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improved proof
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Martin Thoma
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@ -431,82 +431,48 @@ $b, c, d \in \mdr$ be a function.
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\end{theorem}
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\begin{proof}
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Let $g : \mdr \rightarrow \mdr$ be a polynomial of degree 5
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\[g(x) = \tilde{a} x^5 + \tilde{b} x^4 + \tilde{c} x^3 + \tilde{d} x^2 + \tilde{e} x + \tilde{f}\]
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with $\tilde{a} \in \mdr_{> 0},\; \tilde{b} \in \mdr \setminus \Set{0}$ and $\tilde{c}, \tilde{d}, \tilde{e}, \tilde{f} \in \mdr$.
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Then, according to the Abel-Ruffini theorem, the equation
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\[g(x) = 0\]
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cannot be solved algebraicly.
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Suppose you could solve the closest point problem for arbitrary
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cubic functions $f = ax^3 + bx^2 + cx + d$ and arbitrary points $P = (x_P, y_P)$.
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%But lets define $a := \frac{\sqrt{\tilde{a}}}{3}$, $b := \frac{\tilde{b}}{5a}$,
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%$c : = \frac{\frac{\tilde{c}}{2} - b^2}{2a}$, $d := \frac{\tilde{d}}{3} - bc + $
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Then you could solve the following problem for $x$:
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\begin{align}
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0 &\stackrel{!}{=} \left ((d_{P,f}(x))^2 \right )'
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&=-2 x_p + 2x -2y_p(f(x))' + (f(x)^2)'\\
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&= 2 f(x) \cdot f'(x) - 2 y_p f'(x) + 2x - 2 x_p\\
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&= f(x) \cdot f'(x) - y_p f'(x) + x - x_p\\
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&= \underbrace{f'(x) \cdot \left (f(x) - y_p \right )}_{\text{Polynomial of degree 5}} + x - x_p
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\end{align}
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So you can find $a, b, c, d, x_p, y_p$ such that
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General algebraic equations of degree 5 don't have a solution formula.\footnote{TODO: Quelle}
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Although here seems to be more structure, the resulting algebraic
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equation can be almost any polynomial of degree 5:\footnote{Thanks to Peter Košinár on \href{http://math.stackexchange.com/a/584814/6876}{math.stackexchange.com} for this one}
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\begin{align}
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g(x) &= \underbrace{3 a^2}_{= \tilde{a}} x^5 + \underbrace{5ab}_{= \tilde{b}}x^4 + \underbrace{2(2ac + b^2 )}_{= \tilde{c}}x^3 &+& \underbrace{3(ad+bc-ay_p)}_{= \tilde{d}} x^2 \\
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& &+& \underbrace{(2 b d+c^2+1-2 b y_p)}_{= \tilde{e}}x+\underbrace{c d-c y_p-x_p}_{= \tilde{f}}\\
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&= f'(x) \cdot \left (f(x) - y_p \right ) + (x - x_p)\\
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&= f(x) \cdot f'(x) - y_p f'(x) + x - x_p
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\end{align}
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\begin{align}
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0 &\stackrel{!}{=} f'(x) \cdot \left (f(x) - y_p \right ) + (x - x_p)\\
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&= \underbrace{3 a^2}_{= \tilde{a}} x^5 + \underbrace{5ab}_{\tilde{b}}x^4 + \underbrace{2(2ac + b^2 )}_{=: \tilde{c}}x^3 &+& \underbrace{3(ad+bc-ay_p)}_{\tilde{d}} x^2 \\
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& &+& \underbrace{(2 b d+c^2+1-2 b y_p)}_{=: \tilde{e}}x+\underbrace{c d-c y_p-x_p}_{=: \tilde{f}}\\
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0 &\stackrel{!}{=} \tilde{a}x^5 + \tilde{b}x^4 + \tilde{c}x^3 + \tilde{d}x^2 + \tilde{e}x + \tilde{f}
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\end{align}
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And
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\begin{align}
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g(x) &\stackrel{!}{=}0\\
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\Leftrightarrow 0 &\stackrel{!}{=} 2 f(x) \cdot f'(x) - 2 y_p f'(x) + 2x - 2 x_p\\
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&= -2 x_p + 2x -2y_p(f(x))' + (f(x)^2)'\\
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&= \left ((x-x_p)^2 \right )' + \left ( (f(x) - y_p)^2 \right )'\\
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&= \left ((x-x_p)^2 + (f(x) - y_p)^2 \right )'\\
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&= (d_{P,f}(x)^2)'
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\end{align}
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\begin{enumerate}
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\item For any coefficient $\tilde{a} \in \mdr_{> 0}$ of $x^5$ we can choose $a$ such that we get $\tilde{a}$.
