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.

Notice I did not use a derivative anywhere like most numerical methods do.

Also the method can probably be extended nicely to all interpretations of hyperoperators.

Extending towards complex numbers however seems difficult.

(unless we split up by variables for real and imag part )

Can zeration inprove this algoritm ?

regards

tommy1729

" So he can probably do math better than most cows , so what ? "

Ullrich mocking me on sci.math.

" Truth is that what does not go away when you stop believing in it "

tommy1729

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.

Notice I did not use a derivative anywhere like most numerical methods do.

Also the method can probably be extended nicely to all interpretations of hyperoperators.

Extending towards complex numbers however seems difficult.

(unless we split up by variables for real and imag part )

Can zeration inprove this algoritm ?

regards

tommy1729

" So he can probably do math better than most cows , so what ? "

Ullrich mocking me on sci.math.

" Truth is that what does not go away when you stop believing in it "

tommy1729