<s>[INST] Proving that $f$ verifies $f(1−x) + f(x) = 1$ I have a function which is defined for $x \in (0,1)$ and for $p>1$ with the expression
\begin{align*} 
f(x) = \sum_{k=0}^{p-1} \frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}\left(\prod_{i=k-p+1, i\neq0}^{k+p} (x+k-i)\right) 
\end{align*}
I would like to show that $f(1-x) + f(x) = 1$.
What I have done so far
With a little bit of algebra I have
\begin{align*} 
f(x) 
&= \frac{1}{(2p-1)!}\sum_{k=0}^{p-1} (-1)^{p+k} \binom{2p-1}{p+k} \left(\prod_{i=k-p+1, i\neq0}^{k+p} (x+k-i)\right)
&& \text{using }\binom{n}{m} = \frac{n!}{m!(n-m)!}
\\
&= \frac{1}{(2p-1)!}\sum_{k=0}^{p-1} (-1)^{p+k} \binom{2p-1}{p+k} \left(\prod_{j=1-p, j\neq-k}^{p} (x-j)\right) 
&& \text{using the change of indices j=i-k} 
\\
&= \frac{1}{(2p-1)!}\sum_{k=0}^{p-1} (-1)^{p+k} \binom{2p-1}{p+k} \left(\prod_{j=1-p}^{p} (x-j)\right) \frac{1}{x+k}
&& \text{completing the product} 
\\
&= \frac{1}{(2p-1)!} \left(\prod_{j=1-p}^{p} (x-j)\right) \left(\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{x+k}\binom{2p-1}{p+k}\right)
&& \text{factoring the product from the sum} 
\end{align*}
Then I introduce the functions
\begin{align*} 
F(x) &= (2p-1)! f(x)  = G(x) H(x)\\
G(x) &= \prod_{j=1-p}^{p} (x-j) \\
H(x) &= \sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{x+k}\binom{2p-1}{p+k}
\end{align*}
So the initial question reduces to proving that $F(1-x) + F(x) = (2p-1)!$
Next, playing with the indices of the product I have been able to show that $G(1-x) = G(x)$.
So that now I only need to show $G(x) \left( H(1-x) + H(x)\right) = (2p-1)!$
EDIT:
I have just been able to simplify $H(1-x) + H(x)$ as follows :
\begin{align*} 
H(x) 
&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{x+k}\binom{2p-1}{p+k} 
\\
&=\sum_{s=p}^{2p-1}\frac{(-1)^{s}}{x-p+s}\binom{2p-1}{s}
&& \text{using a change of index } s=p+k 
\end{align*}
And
\begin{align*} 
H(1-x) 
&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{1-x+k}\binom{2p-1}{p+k} 
\\
&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{1-x+k}\binom{2p-1}{p-1-k}
&& \text{using } \binom{2p-1}{p+k} = \binom{2p-1}{p-1-k} 
\\
&=\sum_{s=0}^{p-1}\frac{(-1)^{s}}{x-p+s}\binom{2p-1}{s}
&& \text{using a change of index } s=p-1-k 
\end{align*}
So finally
\begin{align*} 
H(1-x) + H(x) 
&=\sum_{s=0}^{2p-1}\frac{(-1)^{s}}{x-p+s}\binom{2p-1}{s}
\end{align*}
EDIT2:
I have just made the same work on $G(x)$ as follows
\begin{align*} 
G(x) 
&=\prod_{j=1-p}^{p} (x-j) \\
&=\prod_{s=0}^{2p-1} (x-p+s)
&& \text{using a change of index } s=p-i 
\end{align*}

Now I think the constant $(2p-1)!$ in the relation $F(1-x)+F(x)=(2p-1)!$ comes from the binomial factor in $H(1-x) + H(x)$ but I don't see how to prove that...
Any lead ?
 [/INST] I have actually found a way to solve the question from EDIT1 and EDIT2
So
\begin{align*} 
F(1-x) + F(x)
&= G(x) \Big( H(1-x) + H(x) \Big) \\
&= \Bigg(\prod_{s=0}^{2p-1}(x-p+s)\Bigg)\Bigg(\sum_{k=0}^{2p-1}\frac{(-1)^{k}}{x-p+k}\binom{2p-1}{k}\Bigg)\\
&= \sum_{k=0}^{2p-1} (-1)^{k}\binom{2p-1}{k}\Bigg(\prod_{s=0, s \neq k}^{2p-1}(x-p+s)\Bigg)
\end{align*}
This the (2p-1)-th order finite difference scheme with step $h=1$ applied to the function
\begin{align*} 
g(x) = \Bigg( \prod_{s=0, s \neq k}^{2p-1}(x-p) \Bigg) = (x-p)^{2p-1}
\end{align*}
So clearly this would mean that :
\begin{align*} 
F(1-x) + F(x) 
&= \Delta_{h}^{2p-1} (x-p)^{2p-1} \\
&= \frac{d^{2p-1}}{dx^{2p-1}} (x-p)^{2p-1} \\
&= (2p-1)!
\end{align*}
Hot point to justify in the conclusion :
So we just need now to justify the passage from $\Delta_{h}^{2p-1}(x-p)^{2p-1}$ to $\frac{d^{2p-1}}{dx^{2p-1}} (x-p)^{2p-1}$.
Is there a property that says the nth-order finite difference scheme is exact on polynomials ?
Found a justification here :
https://hal.archives-ouvertes.fr/hal-01350976/file/on_the_link_between_finite_difference_and_derivative_of_polynomials_kolosov_petro_2017.pdf
</s>