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First, even though I think this is a fun question, it's not really research mathematics and I'm not sure it belongs on mathoverflow. (You know that some really smart people answer questions on math.stackexchange, right?) As was noted in Robert's answer, one is investigating the sequence $x_{n+1} = | \log(x_n)|$, which makes sense for all $x_0$ outside some countable set. Moreover, the periodic points will (surely) be transcendental, and it will be impossible to prove this fact (except for periods of length $1$). So what should one expect for the number of $+$ and $-$ signs? Here is a heuristic description of what happens, with the caveat that I have made no attempt to be rigorous (to do so, I would compare this with the probability theory of the Gauss–Kuzmin distribution).

First, choose $x_0$ in $[0,\infty]$ according to some probability measure $f_0(x)$, with cumulative probability distribution $F_0(x)$. Let $f_n(x)$ be the distribution of $x_n$. By definition, $f_n(x)$ must satisfy the following equation:

$$\int^{b}_{a} f_n(x) dx = \int_{e^a}^{e^b} f_{n-1}(x) dx + \int_{e^{-b}}^{e^{-a}} f_{n-1}(x) dx$$ for all $b \ge a \ge 0$. If $F_n(x) = \int^{x}_{0}f_n(t)dt$ is the cumulative distribution function of $x_n$, then this equation becomes: $$F_n(b) - F_n(a) = F_{n-1}(e^b) - F_{n-1}(e^a) + F_{n-1}(e^{-a}) - F_{n-1}(e^{-b}).$$ Letting $a = 0$, one obtains: $$F_n(z) = F_{n-1}(e^z) - F_{n-1}(e^{-z}).$$ Given some basic assumption on $F_0(z)$, the sequence of functions $F_n(z)$ converges to the unique increasing function $F(z)$ such that $$F(z) = F(e^z) - F(e^{-z}).$$ The independence of $F(z)$ on $F_0(z)$ implies that this function should describe the cumulative distribution function of $x_n$ for $n$ sufficiently large for almost all initial values $x_0$. By choosing random functions $F_0(z)$, one can estimate that $$F(1) \simeq 0.6518\ldots$$ Since the sign in the exponential is determined by whether $x_n > 1$ or not, this implies that the ratio of $-$ signs to $+$ signs (for almost all initial values, which presumably includes $x_0 = 2$) is roughly $1.872$ to $1$. This seems to confirm what you observed experimentally. Moreover, since the function $F(z)$ is strictly increasing, it follows by Kolmogorov's zero-one law that, for almost all initial values $x_0$, that there are arbitrarily long runs of $+$ signs, $-$ signs, etc.

Edit: To make this completely rigorous, one can make $\mathrm{R}^{+}$ a compact measure space given by the measure specified by $\mu([a,b]) = F(b) - F(a)$. The function $T:=|\log(x)|$ (modified so that $T(1) = 1$) is then measure preserving and (as is relatively easy to check) ergodic. The claims then follow from the Birkhoff Ergodic theorem, for suitable choices of test function $f$ (like the step function which is zero for $x < 1$ and one for $x > 1$). (BTW, I may have added a few more decimal digits above than was really justified.)

First, even though I think this is a fun question, it's not really research mathematics and I'm not sure it belongs on mathoverflow. (You know that some really smart people answer questions on math.stackexchange, right?) As was noted in Robert's answer, one is investigating the sequence $x_{n+1} = | \log(x_n)|$, which makes sense for all $x_0$ outside some countable set. Moreover, the periodic points will (surely) be transcendental, and it will be impossible to prove this fact (except for periods of length $1$). So what should one expect for the number of $+$ and $-$ signs? Here is a heuristic description of what happens, with the caveat that I have made no attempt to be rigorous (to do so, I would compare this with the probability theory of the Gauss–Kuzmin distribution).

First, choose $x_0$ in $[0,\infty]$ according to some probability measure $f_0(x)$, with cumulative probability distribution $F_0(x)$. Let $f_n(x)$ be the distribution of $x_n$. By definition, $f_n(x)$ must satisfy the following equation:

$$\int^{b}_{a} f_n(x) dx = \int_{e^a}^{e^b} f_{n-1}(x) dx + \int_{e^{-b}}^{e^{-a}} f_{n-1}(x) dx$$ for all $b \ge a \ge 0$. If $F_n(x) = \int^{x}_{0}f_n(t)dt$ is the cumulative distribution function of $x_n$, then this equation becomes: $$F_n(b) - F_n(a) = F_{n-1}(e^b) - F_{n-1}(e^a) + F_{n-1}(e^{-a}) - F_{n-1}(e^{-b}).$$ Letting $a = 0$, one obtains: $$F_n(z) = F_{n-1}(e^z) - F_{n-1}(e^{-z}).$$ Given some basic assumption on $F_0(z)$, the sequence of functions $F_n(z)$ converges to the unique increasing function $F(z)$ such that $$F(z) = F(e^z) - F(e^{-z}).$$ The independence of $F(z)$ on $F_0(z)$ implies that this function should describe the cumulative distribution function of $x_n$ for $n$ sufficiently large for almost all initial values $x_0$. By choosing random functions $F_0(z)$, one can estimate that $$F(1) \simeq 0.6518\ldots$$ Since the sign in the exponential is determined by whether $x_n > 1$ or not, this implies that the ratio of $-$ signs to $+$ signs (for almost all initial values, which presumably includes $x_0 = 2$) is roughly $1.872$ to $1$. This seems to confirm what you observed experimentally. Moreover, since the function $F(z)$ is strictly increasing, it follows by Kolmogorov's zero-one law that, for almost all initial values $x_0$, that there are arbitrarily long runs of $+$ signs, $-$ signs, etc.

