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 A more consistent definition of tetration of tetration for rational exponents andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/21/2007, 07:19 AM $f(b, t) = \exp_b^{t+2}(-1)$ describes the function your assumption produces, i.e.: $f(b, 0) = \exp_b^{2}(-1) = b^{b^{-1}} = b^{1/b}$ $f(b, 1) = \exp_b^{3}(-1) = b^{b^{b^{-1}}} = b^{b^{1/b}}$ $f(b, 2) = \exp_b^{4}(-1) = b^{b^{b^{b^{-1}}}} = b^{b^{b^{1/b}}}$ and so on, while still allowing $f(f(b, n), 1/n) = f(f(b, 1/n), n) = b$ and retaining the property $\lim_{n \rightarrow \infty}f(b, n) = {}^{\infty}{b}$. By the term orbit I'm refering to a term from dynamical systems where, given a point x, the sequence {x, f(x), f(f(x)), ...} is referred to as the orbit of f from x which is a way of referring to iteration without referring to the t in $f^t(x)$. By using it this way, though, I'm slightly misusing it, since its a sequence, and not a function. Here I'm using it as a function $f \ :\ t \rightarrow f^t(x)$, sorry if it was confusing. One of the reasons why I like the term 'orbit' so much is that it pairs nicely with iterate which, given t, is a function $f \ : \ x \rightarrow f^t(x)$. I've discussed these terms here as well. Andrew Robbins UVIR Junior Fellow Posts: 16 Threads: 2 Joined: Aug 2007 10/21/2007, 10:47 PM andydude Wrote:$f(b, t) = \exp_b^{t+2}(-1)$ describes the function your assumption produces, i.e.: $f(b, 0) = \exp_b^{2}(-1) = b^{b^{-1}} = b^{1/b}$ $f(b, 1) = \exp_b^{3}(-1) = b^{b^{b^{-1}}} = b^{b^{1/b}}$ $f(b, 2) = \exp_b^{4}(-1) = b^{b^{b^{b^{-1}}}} = b^{b^{b^{1/b}}}$ and so on, while still allowing $f(f(b, n), 1/n) = f(f(b, 1/n), n) = b$ and retaining the property $\lim_{n \rightarrow \infty}f(b, n) = {}^{\infty}{b}$. By the term orbit I'm refering to a term from dynamical systems where, given a point x, the sequence {x, f(x), f(f(x)), ...} is referred to as the orbit of f from x which is a way of referring to iteration without referring to the t in $f^t(x)$. By using it this way, though, I'm slightly misusing it, since its a sequence, and not a function. Here I'm using it as a function $f \ :\ t \rightarrow f^t(x)$, sorry if it was confusing. One of the reasons why I like the term 'orbit' so much is that it pairs nicely with iterate which, given t, is a function $f \ : \ x \rightarrow f^t(x)$. I've discussed these terms here as well. Andrew Robbins Ok, I see. Still a pathological function though. It cannot be made continuous at 1. If we define ${^0}e=e^{1/e}$ to conform with the limit of the tetraroots towards zero, then from the functional equation of tetration we must have: ${^1}e=e^{e^{1/e}}$. However, the limit from the left of y=1 can be found by using the following: $lim_{n \to \infty}{^{(10^n-1)/10^n}}e=y \Leftrightarrow\\ e=lim_{n\to\infty}y^{^{1/(10^n-1)}y} \Leftrightarrow\\ e=y^{y^{1/y}}$ Solving the latter with Maple, one gets $y=2.025415088 \neq {^1}e=2.71828...$, so there is a discontinuity at y=1. In short, no matter how this function is defined at y=0 and y=1, it cannot be made continuous simultaneously at y=0 and y=1. I prefer it defined as: ${^0}e=1$ ${^1}e=e$ ${^2}e=e^e$ $\cdots$ and let it do as it pleases in between, even if it ends up discontinuous. « Next Oldest | Next Newest »

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