Lagrange interpolation (Examples and Proof) (2024)

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 thatif $L_{i}(x)$ 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.

Answer
Lagrange's interpolation formula that gives the linear function passing through two points

is

In this example,

We obtain

Answer
Lagrange's interpolation formula that gives the quadratic function passing through three points

is

In this example,

We obtain

Example 3: Cubic interpolation

Let $f (x)$ be a function that passes through four points

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 $(2)$) isa system of $n+1$ linear equations with $n+1$ unknowns.

$X$ is non-singular

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 thatall $( x_{j}- x_{i} )$ 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.

Conclusion

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.