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Zeration = inconsistant ?
#21
(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 can be solved easily by



(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
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