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+--- Thread: Zeration = inconsistant ? (/showthread.php?tid=926)
RE: Zeration = inconsistant ? - MphLee - 10/05/2014
(10/04/2014, 10:20 PM)tommy1729 Wrote: Example :Really interesting and ingenious but I'm a donkey with this kind of math. In other words, I dont see why you should use this sophisticated method when the equation \( x \odot_e^{a}x=b \) can be solved easily by
Solve
exp^[2]( 2 ln^[2](x) ) = 7
This is equivalent to
x^ln(x) = 7.
Take some real a,b >= exp(1).
a_0 = a^{ 1 / ln(b) }
b_0 = 7^{ ln(b) / ln(a) }
replacement rules :
---
a ' = a * 7^{ ln(b) }
b ' = a * b
---
a_1 = a ' ^{ 1/ln(b ') }
b_1 = 7^{ ln(b ') / ln(a ') }
repeat forever
lim n-> oo
a_n/b_n = x.
this gives x = exp( sqrt( ln(7) ) ) as it should.
Numerically we get x = 4.0348084730118923250275859453110072467762717139110...
Notice 1/x is also a solution.
If we take 0 < a,b < 1/exp(1) ( =exp(-1) ) we probably achieve that.
This numerical algorithm can probably be improved with adding some + operators at the right places ...
Still investigating.
\( x=b\oslash_e^{1+a} {\exp}^{\circ a}(2)={\exp}^{\circ a}(\frac{ln^{\circ a}(b)}{2}) \)
(10/04/2014, 10:20 PM)tommy1729 Wrote: Also the method can probably be extended nicely to all interpretations of hyperoperators.How?You are talking about every sequence Hyperoperations sequence??
(10/04/2014, 10:20 PM)tommy1729 Wrote: Can zeration inprove this algoritm ?I have no idea