Proving that $f$ verifies $f(1−x) + f(x) = 1$

Question

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 ?

Answer 1

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

Answer 2

Here we have a partial fraction decomposition in disguise.

We obtain \begin{align*} \color{blue}{f(x)}&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{(p-1-k)!(p+k)!} \prod_{\substack{j=k-p+1\\j\neq 0}}^{k+p}(x+k-j)\\ &=\prod_{j=-p+1}^{p}(x-j)\sum_{k=0}^{p-1} \frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}\,\frac{1}{x+k}\tag{1}\\ &\color{blue}{=\prod_{j=-p}^{p-1}(x+j)\sum_{k=0}^{p-1} \frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}\,\frac{1}{x+k}}\tag{2}\\ \end{align*}

Comment:

• In (1) we expand the summands with $(x+k)/(x+k)$ and shift the product by $k$ to start with $j=-p+1$. This way the product can be factored out from the sum.

• In (2) we substitute $j$ with $-j$ in the product.

We start with (2) and consider $f(1-x)$. We obtain \begin{align*} \color{blue}{f(1-x)}&=\prod_{j=-p}^{p-1}(1-x+j)\sum_{k=0}^{p-1} \frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}\,\frac{1}{1-x+k}\\ &=\prod_{j=-p}^{p-1}(x-j-1)\sum_{k=0}^{p-1} \frac{(-1)^{k-1}}{k!(2p-1-k)!}\,\frac{1}{-x+p-k}\tag{3}\\ &=\prod_{j=-p+1}^{p}(x+j-1)\sum_{k=0}^{p-1} \frac{(-1)^{k}}{k!(2p-1-k)!}\,\frac{1}{x+k-p}\tag{4}\\ &\,\,\color{blue}{=\prod_{j=-p}^{p-1}(x+j)\sum_{k=-p}^{-1} \frac{(-1)^{p+k}}{(p+k)!(p-1-k)!}\,\frac{1}{x+k}}\tag{5}\\ \end{align*}

Comment:

• In (3) we factor out $(-1)^{2p}=1$ from the product. We also change the order of summation $k\to p-1-k$.

• In (4) we substitute $j$ with $-j$ in the product and we expand numerator and denominator of the summands with $-1$.

• In (5) we shift the index $j$ by one to start with $j=-p$. We also shift the index $k$ by $p$ to start with $k=-p$.

We can now sum up (2) and (5) and get \begin{align*} \color{blue}{f(x)+f(1-x)}&\color{blue}{=\prod_{j=-p}^{p-1}(x+j)\sum_{k=-p}^{p-1} \frac{(-1)^{p+k}}{(p+k)!(p-1-k)!}\,\frac{1}{x+k}}\tag{6} \end{align*}

The right-hand side of (6) looks like a partial fraction decomposition. Indeed, we consider a function \begin{align*} g(x)=\frac{1}{\prod_{j=-p}^{p-1}(x+j)} \end{align*} The function $g(x)$ is a rational function with $2p$ pairwise different simple poles. Therefore we can find a representation \begin{align*} g(x)=\frac{1}{\prod_{j=-p}^{p-1}(x+j)}=\sum_{k=-p}^{p-1}\frac{a_k}{x+k}\qquad\qquad a_k\in\mathbb{R} \end{align*}

Multiplication with the denominator of $g(x)$ gives \begin{align*} \color{blue}{1=\prod_{j=-p}^{p-1}(x+j)\sum_{k=-p}^{p-1}\frac{a_k}{x+k}}\tag{7} \end{align*}

We can now evaluate (7) at $x=-k_0, -p\leq k_0\leq p-1$. We obtain \begin{align*} 1&=\left.\prod_{j=-p}^{p-1}(x+j)\sum_{k=-p}^{p-1}\frac{a_k}{x+k}\right|_{x=-k_0}\\ &=\prod_{\substack{j=-p\\j\neq k_0}}^{p-1}\left(-k_0+j\right)a_{k_0}\\ a_{k_0}&=\frac{1}{\prod_{\substack{j=-p\\j\neq k_0}}^{p-1}(j-k_0)}\\ &=\frac{1}{\prod_{j=-p}^{k_0-1}(j-k_0)}\,\frac{1}{\prod_{j=k_0+1}^{p-1}(j-k_0)}\\ &=\frac{(-1)^{p+k_0}}{(p+k_0)!}\,\frac{1}{\left(p-1-k_0\right)!} \end{align*} The coefficients $a_{k_0}$ can now be put into (7) and we obtain expression (6). We conclude \begin{align*} \color{blue}{f(x)+f(1-x)=1} \end{align*} and the claim follows.