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 simple base conversion formula for tetration JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 09/22/2011, 07:41 PM (This post was last modified: 09/22/2011, 07:44 PM by JmsNxn.) Well I finally decided to look a little deeper at my suggested base conversion, and to my surprise, the formula just popped out at me. given the definition $k \ge 0$: $a\,\,\bigtriangleup_{-k}^e\,\, b = \ln^{\circ k}(\exp^{\circ k}(a) + \exp^{\circ k}(b))$ we see instantly: $e^x\,\,\bigtriangleup_{-k}^e\,\,e^y = e^{x\,\,\bigtriangleup_{-k-1}^e\,\,y}$ which works for k = 0 as well since $\bigtriangleup_{1}^e$ is multiplication. It's easy now to generate a formula which works for base conversion using these operators. The notation for such is very cumbersome however, therefore I'll write it out step by step for 2, 3, 4. $b^b = e^{\ln(b) \cdot e^{\ln(b)}} = e^{e^{\ln(b) + \ln(\ln(b))}} = \,\,^{2+\text{slog}_e(\ln(b) + \ln(\ln(b)))}e$ $b^{b^b} = e^{\ln(b) \cdot e^{\ln(b) \cdot e^{\ln(b)}}} = e^{e^{e^{\ln(b) + \ln(\ln(b))\, \bigtriangleup_{-1}^e \,\ln(\ln(\ln(b)))}}}= \,\,^{3+\text{slog}_e(\ln(b) + \ln(\ln(b)) \bigtriangleup_{-1}^e \ln(\ln(\ln(b))))}e$ and which I think the process becomes obvious by this point $b^{b^{b^{b}}} = \,\,^{4 + \text{slog}_e(\ln(b) + \ln(\ln(b)) \,\bigtriangleup_{-1}^e\, \ln(\ln(\ln(b)))\,\bigtriangleup_{-2}^e\,\ln(\ln(\ln(\ln(b)))))} e$ and which by generality becomes: $^k b = \,\,^{k + \text{slog}_e(\ln(b) + \ln^{\circ 2}(b) \,\bigtriangleup_{-1}^e\,\ln^{\circ 3}(b)\,\bigtriangleup_{-2}^e\,\ln^{\circ 4}(b)\,...\,\bigtriangleup_{-k+1}\,\ln^{\circ k-1}(b)\,\bigtriangleup_{-k+2}^e\,\ln^{\circ k}(b))}e$ where we are sure to evaluate the highest operator first. this formula is very easily generalized to include conversion from base $b > \eta$ to any base $c > \eta$ « Next Oldest | Next Newest »

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