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added comment 'symmetry axis'

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Martin Thoma 2013-12-06 00:37:04 +01:00
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documents/math-minimal-distance-to-cubic-function/math-minimal-distance-to-cubic-function.tex
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@ -356,7 +356,7 @@ For all other points $P = (0, w)$, there are exactly two minima $x_{1,2} = \pm \
So the solution is given by So the solution is given by
\begin{align*} \begin{align*}
x_S &:= - \frac{b}{2a}\\ x_S &:= - \frac{b}{2a} \;\;\;\;\; \text{(the symmetry axis)}\\
\underset{x\in\mdr}{\arg \min d_{P,f}(x)} &= \begin{cases} \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} \\ 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} \\
x_2 = -\sqrt{a (y_p + \frac{b^2}{4a} - c) - \frac{1}{2}} + x_S\\ x_2 = -\sqrt{a (y_p + \frac{b^2}{4a} - c) - \frac{1}{2}} + x_S\\