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+--- Thread: Change of base formula using logarithmic semi operators (/showthread.php?tid=674)
Change of base formula using logarithmic semi operators - JmsNxn - 07/05/2011
Well, to start off, I'll give the formula for base conversion and then I'll give the motivation for the definition. Knowledge of Logarithmic semi-operators is required:
http://math.eretrandre.org/tetrationforum/showthread.php?tid=653
First we require definition of a new operator:
\( x\,\,\sum_{n=0}^{R} \bigtriangleup_{n}\,\,f(n) = ((x + f(0))\,\,\bigtriangleup_{1} f(1)) \,\,\bigtriangleup_{2} f(2))...\bigtriangleup_{R}\,\,f( R ) \)
which is basically the summation operator but with alternating operators each time. The bracketing is important. We evaluate the lower operators first.
With this, the change of base formula is given as:
\( x \in C;t \in R,\,\,t \ge 1 \)
\( \exp_b^{\circ t}(x) = \exp_e^{\circ t}(x \sum_{n=0}^{t-1} \bigtriangleup_{-n+1}^e\,\,\ln^{\circ n+1}(b)) \)
which I'll evaluate for 2 and 3, and hopefully you'll see the pattern. It's very simple. Remembering that for \( p \le 0 \)
\( e^{x\,\,\bigtriangleup_{p}^e\,\,y} = e^x \,\, \bigtriangleup_{p+1}^e\,\,e^y \)
\( e^{e^{ln(b)\cdot x + \ln^{\circ 2}(b)} = e^{ln(b)\cdot e^{\ln(b) \cdot x} = b^{b^x} \)
\( e^{e^{e^{(x\cdot \ln(b) + \ln^{\circ 2}(b))\,\,\bigtriangleup_{-1}^e\,\,\ln^{\circ 3}(b)}}} = e^{e^{e^{(x\cdot \ln(b) + \ln^{\circ 2}(b))} + e^{\ln^{\circ 3}(b)}}} = b^{b^{b^x}} \)
so on and so forth.
And for iterated logarithms, the base conversion formula is given as:
\( \log_b^{\circ t}(x) = \ln^{\circ t}(x) \sum_{n=0}^{t-1} \bigtriangledown_{n-t+2}^e \ln^{\circ t-n}(b) \)
I'll leave it to you to figure out how I got that. It's essentially the exact same concept.
The interesting thing is that we get an odd continuum sum-like object for defining tetration:
\( \text{sexp}_b(z) = \,\,^zb = \exp^{\circ z}(1 \sum_{n=0}^{z-1} \bigtriangleup_{-n+1}\,\,\ln^{\circ n+1}(b)) \)
which for natural arguments becomes:
\( \text{sexp}_b(k) = \,\,^kb = \exp^{\circ k}(1 \sum_{n=0}^{k-1} \bigtriangleup_{-n+1}\,\,\ln^{\circ n+1}(b)) = \exp^{\circ k}(\ln(b) + \ln^{\circ 2}(b)\,\, \bigtriangleup_{-1}^{e}\,\, \ln^{\circ 3}(b) \,\,\bigtriangleup_{-2}^e\,\,...\ln^{\circ k}(b)) \)
Now to just solidify it a bit. Given by gottfried's result here:
http://math.eretrandre.org/tetrationforum/showthread.php?tid=653&page=3
which he also got from mike3
the sequence \( a_n \)
\( a_1 = \ln(b)\\a_2 = \ln(\ln(b^b))\\a_3 = \ln(\ln(\ln(b^{b^b}))) \)
converges to a set value, and is reexpressable as:
\( a_2 = \ln(b) + \ln^{\circ 2}(b) \)
\( a_3 = (\ln(b) + \ln^{\circ 2}(b))\bigtriangleup_{-1}^e \ln^{\circ 3}(b) \)
which of course, looks a lot like our change of base formula when x is 1... So putting them together we get:
\( \text{sexp}_b(z) = \exp^{\circ z}(\ln^{\circ z}(\exp_b^{\circ z}(1)) \)
which is like coming around full circle so we can be sure this method works.
