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Piecewise Interpolation: Piecewise Linear Spline Interpolation

Piecewise Linear (Piecewise Affine) Spline Interpolation

Given a set of $k$ data points $(x_1,y_1),(x_2,y_2),\cdots, (x_k,y_k)$ , a piecewise linear (piecewise affine) spline can be defined as:

$y=\begin{cases} a_1x+b_1,&x\in [x_1, x_2) \\ a_2x+b_2,&x\in [x_2, x_3) \\ \vdots & \vdots \\ a_{k-1}x+b_{k-1},&x\in [x_{k-1}, x_k] \end{cases}$

The $k$ data points have $k-1$ intervals. The linear function for each interval is defined using two coefficients, and therefore, we need to find $2(k-1)$ coefficients $a_1,b_1,a_2,b_2,\cdots,a_{k-1},b_{k-1}$ . The coefficients $a_i$ and $b_i$ can be found by solving the two equations:

$\begin{split} y_i&=a_ix_i + b_i\\ y_{i+1}&=a_ix_{i+1}+b_i \end{split}$

Therefore:

$\begin{split} a_i&=\frac{y_i-y_{i+1}}{x_i-x_{i+1}}\\ b_i&=\frac{y_{i+1}x_i-y_ix_{i+1}}{x_i-x_{i+1}} \end{split}$

Example

Consider the following 11 data points: (-1.00,0.038),(-0.80,0.058),(-0.60,0.100),(-0.4,0.200),(-0.20,0.500),(0.00,1.000),(0.20,0.500),(0.40,0.200),(0.60,0.100),(0.80,0.0580),(1.00,0.038) which were generated using the Runge function. Find the explicit representation of the linear spline interpolating function for these data points.

Solution

There are 11 data points surrounding 10 intervals. For each interval $i$ , a set of coefficients $a_i$ and $b_i$ are required for the linear interpolation. For the first interval, the coefficients are:

$\begin{split} a_1&=\frac{y_1-y_2}{x_1-x_2}=\frac{0.038-0.058}{-1-(-0.8)}=0.1\\ b_1&=\frac{y_2x_1-y_1x_2}{x_1-x_2}=\frac{0.058(-1)-0.038(-0.8)}{-1-(-0.8)}=0.138 \end{split}$

Similarly:

$\begin{split} a_2&=\frac{y_2-y_3}{x_2-x_3}=\frac{0.058-0.1}{-0.8-(-0.6)}=0.21\\ b_2&=\frac{y_3x_2-y_2x_3}{x_2-x_3}=\frac{0.1(-0.8)-0.058(-0.6)}{-0.8-(-0.6)}=0.226 \end{split}$

Repeating for the other coefficients, the required explicit equation has the form:

$y=\begin{cases} 0.1x+0.138,&x\in [-1, -0.8) \\ 0.21x+0.226,&x\in [-0.8, -0.6) \\ 0.5x+0.4,&x\in [-0.6, -0.4)\\ 1.5x+0.8,&x\in [-0.4, -0.2)\\2.5x+1,&x\in [-0.2, 0.0)\\-2.5x+1,&x\in [0.0, 0.2)\\-1.5x+0.8,&x\in [0.2, 0.4)\\-0.5x+0.4,&x\in [0.4, 0.6)\\-0.21x+0.226,&x\in [0.6, 0.8)\\-0.1x+0.138,&x\in [0.8, 1.0] \end{cases}$

Notice that for the procedure to work, the $x$ components of the data points have to satisfy: $x_1<x_2<\cdots<x_{k}$ . The following Mathematica code utilizes a user defined procedure “Spline1” that creates the required piecewise linear function.

View Mathematica Code

Data = {{-1, 0.038}, {-0.8, 0.058}, {-0.60, 0.10}, {-0.4,0.20}, {-0.2, 0.50}, {0, 1}, {0.2, 0.5}, {0.4, 0.2}, {0.6,0.1}, {0.8, 0.058}, {1, 0.038}};
Spline1[Data_] := (
k1 = Length[Data] - 1;  
atable = Table[(Data[[i, 2]] - Data[[i + 1, 2]])/(Data[[i, 1]] -Data[[i + 1, 1]]), {i, 1, k1}]; 
btable = Table[(Data[[i + 1, 2]]*Data[[i, 1]] -Data[[i, 2]]*Data[[i + 1, 1]])/(Data[[i, 1]] - Data[[i + 1, 1]]), {i, 1, k1}]; 
pf = Table[{atable[[i]] x + btable[[i]],Data[[i, 1]] <= x <= Data[[i + 1, 1]]}, {i, 1, k1}]; 
Piecewise[pf]
)
y = Spline1[Data]
a = Plot[{y, yactual}, {x, -1, 1}, Epilog -> {PointSize[Large], Point[Data]}, AxesLabel -> {"x", "y1"}, ImageSize -> Medium, PlotRange -> All,  PlotLegends -> {"y1", "yactual"}, PlotLabel -> "Linear splines"]

View Python Code

import numpy as np
import matplotlib.pyplot as plt
Data = [[-1, 0.038], [-0.8, 0.058], [-0.60, 0.10], [-0.4,0.20], [-0.2, 0.50], [0, 1], [0.2, 0.5], [0.4, 0.2], [0.6, 0.1], [0.8, 0.058], [1, 0.038]]
def Spline1(x, Data):
  k1 = len(Data) - 1
  atable = [(Data[i][1] - Data[i + 1][1])/(Data[i][0] - Data[i + 1][0]) for i in range(k1)] 
  btable = [(Data[i + 1][1]*Data[i][0] - Data[i][1]*Data[i + 1][0])/(Data[i][0] - Data[i + 1][0]) for i in range(k1)] 
  display([['{:15}'.format(str(round(atable[i],3))+'x + '+str(round(btable[i],3))),
            '{:15}'.format(str(round(Data[i][0],3))+' <=x<= '+str(round(Data[i + 1][0],3)))] for i in range(k1)])
  return np.piecewise(x, [(x >= Data[i][0])&(x <= Data[i + 1][0]) for i in range(k1)],
                         [lambda x, j=i: atable[j]*x + btable[j] for i in range(k1)])
x = np.arange(-1,1,0.01)
y = Spline1(x, Data)
yactual = 1/(1 + 25*x**2)
plt.plot(x,yactual, label="yactual")
plt.plot(x, y, label="y1")
plt.scatter([point[0] for point in Data],[point[1] for point in Data], c='k')
plt.title("Linear splines")
plt.xlabel("x"); plt.ylabel("y1")
plt.legend(); plt.grid(); plt.show()

MATLAB files: File 1 (ex8_1.m) File 2 (spline1.m)

Open Educational Resources

Piecewise Interpolation: Piecewise Linear Spline Interpolation

Piecewise Linear (Piecewise Affine) Spline Interpolation

Example

Solution

Lecture Video

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