CHAPTER 3: NUMERICAL ALGORITHMS WITH MATLAB

Lecture 3.3: Piecewise polynomial interpolation

Piecewise interpolation:

Problem: Given a set of (n+1) data points:

(x1,y1); (x2,y2); …; (xn,yn); (xn+1,yn+1)

Find a smooth function y = S(x) that connects the given set of (n+1) data points. The function y = S(x) consists of low-order polynomials y = Sj(x) that connect two consecutive data points: (xj,yj) and (xj+1,yj+1).

Example: Given a set of seven data points for a voltage-current characteristic of a zener diode

 Voltage -1 0 1.27 2.55 3.82 4.92 5.02 Current -14.58 0 0 0 0 0.88 11.17

If the seven points are connected by a polynomial y = P6(x) of order 6, the continuous polynomial interpolation is not good because of large swings of the interpolating polynomial between the data points:

The polynomial y = P6(x) of order 6 may have five extremal points (maxima and minima). Indeed, all five extrema are seen on the picture above. This is called the polynomial wiggle. If n is large, interpolating polynomials show large errors and are rarely used for graphical visualization of curves. Instead, the piecewise interpolation made of polynomials of lower order is used to represent a sufficiently smooth curve y = S(x):

·         interp1: applies a piecewise interpolation between given set of data points

x = [ -1,-.866,-.5,0,0.5,0.866,1,1.0402,1.15,1.3,1.54,1.828,2.1736,2.5883,3.086];

y = [ 0,-0.25,-0.433,-0.5,-0.433,-0.25,0,0.15,0.2598,0.3,0.3,0.3,0.3,0.3,0.3];

xInt = -1 : 0.01 : 3; yInt = interp1(x,y,xInt); % default linear interpolation

plot(x,y,'b*',xInt,yInt,'g'); axis([-1,3,-1,1]);

yInt = interp1(x,y,xInt,'spline'); % piecewise cubic spline interpolation

plot(x,y,'b*',xInt,yInt,'g'); axis([-1,3,-1,1]);

yInt = interp1(x,y,xInt,'cubic'); % piecewise cubic Hermite interpolation

plot(x,y,'b*',xInt,yInt,'g'); axis([-1,3,-1,1]);

·         ppval: computes values of the cubic spline and Hermite interpolants in their pp-representations

y1 = ppval(spline(x,y),-0.25)

% computes the value of y at x = -0.25 of the cubic spline interpolant

y2 = ppval(pchip(x,y),-0.25)

% computes the value of y at x = -0.25 of the cubic Hermite interpolant

y1 =   -0.4721

y2 =   -0.4801

·         unmkpp: extracts breaks, coefficients, number of pieces, and order from the cubic spline and Hermite interpolants in their pp-representations

x = [0,1,2,3]; y = humps(x);

[P,R,n,m] = unmkpp(spline(x,y))

% P is the vector of x containing break points (length: n+1)

% R is the matrix of coefficients of cubic spline interpolants (size: n-by-m)

% n – number of pieces of the interpolants (length(x)-1)

% m – number of coefficients in each piece (order of the polynomial + 1)

% Example: n = 3 and m = 4

% Sj(x) = aj + bj*(x-xj) + cj*(x-xj)^2 + dj*(x-xj)^3 – piece between [xj,xj+1]

% aj = R(:,4); bj = R(:,3); cj = R(:,2); dj = R(:,1);

P =    0     1     2     3

R =       8.6251  -41.7147   43.9131    5.1765

8.6251  -15.8394  -13.6409   16.0000

8.6251   10.0360  -19.4443   -4.8552

n =     3

m =     4

x = [0,1,2,3]; y = humps(x);

[P,R,n,m] = unmkpp(pchip(x,y))

P =    0     1     2     3

R =       5.0158  -20.8552   26.6629    5.1765

40.2008  -61.0560         0   16.0000

0.0567    0.6698   -1.5096   -4.8552

n =     3

m =     4

·         mkpp: outputs a pp-representation of a piecewise polynomial from its breaks and coefficients.

x = [0,1,2,3]; y = humps(x)

[P,R,n,m] = unmkpp(spline(x,y));

SS = mkpp(P,R) % the same as SS = spline(x,y)

yy = ppval(SS,x) % evaluation of pp-representation of a polynomial

y =    5.1765   16.0000   -4.8552   -5.6383

SS =     form: 'pp'

breaks: [0 1 2 3]

coefs: [3x4 double]

pieces: 3

order: 4

dim: 1

yy =    5.1765   16.0000   -4.8552   -5.6383

Remark: piecewise representation of cubic spline interpolants is useful for semi-analytical computations of derivatives and integrals of the interpolating polynomials. The piecewise representations can be used for piecewise polynomial interpolations of any order.