RE: Change of base formula using logarithmic semi operators - sheldonison - 07/06/2011
(07/05/2011, 11:44 PM)JmsNxn Wrote: .....
Now to just solidify it a bit. Given by gottfried's result here:
http://math.eretrandre.org/tetrationforum/showthread.php?tid=653&page=3
which he also got from mike3
the sequence \( a_n \)
\( a_1 = \ln(b)\\a_2 = \ln(\ln(b^b))\\a_3 = \ln(\ln(\ln(b^{b^b}))) \)
converges to a set value, and is reexpressable as:
\( a_2 = \ln(b) + \ln^{\circ 2}(b) \)
\( a_3 = (\ln(b) + \ln^{\circ 2}(b))\bigtriangleup_{-1}^e \ln^{\circ 3}(b) \)
which of course, looks a lot like our change of base formula when x is 1... So putting them together we get:
\( \text{sexp}_b(z) = \exp^{\circ z}(\ln^{\circ z}(\exp_b^{\circ z}(1)) \)
which is like coming around full circle so we can be sure this method works.
I'm not sure if this is the same thing or not, but earlier we looked at defining sexp_e(z) (or any other base) from cheta(z), where "k" is a constant chosen so the limit for sexp_e(0)=1. Here \( \text{sexp}_\eta \) and \( \text{slog}_\eta \) refer to the upper super exponential for base \( \eta \), cheta(z) and its inverse.
\( \text{sexpBaseChange}_e(z) = \lim_{n \to\infty}\log^{[n]}\text{sexp}_\eta(z+k+n) \)
\( k=\lim_{n \to \infty} {\text{slog}_\eta(\exp^{[n]}(1))-n \)
I'm not sure if this is the same as what you're getting at or not. We call this definition of sexp, the base change function, which is \( C_\infty \) at the real axis and, see this post, it is conjectured to be nowhere analytic. The base change function has many very interesting properties, in that at the real axis, there is a small 1-periodic wobble connecting it to the "correct" sexp function.
- Sheldon
RE: Change of base formula using logarithmic semi operators - JmsNxn - 07/07/2011
Oh no, this isn't that at all
This is just a simple formula to change the base of a power tower for natural integers.
however, this does provide an interesting continuum like sum for:
\( f(z) = x \cdot \ln(b) + \ln^{\circ 2}(b)\,\,\bigtriangleup_{-1}^e\,\,\ln^{\circ 3}(b)\,\,\bigtriangleup_{-2}^e\,\,\ln^{\circ 4}(b)...\,\,\bigtriangleup_{2-z}^e\,\,\ln^{\circ z}(b) \)
where f(z) is given by:
\( f(z) = \exp_e^{\circ -z}(\exp_b^{\circ z}(x)) \)
mike noted that as z goes to infinity, f(z) converges if x=1 and b=3.
I've seen these alternating operators before. They seem very intimidating, but I feel like there's something powerful about 'em.
RE: Change of base formula using logarithmic semi operators - sheldonison - 07/07/2011
(07/07/2011, 03:56 AM)JmsNxn Wrote: Oh no, this isn't that at allSo, what you have is a way to evaluate the tower of exponents and logarithms. I have to study it more. I know the short-cut for \( \ln(\ln(b^b)) \), but I don't know any short-cuts for taller towers, like \( \ln(\ln(\ln(b^{b^b})) \). That tower sequence is very similar to the base change constant. If you have an alternative equation to evaluate higher towers, that would be very interesting. Could you write out the equation for \( a_3 \)? As you noted, in the limit, this sequence from your earlier post converges.
This is just a simple formula to change the base of a power tower for natural integers.
however, this does provide an interesting continuum like sum for:
\( f(z) = x \cdot \ln(b) + \ln^{\circ 2}(b)\,\,\bigtriangleup_{-1}^e\,\,\ln^{\circ 3}(b)\,\,\bigtriangleup_{-2}^e\,\,\ln^{\circ 4}(b)...\,\,\bigtriangleup_{2-z}^e\,\,\ln^{\circ z}(b) \)
where f(z) is given by:
\( f(z) = \exp_e^{\circ -z}(\exp_b^{\circ z}(x)) \)
mike noted that as z goes to infinity, f(z) converges if x=1 and b=3.
