I was looking at tetration and considering its little taylor series, given by:
(1) = \bigtriangleup \sum_{n=0}^{\infty} a_n + n(x \bigtriangledown l) - ln(n!))
where:
= f(0) \bigtriangleup f(1) \bigtriangleup...{f®})
and:
)
)
specifically, what I was trying to do is find a relationship between the coefficients of tetrations' little taylor series and normal taylor series. We'll say:
 = \sum_{n=0}^{\infty} b_n \frac{(x-l)^n}{n!})
so I'm looking for
as an expression of 
or vice versa.
So I started off by looking at (1):
 = ln(\sum_{n=0}^{\infty} e^{a_n + n(x \bigtriangledown l) - ln(n!)}))
 = \sum_{n=0}^{\infty} e^{a_n} \frac{(e^x - e^l)^n}{n!})
and now let this equal our formula for sexp(x+1) using or the normal taylor series.
^n}{n!} = \sum_{d=0}^{\infty} b_d \frac{(x+1-l)^d}{d!})
and subtract the right hand side from the left hand side
^n - b_n (x+1-l)^n}{n!} = 0)
and now, since x is essentially arbitrary, let x = l to give:
but this contradicts the original Taylor series expansion
which states:
)
Any help would be greatly appreciated, thanks.
The only solution I have is
(1)
where:
and:
specifically, what I was trying to do is find a relationship between the coefficients of tetrations' little taylor series and normal taylor series. We'll say:
so I'm looking for
So I started off by looking at (1):
and now let this equal our formula for sexp(x+1) using
and subtract the right hand side from the left hand side
and now, since x is essentially arbitrary, let x = l to give:
but this contradicts the original Taylor series expansion
which states:
Any help would be greatly appreciated, thanks.
The only solution I have is
