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/tetrationforu...hp?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/tetrationforu...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.
http://math.eretrandre.org/tetrationforu...hp?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/tetrationforu...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.