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\item For any coefficient $\tilde{b} \in \mdr \setminus \Set{0}$ of $x^4$ we can choose $b$ such that we get $\tilde{b}$.
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\item With $c$, we can get any value of $\tilde{c} \in \mdr$.
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\item With $d$, we can get any value of $\tilde{d} \in \mdr$.
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\item With $y_p$, we can get any value of $\tilde{e} \in \mdr$.
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\item With $x_p$, we can get any value of $\tilde{f} \in \mdr$.
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\end{enumerate}
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So the problem of finding a closest point $(x, f(x))$ on a
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cubic function $f$ to $P$ is essentially the same as finding
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a root of a polynomial function of degree 5. As this cannot
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be solved algebraicly, the problem of finding such a point
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can also not be solved algebraicly.$\qed$
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The first restriction guaratees that we have a polynomial of
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degree 5. The second one is necessary, to get a high range of
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$\tilde{e}$.
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This means, that there is no solution formula for the problem of
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finding the closest points on a cubic function to a given point,
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because if there was one, you could use this formula for finding
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roots of polynomials of degree 5. $\qed$
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\end{proof}
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\todo[inline]{Start with theorem that this problem is not solvable
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with analytics only. Use a general 5th degree function and show
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that it can be mapped to a $f$ and $P$ instance.}
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When you want to calculate points with minimal distance, you can
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take the same approach as in Equation \ref{eq:minimizingFirstDerivative}:
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\begin{align}
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0 &\stackrel{!}{=} -2 x_p + 2x -2y_p(f(x))' + (f(x)^2)'\\
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&= 2 f(x) \cdot f'(x) - 2 y_p f'(x) + 2x - 2 x_p\\
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&= f(x) \cdot f'(x) - y_p f'(x) + x - x_p\\
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&= \underbrace{f'(x) \cdot \left (f(x) - y_p \right )}_{\text{Polynomial of degree 5}} + x - x_p
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\end{align}
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General algebraic equations of degree 5 don't have a solution formula.\footnote{TODO: Quelle}
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Although here seems to be more structure, the resulting algebraic
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equation can be almost any polynomial of degree 5:\footnote{Thanks to Peter Košinár on \href{http://math.stackexchange.com/a/584814/6876}{math.stackexchange.com} for this one}
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\begin{align}
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0 &\stackrel{!}{=} f'(x) \cdot \left (f(x) - y_p \right ) + (x - x_p)\\
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&= \underbrace{3 a^2}_{= \tilde{a}} x^5 + \underbrace{5ab}_{\tilde{b}}x^4 + \underbrace{2(2ac + b^2 )}_{=: \tilde{c}}x^3 &+& \underbrace{3(ad+bc-ay_p)}_{\tilde{d}} x^2 \\
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& &+& \underbrace{(2 b d+c^2+1-2 b y_p)}_{=: \tilde{e}}x+\underbrace{c d-c y_p-x_p}_{=: \tilde{f}}\\
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0 &\stackrel{!}{=} \tilde{a}x^5 + \tilde{b}x^4 + \tilde{c}x^3 + \tilde{d}x^2 + \tilde{e}x + \tilde{f}
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\end{align}
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\begin{enumerate}
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\item With $a$, we can get any value of $\tilde{a} \in \mdr \setminus \Set{0}$.
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\item With $b$, we can get any value of $\tilde{b} \in \mdr \setminus \Set{0}$.
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\item With $c$, we can get any value of $\tilde{c} \in \mdr$.
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\item With $d$, we can get any value of $\tilde{d} \in \mdr$.
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\item With $y_p$, we can get any value of $\tilde{e} \in \mdr$.
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\item With $x_p$, we can get any value of $\tilde{f} \in \mdr$.
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\end{enumerate}
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The first restriction only guaratees that we have a polynomial of
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degree 5. The second one is necessary, to get a high range of
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$\tilde{e}$.
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This means, that there is no solution formula for the problem of
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finding the closest points on a cubic function to a given point.
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\subsection{Another approach}
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Just like we moved the function $f$ and the point to get in a
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