04/27/2014, 08:02 PM
A simple yet unsolved equation for slog(z) ?
For z with Re(z)>>1 and z not equal to exp^[m](0) for an integer m and an appropriate real a_0 :
slog(z) = a_0 + (slog(z)/1)^(1-slog(z)) ln^[1](z) + (slog(z)/2)^(2-slog(z)) ln^[2](z)+ ... + (slog(z)/n)^(n-slog(z)) ln^[n](z)
where n is going to +oo.
**
I considered using arc2sinh instead of log or sarc2sinh instead of slog( sarc2sinh the slog analogue of the super of 2sinh )
**
I want my slog to be analytic but wonder if this is possible ?
**
Because this feels again a bit like the base change or my 2sinh method again
**
It comes naturally to consider the integral case instead of the infinite sum.
However my preference for the sum is a fact because otherwise we would need ln^[x] for real x , and that requires tetration. Imho I bet that is a higher level of self-reference and therefore harder or inconsistant. This is just a guess though.
For those who prefer integral transforms I guess they are very intrested.
regards
tommy1729
For z with Re(z)>>1 and z not equal to exp^[m](0) for an integer m and an appropriate real a_0 :
slog(z) = a_0 + (slog(z)/1)^(1-slog(z)) ln^[1](z) + (slog(z)/2)^(2-slog(z)) ln^[2](z)+ ... + (slog(z)/n)^(n-slog(z)) ln^[n](z)
where n is going to +oo.
**
I considered using arc2sinh instead of log or sarc2sinh instead of slog( sarc2sinh the slog analogue of the super of 2sinh )
**
I want my slog to be analytic but wonder if this is possible ?
**
Because this feels again a bit like the base change or my 2sinh method again
**
It comes naturally to consider the integral case instead of the infinite sum.
However my preference for the sum is a fact because otherwise we would need ln^[x] for real x , and that requires tetration. Imho I bet that is a higher level of self-reference and therefore harder or inconsistant. This is just a guess though.
For those who prefer integral transforms I guess they are very intrested.
regards
tommy1729
