Editing Bicubic interpolation (section)

==Extension to rectilinear grids==

Often, applications call for bicubic interpolation using data on a rectilinear grid, rather than the unit square. In this case, the identities for <math>p_x, p_y,</math> and <math>p_{xy}</math> become
<math display="block">p_x(x,y) = \textstyle \sum\limits_{i=1}^3 \sum\limits_{j=0}^3 \frac{a_{ij} i x^{i-1} y^j}{\Delta x},</math>
<math display="block">p_y(x,y) = \textstyle \sum\limits_{i=0}^3 \sum\limits_{j=1}^3 \frac{a_{ij} x^i j y^{j-1}}{\Delta y},</math>
<math display="block">p_{xy}(x,y) = \textstyle \sum\limits_{i=1}^3 \sum\limits_{j=1}^3 \frac{a_{ij} i x^{i-1} j y^{j-1}}{\Delta x \Delta y},</math>
where <math>\Delta x</math> is the <math>x</math> spacing of the cell containing the point <math>(x,y)</math> and similar for <math>\Delta y</math>.
In this case, the most practical approach to computing the coefficients <math>\alpha</math> is to let
<math display="block">x=\left[\begin{smallmatrix}f(0,0)&f(1,0)&f(0,1)&f(1,1)&\Delta x f_x(0,0)&\Delta xf_x(1,0)&\Delta x f_x(0,1)&\Delta x f_x(1,1)&\Delta y f_y(0,0)&\Delta y f_y(1,0)&\Delta y f_y(0,1)&\Delta y f_y(1,1)&\Delta x \Delta y f_{xy}(0,0)&\Delta x \Delta y f_{xy}(1,0)&\Delta x \Delta y f_{xy}(0,1)&\Delta x \Delta y f_{xy}(1,1)\end{smallmatrix}\right]^T,</math>
then to solve <math>\alpha=A^{-1}x</math> with <math>A</math> as before. Next, the normalized interpolating variables are computed as
<math display="block">\begin{align}
\overline{x} &= \frac{x-x_0}{x_1-x_0}, \\
\overline{y} &= \frac{y-y_0}{y_1-y_0}
\end{align}</math>
where <math>x_0, x_1, y_0,</math> and <math>y_1</math> are the <math>x</math> and <math>y</math> coordinates of the grid points surrounding the point <math>(x,y)</math>. Then, the interpolating surface becomes
<math display="block">p(x,y) = \sum\limits_{i=0}^3 \sum_{j=0}^3 a_{ij} {\overline{x}}^i {\overline{y}}^j.</math>