I've seen these alternating operators before. They seem very intimidating, but I feel like there's something powerful about 'em.
Quote:\( a_1 = \ln(b)\\a_2 = \ln(\ln(b^b))\\a_3 = \ln(\ln(\ln(b^{b^b}))) \)
Let us call the value the sequence conveges to \( a_n \). If you take the equation, \( k=\text{slog}_e(a_n) \), then it is the same as this limit, for the constant "k" for the base change equation,
\( k=\lim_{n \to \infty} {\text{slog}_e(\exp_b^{[n]}(1))-n=\lim_{n \to \infty} {\text{slog}_e(\log_e^{[n]} (\exp_b^{[n]}(1))) \).
This is the "k" constant for:
\( \text{sexpBaseChange}_b(z) = \lim_{n \to\infty}\log_b^{[n]}\text{sexp}_e(z+k+n) \)
For large enough numbers, k is an approximation for how the delta for how the two different sexp bases would represent large numbers. You might expect that the "k" constant represented how to convert from one sexp base to another, but the value is only approximate, and in fact has a 1-cyclic period. For base e, and base eta, the 1-cyclic function k(z) function is approximately 0.5835+/-0.0004
\( k(z)=\lim_{n \to \infty} {\text{slog}_\eta(\text{sexp}_e(z+n))-n \)
- Sheldon
RE: Change of base formula using logarithmic semi operators - JmsNxn - 07/08/2011
(07/07/2011, 05:02 AM)sheldonison Wrote: So, what you have is a way to evaluate the tower of exponents and logarithms. I have to study it more. I know the short-cut for \( \ln(\ln(b^b)) \), but I don't know any short-cuts for taller towers, like \( \ln(\ln(\ln(b^{b^b})) \). That tower sequence is very similar to the base change constant. If you have an alternative equation to evaluate higher towers, that would be very interesting. Could you write out the equation for \( a_3 \)? As you noted, in the limit, this sequence from your earlier post converges.
\( a_3 = (\ln(b) + \ln^{\circ 2}(b)) \,\,\bigtriangleup_{-1}^e\,\,\ln^{\circ 3}(b) \)
\( a_4 = ((\ln(b) + \ln^{\circ 2}(b)) \,\,\bigtriangleup_{-1}^e\,\,\ln^{\circ 3}(b))\,\,\bigtriangleup_{-2}^e \,\,\ln^{\circ 4}(b) \)
where
\( p \le 0 \)
\( a\,\,\bigtriangleup_p^e\,\,b = \exp_e^{\circ p}(\exp_e^{\circ -p}(a) + \exp_e^{\circ -p}(b)) \)
Quote:Let us call the value the sequence conveges to \( a_n \). If you take the equation, \( k=\text{slog}_e(a_n) \), then it is the same as this limit, for the constant "k" for the base change equation,
\( k=\lim_{n \to \infty} {\text{slog}_e(\exp_b^{[n]}(1))-n=\lim_{n \to \infty} {\text{slog}_e(\log_e^{[n]} (\exp_b^{[n]}(1))) \).
This is the "k" constant for:
\( \text{sexpBaseChange}_b(z) = \lim_{n \to\infty}\log_b^{[n]}\text{sexp}_e(z+k+n) \)
For large enough numbers, k is an approximation for how the delta for how the two different sexp bases would represent large numbers. You might expect that the "k" constant represented how to convert from one sexp base to another, but the value is only approximate, and in fact has a 1-cyclic period. For base e, and base eta, the 1-cyclic function k(z) function is approximately 0.5835+/-0.0004
\( k(z)=\lim_{n \to \infty} {\text{slog}_\eta(\text{sexp}_e(z+n))-n \)
- Sheldon
that's really quite interesting