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:

$$ \left( \sum\limits_{n=0}^{N} \binom{N}{n} \frac{(-1)^n}{2n+1} \right)^{-1} (1-x^2)^{N} = \sum_{n=0}^{N} \binom{-N-1}{n}\binom{2N+1}{N-n} (N+n+1) \left(\tfrac12(x+1)\right)^{N+n} $$

  • 1
    Your question is to prove the identity $\mathrm{S}_N\left(1-x\right) = 1 - \mathrm{S}_N\left(x\right)$, right? (Proving that $\mathrm{S}_N\left(x\right)$ has its first $N-1$ derivatives at $0$ vanish is obvious; now, once the identity is proven, it will follow that the same holds for the first $N-1$ derivatives at $1$.) – darij grinberg Apr 27 '17 at 18:26
  • 1
    no @darijgrinberg, not precisely. my question as stated is to prove this identity: $$\frac{\sum\limits_{n=0}^{N-1} \binom{N-1}{n} \frac{(-1)^n}{2n+1} x^{2n+1}}{\sum\limits_{n=0}^{N-1} \binom{N-1}{n} \frac{(-1)^n}{2n+1}} = 2\sum\limits_{n=0}^{N-1} \binom{-N}{n}\binom{2N-1}{N-n-1} \big(\tfrac12(x+1)\big)^{N+n}-1$$ primarily. but i am happy if this identity is proven: $$\frac{(1-x^2)^{N-1}}{\sum\limits_{n=0}^{N-1} \binom{N-1}{n} \frac{(-1)^n}{2n+1}} = \sum_{n=0}^{N-1} \binom{-N}{n}\binom{2N-1}{N-n-1} (N+n) \left(\tfrac12(x+1)\right)^{N+n-1} $$ but i am happy to get a UMN prof's help. ;-) – robert bristow-johnson Apr 27 '17 at 21:34
  • and @darijgrinberg, i understand your point to the extent that if $$\operatorname{S}_N(1-x)=1-\operatorname{S}_N(x)$$ then it's true that $$ \operatorname{S}_N(\tfrac12-x)=1-\operatorname{S}_N(\tfrac12+x)$$ and that will show that $\operatorname{R}_N(-x)=-\operatorname{R}_N(x)$ which is an important property of my odd-symmetric Smoothstep derived independently. i want to somehow make sure that the two polynomials have equivalent coefficients, when $x$ is correctly scaled and offset between the two definitions. – robert bristow-johnson Apr 27 '17 at 21:48
up vote 7 down vote accepted
+300

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*}

  • 1
    @robertbristow-johnson: Ok, I see. You might find table 174 (p. 174) of Concrete Mathematics helpful. – Markus Scheuer Apr 28 '17 at 16:45
  • 1
    @robertbristow-johnson: With respect to your first comment of the last triple: Good observation! This drastically simplifies the proof. :-) – Markus Scheuer Apr 29 '17 at 6:11
  • 1
    @robertbristow-johnson: Many thanks for accepting my answer and granting the bounty! :-) – Markus Scheuer Apr 29 '17 at 19:46
  • 1
    i had originally defined $N$ as you do here, but the dumb Wikipedia article wanted to start off with $\operatorname{S}_1$ rather than $\operatorname{S}_0$ and in the math.SE i wanted to be consistent with the Wikipedia definition. this counting from 0 vs. counting from 1 debate is old but i think Dijkstra has nailed it. i just like consistency wherever possible. – robert bristow-johnson Apr 29 '17 at 20:11
  • 2
    This looks like an anagram of a really nice proof, Markus, but I really wouldn't mind seeing it unscrambled :) For example, it took me a while to see that your (1), (2) and (3) seemingly refer to the equality of whatever-is-on-that-line to the right hand side (not to the left hand side). It is generally better to write proofs bottom-to-top unless it's really clear how you are proceeding. – darij grinberg Apr 30 '17 at 2:26

Here is shown the development of the odd-symmetric Smoothstep function:

$$\operatorname{R}_N(x) = \begin{cases} -1 & x < -1 \\ \frac{1}{f_N} \int\limits_{0}^{x} \big(1 - u^2 \big)^{N} \ du \quad & -1 \le x \le 1 \\ +1 & 1 < x \\ \end{cases}$$

where $x \in \mathbb{R}, \ N\ge0 \in \mathbb{Z},$ and $\tfrac{1}{f_N}$ is a scaling constant judiciously chosen so that $\operatorname{R}_N(x)$ is continuous and $\operatorname{R}_N(\pm 1) = \pm 1$.

$$ f_N = \int\limits_{0}^{1} \big(1 - u^2 \big)^{N} \ du $$

Since the integrand is positive for $|u| < 1$, this is a monotonic increasing function. It is also clear that odd-symmetry prevails:

$$ \operatorname{R}_N(-x) = -\operatorname{R}_N(x) \qquad \forall x \in \mathbb{R}, \ N\ge0 \in \mathbb{Z} $$