Linear interpolation:

Two consecutive data points (xj,yj) and (xj+1,yj+1) can be connected by a straight line:

y = Sj(x) = yj + ( x - xj ) = aj + bj ( x - xj )

The linear interpolants y = Sj(x) for 1  j n connect all (n+1) data points but they do not have continuous first derivatives at the data points. The piecewise linear interpolation is used in plotting of functions y = f(x) from a given set of data points.

% computation of coefficients of linear interpolants from a given data set [x,y]

x = -1 : 0.25 : 3; y = humps(x);

n = length(y)-1; a = y(1:n); b = (y(2:n+1)-y(1:n))./(x(2:n+1)-x(1:n));

% evaluation of linear interpolants at data points xInt

xInt = -1 : 0.01 : 3;

for j = 1 : length(xInt)

if xInt(j) ~= x(n+1)

iInt(j) = sum(x <= xInt(j));

else

iInt(j) = n;

end

end

yInt = a(iInt) + b(iInt).*(xInt-x(iInt));

yEx = humps(xInt);

plot(x,y,'b*',xInt,yInt,'g',xInt,yEx,'r:'); axis([-1,3,-10,100]);

Cubic spline interpolation:

Two consecutive data points (xj,yj) and (xj+1,yj+1) can be connected by a cubic polynomial:

y = Sj(x) = aj + bj (x - xj ) +cj (x - xj)2 + dj (x - xj)3

The cubic spline interpolants y = Sj(x) for 1  j n connect all (n+1) data points and they have continuous first and second derivatives at the data points. Cubic spline interpolants provide the smoothest solution of the interpolation problem and are suitable for plotting of smooth functions y = f(x) from a given set of data points.

Problem: Find coefficients [aj,bj,cj,dj] of the cubic spline interpolants from the conditions:

continuity of functions:                     Sj(xj) = yj; Sj(xj+1) = yj+1

continuity of slopes:                         S'j(xj) = S'j-1(xj)

continuity of second derivatives:      S''j(xj) = S''j-1(xj)

There are (2n) conditions for continuity of functions, (n-1) conditions for continuity of slopes and (n-1) conditions for continuity of second derivatives, i.e. the total number of conditions is (4n – 2). However, there are (4n) unknown coefficients [aj,bj,cj,dj]. Therefore, two more conditions should be added to the problem. These two conditions are set at the end points as the natural spline conditions:

S''1(x1) = 0;    S''n(xn+1) = 0

Solution: Denote mj = S''j(xj) and hj = xj+1 - xj, then two coefficients are found from continuity of second derivatives:

cj =       and     dj =

Two coefficients are found from continuity of functions:

aj = yj      and     bj =  - hj

Thus, all coefficients are parameterized by values of mj. Those values are found from continuity of slopes at the (n-1) interior points. The latter conditions result in a system of (n-1) equations:

hj-1 mj-1 + 2 (hj-1 + hj) mj + hj mj+1 = 6  - ;            j = 2,…,n

The boundary values for natural splines: m1 = 0 and mn+1 = 0.

The boundary values for not-a-knot (cubic runout) splines:

m1 = (1 + )m2m3  and  mn+1 = (1 + )mnmn-1

% computation of coefficients of cubic spline interpolants from a set [x,y]

x = -1 : 0.25 : 3;      % data partition

y = humps(x);           % function values

n = length(x)-1;

h = x(2:n+1)-x(1:n); % vector of step sizes of the partition of x

A = 2*diag(h(1:n-1)) + 2*diag(h(2:n)) + diag(h(2:n-1),1) + diag(h(2:n-1),-1);

b = 6*((y(3:n+1)-y(2:n))./h(2:n)-(y(2:n)-y(1:n-1))./h(1:n-1));

m = A\b';

m = [0;m;0]'; % solving the linear system for natural splines

a = y(1:n);

b = (y(2:n+1)-y(1:n))./h(1:n)-h(1:n).*(m(2:n+1)+2*m(1:n))/6;

c = 0.5*m(1:n);

d = (m(2:n+1)-m(1:n))./(6*h(1:n));

