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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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documents/math-minimal-distance-to-cubic-function/constant-functions.tex
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@ -116,3 +116,11 @@ The point with minimum distance can be found by:
\Set{a} &\text{if } S_0(f,P) \ni x_P < a\\
\Set{b} &\text{if } S_0(f,P) \ni x_P > 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}