誤植

Errata for 1st version( 2022/09/13 ).

P.50 Equation (3.3)

$$h(x) = \left\{ \begin{array}{ll} \text{mix}(0.3, 0.9, c(x+1)) = 0.9 + 1.8x^2 + 1.2x^3 & (-1 \leq x \leq 0)\\ \text{mix}(0.9, 0.6, c(x)) = 0.9 - 0.9x^2 + 0.6x^3 & (0 \leq x \leq 1) \end{array} \right.$$

->

$$h(x) = \left\{ \begin{array}{ll} \text{mix}(0.3, 0.9, c(x+1)) = 0.9 - 1.8x^2 - 1.2x^3 & (-1 \leq x \leq 0)\\ \text{mix}(0.9, 0.6, c(x)) = 0.9 - 0.9x^2 + 0.6x^3 & (0 \leq x \leq 1) \end{array} \right.$$

P.51 left differential coefficient

$$\frac{h( - \varepsilon) - h(0)}{\varepsilon} = \frac{ 1.8 \varepsilon ^ 2 - 1.2 \varepsilon ^ 3 }{ \varepsilon } \rightarrow 0$$

->

$$\frac{h( - \varepsilon) - h(0)}{\varepsilon} = \frac{ -1.8 \varepsilon ^ 2 + 1.2 \varepsilon ^ 3 }{ \varepsilon } \rightarrow 0$$

P.51 derived function

$$h'(x) = \left\{ \begin{array}{ll} 3.6x(1+x) & (-1 \leq x \leq 0)\\ 1.8x(-1+x) & (0 \leq x \leq 1) \end{array} \right.$$

->

$$h'(x) = \left\{ \begin{array}{ll} -3.6x(1+x) & (-1 \leq x \leq 0)\\ 1.8x(-1+x) & (0 \leq x \leq 1) \end{array} \right.$$

P.52 second derived function

$$h''(x) = \left\{ \begin{array}{ll} 3.6+7.2x & (-1 \leq x \lt 0)\\ -1.8+3.6x & (0 \lt x \leq 1) \end{array} \right.$$

->

$$h''(x) = \left\{ \begin{array}{ll} -3.6-7.2x & (-1 \leq x \lt 0)\\ -1.8+3.6x & (0 \lt x \leq 1) \end{array} \right.$$

P.131 コード 9.3

This is not strictly a typo but if you want to conform to the text, this expression might be better.

float smin(float a, float b, float k){
  float h = clamp(0.5 + 0.5 * (b - a) / k, 0.0, 1.0);
  return mix(b, a, h) - k * h * (1.0 - h); 
}

->

float smin(float a, float b, float k){
  float h = clamp(0.5 - 0.5 * (b - a) / k, 0.0, 1.0);
  return mix(a, b, h) - k * h * (1.0 - h); 
}