% evaluation of cubic spline interpolants at data points xInt

xInt = -1 : 0.01 : 3;       % another partition with better resolution

for j = 1 : length(xInt)

if xInt(j) ~= x(n+1)

iInt(j) = sum(x <= xInt(j));

else

iInt(j) = n;

end

end

yInt = a(iInt) + b(iInt).*(xInt-x(iInt)) + c(iInt).*(xInt - x(iInt)).^2 + d(iInt).*(xInt - x(iInt)).^3;

yEx = humps(xInt);

plot(x,y,'b*',xInt,yInt,'g',xInt,yEx,'r:'); axis([-1,3,-10,100]);

Cubic Hermite interpolation:

Two consecutive data points (xj,yj) and (xj+1,yj+1) can be connected by a cubic Hermite polynomial:

y = Sj(x) = aj + bj (x - xj ) +cj (x - xj)2 + dj (x - xj)2 (x - xj+1)

The cubic Hermite interpolants y = Sj(x) for 1  j n connect the data points (xj,yj) with the prescribed slopes sj = y'j at each data point. Therefore, the cubic Hermite interpolants have continuous first derivatives at the data points. However, they do not have continuous second derivatives at the data points. The cubic Hermite interpolation may be too crude in graphical applications, because the human eye can detect discontinuities in the second derivative (curvature).

Problem: Find coefficients [aj,bj,cj,dj] of the cubic Hermite interpolants from the conditions:

continuity of functions:                     Sj(xj) = yj; Sj(xj+1) = yj+1

continuity of slopes:                         S'j(xj) = sj; S'j-1(xj) = sj+1

There are (2n) conditions for continuity of functions and (2n) conditions for continuity of slopes, i.e. the total number of conditions is (4n). There are (4n) unknown coefficients [aj,bj,cj,dj]. Therefore, the problem has a unique solution.

Solution: At the left-end point x = xj, we find two coefficients:

aj = yj      and     bj = sj.

At the right-end point x = xj+1, we find a system of equations for cj,dj that has a solution:

cj =   - sj    and   dj = sj+1 + sj - 2

% computation of coefficients of cubic Hermite interpolants from a set [x,y]

x = 0:0.125:1;               % data partition

y = cos(10*pi*x);         % function values

s = -10*pi*sin(10*pi*x);     % slope values

n = length(x)-1;

h = x(2:n+1)-x(1:n); % vector of step sizes of the partition of x

a = y(1:n);

b = s(1:n);

c = ((y(2:n+1)-y(1:n))./h(1:n)-s(1:n))./h(1:n);

d = (s(2:n+1)+s(1:n)-2*(y(2:n+1)-y(1:n))./h(1:n))./(h(1:n).^2);

% evaluation of cubic Hermite interpolants at data points xInt

xInt = 0 : 0.01 : 1;       % another partition with better resolution

for j = 1 : length(xInt)

if xInt(j) ~= x(n+1)

iInt(j) = sum(x <= xInt(j));

else

iInt(j) = n;

end

end

yInt = a(iInt) + b(iInt).*(xInt-x(iInt)) + c(iInt).*(xInt - x(iInt)).^2 + d(iInt).*(xInt - x(iInt)).^2.*(xInt-x(iInt+1));

yEx = cos(10*pi*xInt);

plot(x,y,'b*',xInt,yInt,'g',xInt,yEx,'r:');

The example above is difficult for interpolation since the function oscillates rapidly and too few data points are given. However, the cubic Hermite interpolation resembles the behaviour of the function. For comparison, below is the result of the cubic spline interpolation on the same data set. Without information about slopes of function at the given data points, the cubic spline interpolation works worse to display the function.

x = 0:0.125:1; y = cos(10*pi*x); n = length(x)-1; h = x(2:n+1)-x(1:n);

A = 2*diag(h(1:n-1)) + 2*diag(h(2:n)) + diag(h(2:n-1),1) + diag(h(2:n-1),-1);

b = 6*((y(3:n+1)-y(2:n))./h(2:n)-(y(2:n)-y(1:n-1))./h(1:n-1));

m = A\b'; m = [0;m;0]'; a = y(1:n);

b = (y(2:n+1)-y(1:n))./h(1:n)-h(1:n).*(m(2:n+1)+2*m(1:n))/6;

c = 0.5*m(1:n); d = (m(2:n+1)-m(1:n))./(6*h(1:n));

xInt = 0 : 0.01 : 1;       % another partition with better resolution

yInt = a(iInt) + b(iInt).*(xInt-x(iInt)) + c(iInt).*(xInt - x(iInt)).^2 + d(iInt).*(xInt - x(iInt)).^3;

yEx = cos(10*pi*xInt);

plot(x,y,'b*',xInt,yInt,'g',xInt,yEx,'r:');