Ops ... sorry! Once again!
the same situation is found in the definition of h(b), inverse of b = h^(1/h). In fact, we know that the "classical" solution is:
if: b = h^(1/h)
then: h = - W(-lnb)/lnb
We also know that, if we only take the W(0) branch of the Lambert Function, we only get the lower branch of h(b), which is not sufficient. Therefore, I propose to use the plog(z) operator, always meaning: plog(z) = W(0,z) OR W(-1,z), and write:
h(b) = plog(-lnb)/(-lnb) = b#(+oo)
See the attachment.
We can see (as we also know) that:
h(1/a) . ssqrt(a) = h(a) . ssqrt(1/a) = 1
GFR
the same situation is found in the definition of h(b), inverse of b = h^(1/h). In fact, we know that the "classical" solution is:
if: b = h^(1/h)
then: h = - W(-lnb)/lnb
We also know that, if we only take the W(0) branch of the Lambert Function, we only get the lower branch of h(b), which is not sufficient. Therefore, I propose to use the plog(z) operator, always meaning: plog(z) = W(0,z) OR W(-1,z), and write:
h(b) = plog(-lnb)/(-lnb) = b#(+oo)
See the attachment.
We can see (as we also know) that:
h(1/a) . ssqrt(a) = h(a) . ssqrt(1/a) = 1
GFR