TITLE: Smoothstep sigmoid-like function: Can anyone prove this relation?
QUESTION [11 upvotes]: The Smoothstep sigmoid-like function is defined as the polynomial
$$ \begin{align}
\operatorname{S}_N(x) &= x^{N+1} \sum_{n=0}^{N} \binom{N+n}{n} \binom{2N+1}{N-n} (-x)^{n} \qquad N \in \mathbb{Z} \ge 0 \\
                      &= \sum_{n=0}^{N}  (-1)^n \binom{N+n}{n} \binom{2N+1}{N-n} x^{N+n+1} \\
                      &= \sum_{n=0}^{N}        \binom{-N-1}{n} \binom{2N+1}{N-n} x^{N+n+1} \\
\end{align} $$
The first 7 examples are:
$$\begin{align}
\operatorname{S}_0(x) &= x \\
\operatorname{S}_1(x) &= -2x^3 + 3x^2 \\
\operatorname{S}_2(x) &= 6x^5 - 15x^4 + 10x^3 \\
\operatorname{S}_3(x) &= -20x^7 + 70x^6 - 84x^5 + 35x^4 \\
\operatorname{S}_4(x) &= 70x^9 - 315x^8 + 540x^7 - 420x^6 + 126x^5 \\
\operatorname{S}_5(x) &= -252x^{11} + 1386x^{10} - 3080x^9 + 3465x^8 - 1980x^7 + 462x^6 \\
\operatorname{S}_6(x) &= 924x^{13} - 6006x^{12} + 16380x^{11} - 24024x^{10} + 20020x^9 - 9009x^8 + 1716x^7 \\
\\
\end{align} $$
It is purported, for all non-negative integer $N$, that $\operatorname{S}_N(0) = 0$ and $\operatorname{S}_N(1) = 1$ and, at those two points, as many derivatives equal zero as possible.  I think it is also purported that $\operatorname{S}_N(\tfrac12) = \tfrac12$ and that this polynomial display odd-symmetry about the point at $x=\tfrac12$
If we define a linearly-scaled and offset version of the Smoothstep polynomial as:
$$ \operatorname{R}_N(x) = 2\operatorname{S}_N\left( \tfrac12(x+1) \right) - 1  $$
Then this means that $\operatorname{R}_N(-1) = -1$, $\operatorname{R}_N(1) = 1$, and as many derivatives as possible at those two points are zero.  And we see that $\operatorname{R}_N(0) = 0$ and that  odd-symmetry exists: $\operatorname{R}_N(-x) = -\operatorname{R}_N(x)$
Can anyone show, with the least amount of pain possible, that the derivative of $\operatorname{R}_N(x)$ becomes
$$\begin{align}
\operatorname{R}^{'}_{N}(x) &= \operatorname{S}^{'}_{N}\left( \tfrac12(x+1) \right)  \\
&= \left( \sum\limits_{n=0}^{N} \frac{N!}{n! (N-n)!} \frac{(-1)^n}{2n+1} \right)^{-1} (1-x^2)^{N} \qquad ? \\
\end{align}$$
This is not homework (I haven't been in skool since the early '80s). This DSP question and answer show the previous work I have done with this.  It's just a little bit bitchy and I am not sure the least painful way to go about doing this.
I guess that I am trying to show that
$$\begin{align}
\operatorname{S}^{'}_{N}\left( \tfrac12(x+1) \right) &= \sum_{n=0}^{N} \binom{-N-1}{n}\binom{2N+1}{N-n}(N+n+1) \left(\tfrac12(x+1)\right)^{N+n}  \\
&= \left( \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} \frac{(-1)^n}{2n+1} \right)^{-1} (1-x^2)^{N} \\
\end{align}$$
This really looks like a copulating female canine to me.

