(B6) - Monotonic Cubics

As we physically push an analog-trigger downwards from resting position to fully depressed, we expect to observe a continuously increasing input from $0 \rightarrow 1$. If we are applying a sensitivity curve to our raw-input, we should make sure that our curve does not break this property. If we denote our curve by $s: [0,1] \rightarrow [0,1]$, we want $s$ to be continuous, surjective (onto), and monotone-increasing. In my recent post about analog inputs, I discuss how to define $s$ by a cubic polynomial.

$$s(x) = c_3 x^3 + c_2 x^2 + c_1 x + c_0$$

Given that $s$ is a polynomial, our constraints are equivalent to ensuring $s(0)=0$, $s(1)=1$, and that there are no local extrema in the interval $[0,1]$. We seek to describe the coefficients $c_i$ that satisfy these constraints. In the post I simply stated the results. In this blurb we will perform the derivation.

Generally speaking, an $n\mathtt{th}$ degree polynomial will uniquely fit $n+1$ given points. Since we are only specifying two points $(0,0)$ and $(1,1)$, we expect two degrees of freedom in our coefficients (constrained by monotonicity). Indeed, our fixed points imply

$$c_0 = 0 \\ c_3 + c_2 + c_1 = 1$$

As a matter of choice, we will take $c_1 = 1- c_3 - c_2$ and try to find the region in the $c_3 c_2$-plane that yields monotonicity. A curve takes on local extrema only when its slope is zero and its change in slope is nonzero, that is when $s'(x) = 0$ and $s''(x) \neq 0$. Plugging in $c_1$ and taking the derivatives we have

$$\newcommand{\uminus}{\raise.17ex\hbox{$\scriptstyle{-}$}} \begin{array}{lll} s(x) \; &= & \; c_3 \, x^3 + c_2 \, x^2 + (1-c_3-c_2) \, x \\ s'(x) \; &= & \; 3 c_3 \, x^2 + 2 c_2 \, x + (1 - c_3 - c_2) \\ s''(x) \; &= & \; 2 \, (3 c_3 \, x + c_2) \end{array}$$

$$\begin{align} s'(x) = 0 \quad \quad & \text{when} \quad \quad x_{\pm} := \frac{\uminus c_2 \pm \sqrt{c_2^2 + 3c_3c_2 + 3c_3\,(c_3-1)}}{3\,c_3} \\ \\ s''(x) = 0 \quad \quad & \text{when} \quad \quad x = \frac{\uminus c_2}{3\,c_3} \end{align}$$

Letting $\theta$ denote the portion inside the $\mathtt{sqrt}$, that is $\theta = c_2^2 - 3 c_3 c_1$, we notice that $\theta = 0$ gives us an inflection while $\theta < 0$ means $s$ has no real stationary points (or extrema). So the region $\theta \leq 0$ is allowed and is shown below.

The blue region is bounded by $\theta = 0$, explicitly represented by two curves $\gamma_\pm = \tfrac{1}{2} \left(\uminus 3c_3 \pm \sqrt{3c_3 ( 4- c_3)}\right)$, where $\gamma_+$ is shown in pink and $\gamma_-$ is shown in yellow. These curves meet with vertical slope at (0,0) and (4,-6). Before we continue we realize that we must have a nonnegative slope at $x=0$ and $x=1$. This bounds the allowable $c_2 c_3$-region by the two lines shown below.

Indeed the $\theta \leq 0$ falls within such bounds. When $\theta > 0$ our curve $s$ will have two extremum at $x_\pm$. Our coefficients $c_2, c_3$ are allowed when neither $x_\pm$ lie within $(0,1)$. This can be broken up into the following regions.

$$\begin{array}{llll} \mathtt{(i)}  \quad & (c_3 < 0) \; \cap  \; (x_- \leq 0) \\ \mathtt{(ii)}  \quad & (c_3 < 0) \; \cap  \; (x_+ \geq 1) \\ \mathtt{(iii)}  \quad & (c_3 < 0) \; \cap  \; (x_+ \leq 0) \; \cap \; (x_- \geq 1) \\ \mathtt{(iv)}  \quad & (c_3 > 0) \; \cap  \; (x_+ \leq 0) \\ \mathtt{(v)}  \quad & (c_3 > 0) \; \cap  \; (x_- \geq 1) \\ \mathtt{(vi)}  \quad & (c_3 > 0) \; \cap  \; (x_- \leq 0) \; \cap \; (x_+ \geq 1) \\ \end{array}$$

The regions $\mathtt{(i)} = \mathtt{(ii)} = \mathtt{(vi)} = \emptyset$. Putting this all together we have

$$\begin{align} c_1 \; & = \; 1-c_3-c_2 \\ \\ c_2 \; & \in \; \left\{ \begin{array}{lll} [\uminus 1 - 2 c_3, 1-c_3] & c_3 \leq 1 \\ [\gamma_- , \gamma_+] & c_3 > 1 \end{array} \right. \\ \\ c_3 \; & \in \; [\uminus 2,4] \\ \\ \gamma_\pm \; & = \; \tfrac{1}{2} \left(\uminus 3c_3 \pm \sqrt{3c_3 ( 4- c_3)}\,\right) \end{align}$$

I was interested in what the three dimensional region looks like for quartics. Even with a symbolic algebra manipulator the equations quickly became unwieldy (roots of a cubic are not so simple), so the following scatterplot is empirical. The resulting region in $c_4 c_3 c_2$-space looks like a twisted surfboard.

The approximate bounds are $c_4 \in [\uminus 9,9]$, $c_3 \in [\uminus 16.48,20.39]$, $c_2 \in [\uminus 16.11, 9]$, $c_1 \in [0,6]$. The white specks are the projected "shadow" onto the graph's walls. The cross-sections $c_4 = \alpha$ are shown below. The purple region $c_4=0$ is exactly what we had found for cubics. Because the coefficients $c_i$ are related by the relation $c_1 + c_2 + c_3 + c_4 = 1$, we get similar contours for each.