Martin Thoma 11 vuotta sitten
vanhempi
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ba879e5ae8

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documents/math-minimal-distance-to-cubic-function/math-minimal-distance-to-cubic-function.pdf


+ 2 - 1
documents/math-minimal-distance-to-cubic-function/math-minimal-distance-to-cubic-function.tex

@@ -402,7 +402,8 @@ $t$:
 So the solution is given by
 \begin{align*}
 x_S &:= - \frac{b}{2a} \;\;\;\;\; \text{(the symmetry axis)}\\
-w &:= y_P+\frac{b^2}{4a}-c \;\;\; \text{ and } \;\;\; \alpha := \frac{(1- 2 aw)}{2 a^2} \;\;\;\text{ and }\;\;\; \beta := \frac{-z}{2 a^2}\\
+w &:= y_P+\frac{b^2}{4a}-c \;\;\; \text{ and } \;\;\; z := x_P+\frac{b}{2a}\\
+\alpha &:= \frac{(1- 2 aw)}{2 a^2} \;\;\;\text{ and }\;\;\; \beta := \frac{-z}{2 a^2}\\
 t &:= \sqrt[3]{\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)} -9\beta}\\
 \underset{x\in\mdr}{\arg \min d_{P,f}(x)} &= \begin{cases}
      x_1 = +\sqrt{a (y_p + \frac{b^2}{4a} - c) - \frac{1}{2}} + x_S \text{ and }   &\text{if } x_P = x_S \text{ and } y_p + \frac{b^2}{4a} - c >  \frac{1}{2a} \\