Lagrange’s interpolation formula: Theoreme and Example I would like to know where they come up with the formula of Lagrange interpolation (Lagrange’s interpolation formula),Lagrange_polynomial because I did some research, but I find a different definition of Lagrange interpolation. Beside would someone explain and elaborate more how they apply the formula 5.1 in example 5.6. because I don't know how to apply it. for example, what is $c_{i1}$
Thanks in advanced.
Theorem $5.5 ([26])$.
Let $f : k^{r} \to k$ be any function on a finite field $k$. Then there exists a unique polynomial $g : k^{r} \to k$, such that $\forall x \in k^{r}, f(x)=g(x).$
Any such mapping over a finite field can be described by a unique polynomial.
Using Lagrange interpolation, we can easily determine the polynomial. Let $f: k^{r}\to k$ be any function on $k$. Then
$$g(x)=\sum_{\left(c_{i1},\ldots,c_{ir}\right)\in k^{r}}f\left(c_{i1},\ldots,c_{ir} \right)\prod_{j=1}^{r}\left(1-\left(x_{j}-c_{ij} \right)^{p-1} \right)\quad \mbox{(5.1)}$$
is the unique polynomial that defines the same mapping as $f$.
Example $5.6$
Suppose $k=\mathbb{F}_{3},r=2,$ and the mapping $f$ is defined on $\mathbb{F}_{3}^{2}=\{0,1,2\}\times\{0,1,2\}$ as follows:
$$f(0,0)=0,$$
$$f(0,1)=1,$$
$$f(0,2)=2,$$
$$f(1,0)=1,$$
$$f(1,1)=2,$$
$$f(1,2)=0,$$
$$f(2,0)=2,$$
$$f(2,1)=0,$$
$$f(2,2)=1.$$
Then the polynomial $g$ that defines the same mapping as $f$ is constructed as follows:
$$\begin{align*}
g(x,y)&=0\\
&+1\left((1-x^{2})(1-(y-1)^{2}) \right)\\
&+2\left((1-x^{2})(1-(y-2)^{2}) \right)\\
&+1\left((1-(x-1)^{2})(1-y^{2}) \right)\\
&+2\left((1-(x-1)^{2})(1-(y-1)^{2}) \right)\\
&+0\\
&+2\left((1-(x-2)^{2})(1-y^{2}) \right)\\
&+0\\
&+1\left((1-(x-2)^{2})(1-(y-2)^{2})\right)\\
&=x+y.
\end{align*}$$
Reference:
Mathematical Concepts and Methods in Modern Biology 1st Edition Using Modern Discrete Models

 A: You would do well to study the idea of Lagrange interpolation. There is a unique polynomial of the minimal reasonable degree fitting data. In your setting, $t^4=t$ so any of $x+y,x^4+y,x+y^4,x^4+y^4$ or $(x+y)^4=x^4+x^3y+xy^4+x^4$ agree at all points.
Here are a few observations about your example: Consider the $9$ function values you specify, but as integers. The standard Lagrange interpolation would be
$$\begin{align*}
g(x,y)&=0\\
&+1 \frac {x \left( x-2 \right) 
 \left( y-2 \right)  \left( y-1 \right) }{-2}
\\
&+2\frac {x \left( x-1
 \right)  \left( y-2 \right)  \left( y-1 \right) }{4}
\\
&+1\frac {y
 \left( x-2 \right)  \left( x-1 \right)  \left( y-2 \right) }{-2}
\\
&+2\frac{xy \left( x-2 \right)  \left( y-2 \right)}{1} \\
&+0\\
&+2\frac {y \left( x-2 \right) 
 \left( x-1 \right)  \left( y-1 \right) }{4}
\\
&+0\\
&+1\frac {xy \left( x-1
 \right)  \left( y-1 \right) }{4}
\\
&=\frac{9\,{x}^{2}{y}^{2}-15\,{x}^{2}y-15\,x{y}^{2}+21\,xy+4\,x+4\,y}{4}.
\end{align*}$$
And that is the unique polynomial of that degree over the reals, which does that. Then, $\mod 3,$ we have $4=1$ and the other coefficients are $0$, so indeed, $x+y.$
What you (and your book) wrote, treated as over the integers, comes out $9\,{x}^{2}{y}^{2}-18\,{x}^{2}y-18\,x{y}^{2}+6\,{x}^{2}+24\,xy+6\,{y}^{
2}-2\,x-2\,y-3
$
The values at the appropriate integer points are $$[0, 0, -3], [1, 0, 1], [2, 0, 17], [0, 1, 1], [1, 1, 2], [2, 1, -3], [0, 2, 17], [1, 2, -3], [2, 2, -11]$$ but again,$\mod 3,$ everything comes out fine.
