Spline interpolation

For a broader coverage related to this topic, see Spline (mathematics).

In the mathematical field of numerical analysis, spline interpolation is a form of interpolation where the interpolant is a special type of piecewise polynomial called a spline. Spline interpolation is often preferred over polynomial interpolation because the interpolation error can be made small even when using low degree polynomials for the spline. Spline interpolation avoids the problem of Runge's phenomenon, in which oscillation can occur between points when interpolating using high degree polynomials.

Introduction

Originally, spline was a term for elastic rulers that were bent to pass through a number of predefined points ("knots"). These were used to make technical drawings for shipbuilding and construction by hand, as illustrated by Figure 1.

Figure 1: Interpolation with cubic splines between eight points. Hand-drawn technical drawings were made for shipbuilding etc. using flexible rulers that were bent to follow pre-defined points

The approach to mathematically model the shape of such elastic rulers fixed by $n + 1$ knots $\left\{(x_{i},y_{i}):i=0,1,\cdots ,n\right\}$ is to interpolate between all the pairs of knots $(x_{{i-1}},y_{{i-1}})$ and $(x_{i},y_{i})$ with polynomials $y=q_{i}(x),i=1,2,\cdots ,n$ .

The curvature of a curve $y=f(x)$ is given by:

\kappa= \frac{y''}{(1+y'^2)^{3/2}}

As the spline will take a shape that minimizes the bending (under the constraint of passing through all knots) both $y'$ and $y''$ will be continuous everywhere and at the knots. To achieve this one must have that

{\begin{cases}q'_{i}(x_{i})=q'_{{i+1}}(x_{i})\\q''_{i}(x_{i})=q''_{{i+1}}(x_{i})\end{cases}}\qquad 1\leq i\leq n-1

This can only be achieved if polynomials of degree 3 or higher are used. The classical approach is to use polynomials of degree 3 — the case of cubic splines.

Algorithm to find the interpolating cubic spline

A third order polynomial $q(x)$ for which

q(x_1)=y_1

q(x_2)=y_2

q'(x_1)=k_1

q'(x_2)=k_2

can be written in the symmetrical form

$q=(1-t)y_{1}+ty_{2}+t(1-t)(a(1-t)+bt)$

(1)

where

$t={\frac {x-x_{1}}{x_{2}-x_{1}}},$

(2)

$a=k_{1}(x_{2}-x_{1})-(y_{2}-y_{1}),$

(3)

$b=-k_2 (x_2 - x_1)+(y_2 - y_1).$

(4)

q'={\frac {dq}{dx}}={\frac {dq}{dt}}{\frac {dt}{dx}}={\frac {dq}{dt}}{\frac {1}{x_{2}-x_{1}}}

one gets that:

$q'={\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}+(1-2t){\frac {a(1-t)+bt}{x_{2}-x_{1}}}+t(1-t){\frac {b-a}{x_{2}-x_{1}}},$

(5)

$q''=2\frac {b-2a+(a-b)3t}{{(x_2-x_1)}^2}.$

(6)

Setting $x = x 1$ and $x = x 2$ respectively in equations (5) and (6) one gets from (2) that indeed first derivatives $q' (x 1) = k 1$ and $q' (x 2) = k 2$ and also second derivatives

$q''(x_1)=2\frac {b-2a}{{(x_2-x_1)}^2}$

(7)

$q''(x_2)=2\frac {a-2b}{{(x_2-x_1)}^2}$

(8)

If now $(x i, y i), i = 0, 1, ..., n$ are $n + 1$ points and

$q_{i}=(1-t)y_{{i-1}}+ty_{i}+t(1-t)(a_{i}(1-t)+b_{i}t)$

(9)

where i = 1, 2, ..., n and $t={\tfrac {x-x_{{i-1}}}{x_{{i}}-x_{{i-1}}}}$ are n third degree polynomials interpolating $y$ in the interval $x i -1 \leq x \leq x i$ for i = 1, ..., n such that $q' i (x i) = q' i +1 (x i)$ for i = 1, ..., n−1 then the n polynomials together define a differentiable function in the interval $x 0 \leq x \leq x n$ and

$a_i=k_{i-1}(x_i-x_{i-1})-(y_i - y_{i-1})$

(10)

$b_i=-k_i(x_i-x_{i-1})+(y_i - y_{i-1})$

(11)

for i = 1, ..., n where

$k_0=q_1'(x_0)$

(12)

$k_{i}=q_{i}'(x_{i})=q_{{i+1}}'(x_{i})\qquad i=1,\dotsc ,n-1$

(13)

$k_n=q_n'(x_n)$

(14)

If the sequence $k 0, k 1, ..., k n$ is such that, in addition, $q'' i (x i) = q'' i +1 (x i)$ holds for i = 1, ..., n-1, then the resulting function will even have a continuous second derivative.

From (7), (8), (10) and (11) follows that this is the case if and only if

${\frac {k_{{i-1}}}{x_{i}-x_{{i-1}}}}+\left({\frac {1}{x_{i}-x_{{i-1}}}}+{\frac {1}{{x_{{i+1}}-x_{i}}}}\right)2k_{i}+{\frac {k_{{i+1}}}{{x_{{i+1}}-x_{i}}}}=3\left({\frac {y_{i}-y_{{i-1}}}{{(x_{i}-x_{{i-1}})}^{2}}}+{\frac {y_{{i+1}}-y_{i}}{{(x_{{i+1}}-x_{i})}^{2}}}\right)$

(15)

for i = 1, ..., n-1. The relations (15) are $n - 1$ linear equations for the $n + 1$ values $k 0, k 1, ..., k n$ .

