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A(z) = exp( A(z-1) - exp(-z) ) ?
#2
So for instance 

A(z) = exp( A(z-1) - exp(-z) )

We use A(z) = exp( - exp(-z) + exp( - exp(-z+1) + exp( - exp(-z+2) +  ... 

We define the functional inverse of A(z) := C(z).

Now we can go 

James/sheldon method :
for lim ln(ln(...* t times * A(z+t)...) similar to the phi method or - as i prefer - 

tommy method :
tet(v + slog(z) + p) = lim ln^[t]( A( C(exp^[t](z)) +v) )

for some constant p and 0<v<2.

both are C^oo and maybe analytic.
I conjecture my method is.

ARE both equal ??





regards

tommy1729
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Messages In This Thread
A(z) = exp( A(z-1) - exp(-z) ) ? - by tommy1729 - 03/03/2021, 12:26 AM
RE: A(z) = exp( A(z-1) - exp(-z) ) ? - by tommy1729 - 03/03/2021, 01:26 PM



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