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Rational operators (a {t} b); a,b > e solved
#1
Well alas, logarithmic semi operators have paid off and have given a beautiful smooth curve over domain . This solution for rational operators is given by :



Which extends the ackerman function to domain real (given the restrictions provided).
the upper superfunction of is used (i.e: the cheta function).


Logarithmic semi-operators contain infinite rings and infinite abelian groups. In so far as {t} and {t-1} always form a ring and {t-1} is always an abelian group (therefore any operator greater than {1} is not commutative and is not abelian). There is an identity function S(t), however its values occur below e and are therefore still unknown for operators less than {1} (except at negative integers where it is a variant of infinity (therefore difficult to play with) and at 0 where it is 0). Greater than {1} operators have identity 1.

The difficulty is, if we use the lower superfunction of to define values less than e we get a hump in the middle of our transformation from . Therefore we have difficulty in defining an inverse for rational exponentiation. however, we still have a piecewise formula:



therefore rational roots, the inverse of rational exponentiation is defined so long as and .
rational division and rational subtraction is possible if and .

Here are some graphs, I'm sorry about their poor quality but I'm rather new to pari-gp so I don't know how to draw graphs using it. I'm stuck using python right now. Nonetheless here are the graphs.

the window for these ones is xmin = -1, xmax = 2, ymin = 0, ymax = 100

[Image: 3x4.png][Image: 4x3.png]
[Image: 5x3.png]

If there's any transformation someone would like to see specifically, please just ask me. I wanted to do the transformation of as we slowly raise t, but the graph doesn't look too good since x > e.

Some numerical values:

(I know I'm not supposed to be able to calculate the second one, but that's the power of recursion)

I'm very excited by this, I wonder if anyone has any questions comments?

for more on rational operators in general, see the identities they follow on this thread http://math.eretrandre.org/tetrationforu...hp?tid=546

thanks, James

PS: thanks go to Sheldon for the taylor series approximations of cheta and its inverse which allowed for the calculations.
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Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/06/2011, 02:45 AM

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