Question

i 0) 2) Interolae a quadelgn 0.6 1.3 1.7 2.1 1.2 2.4 3.0 4.2

Answer 1

% This code solves for quadratic interpolation
x = [1.2 2.4 3.0 4.2];
y = [06 1.3 1.7 2.1];
N = length(x)-1;
% The unknowns are 3*N with ao=0 "Linear Spline"
% Filling Matrix from point matching
V = [0;zeros(2*N,1);zeros(N-1,1)];
Z = zeros(length(V),length(V));
j=1;
f=1;
for i=2:2:2*N
Z(i,f:f+2) = [x(j)^2 x(j) 1];
V(i) = y(j);
j = j+1;
Z(i+1,f:f+2) = [x(j)^2 x(j) 1];
V(i+1) = y(j);
f = f+3;
end
% Filling Matrix from smoothing condition
j=1;
l=2;
for i=2*N+2:3*N
  
Z(i,j:j+1) = [2*x(l) 1];
Z(i,j+3:j+4) = [-2*x(l) -1];
j = j+3;
l = l+1;
end
% Adjusting the value of a1 to be zero "Linear Spline"
Z(1,1)=1;
% Inverting and obtaining the coeffiecients, Plotting
Coeff = Z\V;
j=1;
hold on;
for i=1:N
curve=@(l) Coeff(j)*l.^2+Coeff(j+1)*l+Coeff(j+2);
ezplot(curve,[x(i),x(i+1)]);
hndl=get(gca,'Children');
set(hndl,'LineWidth',2);
hold on
j=j+3;
end
scatter(x,y,50,'r','filled')
grid on;
xlim([min(x)-2 max(x)+2]);
ylim([min(y)-2 max(y)+2]);
xlabel('x');
ylabel('y');
title('Quadratic Spline')

Quadratic Spline