Inverse distance weighting

Builds on Multivariate interpolation Metric space Euclidean distance

Definition

Let $(X, \delta)$ be a metric space, usually $\R^d$ with the Euclidean distance. Let $x_1, \dots, x_N \in X$ be pairwise distinct, let $u_1, \dots, u_N \in \R$ , and let $p > 0$ . For $x \in X$ and $i = 1, \dots, N$ with $x \ne x_i$ , put $w_i(x) = \delta(x, x_i)^{-p}$ . The inverse distance weighting interpolant of the data is the function $u \colon X \to \R$ given as follows. If $x$ is not one of the $x_i$ , then $u(x) = \frac{\sum_{i=1}^{N} w_i(x) \, u_i}{\sum_{i=1}^{N} w_i(x)}.$ At the data points, $u(x_i) = u_i$ for $i = 1, \dots, N$ .

Remarks

The interpolant reproduces the data by definition, and it is continuous at each $x_i$ , because $w_i(x) \to \infty$ as $x \to x_i$ while the other weights stay bounded. Each value $u(x)$ is a convex combination of the $u_i$ , so $\min_i u_i \le u(x) \le \max_i u_i$ . This scheme is also called Shepard's method. It is the case of constant fits with weight $r^{-p}$ in moving least squares.

Example: on $\R$ with $p = 2$ , data $u_1 = 10$ at $x_1 = 0$ and $u_2 = 40$ at $x_2 = 3$ , at $x = 1$ the weights are $1$ and $1/4$ . So $u(1) = \frac{10 + 40/4}{1 + 1/4} = 16$ , which lies between $10$ and $40$ .

Non-example: the plain average $\sum_i u_i / N$ with equal weights does not depend on the distances and does not reproduce the data; for the same data it gives $25 \ne 10$ at $x_1$ .

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Tag	`55V8WAJXXJ`
URI	`m2://55V8WAJXXJ`
Kind	definition
Writing	W0 stub
MSC	65D05, 41A05, 41A63
Defines	Inverse distance weighting
Uses	Multivariate interpolation, Metric space, Euclidean distance

Inverse distance weighting

Definition

Remarks

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