First let me recapitulate the method from my point of view.
We start with as usual with a function
and want to obtain a superfunction 
of it, i.e. a function that satisfies
(1)=f(\sigma(z)))
.
Now by differentiating once we get
=\sigma'(z) f'(\sigma(z)))
expanding twice:
=\sigma'(z) f'(\sigma(z)) f'(\sigma(z+1)))
and generally by induction:
=\sigma'(z) \prod_{k=0}^{n-1} f'(\sigma(z+k)))
making
our new variable:
}{\sigma'(z_0)}= \prod_{k=0}^{z-1} f'(\sigma(z_0+k)))
To extend the product to non-integer
we have a look at the sum
(2)-\ln\sigma'(z_0)= \sum_{k=0}^{z-1} \ln f'(\sigma(z_0+k)))
Extending the sum to non-integer boundaries
The sum can extended to non-integer values via Faulhaber's formula which provides the sum for monomials:
 - \varphi_{p+1}(n_0)}{p+1})
,
where = \sum_{k=0}^{p} \left(p\\k\right) B_k x^{p-k})
is the 
-th Bernoulli polynomial with 
the Bernoulli numbers.
The sum operator is linear for natural boundaries so one can extend Faulhaber's formula to linear combinations of monomials (also called polynomials
), and finally to powerseries. So for non-integer we define for
=\sum_{n=0}^\infty f_n z^n)
the extended sum  := \sum_{n=0}^\infty f_n \frac{\varphi_{n+1}(z+1)-\varphi_{n+1}(z_0)}{n+1})
.
To simplify the matter we can define the antidifference or indefinite sum:
 :=\sum_{n=0}^\infty f_n \frac{\varphi_{n+1}(z)}{n+1})
and get
 = \Delta^{-1}[f](z+1) - \Delta^{-1}[f](z_0))
Extended sum and shifts
We hope that the extended sum behaves like a normal sum with integer boundaries, e.g. we want to have
(3) = \sum_{\zeta=a+c}^{b+c} f(\zeta))
.
We show it for polynomials. Let be a polynomial of degree 
, then we know that (3) is satisfied for all integers 
and both sides are polynomials of maximal degree 
in . But equality at different points already implies the equality of both polynomials.
Hence not only for integer but for every the equality holds.
This than carries over to powerseries (as limit of a polynomial sequence) if no convergence issues arise.
This property lets us formulate (2) in a much nicer way by applying (3) with
and afterwards setting 
:
(4)- \ln\sigma'(z_0)=\sum_{\zeta=z_0}^{z-1} \ln f'(\sigma(\zeta)))
Uniqueness of extended sum superfunction
Sketch of proof:
Let
beside be another at 
analytic super-exponential (1) that satisfies (2). We further assume and to be invertible at and =\rho(z_0))
. We set 
in a vicinity of , so =\sigma(\delta(z)))
there.
=z_0)
and by (1) =\delta(z)+n)
so =z_0+n)
for integer 
.
So put into (4), with =\sigma'(\delta(z)) \delta'(z))
:
)- \ln\sigma'(z_0)+\ln\delta'(z) =\sum_{\zeta=z_0}^{z-1} \ln f'(\sigma(\delta(\zeta))))
the left difference can be replaced with (4), knowing:
(5)}^{\delta(z)-1} \ln f'(\sigma(\zeta))+\ln\delta'(z) =\sum_{\zeta=z_0}^{z-1} \ln f'(\sigma(\delta(\zeta))))
Now consider the forward difference
 (z) = h(\delta(z+1))-h(\delta(z)) = h(\delta(z)+1)-h(\delta(z)) = \Delta(h)\circ \delta(z))
.
Hence
 = \Delta^{-1}(h) \circ \delta)
for any 
up to perhaps a constant, so
) = \sum_{\zeta=\delta(z_0)}^{\delta(z)-1} h(\zeta))
.
Now we apply this to (5):
 =0)
=1)
=a z + b)
together with and =z_0+1)
this gives 
and 
, so =z)
and =\rho(z))
.
We start with as usual with a function
(1)
Now by differentiating once we get
expanding twice:
and generally by induction:
making
To extend the product to non-integer
(2)
Extending the sum to non-integer boundaries
The sum can extended to non-integer values
where
The sum operator is linear for natural boundaries so one can extend Faulhaber's formula to linear combinations of monomials (also called polynomials
), and finally to powerseries. So for non-integer To simplify the matter we can define the antidifference or indefinite sum:
and get
Extended sum and shifts
We hope that the extended sum behaves like a normal sum with integer boundaries, e.g. we want to have
(3)
We show it for polynomials. Let
Hence not only for integer
This than carries over to powerseries (as limit of a polynomial sequence) if no convergence issues arise.
This property lets us formulate (2) in a much nicer way by applying (3) with
(4)
Uniqueness of extended sum superfunction
Sketch of proof:
Let
So put
the left difference can be replaced with (4), knowing
(5)
Now consider the forward difference
Hence
Now we apply this to (5):
together with