For the elastic rulers being the model for the spline interpolation one has that to the left of the left-most "knot" and to the right of the right-most "knot" the ruler can move freely and will therefore take the form of a straight line with $q'' = 0$ . As $q''$ should be a continuous function of $x$ one gets that for "Natural Splines" one in addition to the $n - 1$ linear equations (15) should have that

q''_{1}(x_{0})=2{\frac {3(y_{1}-y_{0})-(k_{1}+2k_{0})(x_{1}-x_{0})}{{(x_{1}-x_{0})}^{2}}}=0,

q''_{n}(x_{n})=-2{\frac {3(y_{n}-y_{{n-1}})-(2k_{n}+k_{{n-1}})(x_{n}-x_{{n-1}})}{{(x_{n}-x_{{n-1}})}^{2}}}=0,

i.e. that

${\frac {2}{x_{1}-x_{0}}}k_{0}+{\frac {1}{x_{1}-x_{0}}}k_{1}=3{\frac {y_{1}-y_{0}}{(x_{1}-x_{0})^{2}}},$

(16)

${\frac {1}{x_{n}-x_{{n-1}}}}k_{{n-1}}+{\frac {2}{x_{n}-x_{{n-1}}}}k_{n}=3{\frac {y_{n}-y_{{n-1}}}{(x_{n}-x_{{n-1}})^{2}}}.$

(17)

Eventually, (15) together with (16) and (17) constitute $n + 1$ linear equations that uniquely define the $n + 1$ parameters $k 0, k 1, ..., k n$ .

There exist other end conditions: "Clamped spline", that specifies the slope at the ends of the spline, and the popular "not-a-knot spline", that requires that the third derivative is also continuous at the $x 1$ and $x N -1$ points. For the "not-a-knot" spline, the additional equations will read:

q'''_1(x_1) =q'''_2(x_1) \Rightarrow \frac{1}{\Delta x_1^2} k_0 + \left( \frac{1}{\Delta x_1^2} - \frac{1}{\Delta x_2^2} \right) k_1 - \frac{1}{\Delta x_2^2} k_2 = 2 \left( \frac{\Delta y_1}{\Delta x_1^3} - \frac{\Delta y_2}{\Delta x_2^3} \right)

q'''_{n-1}(x_{n-1}) =q'''_n(x_{n-1}) \Rightarrow \frac{1}{\Delta x_{n-1}^2} k_{n-2} + \left( \frac{1}{\Delta x_{n-1}^2} - \frac{1}{\Delta x_n^2} \right) k_{n-1} - \frac{1}{\Delta x_n^2} k_n = 2\left( \frac{\Delta y_{n-1} }{\Delta x_{n-1}^3 }- \frac{ \Delta y_n}{ \Delta x_n^3 } \right)

where $\Delta x_{i}=x_{i}-x_{{i-1}},\Delta y_{i}=y_{i}-y_{{i-1}}$ .

Example

In case of three points the values for $k_0,k_1,k_2$ are found by solving the tridiagonal linear equation system

\begin{bmatrix} a_{11} & a_{12} & 0 \\ a_{21} & a_{22} & a_{23} \\ 0 & a_{32} & a_{33} \\ \end{bmatrix} \begin{bmatrix} k_0 \\ k_1 \\ k_2 \\ \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \\ \end{bmatrix}

with

a_{11}=\frac{2}{x_1-x_0}

a_{12}=\frac{1}{x_1-x_0}

a_{21}=\frac{1}{x_1-x_0}

a_{22}=2\ \left(\frac {1}{x_1-x_0}+ \frac {1}{{x_2-x_1}}\right)

a_{23}=\frac {1}{{x_2-x_1}}

a_{32}=\frac{1}{x_2-x_1}

a_{33}=\frac{2}{x_2-x_1}

b_1=3\ \frac{y_1-y_0}{(x_1-x_0)^2}

b_2=3\ \left(\frac {y_1 - y_0}{{(x_1-x_0)}^2}+\frac {y_2 - y_1}{{(x_2-x_1)}^2}\right)

b_3=3\ \frac{y_2-y_1}{(x_2-x_1)^2}

For the three points

(-1,0.5)\ ,\ (0,0)\ ,\ (3,3)

one gets that

k_0=-0.6875\ ,\ k_1=-0.1250\ ,\ k_2=1.5625

and from (10) and (11) that

a_1= k_0(x_1-x_0)-(y_1 - y_0)=-0.1875

b_1=-k_1(x_1-x_0)+(y_1 - y_0)=-0.3750

a_2= k_1(x_2-x_1)-(y_2 - y_1)=-3.3750

b_2=-k_2(x_2-x_1)+(y_2 - y_1)=-1.6875

In Figure 2 the spline function consisting of the two cubic polynomials $q_1(x)$ and $q_2(x)$ given by (9) is displayed

Computer code

TinySpline: Open source C-library for splines which implements cubic spline interpolation

External links

Cubic Spline Interpolation Online Calculation and Visualization Tool (with JavaScript source code)
Hazewinkel, Michiel, ed. (2001), "Spline interpolation", Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4
Dynamic cubic splines with JSXGraph
Lectures on the theory and practice of spline interpolation
Paper which explains step by step how cubic spline interpolation is done, but only for equidistant knots.
Numerical Recipes in C, Go to Chapter 3 Section 3-3.
A note on cubic splines

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