Suppose that you are given a data set D= x 1 y 1 … x m y m D x 1 y 1 … x m y m in R2 2 and that you must find a smooth curve to these points. There are many solutions to such a problem. Most simply, you could "eyeball it" and draw in a curve that looks good; however, this solution is too subjective, not very precise. You could just connect the dots, but this will probably not give a smooth line. You could fit a polynomial, but this fitted polynomial may wiggle too much. Or you could use splines.

Splines are piecewise polynomials with pieces smoothly connected together. The joining points of the polynomial pieces are called knots. Knots do not have to be evenly spaced. When each segment of a spline is a polynomial of degree nn, we say that the spline is a spline of degree n.

We need to add some constraints to ensure smoothness. For a spline of degree nn, we require that the spline has continuous derivatives up to order n−1 n 1 at each of the knots, i.e. a spline of degree nn is in Cn−1 C n 1 .

Remember our problem of fitting a curve to the data? There are generally two ways of approaching this problem: interpolation and smoothing.

We can fit a spline to interpolate the data; i.e. ∀i,i∈1…m:f⁢ x i = y i i i 1 … m f x i y i

We can fit a smoothing spline: i.e. find ff to minimize 1n⁢∑i=1n y i −f⁢ x i 2+λ⁢∫ x 1 x m fmx⁢u2du 1 n i 1 n y i f x i 2 λ u x 1 x m f x m u 2 The first term measures the closeness of the fitted function to the data, while the second penalizes the curvature in the function. λλ established the trade off between the two. For 0<λ<∞ 0 λ , this constraint is minimized by a natural spline of degree m+1 m 1 ( Schoenberg). If λ=0 λ 0 , then ff can be any function which interpolates the data. If λ=∞ λ and m+2 m 2 , then this is the simple least squares line fit, since no second derivative can be tolerated.

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This work is licensed by Alena Scott under a Creative Commons Attribution License (CC-BY 1.0), and is an Open Educational Resource.

Last edited by Elizabeth Gregory on Jun 7, 2005 3:28 pm -0500.