Lagrange’s interpolation formula: Theoreme and Example [closed]

Question

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

Answer 1

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.