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first part of solution formula is ready

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Martin Thoma 2013-12-06 00:30:05 +01:00
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documents/math-minimal-distance-to-cubic-function/math-minimal-distance-to-cubic-function.tex
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@ -334,12 +334,12 @@ Because:
Then move $f_1$ and $P_1$ by $\frac{b^2}{4a}-c$ in $y$ direction. You get:
\[f_2(x) = ax^2\;\;\;\text{ and }\;\;\; P_2 = \left (x_p+\frac{b}{2a},\;\; y_p+\frac{b^2}{4a}-c \right )\]
As $f(x) = ax^2$ is symmetric to the $y$ axis, only points
As $f_2(x) = ax^2$ is symmetric to the $y$ axis, only points
$P = (0, w)$ could possilby have three minima.
Then compute:
\begin{align}
d_{P,f}(x) &= \sqrt{(x-x_{P})^2 + (f(x)-w)^2}\\
d_{P,{f_2}}(x) &= \sqrt{(x-x_{P})^2 + (f(x)-w)^2}\\
&= \sqrt{x^2 + (ax^2-w)^2}\\
&= \sqrt{x^2 + a^2 x^4-2aw x^2+w^2}\\
&= \sqrt{a^2 x^4 + (1-2aw) x^2 + w^2}\\
@ -347,16 +347,23 @@ Then compute:
&= \sqrt{\left (a^2 x^2 + \nicefrac{1}{2}-a w \right )^2 + (w^2 - (1-2 a w)^2)}\\
\end{align}
For $w \leq \nicefrac{1}{2a}$ you only have $x = 0$ as a minimum.
For all other points $P = (0, w)$, there are exactly two minima.
The term
\[a^2 x^2 + (\nicefrac{1}{2}-a w)\]
should get as close to $0$ as possilbe when we want to minimize
$d_{P,{f_2}}$. For $w \leq \nicefrac{1}{2a}$ you only have $x = 0$ as a minimum.
For all other points $P = (0, w)$, there are exactly two minima $x_{1,2} = \pm \sqrt{aw - \nicefrac{1}{2}}$.
So the solution is given by
\[\underset{x\in\mdr}{\arg \min d_{P,f}(x)} = \begin{cases}
x_1 = todo \text{ and } x_2 = todo &\text{if } x_P = - \frac{b}{2a} \text{ and } y_p + \frac{b^2}{4a} - c > \frac{1}{2a} \\
x = todo &\text{if } x_P = - \frac{b}{2a} \text{ and } y_p + \frac{b^2}{4a} - c \leq \frac{1}{2a} \\
x = todo &\text{if } x_P \neq - \frac{b}{2a}
\end{cases}\]
\begin{align*}
x_S &:= - \frac{b}{2a}\\
\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_2 = -\sqrt{a (y_p + \frac{b^2}{4a} - c) - \frac{1}{2}} + x_S\\
x_1 = x_S &\text{if } x_P = x_S \text{ and } y_p + \frac{b^2}{4a} - c \leq \frac{1}{2a} \\
x_1 = todo &\text{if } x_P \neq x_S
\end{cases}
\end{align*}
\clearpage
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