Lagrange interpolation

Proof
  Let $$ \tag{1} $$ be $n+1$ points on a real function $f(x)$, where Let $p(x)$ be the nth degree polynomial passing through $(1)$. $p(x)$ satisfies $$ \tag{2} $$ (See fig. below).
  We put $p(x)$ as $$ \tag{3} $$ , where $L_{i}(x)$ $(i=0,1, \cdots,n)$ are nth degree polynomials. (The specific form of $L_{i}(x)$ has not been determined at this point.)
  If $L_{i}(x)$ satisfy , $p(x_{0}) = f(x_{0}) $. If $L_{i}(x)$ satisfy $p(x_{1}) = f(x_{1}) $. In the same way, we see that if $L_{i}(x)$ satisfy, $$ \tag{4} $$ , $p(x_{j}) = f(x_{j})$ $(j=0,1,\cdots,n)$.
  In $(4)$, $L_{0}(x_{i}) = 0$ for $i =1, 2, \cdots, n$. By the factor theorem, $L_{0} (x)$ can be expressed as , where $\alpha_{0} $ is a constant. By $L_{0}(x_{0}) = 1$, it is derived as Therefore we obtain
  In a similar way, since $L_{1}(x_{i}) = 0$ for $i =0,2,3\cdots,n$, by the factor theorem, we obtain $L_{1}(x)$ can be written as , where $\alpha_{1} $ is a constant. By $L_{1}(x_{1}) = 1$, it is derived as . Therefore we obtain By repeating the same discussion, we can derive $L_{j}(x)$ for $j = 0,1,\cdots, n$ as . This expression can be written by the symbol $\prod$ as $$ \tag{5} $$
  Substituing $(5)$ into $(3)$, we have It is clear that this function passes through points $(1)$ (that is, it satisfies $(2)$), since $L_{j}(x)$ satisfies $(4)$.
  Approximating the original function $f (x)$ with the polynomial function $p (x)$ defined in this way, that is, is called Lagrange interpolation.

Proof  

Problems solving a system of linear equations
  Let $f(x)$ be a polynomial of degree $n$ defined as , and that passes through $n+1$ different points, . We have $$ \tag{2} $$
  Let $X$ be an $(n+1) \times (n+1)$ matrix, and $\mathbf{a}$ and $\mathbf{y}$ be $n$ dimensional vectors defined as . Equations $(2)$ can be written as $$ \tag{3} $$
  Equation $(3)$ ( or $(2)$) is a system of $n+1$ linear equations with $n+1$ unknowns.   In order for the function $f (x)$ to be unique, each coefficient $a_{0}, a_{1}, \cdots, a_{n}$ must be unique. To be so, the solution of system of linear equations $(2)$, that is $\mathbf{a}$, must be unique.

  A necessary and sufficient condition for the system of linear equations whose coefficient matrix is a square matrix to have a single solution is that the coefficient matrix is a non-singular matrix (a matrix having and inverse matrix). Therefore, if it is shown that the coefficient matrix of $(3)$ is a non-singular matrix, it means that the solution of $(3)$ is unique.
  Let us focus on the coefficient matrix $X$. The transpose matrix of $X$ is a Vandermonde matrix It is known that the determinant of the Vandermonde matrix is given as , where $\prod_{1 \leq i < j \leq n}$ means that all $( x_{j}- x_{i} )$ are multiplied if $1 \leq i < j \leq n$. Specifically,
  In our discussion, every $x_{i}$ is different. If $i \neq j$, $x_{i} \neq x_{j}$. We have Generally, the determinant of the transposed matrix is equal to the determinant of the original matrix. We have Since a matrix whose determinant is not $0$ is a non-singular matrix, $X$ is shown to be a non-singular matrix.   As described above, the coefficient matrix $ X $ of the system of linear equations $(3)$ is a non-singular matrix and therefore has the unique solution. Solving $(2)$ gives the unique coeffient $a_{0}, a_{1} \cdots, a_{n}$. The function of $f(x)$ is uniquely determined. Therefore, a function that passes through different $n+1$ points is unique.