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moved argument to the correct location; consistency for titles; added proof to theorem (only two solutions for quadratic problem)

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Martin Thoma 2013-12-16 11:56:21 +01:00
parent dc82db6336
commit bdf8bbca3d
5 changed files with 40 additions and 11 deletions
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10
documents/math-minimal-distance-to-cubic-function/linear-functions.tex
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@ -29,6 +29,8 @@ $t \in \mdr$ be a linear function.
\addplot[domain=-5:5, thick,samples=50, red] {0.5*x};
\addplot[domain=-5:5, thick,samples=50, blue] {-2*x+6};
\addplot[black, mark = *, nodes near coords=$P$,every node near coord/.style={anchor=225}] coordinates {(2, 2)};
\addplot[blue, nodes near coords=$f_\bot$,every node near coord/.style={anchor=225}] coordinates {(1.5, 3)};
\addplot[red, nodes near coords=$f$,every node near coord/.style={anchor=225}] coordinates {(0.9, 0.5)};
\addlegendentry{$f(x)=\frac{1}{2}x$}
\addlegendentry{$f_\bot(x)=-2x+6$}
\end{axis}
@ -115,10 +117,4 @@ The point with minimum distance can be found by:
\Set{b} &\text{if } S_1(f, P) \ni x > b
\end{cases}\]
Because:
\begin{align}
\underset{x\in[a,b]}{\arg \min d_{P,f}(x)} &= \underset{x\in[a,b]}{\arg \min d_{P,f}(x)^2}\\
&=\underset{x\in[a,b]}{\arg \min} x^2 - 2x_P x + (x_P^2 + y_P^2 - 2 y_P c + c^2)\\
&=\underset{x\in[a,b]}{\arg \min} x^2 - 2 x_P x + x_P^2\\
&=\underset{x\in[a,b]}{\arg \min} (x-x_P)^2\\
\end{align}
\todo[inline]{argument? proof?}