Smoothstep

Starting with a generic third-order polynomial function and its first derivative:

${\displaystyle {\begin{alignedat}{9}\operatorname {S} _{1}(x)&&\;=\;&&a_{3}x^{3}&&\;+\;&&a_{2}x^{2}&&\;+\;&&a_{1}x&&\;+\;&&a_{0},&\\\operatorname {S} _{1}'(x)&&\;=\;&&3a_{3}x^{2}&&\;+\;&&2a_{2}x&&\;+\;&&a_{1}.&\end{alignedat}}}$

Applying the desired values for the function at both endpoints:

${\displaystyle {\begin{alignedat}{13}\operatorname {S} _{1}(0)&&\;=\;&&0\quad &&\Rightarrow &&\quad 0\;&&+&&\;0\;&&+&&\;0\;&&+&&\;a_{0}&&\;=\;&&0,&\\\operatorname {S} _{1}(1)&&\;=\;&&1\quad &&\Rightarrow &&\quad a_{3}\;&&+&&\;a_{2}\;&&+&&\;a_{1}\;&&+&&\;a_{0}&&\;=\;&&1.&\end{alignedat}}}$

Applying the desired values for the first derivative of the function at both endpoints:

${\displaystyle {\begin{alignedat}{11}\operatorname {S} _{1}'(0)&&\;=\;&&0\quad &&\Rightarrow &&\quad 0\;&&+&&\;0\;&&+&&\;a_{1}\;&&=\;&&0,&\\\operatorname {S} _{1}'(1)&&\;=\;&&0\quad &&\Rightarrow &&\quad 3a_{3}\;&&+&&\;2a_{2}\;&&+&&\;a_{1}\;&&=\;&&0.&\end{alignedat}}}$

Solving the system of 4 unknowns formed by the last 4 equations result in the values of the polynomial coefficients:

${\displaystyle a_{0}=0,\quad a_{1}=0,\quad a_{2}=3,\quad a_{3}=-2.}$

This results in the third-order "smoothstep" function:

${\displaystyle \operatorname {S} _{1}(x)=-2x^{3}+3x^{2}.}$

Starting with a generic fifth-order polynomial function, its first derivative and its second derivative:

${\displaystyle {\begin{alignedat}{13}\operatorname {S} _{2}(x)&&\;=\;&&a_{5}x^{5}&&\;+\;&&a_{4}x^{4}&&\;+\;&&a_{3}x^{3}&&\;+\;&&a_{2}x^{2}&&\;+\;&&a_{1}x&&\;+\;&&a_{0},&\\\operatorname {S} _{2}'(x)&&\;=\;&&5a_{5}x^{4}&&\;+\;&&4a_{4}x^{3}&&\;+\;&&3a_{3}x^{2}&&\;+\;&&2a_{2}x&&\;+\;&&a_{1},&\\\operatorname {S} _{2}''(x)&&\;=\;&&20a_{5}x^{3}&&\;+\;&&12a_{4}x^{2}&&\;+\;&&6a_{3}x&&\;+\;&&2a_{2}.&\end{alignedat}}}$

Applying the desired values for the function at both endpoints:

${\displaystyle {\begin{alignedat}{17}\operatorname {S} _{2}(0)&&\;=\;&&0\;\;\;\;\;&&\Rightarrow &&\;\;\;\;\;0\;&&+&&\;0\;&&+&&\;0\;&&+&&\;0\;&&+&&\;0\;&&+&&\;a_{0}&&\;=\;&&0,&\\\operatorname {S} _{2}(1)&&\;=\;&&1\;\;\;\;\;&&\Rightarrow &&\;\;\;\;\;a_{5}\;&&+&&\;a_{4}\;&&+&&\;a_{3}\;&&+&&\;a_{2}\;&&+&&\;a_{1}\;&&+&&\;a_{0}&&\;=\;&&1.&\end{alignedat}}}$

Applying the desired values for the first derivative of the function at both endpoints:

${\displaystyle {\begin{alignedat}{15}\operatorname {S} _{2}'(0)&&\;=\;&&0\;\;\;\;&&\Rightarrow &&\;\;\;\;0\;&&+&&\;0\;&&+&&\;0\;&&+&&\;0\;&&+&&\;a_{1}\;&&=\;&&0,&\\\operatorname {S} _{2}'(1)&&\;=\;&&0\;\;\;\;&&\Rightarrow &&\;\;\;\;5a_{5}\;&&+&&\;4a_{4}\;&&+&&\;3a_{3}\;&&+&&\;2a_{2}\;&&+&&\;a_{1}\;&&=\;&&0.&\end{alignedat}}}$

Applying the desired values for the second derivative of the function at both endpoints:

