10/05/2014, 03:36 PM

(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

MSE MphLee

Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)

S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)