shall construct the first function $x\rightarrow d^2_{M(\beta)0}(x,y)$ for arbitrary multiindex $\beta$. The other functions can be constructed completely analogously. We start with the $L^p$-approximation. $$d_{N0}^2(x,y)=d^ 2_{A^ {-1}(y)}(x,y)+\sum_{j=1}^Nc_{j0}\Pi_{r=1}^j\Pi_{l=1}^n(x_l-x^r_l)^2d^ 2_{A^ {-1}(x_j)}(x,y),$$ where the numbers $c_{j1}$ have been determined according to section 4.1.. Next we define $d^2_{M(\beta)0}(x,x^N)$ for multiindices of order $|\beta|=3$. Let $\beta^0,\cdots, \beta^k,\cdots ,\beta^{R}$ a list of multiindices of order $3$. The length $R$ of this list is dependent of the dimension $n$ of course. Start with $\beta^0=(\beta^0_1,\cdots,\beta^0_n)$ and let $\gamma^0$ be an multiindex with $|\gamma|=2$ such that $\beta^0-\gamma =1_i$ for some index $i$. Define (recall that $x^0=y$) $$d^2_{\beta^0 0}(x,y)=d^2_{N0}(x,y)+\frac{1}{\beta^0!}c_{\beta^0}^0(x-y)^{\beta^0}.$$ Then plug $d^2_{\beta^0 0}(x,y)$ into the equation $$\label{beta0}
\begin{array}{ll}
\partial^{(\beta^0-\gamma^0)}_xd^2(x,y)=\partial^{(\beta^0-\gamma^0)}_x\left(\frac{1}{4}\sum_{ij}a_{ij}(x)\frac{\partial d^2}{\partial x_i}\frac{\partial d^2}{\partial x_j} \right),
\end{array}$$ evaluate at $x=x^0=y$ and solve for the real number $c_{\beta^0}^0$. Then proceed recursively: having defined the function $x\rightarrow d^2_{\beta^0 (k-1)}(x,y)$ define $$d^2_{\beta^0 k}(x,y)=d^2_{\beta^0 (k-1)}(x,y)+c_{\beta^0}^k\Pi_{l=0}^{k-1}(x-x^l)^{\beta^0+{\bf 1}}\frac{1}{\beta^0!}(x-x^k)^{\beta^0},$$ where ${\bf 1}=(1,1,\cdots,1)$. Then plug $d^2_{\beta^0 k}(x,y)$ into the equation , evaluate at $y$, and solve for $c_{\beta^0}^k$. When $k=N$ we have got the approximation $$d^2_{\beta^0 N}(x,y)=d^2_{N0}(x,y)+\sum_{k=0}^Nc_{\beta^0}^k\Pi_{l=0}^{k-1}(x-x^l)^{\beta^0+{\bf 1}}\frac{1}{\beta^0!}(x-x^k)^{\beta^0}.$$ with $N+1$ real numbers $c_{\beta^0}^k$ for $0\leq k\leq N$ determined recursively. Note that the function $x\rightarrow d^2_{\beta^0 k}(x,y)$ satisfies the equations and at all interpolation points $x^0,\cdots,x^N$. Then we take the next 