First let me recapitulate the method from my point of view.
We start with as usual with a function \( f \) and want to obtain a superfunction \( \sigma \) of it, i.e. a function that satisfies
(1) \( \sigma(z+1)=f(\sigma(z)) \).
Now by differentiating once we get
\( \sigma'(z+1)=\sigma'(z) f'(\sigma(z)) \)
expanding twice:
\( \sigma'(z+2)=\sigma'(z) f'(\sigma(z)) f'(\sigma(z+1)) \)
and generally by induction:
\( \sigma'(z+n)=\sigma'(z) \prod_{k=0}^{n-1} f'(\sigma(z+k)) \)
making \( z=n \) our new variable:
\( \frac{\sigma'(z_0+z)}{\sigma'(z_0)}= \prod_{k=0}^{z-1} f'(\sigma(z_0+k)) \)
To extend the product to non-integer \( z \) we have a look at the sum
(2) \( \ln\sigma'(z_0+z)-\ln\sigma'(z_0)= \sum_{k=0}^{z-1} \ln f'(\sigma(z_0+k)) \)
Extending the sum to non-integer boundaries
The sum can extended to non-integer values \( z \) via Faulhaber's formula which provides the sum for monomials:
\( \sum_{k=n_0}^{n} k^p = \frac{\varphi_{p+1}(n+1) - \varphi_{p+1}(n_0)}{p+1} \),
where \( \varphi_{p}(x) = \sum_{k=0}^{p} \left(p\\k\right) B_k x^{p-k} \) is the \( p \)-th Bernoulli polynomial with \( B_k \) 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 \( z \) we define for
\( f(z)=\sum_{n=0}^\infty f_n z^n \) the extended sum \( \sum_{\zeta = z_0}^{z} f(\zeta) := \sum_{n=0}^\infty f_n \frac{\varphi_{n+1}(z+1)-\varphi_{n+1}(z_0)}{n+1} \).
To simplify the matter we can define the antidifference or indefinite sum:
\( \Delta^{-1}[f](z) :=\sum_{n=0}^\infty f_n \frac{\varphi_{n+1}(z)}{n+1} \)
and get
\( \sum_{\zeta = z_0}^{z} f(\zeta) = \Delta^{-1}[f](z+1) - \Delta^{-1}[f](z_0) \)
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) \( \sum_{\zeta=a}^b f(\zeta+c) = \sum_{\zeta=a+c}^{b+c} f(\zeta) \).
We show it for polynomials. Let \( p \) be a polynomial of degree \( N \), then we know that (3) is satisfied for all integers \( c \) and both sides are polynomials of maximal degree \( N+1 \) in \( c \). But equality at \( N+1 \) different points already implies the equality of both polynomials.
Hence not only for integer \( c \) but for every \( c \) 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 \( c=z_0 \) and afterwards setting \( z:=z+z_0 \):
(4) \( \ln\sigma'(z)- \ln\sigma'(z_0)=\sum_{\zeta=z_0}^{z-1} \ln f'(\sigma(\zeta)) \)
Uniqueness of extended sum superfunction
Sketch of proof:
Let \( \rho \) beside \( \sigma \) be another at \( z_0 \) analytic super-exponential (1) that satisfies (2). We further assume \( \sigma \) and \( \rho \) to be invertible at \( z_0 \) and \( \sigma(z_0)=\rho(z_0) \). We set \( \delta=\sigma^{-1}\circ \rho \) in a vicinity of \( z_0 \), so \( \rho(z)=\sigma(\delta(z)) \) there.
\( \delta(z_0)=z_0 \) and by (1) \( \delta(z+n)=\delta(z)+n \) so \( \delta(z_0+n)=z_0+n \) for integer \( n \).
So put \( \rho \) into (4), with \( \rho'(z)=\sigma'(\delta(z)) \delta'(z) \):
\( \ln\sigma'(\delta(z))- \ln\sigma'(z_0)+\ln\delta'(z) =\sum_{\zeta=z_0}^{z-1} \ln f'(\sigma(\delta(\zeta))) \)
the left difference can be replaced with (4), knowing \( \delta(z_0)=z_0 \):
(5) \( \sum_{\zeta=\delta(z_0)}^{\delta(z)-1} \ln f'(\sigma(\zeta))+\ln\delta'(z) =\sum_{\zeta=z_0}^{z-1} \ln f'(\sigma(\delta(\zeta))) \)
Now consider the forward difference
\( \Delta (h\circ \delta) (z) = h(\delta(z+1))-h(\delta(z)) = h(\delta(z)+1)-h(\delta(z)) = \Delta(h)\circ \delta(z) \).
Hence
\( \Delta^{-1} (h\circ \delta) = \Delta^{-1}(h) \circ \delta \) for any \( h \) up to perhaps a constant, so
\( \sum_{\zeta=z_0}^{z-1} h(\delta(\zeta)) = \sum_{\zeta=\delta(z_0)}^{\delta(z)-1} h(\zeta) \).
Now we apply this to (5):
\( \ln\delta'(z) =0 \)
\( \delta'(z)=1 \)
\( \delta(z)=a z + b \)
together with \( \delta(z_0)=z_0 \) and \( \delta(z_0+1)=z_0+1 \) this gives \( a=1 \) and \( b=0 \), so \( \delta(z)=z \) and \( \sigma(z)=\rho(z) \).
