Sorry for being little out of touch, to me "upping " things one step looks very interesting in general ( because of log-Poisson distribution):
\( e^{\lambda^k /k!*e^{-\lambda} \)
Can we not replace \( \lambda^k/k! \) by any function \( f(\lambda) \) which has a powerseries expansion, as that expansion would usually contain something like \( c* \lambda^k/k! \) ? Of course, coefficients at powers of \( \lambda \) can be also anything else. I can only see so far that terms of powerseries expansion has to be put into "double exponentiation operator" one by one:
\( e^{f_k (\lambda)*e^{-\lambda}} \)
Then the summation of terms of powerseries has to happen in exponent, and than we end up with :
\( e^{f(\lambda)*e^{-\lambda}} \)
Perhaps than we can replace \( f(\lambda) = f (T(z)) \) and see what happens. In case \( f( \lambda) = \lambda =T(z) \) we get
\( e^{T(z)*e^{-T(z)}=e^z \)
\( T(z) \) is Euler tree function.
if now \( z=\ln w \) then
\( e^{T(\ln w) * e^{-T(\ln w)} = e^{\ln w}= w \)
\( T(\ln w) = h ( w) * \ln w \) so
\( e^{h(w)*\ln w*e^{-(h(w)*\ln w) }} = w \)
Just an idea.
Ivars
\( e^{\lambda^k /k!*e^{-\lambda} \)
Can we not replace \( \lambda^k/k! \) by any function \( f(\lambda) \) which has a powerseries expansion, as that expansion would usually contain something like \( c* \lambda^k/k! \) ? Of course, coefficients at powers of \( \lambda \) can be also anything else. I can only see so far that terms of powerseries expansion has to be put into "double exponentiation operator" one by one:
\( e^{f_k (\lambda)*e^{-\lambda}} \)
Then the summation of terms of powerseries has to happen in exponent, and than we end up with :
\( e^{f(\lambda)*e^{-\lambda}} \)
Perhaps than we can replace \( f(\lambda) = f (T(z)) \) and see what happens. In case \( f( \lambda) = \lambda =T(z) \) we get
\( e^{T(z)*e^{-T(z)}=e^z \)
\( T(z) \) is Euler tree function.
if now \( z=\ln w \) then
\( e^{T(\ln w) * e^{-T(\ln w)} = e^{\ln w}= w \)
\( T(\ln w) = h ( w) * \ln w \) so
\( e^{h(w)*\ln w*e^{-(h(w)*\ln w) }} = w \)
Just an idea.
Ivars
