus, to show that $\range(T)=\left\{ \frac{2k-1}{3^{n-1}}:k,n\in \mathbb{N} \right\} $, assuming that $\bigcup_{d=1}^{\infty} \range(T_d) =\left\{ \frac{2k-1}{3^{n-1}}:k,n\in \mathbb{N} \right\}$, all that remains to show is the opposite inclusion, i.e. $$\range(T) \subset \bigcup_{d=1}^{\infty} \range(T_d). $$ So let $y \in \range(T)$, show that there must exist some $d$ such that $y=T(x_1,\dots, x_d, 2, 2, \dots)=T_d(x_1, \dots, x_d)$, conclude that $y \in \range(T_d) \subset \bigcup_{d=1}^{\infty} \range(T_d)$ and thus that $\range(T) \subset \bigcup_{d=1}^{\infty} \range(T_d)$.

Since $\range(T) \subset \bigcup_{d=1}^{\infty} \range(T_d)$ and $\bigcup_{d=1}^{\infty} \range(T_d) \subset \range(T)$, one would have $\range(T)=\bigcup_{d=1}^{\infty} \range(T_d)$ and you would be done.

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    $\begingroup$ It's the $range(T_d)$ I'm struggling with. $\endgroup$ – Ben Crossley Oct 27 '16 at 8:16
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    $\begingroup$ Showing that $Range(T_{d+1}) \supset Range(T_{d})$ is very easy. Since: $T_{d+1}(x_1,\dots,x_d,2) = T_d(x_1,\dots,x_d)$ $\endgroup$ – Ben Crossley Oct 27 '16 at 8:18
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    $\begingroup$ Any ideas how I could go about finding $\range(T_d)$? $\endgroup$ – Ben Crossley Oct 27 '16 at 8:20
  • $\begingroup$ @BenCrossley-hobbyist To be honest I don't quite really understand the definition of the function very well or its motivation. If nothing else, one could say that: $$\range(T_d) = \left\{ n\in \mathbb{N}: n= \frac{2^{x_1+\dots+x_d}}{3^d}-\frac{1}{3^d}\sum_{a=1}^{d-1} 3^{d-a-1}\times2^{x_0+\dots+x_a} \ \text{for some }(x_1, \dots, x_d)\in \mathbb{N}^d \right\}$$ but my guess is that this obviously isn't very helpful for you. I guess the only comment I can think of is that a value for the left term determines all possible values for the right term, so first determine all possible values for $\endgroup$ – Chill2Macht Oct 27 '16 at 10:04
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    $\begingroup$ That is essentially how I got these formulas in the first place. Motivation: Proving this proves the Collatz conjecture. I've been working on it for several months and this is the final step. If my formula produces every odd number, then the conjecture is true, if it doesn't the conjecture is false. $\endgroup$ – Ben Crossley Oct 27 '16 at 10:17

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