Natural cubic spline interpolation using MATLAB
MATLAB Program:
%
Natural cubic spline interpolation
%
Find the approximate value of f(1.5) from
%
(x,y)= (0,1), (1,e), (2,e^2) & (3,e^3).
n = input('Enter
n for (n+1) nodes, n: ');
x = zeros(1,n+1);
a = zeros(1,n+1);
for i = 0:n
fprintf('Enter
x(%d) and f(x(%d)) on separate lines:
\n', i, i);
x(i+1)
= input(' ');
a(i+1) = input(' ');
end
m = n - 1;
h = zeros(1,m+1);
for i = 0:m
h(i+1) = x(i+2) - x(i+1);
end
xa = zeros(1,m+1);
for i = 1:m
xa(i+1) =
3.0*(a(i+2)*h(i)-a(i+1)*(x(i+2)-x(i))+a(i)*h(i+1))/(h(i+1)*h(i));
end
xl = zeros(1,n+1);
xu = zeros(1,n+1);
xz = zeros(1,n+1);
xl(1) = 1;
xu(1) = 0;
xz(1) = 0;
for i = 1:m
xl(i+1) = 2*(x(i+2)-x(i))-h(i)*xu(i);
xu(i+1) = h(i+1)/xl(i+1);
xz(i+1) = (xa(i+1)-h(i)*xz(i))/xl(i+1);
end
xl(n+1) = 1;
xz(n+1) = 0;
b = zeros(1,n+1);
c = zeros(1,n+1);
d = zeros(1,n+1);
c(n+1) = xz(n+1);
for i = 0:m
j = m-i;
c(j+1) = xz(j+1)-xu(j+1)*c(j+2);
b(j+1) = (a(j+2)-a(j+1))/h(j+1) - h(j+1) *
(c(j+2) + 2.0 * c(j+1)) / 3.0;
d(j+1) = (c(j+2) - c(j+1)) / (3.0 * h(j+1));
end
fprintf('\nThe
numbers x(0), ..., x(n) are:\n');
for i = 0:n
fprintf(' %5.4f',
x(i+1));
end
fprintf('\n\nThe
coefficients of the spline on the subintervals are:\n');
fprintf(' a(i)
b(i) c(i) d(i)\n');
for i = 0:m
fprintf('%11.8f %11.8f
%11.8f %11.8f \n',a(i+1),b(i+1),c(i+1),d(i+1));
end
>> Natural_spline
Enter n for (n+1) nodes, n: 3
Enter x(0) and f(x(0)) on separate lines:
0
1
Enter x(1) and f(x(1)) on separate lines:
1
2.7
Enter x(2) and f(x(2)) on separate lines:
2
7.29
Enter x(3) and f(x(3)) on separate lines:
3
19.68
The numbers x(0), ..., x(n) are:
0.0000 1.0000 2.0000 3.0000
The coefficients of the spline on the subintervals are:
a(i) b(i) c(i) d(i)
1.00000000 1.44933333 0.00000000 0.25066667
2.70000000 2.20133333 0.75200000 1.63666667
7.29000000 8.61533333 5.66200000 -1.88733333
>>
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