This odd-symmetric Smoothstep function is, I believe, directly related to the commonly-defined Smoothstep sigmoid-like function as

$$ \operatorname{R}_N(x) = 2\operatorname{S}_N\big(\tfrac12 (x+1) \big) -1 \qquad -\infty < x < +\infty $$

where

$$\operatorname{S}_N(x) = \begin{cases} 0 & x < 0 \\ \sum\limits_{n=0}^{N} \binom{-N-1}{n}\binom{2N+1}{N-n} x^{N+n+1} \quad & 0 \le x \le 1 \\ 1 & 1 < x \\ \end{cases}$$

for $x \in \mathbb{R}, \ N\ge0 \in \mathbb{Z}$ . This relationship is what I want rigorously proven. (Or, at least, a decent amount of rigor.)

For $|x| < 1$ the 1st derivative of $\operatorname{R}_N(x)$ is

$$ \operatorname{R}_N^{'}(x) = \tfrac{1}{f_N} \big(1 - x^2 \big)^{N} $$

The 2nd derivative of $\operatorname{R}_N(x)$ is

$$ \operatorname{R}_N^{''}(x) = \tfrac{1}{f_N} N \big(1 - x^2 \big)^{N-1} (-2x)$$

The 3rd derivative of $\operatorname{R}_N(x)$ is

$$\begin{align} \operatorname{R}_N^{'''}(x) &= \tfrac{1}{f_N} N(N-1) \big(1 - x^2 \big)^{N-2} (-2x)^2 \ + \ \tfrac{1}{f_N} N \big(1 - x^2 \big)^{N-1} (-2) \\ &= \tfrac{1}{f_N} \big(1 - x^2 \big)^{N-2} \bigg( N(N-1)(-2x)^2 - 2N(1 - x^2) \bigg) \\ \end{align}$$

and, for $n \ge 1$, the $n$th derivative is

$$ \operatorname{R}_N^{(n)}(x) = \tfrac{1}{f_N} \big(1 - x^2 \big)^{N-n+1} \, g_n(x) $$

where $g_n(x)$ is some ($n$-1)th order polynomial function of $x$ and is finite in value. This can be proven inductively by considering the ($n$+1)th derivative:

$$\begin{align} \\ \operatorname{R}_N^{(n+1)}(x) &= \tfrac{d}{dx} \Big( \operatorname{R}_N^{(n)}(x) \Big) \\ &= \tfrac{d}{dx} \Big(\tfrac{1}{f_N} \, \big(1 - x^2 \big)^{N-n+1} \, g_n(x)\Big) \\ &= \tfrac{1}{f_N}(N-n+1)\big(1 - x^2 \big)^{N-n} (-2x) g_n(x) \, + \, \tfrac{1}{f_N} \big(1-x^2 \big)^{N-n+1}g'_n(x) \\ &= \tfrac{1}{f_N} \big(1 - x^2 \big)^{N-n} \Big( (N-n+1)(-2x) g_n(x) \, + \, (1-x^2) g'_n(x) \Big) \\ &= \tfrac{1}{f_N} \big(1 - x^2 \big)^{N-n} \, g_{n+1}(x) \\ \\ \end{align}$$

where $ g_{n+1}(x) = (N-n+1)(-2x) g_n(x) + (1-x^2) g'_n(x) $ .

Because of differentiation, the order of polynomial $g'_n(x)$ is one less than the order of $g_n(x)$, but polynomial $(1-x^2)g'_n(x)$ is one order greater than $g_n(x)$ and so also is $(-2x) g_n(x)$.

When $x = \pm 1$, then the first $N$ derivatives are zero, $$ \operatorname{R}^{(n)}(x) \Bigg|_{x=\pm 1} = 0 \qquad \text{for } 1 \le n \le N $$ making this polynomial maximally flat at $x = \pm 1$.

The integrand is a binomial and can be expressed as a power series using binomial expansion:

$$\begin{align} \big(1 - u^2 \big)^{N} & = \sum\limits_{n=0}^{N} \binom{N}{n} \big(-u^2\big)^n (1)^{N-n} \\ & = \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} \big(-u^2\big)^n (1)^{N-n} \\ & = \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} (-1)^n u^{2n} \\ \end{align}$$

So the integral can be expressed as an integral of a power series:

$$\begin{align} \int\limits_{0}^{x} \big(1 - u^2 \big)^{N} \ du & = \int\limits_{0}^{x} \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} (-1)^n u^{2n} \ du \\ & = \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} (-1)^n \int\limits_{0}^{x} u^{2n} \ du \\ & = \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} \frac{(-1)^n}{2n+1} u^{2n+1} \Bigg|_0^x \\ & = \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} \frac{(-1)^n}{2n+1} x^{2n+1} \\ & = x \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} \frac{(-1)^n}{2n+1} \big(x^2 \big)^n \\ \end{align}$$

When $x = \pm 1$, we get

$$ \int\limits_{0}^{\pm 1} \big(1 - u^2 \big)^{N} \ du = \pm \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} \frac{(-1)^n}{2n+1} $$

This makes the scaler $f_N$ to be

$$ f_N = \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} \frac{(-1)^n}{2n+1} $$

and makes the odd-symmetric Smoothstep function to be:

$$ \operatorname{R}_N(x) = \left( \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} \frac{(-1)^n}{2n+1} \right)^{-1} \ \ \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} \frac{(-1)^n}{2n+1} x^{2n+1} $$

for $|x| \le 1$ .

