Spline interpolation

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In the mathematical subfield 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 preferred over polynomial interpolation because the interpolation error can be made small even when using low degree polynomials for the spline. Thus, spline interpolation avoids the problem of Runge's phenomenon which occurs when using high degree polynomials.

1 Definition
2 Spline interpolant
3 Linear spline interpolation
4 Quadratic spline interpolation
5 Cubic spline interpolation
- 5.1 Minimality of the cubic splines
- 5.2 Interpolation using natural cubic spline
6 Example
- 6.1 Linear spline interpolation
- 6.2 Quadratic spline interpolation
7 See also

[edit] Definition

Given n+1 distinct knots x_i such that

$x_0 < x_1 < \cdots < x_{n-1} < x_n, \,\!$

with n+1 knot values y_i we are trying to find a spline function of degree n

$S(x) := \left\{\begin{matrix} S_0(x) & x \in [x_0, x_1] \\ S_1(x) & x \in [x_1, x_2] \\ \vdots & \vdots \\ S_{n-1}(x) & x \in [x_{n-1}, x_n] \end{matrix}\right.$

where each S_i(x) is a polynomial of degree k.

[edit] Spline interpolant

Using polynomial interpolation, the polynomial of degree n which interpolates the data set is uniquely defined by the data points. The spline of degree n which interpolates the same data set is not uniquely defined, and we have to fill in n-1 additional degrees of freedom to construct a unique spline interpolant.

[edit] Linear spline interpolation

Linear spline interpolation is the simplest form of spline interpolation and is equivalent to linear interpolation. The data points are graphically connected by straight lines. The resultant spline is just a polygon.

Algebraically, each S_i is a linear function constructed as

$S_i(x) = y_i + \frac{y_{i+1}-y_i}{x_{i+1}-x_i}(x-x_i)$

The spline must be continuous at each data point, that is

$S_i(x_i) = S_{i+1}(x_i) \qquad \mbox{ , } i=1,\ldots n -1$

This is the case as we can easily see

$S_{i-1}(x_i) = y_{i-1} + \frac{y_{i}-y_{i-1}}{x_{i}-x_{i-1}}(x_i-x_{i-1}) = y_i$

$S_{i}(x_i) = y_i + \frac{y_{i+1}-y_i}{x_{i+1}-x_i}(x_i-x_i) = y_i$

[edit] Quadratic spline interpolation

The quadratic spline can be constructed as

$S_i(x) = y_i + z_i(x-x_i) + \frac{z_{i+1}-z_i}{2(x_{i+1}-x_i)}(x-x_i)^2$

The coefficients can be found by choosing a $z 0$ and then using the recurrence relation:

$z_{i+1} = -z_i + 2 \frac{y_{i+1}-y_i}{x_{i+1}-x_i}$

[edit] Cubic spline interpolation

For a data set {x_i} of n+1 points, we can construct a cubic spline with n piecewise cubic polynomials between the data points. If

$S(x)=\left\{\begin{matrix} S_0(x),\ x\in[x_0,x_1] \\ S_1(x),\ x\in[x_1,x_2]\\ \cdots \\ S_{n-1}(x),\ x\in[x_{n-1},x_n]\end{matrix}\right.$

represents the spline function interpolating the function f, we require:

the interpolating property, S(x_i)=f(x_i)
the splines to join up, S_i-1(x_i) = S_i(x_i), i=1,...,n-1
twice continuous differentiable, S'_i-1(x_i) = S'_i(x_i) and S''_i-1(x_i) = S''_i(x_i), i=1,...,n-1.

For the n cubic polynomials comprising S, this means to determine these polynomials, we need to determine 4n conditions (since for one polynomial of degree three, there are four conditions on choosing the curve). However, the interpolating property gives us n + 1 conditions, and the conditions on the interior data points give us n + 1 − 2 = n − 1 data points each, summing to 4n − 2 conditions. We require two other conditions, and these can be imposed upon the problem for different reasons.