First, even though I think this is a fun question, it's not really research mathematics and I'm not sure it belongs on mathoverflow. (You know that some really smart people answer questions on math.stackexchange, right?) As was noted in Robert's answer, one is investigating the sequence $x_{n+1} = | \log(x_n)|$, which makes sense for all $x_0$ outside some countable set. Moreover, the periodic points will (surely) be transcendental, and it will be impossible to prove this fact (except for periods of length $1$). So what should one expect for the number of $+$ and $-$ signs? Here is a heuristic description of what happens, with the caveat that I have made no attempt to be rigorous (to do so, I would compare this with the probability theory of the Gauss–Kuzmin distribution).

First, choose $x_0$ in $[0,\infty]$ according to some probability measure $f_0(x)$, with cumulative probability distribution $F_0(x)$. Let $f_n(x)$ be the distribution of $x_n$. By definition, $f_n(x)$ must satisfy the following equation:

$$\int^{b}_{a} f_n(x) dx = \int_{e^a}^{e^b} f_{n-1}(x) dx + \int_{e^{-b}}^{e^{-a}} f_{n-1}(x) dx$$ for all $b \ge a \ge 0$. If $F_n(x) = \int^{x}_{0}f_n(t)dt$ is the cumulative distribution function of $x_n$, then this equation becomes: $$F_n(b) - F_n(a) = F_{n-1}(e^b) - F_{n-1}(e^a) + F_{n-1}(e^{-a}) - F_{n-1}(e^{-b}).$$ Letting $a = 0$, one obtains: $$F_n(z) = F_{n-1}(e^z) - F_{n-1}(e^{-z}).$$ Given some basic assumption on $F_0(z)$, the sequence of functions $F_n(z)$ converges to the unique increasing function $F(z)$ such that $$F(z) = F(e^z) - F(e^{-z}).$$ The independence of $F(z)$ on $F_0(z)$ implies that this function should describe the cumulative distribution function of $x_n$ for $n$ sufficiently large for almost all initial values $x_0$. By choosing random functions $F_0(z)$, one can estimate that $$F(1) \simeq 0.6518\ldots$$ Since the sign in the exponential is determined by whether $x_n > 1$ or not, this implies that the ratio of $-$ signs to $+$ signs (for almost all initial values, which presumably includes $x_0 = 2$) is roughly $1.872$ to $1$. This seems to confirm what you observed experimentally. Moreover, since the function $F(z)$ is strictly increasing, it follows by Kolmogorov's zero-one law that, for almost all initial values $x_0$, that there are arbitrarily long runs of $+$ signs, $-$ signs, etc.

Edit: To make this completely rigorous, one can make $\mathrm{R}^{+}$ a compact measure space given by the measure specified by $\mu([a,b]) = F(b) - F(a)$. The function $T:=|\log(x)|$ (modified so that $T(1) = 1$) is then measure preserving and (as is relatively easy to check) ergodic. The claims then follow from the Birkhoff Ergodic theorem, for suitable choices of test function $f$ (like the step function which is zero for $x < 1$ and one for $x > 1$). (BTW, I may have added a few more decimal digits above than was really justified.)

Post Made Community Wiki
Source Link
user631
user631

First, even though I think this is a fun question, it's not really research mathematics and I'm not sure it belongs on mathoverflow. (You know that some really smart people answer questions on math.stackexchange, right?) As was noted in Robert's answer, one is investigating the sequence $x_{n+1} = | \log(x_n)|$, which makes sense for all $x_0$ outside some countable set. Moreover, the periodic points will (surely) be transcendental, and it will be impossible to prove this fact (except for periods of length $1$). So what should one expect for the number of $+$ and $-$ signs? Here is a heuristic description of what happens, with the caveat that I have made no attempt to be rigorous (to do so, I would compare this with the probability theory of the Gauss–Kuzmin distribution).

First, choose $x_0$ in $[0,\infty]$ according to some probability measure $f_0(x)$, with cumulative probability distribution $F_0(x)$. Let $f_n(x)$ be the distribution of $x_n$. By definition, $f_n(x)$ must satisfy the following equation:

$$\int^{b}_{a} f_n(x) dx = \int_{e^a}^{e^b} f_{n-1}(x) dx + \int_{e^{-b}}^{e^{-a}} f_{n-1}(x) dx$$ for all $b \ge a \ge 0$. If $F_n(x) = \int^{x}_{0}f_n(t)dt$ is the cumulative distribution function of $x_n$, then this equation becomes: $$F_n(b) - F_n(a) = F_{n-1}(e^b) - F_{n-1}(e^a) + F_{n-1}(e^{-a}) - F_{n-1}(e^{-b}).$$ Letting $a = 0$, one obtains: $$F_n(z) = F_{n-1}(e^z) - F_{n-1}(e^{-z}).$$ Given some basic assumption on $F_0(z)$, the sequence of functions $F_n(z)$ converges to the unique increasing function $F(z)$ such that $$F(z) = F(e^z) - F(e^{-z}).$$ The independence of $F(z)$ on $F_0(z)$ implies that this function should describe the cumulative distribution function of $x_n$ for $n$ sufficiently large for almost all initial values $x_0$. By choosing random functions $F_0(z)$, one can estimate that $$F(1) \simeq 0.6518\ldots$$ Since the sign in the exponential is determined by whether $x_n > 1$ or not, this implies that the ratio of $-$ signs to $+$ signs (for almost all initial values, which presumably includes $x_0 = 2$) is roughly $1.872$ to $1$. This seems to confirm what you observed experimentally. Moreover, since the function $F(z)$ is strictly increasing, it follows by Kolmogorov's zero-one law that, for almost all initial values $x_0$, that there are arbitrarily long runs of $+$ signs, $-$ signs, etc.