EDIT:
Here is the clearest way for me to state the question:
Let $x \in \mathbb{R}, \ N \ge 0 \in \mathbb{Z}$.
Define:
$$\begin{align}
 f_N & \triangleq \int\limits_{0}^{1} \big(1 - u^2 \big)^{N} \ du \\
     & = \sum\limits_{n=0}^{N} \binom{N}{n} \frac{(-1)^n}{2n+1} \\
\end{align}$$
Define:
$$ \operatorname{R}_N(x) \triangleq \frac{1}{f_N} \int\limits_{0}^{x} \big(1 - u^2 \big)^{N} \ du $$
Define from the Wikipedia definition of the Smoothstep sigmoid-like function:
$$\operatorname{S}_N(x) \triangleq \sum\limits_{n=0}^{N} \binom{-N-1}{n}\binom{2N+1}{N-n} x^{N+n+1} $$
Prove:
$$ \operatorname{R}_N(x) = 2\operatorname{S}_N\big(\tfrac12 (x+1) \big) - 1 $$
It suffices to prove that their first derivatives of the left-hand and right-hand sides are equal, because we know the left-hand and right-hand sides are equal at $x=-1$.  This means it is sufficient to show that:
$$ \frac{1}{f_N} (1-x^2)^{N} = \frac{d}{du}\,\operatorname{S}_N(u) \Bigg|_{u=\frac12 (x+1)} $$
or
$$ \frac{1}{f_N} (1-x^2)^{N} = \sum_{n=0}^{N} \binom{-N-1}{n}\binom{2N+1}{N-n} (N+n+1) \left(\tfrac12(x+1)\right)^{N+n} $$
or, explicitly:
$$ \left( \sum\limits_{n=0}^{N} \binom{N}{n} \frac{(-1)^n}{2n+1} \right)^{-1} (1-x^2)^{N} = \sum_{n=0}^{N} \binom{-N-1}{n}\binom{2N+1}{N-n} (N+n+1) \left(\tfrac12(x+1)\right)^{N+n} $$

REPLY [7 votes]: We want to show for $N\geq 0$
\begin{align*}
\color{blue}{\left(\sum_{n=0}^N\right.}&\color{blue}{\left.\binom{N}{n}\frac{(-1)^n}{2n+1}\right)^{-1}(1-x^2)^N}\\
&=\color{blue}{\sum_{n=0}^N\binom{-N-1}{n}\binom{2N+1}{N-n}(N+n+1)\left(\frac{1}{2}(x+1)\right)^{N+n}}\tag{1}\\
\end{align*}
We start  with the   RHS and obtain
  \begin{align*}
\sum_{n=0}^N&\binom{-N-1}{n}\binom{2N+1}{N-n}(N+n+1)\left(\frac{1}{2}(x+1)\right)^{N+n}\\
&=\sum_{n=0}^N(-1)^n\binom{N+n}{n}\binom{2N+1}{N+n+1}(N+n+1)\left(\frac{1}{2}(x+1)\right)^{N+n}\tag{2}\\
&=(2N+1)\sum_{n=0}^N(-1)^n\binom{N+n}{n}\binom{2N}{N+n}\left(\frac{1}{2}(x+1)\right)^{N+n}\tag{3}\\
&=(2N+1)\binom{2N}{N}\sum_{n=0}^N(-1)^n\binom{N}{n}\left(\frac{1}{2}(x+1)\right)^{N+n}\tag{4}
\end{align*}

Comment:

In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
In (3) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.
In (4) we use the binomial identity $\binom{p}{m}\binom{m}{q}=\binom{p}{q}\binom{p-q}{m-q}$


With (4) the    claim (1) boils down to show
\begin{align*}
\color{blue}{\left(\sum_{n=0}^N\right.}&\color{blue}{\left.\binom{N}{n}\frac{(-1)^n}{2n+1}\right)^{-1}(1-x^2)^N}\\
&=\color{blue}{(2N+1)\binom{2N}{N}\sum_{n=0}^N(-1)^n\binom{N}{n}\left(\frac{1}{2}(x+1)\right)^{N+n}}\tag{5}\\
\end{align*}

Intermezzo:
We can find a closed formula for the denominator of the LHS. The following is valid:
\begin{align*}
\sum_{n=0}^N\binom{N}{n}\frac{(-1)^n}{2n+1}=\frac{4^N}{2N+1}\binom{2N}{N}^{-1}\tag{6}
\end{align*}
A proof is given in the appendix.

