Last modified: May 19, 2025

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Interpolation

Interpolation is the problem of reconstructing an unknown function from a finite set of exact data pairs

$$ {(x_i,y_i)}_{i=0}^{n},\qquad x_0< x_1<\dots <x_n,\qquad y_i=f(x_i). $$

Given a query point $x$ in the closed interval $[x_0,x_n]$, the task is to compute an interpolant $\hat f(x)$ such that

$$ \hat f(x_i)=y_i \text{for every }i=0,\dots ,n, \qquad\text{and}\qquad \hat f(x)\approx f(x) \text{for }x\in[x_0,x_n]. $$

Because the nodes $x_i$ are distinct, the classical existence-and-uniqueness theorem guarantees that there is a unique algebraic polynomial $P_n$ of degree at most $n$ satisfying the interpolation conditions.

Extrapolation is the computation of $\hat f(x)$ for $x\notin[x_0,x_n]$ and is not covered by the uniqueness theorem; error growth can be dramatic.

Assumptions

Concepts

Method Essential idea Interpolant form Error behavior* Typical use-case
Linear Join successive points by straight segments Piecewise degree-1 $O(h^2)$ Fast previews, computer graphics
Polynomial Single poly of degree $n$ through all nodes Lagrange, Newton, barycentric, etc. $O(h^{n+1})$ but risk of Runge oscillations when $n$ is large Small $n$ on modest intervals
Spline Glue low-degree polynomials with continuity constraints Piecewise degree-$k$ (usually 3) $O(h^{k+1})$ plus good shape control Smooth curves, CAD/CAM, statistics
Radial Basis Function (RBF) Weighted radial kernels centered at nodes $\displaystyle \hat f(x)=\sum_{i=0}^{n} \lambda_i\,\phi\bigl(\lVert x - x_i\rVert\bigr)$ Spectral for smooth $\phi$ Scattered data, high-dimensional grids

Assuming evenly spaced nodes with spacing $h=\max_i (x_{i+1}-x_i)$.

Mathematical Formulation

I. Linear interpolation (piecewise)

For $x\in[x_i,x_{i+1}]$

$$ \hat f(x)=y_i+\frac{y_{i+1}-y_i}{x_{i+1}-x_i}\,(x-x_i). $$

II. Polynomial interpolation (global)

Lagrange form

$$ P_n(x)=\sum_{i=0}^{n} y_i\,\ell_i(x), \qquad \ell_i(x)=\prod_{\substack{j=0 \ j\neq i}}^{n} \frac{x-x_j}{x_i-x_j}. $$

Error bound (Peano form): if $f\in C^{n+1}$, then for some $\xi(x)\in[x_0,x_n]$

$$ f(x)-P_n(x)=\frac{f^{(n+1)}(\xi(x))}{(n+1)!}\prod_{i=0}^{n}(x-x_i). $$

III. Natural cubic spline

Solve the tridiagonal system

$$ h_{i-1}M_{i-1}+2(h_{i-1}+h_i)M_i+h_iM_{i+1}=6 \left(\frac{y_{i+1}-y_i}{h_i}-\frac{y_i-y_{i-1}}{h_{i-1}}\right), $$

with $M_0=M_n=0$, where $h_i=x_{i+1}-x_i$.

Each piece is then

$$ S_i(x)=\frac{M_{i+1}(x-x_i)^3+M_i(x_{i+1}-x)^3}{6h_i} +\frac{y_{i+1}}{h_i}(x-x_i)-\frac{y_i}{h_i}(x_{i+1}-x) -\frac{h_i}{6}\bigl(M_{i+1}-M_i\bigr)(x-x_i). $$

Example

Time (hh) Temp (°C)
9 20
10 22
11 26
12 28
13 30
14 31
15 31

Estimate $T(10.5)$.

I. Linear (segment 10 – 11)

$$ T(10.5)=22+\frac{26-22}{11-10}(10.5-10)=24\text{ °C}. $$

II. Natural cubic spline (all nodes)

Solving the spline system yields second derivatives

$$ (M_0,\dots,M_6)\approx(0,\;2.4,\;3.6,\;0.6,\;-1.2,\;-0.6,\;0). $$

From the $i=1$ piece (10–11 h) one obtains

$$ T(10.5)\approx 24.77\text{ °C}. $$

Because the temperature rise slows near noon, the spline—by “seeing” the curvature—predicts a slightly higher value than the straight-line model.

Advantages

Limitations

Table of Contents

    Interpolation
    1. Assumptions
    2. Concepts
    3. Mathematical Formulation
    4. Example
    5. Advantages
    6. Limitations