05/26/2015, 12:10 PM
The answer appears to be simple.
That is considering we work with the same continuum sum.
Within an analytic region :
Let r > 0.
exp^[1+r](x) = exp^[r](0) + F(exp(x))
where F(0) = 0 and is analytic.
Therefore the continuum sum CS is given by
T_r(x) = CS exp^[1+r](x) = exp^[r](0) x + G(exp(x)).
Therefore T(x) - T(x - 2pi i) = exp^[r](0) * 2 pi i.
regards
tommy1729
That is considering we work with the same continuum sum.
Within an analytic region :
Let r > 0.
exp^[1+r](x) = exp^[r](0) + F(exp(x))
where F(0) = 0 and is analytic.
Therefore the continuum sum CS is given by
T_r(x) = CS exp^[1+r](x) = exp^[r](0) x + G(exp(x)).
Therefore T(x) - T(x - 2pi i) = exp^[r](0) * 2 pi i.
regards
tommy1729