Using (6) and since  $(1-x^2)^N=(1-x)^N(1+x)^N$ the equation (5) can be  simplified to
\begin{align*}
\color{blue}{\frac{1}{2^N}(1-x)^N
=\sum_{n=0}^N\frac{(-1)^n}{2^n}\binom{N}{n}(x+1)^{n}}
\end{align*}
Applying the binomial theorem to the RHS we finally get
\begin{align*}
\sum_{n=0}^N\frac{(-1)^n}{2^n}\binom{N}{n}(x+1)^{n}&=\sum_{n=0}^N\binom{N}{n}\left(-\frac{1+x}{2}\right)^{n}\\
&=\left(1-\frac{1+x}{2}\right)^N\\
&=\frac{1}{2^N}(1-x)^N
\end{align*}
and the claim (1) follows.

We now prove formula (6) following chapter I (problem section, problem 4 with solution) in Combinatorial Identities by John Riordan.

Appendix: The following is valid
  \begin{align*}
\qquad\qquad\sum_{n=0}^N\binom{N}{n}\frac{(-1)^n}{2n+1}=\frac{4^N}{2N+1}\binom{2N}{N}^{-1}\qquad\qquad N\geq 0
\end{align*}
We obtain using the Kronecker delta $\delta_{N,0}$ 
  \begin{align*}
f_N=\sum_{n=0}^N(-1)^n\binom{N}{n}\frac{1}{2n+1}&=\sum_{n=0}^N(-1)^n\binom{N}{n}\left(1-\frac{2n}{2n+1}\right)\\
&=\delta_{N,0}-\sum_{n=0}^N(-1)^n\binom{N}{n}\frac{2n}{2n+1}\tag{7}\\
&=\delta_{N,0}-2N\sum_{n=1}^N(-1)^n\binom{N-1}{n-1}\frac{1}{2n+1}\tag{8}\\
\end{align*}

Comment:

In (7) we apply $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.


We also get
  \begin{align*}
f_N=\sum_{n=0}^N(-1)^n\binom{N}{n}\frac{1}{2n+1}
&=\sum_{n=0}^N(-1)^n\left[\binom{N-1}{n}+\binom{N-1}{n-1}\right]\frac{1}{2n+1}\\
&=f_{N-1}+\sum_{n=1}^N(-1)^n\binom{N-1}{n-1}\frac{1}{2n+1}\tag{9}\\
\end{align*}

Adding (8) and $2N$ times (9) we get
\begin{align*}
\qquad\qquad(2N+1)f_N=2Nf_{N-1}+\delta_{N,0}\qquad\qquad N\geq 0\tag{10}
\end{align*}

Iterating (10) we obtain with $f_0=1$ for $N>0$
  \begin{align*}
f_N&=\frac{2N}{2N+1}f_{N-1}=\frac{(2N)(2N-2)}{(2N+1)(2N-1)}f_{N-2}=\ldots\\
&=\frac{(2N)!!}{(2N+1)!!}\\
&=\frac{(2N)!!(2N)!!}{(2N+1)!}\\
&=\frac{1}{2N+1}\cdot\frac{2^{2N}N!N!}{(2N)!}\\
&=\frac{4^N}{2N+1}\binom{2N}{N}^{-1}
\end{align*}
  and the claim (6) follows.

Here we use double factorials
\begin{align*}
(2N)!!&=(2N)(2N-2)\cdots4\cdot 2\\
(2N+1)!!&=(2N+1)(2N-1)\cdots 3\cdot 1\\
\end{align*}
and the formulae
\begin{align*}
(2N)!&=(2N)!!(2N-1)!!\\
(2N)!!&=2^NN!
\end{align*}