One such choice results in the so-called clamped cubic spline, with

$S'(x_0) = u \,\!$

$S'(x_k) = v \,\!$

for given values u and v.

Alternately, we can set

$S''(x_0) = S''(x_n) = 0 \,\!$ .

resulting in the natural cubic spline. The natural cubic spline is approximately the same curve as created by the spline device.

Amongst all twice continuously differentiable functions, clamped and natural cubic splines yield the least oscillation about the function f which is interpolated.

Another choice gives the periodic cubic spline if

$S(x_0) = S(x_n) \,\!$

$S'(x_0) = S'(x_n) \,\!$

$S''(x_0) = S''(x_n) \,\!$

Another choice gives the complete cubic spline if

$S(x_0) = S(x_n) \,\!$

$S'(x_0) = S'(x_n) \,\!$

$S''(x_0) = f'(x_0),\quad S''(x_n)=f'(x_n) \,\!$

[edit] Minimality of the cubic splines

The cubic spline has a very important variational interpretation, in fact it is the function that minimizes the functional

$J(f)=\int_a^b |f''(x)|^2 dx$ ,

over the function in the Sobolev space $H 2 ([a; b])$ .

The functional $J$ contains an approximation of the total curvature $\left|\frac{f''(x)}{(1+f'(x)^2)^{\frac{3}{2}}}\right|$ of the graph of $f (x)$ and then the spline is the approximation of $f (x)$ with minimal curvature, and then is the more well looking psychologically.

Since the total energy of an elastic strip is proportional to the curvature, the spline is the configuration of minimal energy of an elastic strip constrained to $n$ points. A spline is also an instrument to design based on an elastic strip.

[edit] Interpolation using natural cubic spline

It can be defined as

$S_i(x) = \frac{z_{i+1} (x-x_i)^3 + z_i (x_{i+1}-x)^3}{6h_i} + \left(\frac{y_{i+1}}{h_i} - \frac{h_i}{6} z_{i+1}\right)(x-x_i) + \left(\frac{y_{i}}{h_i} - \frac{h_i}{6} z_i\right) (x_{i+1}-x)$

and

$h_i = x_{i+1} - x_i \,\!$ .

The coefficients can be found by solving this system of equations:

$\begin{align} z_0 &= 0 \\ h_{i-1} z_{i-1} + 2(h_{i-1} + h_i) z_i + h_i z_{i+1} &= 6 \left( \frac{y_{i+1}-y_i}{h_i} - \frac{y_i-y_{i-1}}{h_{i-1}} \right) \\ z_n &= 0 \end{align}$

[edit] Example

[edit] Linear spline interpolation

Consider the problem of finding a linear spline for

$f(x) = e^{-x^2}$

with the following knots

$(x_0,f(x_0)) = (x_0,y_0) = \left(-1,\ e^{-1}\right) \,\!$

$(x_1,f(x_1)) = (x_1,y_1) = \left(-\frac{1}{2},\ e^{-\frac{1}{4}}\right) \,\!$

$(x_2,f(x_2)) = (x_2,y_2) = \left(0,\ 1\right) \,\!$

$(x_3,f(x_3)) = (x_3,y_3) = \left(\frac{1}{2},\ e^{-\frac{1}{4}}\right) \,\!$

$(x_4,f(x_4)) = (x_4,y_4) = \left(1,\ e^{-1}\right) \,\!$

After directly applying the spline formula, we get the following spline:

$S(x) = \left\{\begin{matrix} e^{-1} + 2(e^{-\frac{1}{4}} - e^{-1})(x+1) & x \in [-1, -\frac{1}{2}] \\ e^{-\frac{1}{4}} + 2(1-e^{-\frac{1}{4}})(x+\frac{1}{2}) & x \in [-\frac{1}{2},0] \\ 1 + 2(e^{-\frac{1}{4}}-1)x & x \in [0,\frac{1}{2}] \\ e^{-\frac{1}{4}} + 2(e^{-1} - e^{-\frac{1}{4}})(x-\frac{1}{2}) & x \in [\frac{1}{2},1] \\ \end{matrix}\right.$