We start with as usual with a function \( f \) and want to obtain a superfunction \( \sigma \) of it, i.e. a function that satisfies
(1) \( \sigma(z+1)=f(\sigma(z)) \).
Now by differentiating once we get
\( \sigma'(z+1)=\sigma'(z) f'(\sigma(z)) \)
expanding twice:
\( \sigma'(z+2)=\sigma'(z) f'(\sigma(z)) f'(\sigma(z+1)) \)
and generally by induction:
\( \sigma'(z+n)=\sigma'(z) \prod_{k=0}^{n-1} f'(\sigma(z+k)) \)
making \( z=n \) our new variable:
\( \frac{\sigma'(z_0+z)}{\sigma'(z_0)}= \prod_{k=0}^{z-1} f'(\sigma(z_0+k)) \)
To extend the product to non-integer \( z \) we have a look at the sum
(2) \( \ln\sigma'(z_0+z)-\ln\sigma'(z_0)= \sum_{k=0}^{z-1} \ln f'(\sigma(z_0+k)) \)
Extending the sum to non-integer boundaries
The sum can extended to non-integer values \( z \) via Faulhaber's formula which provides the sum for monomials:
\( \sum_{k=n_0}^{n} k^p = \frac{\varphi_{p+1}(n+1) - \varphi_{p+1}(n_0)}{p+1} \),
where \( \varphi_{p}(x) = \sum_{k=0}^{p} \left(p\\k\right) B_k x^{p-k} \) is the \( p \)-th Bernoulli polynomial with \( B_k \) 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 \( z \) we define for\( f(z)=\sum_{n=0}^\infty f_n z^n \) the extended sum \( \sum_{\zeta = z_0}^{z} f(\zeta) := \sum_{n=0}^\infty f_n \frac{\varphi_{n+1}(z+1)-\varphi_{n+1}(z_0)}{n+1} \).
To simplify the matter we can define the antidifference or indefinite sum:
\( \Delta^{-1}[f](z) :=\sum_{n=0}^\infty f_n \frac{\varphi_{n+1}(z)}{n+1} \)
and get
\( \sum_{\zeta = z_0}^{z} f(\zeta) = \Delta^{-1}[f](z+1) - \Delta^{-1}[f](z_0) \)
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) \( \sum_{\zeta=a}^b f(\zeta+c) = \sum_{\zeta=a+c}^{b+c} f(\zeta) \).
We show it for polynomials. Let \( p \) be a polynomial of degree \( N \), then we know that (3) is satisfied for all integers \( c \) and both sides are polynomials of maximal degree \( N+1 \) in \( c \). But equality at \( N+1 \) different points already implies the equality of both polynomials.
Hence not only for integer \( c \) but for every \( c \) 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 \( c=z_0 \) and afterwards setting \( z:=z+z_0 \):
(4) \( \ln\sigma'(z)- \ln\sigma'(z_0)=\sum_{\zeta=z_0}^{z-1} \ln f'(\sigma(\zeta)) \)
Uniqueness of extended sum superfunction
Sketch of proof:
Let \( \rho \) beside \( \sigma \) be another at \( z_0 \) analytic super-exponential (1) that satisfies (2). We further assume \( \sigma \) and \( \rho \) to be invertible at \( z_0 \) and \( \sigma(z_0)=\rho(z_0) \). We set \( \delta=\sigma^{-1}\circ \rho \) in a vicinity of \( z_0 \), so \( \rho(z)=\sigma(\delta(z)) \) there.
\( \delta(z_0)=z_0 \) and by (1) \( \delta(z+n)=\delta(z)+n \) so \( \delta(z_0+n)=z_0+n \) for integer \( n \).
So put \( \rho \) into (4), with \( \rho'(z)=\sigma'(\delta(z)) \delta'(z) \):
\( \ln\sigma'(\delta(z))- \ln\sigma'(z_0)+\ln\delta'(z) =\sum_{\zeta=z_0}^{z-1} \ln f'(\sigma(\delta(\zeta))) \)
the left difference can be replaced with (4), knowing \( \delta(z_0)=z_0 \):
(5) \( \sum_{\zeta=\delta(z_0)}^{\delta(z)-1} \ln f'(\sigma(\zeta))+\ln\delta'(z) =\sum_{\zeta=z_0}^{z-1} \ln f'(\sigma(\delta(\zeta))) \)
Now consider the forward difference
\( \Delta (h\circ \delta) (z) = h(\delta(z+1))-h(\delta(z)) = h(\delta(z)+1)-h(\delta(z)) = \Delta(h)\circ \delta(z) \).
Hence
\( \Delta^{-1} (h\circ \delta) = \Delta^{-1}(h) \circ \delta \) for any \( h \) up to perhaps a constant, so
\( \sum_{\zeta=z_0}^{z-1} h(\delta(\zeta)) = \sum_{\zeta=\delta(z_0)}^{\delta(z)-1} h(\zeta) \).
Now we apply this to (5):
\( \ln\delta'(z) =0 \)
\( \delta'(z)=1 \)
\( \delta(z)=a z + b \)
together with \( \delta(z_0)=z_0 \) and \( \delta(z_0+1)=z_0+1 \) this gives \( a=1 \) and \( b=0 \), so \( \delta(z)=z \) and \( \sigma(z)=\rho(z) \).