${\displaystyle {\begin{alignedat}{15}\operatorname {S} _{2}''(0)&&\;=\;&&0\;\;\;\;&&\Rightarrow &&\;\;\;\;0\;&&+&&\;0\;&&+&&\;0\;&&+&&\;2a_{2}\;&&=\;&&0,&\\\operatorname {S} _{2}''(1)&&\;=\;&&0\;\;\;\;&&\Rightarrow &&\;\;\;\;20a_{5}\;&&+&&\;12a_{4}\;&&+&&\;6a_{3}\;&&+&&\;2a_{2}\;&&=\;&&0.&\end{alignedat}}}$

Solving the system of 6 unknowns formed by the last 6 equations result in the values of the polynomial coefficients:

${\displaystyle a_{0}=0,\quad a_{1}=0,\quad a_{2}=0,\quad a_{3}=10,\quad a_{4}=-15,\quad a_{5}=6.}$

This results in the fifth-order "smootherstep" function:

${\displaystyle \operatorname {S} _{2}(x)=6x^{5}-15x^{4}+10x^{3}.}$

Applying similar techniques, the 7th-order equation is found to be:

${\displaystyle \operatorname {S} _{3}(x)=-20x^{7}+70x^{6}-84x^{5}+35x^{4}.}$

Smoothstep polynomials are generalized, with 0 ≤ x ≤ 1 as

${\displaystyle {\begin{aligned}\operatorname {S} _{N}(x)&=x^{N+1}\sum _{n=0}^{N}{\binom {N+n}{n}}{\binom {2N+1}{N-n}}(-x)^{n}\qquad N\in \mathbb {N} \\&=\sum _{n=0}^{N}(-1)^{n}{\binom {N+n}{n}}{\binom {2N+1}{N-n}}x^{N+n+1}\\&=\sum _{n=0}^{N}{\binom {-N-1}{n}}{\binom {2N+1}{N-n}}x^{N+n+1},\\\end{aligned}}}$

where N determines the order of the resulting polynomial function, which is 2N + 1. The first seven smoothstep polynomials, with 0 ≤ x ≤ 1, are

${\displaystyle {\begin{aligned}\operatorname {S} _{0}(x)&=x,\\\operatorname {S} _{1}(x)&=-2x^{3}+3x^{2},\\\operatorname {S} _{2}(x)&=6x^{5}-15x^{4}+10x^{3},\\\operatorname {S} _{3}(x)&=-20x^{7}+70x^{6}-84x^{5}+35x^{4},\\\operatorname {S} _{4}(x)&=70x^{9}-315x^{8}+540x^{7}-420x^{6}+126x^{5},\\\operatorname {S} _{5}(x)&=-252x^{11}+1386x^{10}-3080x^{9}+3465x^{8}-1980x^{7}+462x^{6},\\\operatorname {S} _{6}(x)&=924x^{13}-6006x^{12}+16380x^{11}-24024x^{10}+20020x^{9}-9009x^{8}+1716x^{7}.\\\end{aligned}}}$

It can be shown that the smoothstep polynomials ${\displaystyle \operatorname {S} _{N}(x)}$ that transition from 0 to 1 when x transitions from 0 to 1 can be simply mapped to odd-symmetry polynomials

${\displaystyle \operatorname {R} _{N}(x)=\left(\int _{0}^{1}{\big (}1-u^{2}{\big )}^{N}\,du\right)^{-1}\int _{0}^{x}{\big (}1-u^{2}{\big )}^{N}\,du,}$

where

${\displaystyle \operatorname {S} _{N}(x)={\tfrac {1}{2}}\operatorname {R} _{N}(2x-1)+{\tfrac {1}{2}}}$

and

${\displaystyle \operatorname {R} _{N}(-x)=-\operatorname {R} _{N}(x).}$

The argument of R_N(x) is −1 ≤ x ≤ 1 and is appended to the constant −1 on the left and +1 at the right.

An implementation of ${\displaystyle \operatorname {S} _{N}(x)}$ in Javascript:[6]

// Generalized smoothstep
function generalSmoothStep(N, x) {
  x = clamp(x, 0, 1); // x must be equal to or between 0 and 1
  var result = 0;
  for (var n = 0; n <= N; ++n)
    result += pascalTriangle(-N - 1, n) *
              pascalTriangle(2 * N + 1, N - n) *
              Math.pow(x, N + n + 1);
  return result;
}

// Returns binomial coefficient without explicit use of factorials,
// which can't be used with negative integers
function pascalTriangle(a, b) {
  var result = 1; 
  for (var i = 0; i < b; ++i)
    result *= (a - i) / (i + 1);
  return result;
}

function clamp(x, lowerlimit, upperlimit) {
  if (x < lowerlimit)
    x = lowerlimit;
  if (x > upperlimit)
    x = upperlimit;
  return x;
}