ConceptComplete

Polynomial Interpolation

Polynomial interpolation constructs a polynomial passing through given data points, providing a fundamental tool for approximating functions and numerical differentiation and integration.

DefinitionPolynomial Interpolation Problem

Given n+1n+1 distinct points (x0,y0),(x1,y1),…,(xn,yn)(x_0, y_0), (x_1, y_1), \ldots, (x_n, y_n), find a polynomial pn(x)p_n(x) of degree at most nn such that: pn(xi)=yi,i=0,1,…,np_n(x_i) = y_i, \quad i = 0, 1, \ldots, n

Such a polynomial exists and is unique. It is called the interpolating polynomial for the given data.

The existence and uniqueness follow from linear algebra: the system of n+1n+1 equations in n+1n+1 unknowns (polynomial coefficients) has a unique solution when the xix_i are distinct. The Vandermonde matrix has non-zero determinant ∏i<j(xjβˆ’xi)β‰ 0\prod_{i<j}(x_j - x_i) \neq 0.

DefinitionLagrange Form

The Lagrange form of the interpolating polynomial is: pn(x)=βˆ‘i=0nyiLi(x)p_n(x) = \sum_{i=0}^n y_i L_i(x) where the Lagrange basis polynomials are: Li(x)=∏j=0,jβ‰ inxβˆ’xjxiβˆ’xjL_i(x) = \prod_{j=0, j \neq i}^n \frac{x - x_j}{x_i - x_j}

Each Li(x)L_i(x) satisfies Li(xj)=Ξ΄ijL_i(x_j) = \delta_{ij} (Kronecker delta), so pn(xi)=yip_n(x_i) = y_i automatically.

ExampleQuadratic Interpolation

Interpolate (0,1)(0,1), (1,0)(1,0), (2,3)(2,3). The Lagrange basis polynomials are: L0(x)=(xβˆ’1)(xβˆ’2)(0βˆ’1)(0βˆ’2)=12(xβˆ’1)(xβˆ’2)L_0(x) = \frac{(x-1)(x-2)}{(0-1)(0-2)} = \frac{1}{2}(x-1)(x-2) L1(x)=(xβˆ’0)(xβˆ’2)(1βˆ’0)(1βˆ’2)=βˆ’(x)(xβˆ’2)=βˆ’x2+2xL_1(x) = \frac{(x-0)(x-2)}{(1-0)(1-2)} = -(x)(x-2) = -x^2 + 2x L2(x)=(xβˆ’0)(xβˆ’1)(2βˆ’0)(2βˆ’1)=12x(xβˆ’1)L_2(x) = \frac{(x-0)(x-1)}{(2-0)(2-1)} = \frac{1}{2}x(x-1)

The interpolating polynomial is: p2(x)=1β‹…L0(x)+0β‹…L1(x)+3β‹…L2(x)=12(xβˆ’1)(xβˆ’2)+32x(xβˆ’1)p_2(x) = 1 \cdot L_0(x) + 0 \cdot L_1(x) + 3 \cdot L_2(x) = \frac{1}{2}(x-1)(x-2) + \frac{3}{2}x(x-1) =12[x2βˆ’3x+2+3x2βˆ’3x]=2x2βˆ’3x+1= \frac{1}{2}[x^2 - 3x + 2 + 3x^2 - 3x] = 2x^2 - 3x + 1

Verification: p2(0)=1p_2(0) = 1, p2(1)=0p_2(1) = 0, p2(2)=8βˆ’6+1=3p_2(2) = 8 - 6 + 1 = 3. βœ“

Remark

Computational Considerations: The Lagrange form is conceptually simple but computationally expensive (O(n2)O(n^2) operations per evaluation). The Newton divided difference form is more efficient for sequential data addition and uses O(n)O(n) operations per evaluation after O(n2)O(n^2) preprocessing.

Runge Phenomenon: High-degree polynomial interpolation with equally-spaced points can exhibit large oscillations near interval endpoints. Using Chebyshev nodes xi=cos⁑(2i+12(n+1)Ο€)x_i = \cos\left(\frac{2i+1}{2(n+1)}\pi\right) minimizes this effect.

Polynomial interpolation forms the basis for numerical quadrature (integration), differentiation formulas, and piecewise polynomial methods like splines. The interpolation error for smooth functions is ∣f(x)βˆ’pn(x)βˆ£β‰€Mn+1(n+1)!∏i=0n∣xβˆ’xi∣|f(x) - p_n(x)| \leq \frac{M_{n+1}}{(n+1)!} \prod_{i=0}^n |x - x_i| where Mn+1=max⁑∣f(n+1)(ΞΎ)∣M_{n+1} = \max |f^{(n+1)}(\xi)|, showing that error depends on data spacing and function smoothness.