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 Zeration = inconsistant ? MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 10/05/2014, 03:36 PM (10/04/2014, 10:20 PM)tommy1729 Wrote: Example : 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.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 $x=b\oslash_e^{1+a} {\exp}^{\circ a}(2)={\exp}^{\circ a}(\frac{ln^{\circ a}(b)}{2})$ (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 MathStackExchange account:MphLee « Next Oldest | Next Newest »

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