Quadratic functions in two variables

We originally might have the following set of data on the triangle with vertices $(1,1),(3,1),(1.5,3)$: Taking the vertex $(1.0,1.0)$ as reference point we might consider this triangle as the image of the standard triangle with vertices $(0,0),(1,0),(0,1)$ under the linear transformation

${\displaystyle \left(\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \end{array}\right)\hspace{0.5em}=\hspace{0.5em}\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right)+\left(\begin{array}{cc}\hfill 2& \hfill 0.5\\ \hfill 0& \hfill 2\end{array}\right)\left(\begin{array}{c}\hfill {\xi }_1\hfill \\ \hfill {\xi }_2\hfill \end{array}\right)}$

and hence the inverse transformation is

${\displaystyle \left(\begin{array}{c}\hfill {\xi }_1\hfill \\ \hfill {\xi }_2\hfill \end{array}\right)\hspace{0.5em}=\hspace{1em}\left(\begin{array}{cc}\hfill 0.5& \hfill -0.125\\ \hfill 0& \hfill 0.5\end{array}\right)\left(\begin{array}{c}\hfill x_1-1\hfill \\ \hfill x_2-1\hfill \end{array}\right)}$

Now, on the standard triangle we have the data and expressing the quadratic here in $({\xi }_1,{\xi }_2)$ is quite simple: we get the linear system

${\displaystyle \left(\begin{array}{cccccc}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0\\ \hfill 1& \hfill 0.5& \hfill 0& \hfill 0.25& \hfill 0& \hfill 0\\ \hfill 1& \hfill 1& \hfill 0& \hfill 1& \hfill 0& \hfill 0\\ \hfill 1& \hfill 0& \hfill 0.5& \hfill 0& \hfill 0& \hfill 0.25\\ \hfill 1& \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 1\\ \hfill 1& \hfill 0.5& \hfill 0.5& \hfill 0.25& \hfill 0.25& \hfill 0.25\end{array}\right)\left(\begin{array}{c}\hfill f_0\\ \hfill g_1\\ \hfill g_2\\ \hfill h_{11}/2\\ \hfill h_{12}/2\\ \hfill h_{22}/2\end{array}\right)\hspace{0.5em}=\hspace{0.5em}\left(\begin{array}{c}\hfill 1\\ \hfill -2\\ \hfill 0\\ \hfill -1\\ \hfill 2\\ \hfill 1\end{array}\right)}$

with the solution

${\displaystyle q({\xi }_1,{\xi }_2)=1-11{\xi }_1-9{\xi }_2+\backslash tfrac12(20{\xi }_2^2+40{\xi }_1{\xi }_2+20{\xi }_2^2)}$

Expressing the variables ${\xi }_i$ in the variables $x_i$ we get $q$ expressed in these. This technique works in any dimension on any full simplex.

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On 16 Jun 2016, 17:26.