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Rational operators (a {t} b); a,b > e solved
#17
(06/07/2011, 06:59 AM)bo198214 Wrote: You mean "triation". Its greek numbering not latin Smile
Actually I agree with you, in certain contexts I worked with it also seems more suitable to use addition as operation 0. (Even Ackermann did so.)
On the other hand, on different contexts there it seems more preferable to start with 1. E.g. to have the zero-th operation a [0] x = x+1.
And I dont think one can roll back the whole historic naming development, tetration, the forth operation is already ingrained in the minds.

I'm glad you agree with me about centering with addition as zero. I actually originally centered it at zero because Ackermann did.

triation sounds nicer than tertiation.


But besides this, I have something to report.

which converges absolutely for all and .
Interestingly enough L does not change with a and b and seems to be only x dependent. L seems to always be greater than 2.6 but just around it. At first I thought it was a universal constant they were all converging to, but only showed convergence to this value.

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RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/08/2011, 04:54 AM

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