The Smoothstep polynomials (without splicing to the $\operatorname{sgn}(x) = \pm 1$ saturated components) look like

enter image description here

I think the order $2N+1$ starts at 1 and goes to 9 (or $0 \le N \le 4$)

With the saturation attached, the Smoothstep sigmoid-like curves look like

enter image description here

The odd-symmetry Smoothstep function is continuous everywhere and all derivatives, up to the $N$th derivative, are continuous everywhere and the ($N$+1)th derivative and higher are continuous everywhere except where the constant saturation is spliced to the polynomial at $x = \pm 1$.

What I want to be able to prove is that the two definitions of these Smoothstep polynomials are equivalent, given the proper scaling and offset of $x$.

$$\begin{align} \operatorname{R}_N(x) &= 2\operatorname{S}_N\big(\tfrac12(x+1)\big)-1 & -\infty < x < +\infty \\ \\ \frac{\sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} \frac{(-1)^n}{2n+1} x^{2n+1}}{\sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} \frac{(-1)^n}{2n+1}} &= 2 \sum\limits_{n=0}^{N} \binom{-N-1}{n}\binom{2N+1}{N-n} \big(\tfrac12(x+1)\big)^{N+n+1} - 1 & -1 \le x \le 1 \\ \end{align}$$

Because it is not hard to show equality at $x=-1$, to show both sides as equivalent, it suffices to show their derivatives as equivalent (with the proper scaling and offset of $x$). So it suffices to prove that

$$ \frac{(1-x^2)^{N}}{\sum\limits_{n=0}^{N} \frac{N!}{n! (N-n)!} \frac{(-1)^n}{2n+1}} = \sum_{n=0}^{N} \binom{-N-1}{n}\binom{2N+1}{N-n} (N+n+1) \left(\tfrac12(x+1)\right)^{N+n} $$

This is what the bounty is for.

I do not know how much this could help you.

What it seems it that, using $$S_n(x)=\sum_{i=0}^{n-1} \binom{-n}{i} \binom{2 n-1}{n-i-1} x^{n+i}=n \binom{2 n-1}{n-1} B_x(n,n)$$ where appears the incomplete beta function. $$S_n\left(\frac{x+1}2\right)=n \binom{2 n-1}{n-1} B_{\frac{x+1}{2}}(n,n)$$ $$R_n(x)=2 n \binom{2 n-1}{n-1} B_{\frac{x+1}{2}}(n,n)-1 $$ $$S^{'}_n\left(\frac{x+1}2\right)=\frac{\Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi } \, \Gamma (n)}\left(1-x^2\right)^{n-1} $$ $$R^{'}_n(x)=\frac{2 \Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }\, \Gamma (n)}\left(1-x^2\right)^{n-1}$$
I hope and wish no mistakes on my side.

  • thanks Claude. i am wondering where we're getting the "$(1-x^2)^{n-1}$" expression for $\operatorname{S}^{'}_n(\cdot)$ from? $$ $$ and i don't see a subscript $x$ for the beta function in the reference you gave. – robert bristow-johnson Apr 24 '17 at 7:53
  • @robertbristow-johnson. If we consider $B_{a x+b}(n,n)$, the derivative is $a (1-a x-b)^{n-1} (a x+b)^{n-1}$ – Claude Leibovici Apr 24 '17 at 7:59
  • hey Claude, thank you for your contribution. please don't take it badly if i put some of my paltry reputation on a bounty for this. i would really like to see a rigorously constructed answer. and i also do not understand why i am off by a factor of two. i have two polynomials of the same order, and with proper scaling on both $x$ and $y=S_n(x)$, i can map $0\le x\le 1$ and $0\le S_N(x)\le 1$ to $-1\le x\le 1$ and $-1\le 2S_N\left(\tfrac12(x+1)\right)-1\le 1$. – robert bristow-johnson Apr 25 '17 at 2:49
  • Claude, where is this relationship between the Smoothstep and the Incomplete Beta function documented? $$S_N(x)=\sum_{n=0}^{N-1} \binom{-N}{n} \binom{2N-1}{N-n-1} x^{N+n}=N \binom{2N-1}{N-1} B_x(N,N)$$ for $N$ = 1, 2, 3... – robert bristow-johnson Apr 26 '17 at 6:58

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.