I'm trying to write a cubic spline interpolation program. I have written the program but, the graph is not coming out correctly. The spline uses natural boundary conditions(second dervative at start/end node are 0). The code is in Matlab and is shown below,
clear all %Function to Interpolate k = 10; %Number of Support Nodes-1 xs(1) = -1; for j = 1:k xs(j+1) = -1 +2*j/k; %Support Nodes(Equidistant) end; fs = 1./(25.*xs.^2+1); %Support Ordinates x = [-0.99:2/(2*k):0.99]; %Places to Evaluate Function fx = 1./(25.*x.^2+1); %Function Evaluated at x %Cubic Spline Code(Coefficients to Calculate 2nd Derivatives) f(1) = 2*(xs(3)-xs(1)); g(1) = xs(3)-xs(2); r(1) = (6/(xs(3)-xs(2)))*(fs(3)-fs(2)) + (6/(xs(2)-xs(1)))*(fs(1)-fs(2)); e(1) = 0; for i = 2:k-2 e(i) = xs(i+1)-xs(i); f(i) = 2*(xs(i+2)-xs(i)); g(i) = xs(i+2)-xs(i+1); r(i) = (6/(xs(i+2)-xs(i+1)))*(fs(i+2)-fs(i+1)) + ... (6/(xs(i+1)-xs(i)))*(fs(i)-fs(i+1)); end e(k-1) = xs(k)-xs(k-1); f(k-1) = 2*(xs(k+1)-xs(k-1)); r(k-1) = (6/(xs(k+1)-xs(k)))*(fs(k+1)-fs(k)) + ... (6/(xs(k)-xs(k-1)))*(fs(k-1)-fs(k)); %Tridiagonal System i = 1; A = zeros(k-1,k-1); while i = min(x)) error('Outside Range'); end P = zeros(size(length(x),length(x))); i = 1; for Counter = 1:length(x) for j = 1:k-1 a(j) = x(Counter)- xs(j); end i = find(a == min(a(a>=0))); if i == 1 c1 = 0; c2 = xn(1)/6/(xs(2)-xs(1)); c3 = fs(1)/(xs(2)-xs(1)); c4 = fs(2)/(xs(2)-xs(1))-xn(1)*(xs(2)-xs(1))/6; t1 = c1*(xs(2)-x(Counter))^3; t2 = c2*(x(Counter)-xs(1))^3; t3 = c3*(xs(2)-x(Counter)); t4 = c4*(x(Counter)-xs(1)); P(Counter) = t1 +t2 +t3 +t4; else if i
When I run the code, the interpolation function is not symmetric and, it doesn't converge correctly. Can anyone offer any suggestions about problems in my code? Thanks.
I wrote a cubic spline package in Mathematica a long time ago. Here is my translation of that package into Matlab. Note I haven't looked at cubic splines in about 7 years, so I'm basing this off my own documentation. You should check everything I say.
The basic problem is we are given n data points (x(1), y(1)) , ... , (x(n), y(n)) and we wish to calculate a piecewise cubic interpolant. The interpolant is defined as
S(x) = { Sk(x) when x(k)
Here Sk(x) is a cubic polynomial of the form
Sk(x) = sk0 + sk1*(x-x(k)) + sk2*(x-x(k))^2 + sk3*(x-x(k))^3
The properties of the spline are:
- The spline pass through the data point
Sk(x(k)) = y(k) - The spline is continuous at the end-points and thus continuous everywhere in the interpolation interval
Sk(x(k+1)) = Sk+1(x(k+1)) - The spline has continuous first derivative
Sk'(x(k+1)) = Sk+1'(x(k+1)) - The spline has continuous second derivative
Sk''(x(k+1)) = Sk+1''(x(k+1))
To construct a cubic spline from a set of data point we need to solve for the coefficients sk0, sk1, sk2 and sk3 for each of the n-1 cubic polynomials. That is a total of 4*(n-1) = 4*n - 4 unknowns. Property 1 supplies n constraints, and properties 2,3,4 each supply an additional n-2 constraints. Thus we have n + 3*(n-2) = 4*n - 6 constraints and 4*n - 4 unknowns. This leaves two degrees of freedom. We fix these degrees of freedom by setting the second derivative equal to zero at the start and end nodes.
Let m(k) = Sk''(x(k)) , h(k) = x(k+1) - x(k) and d(k) = (y(k+1) - y(k))/h(k). The following three-term recurrence relation holds
h(k-1)*m(k-1) + 2*(h(k-1) + h(k))*m(k) + h(k)*m(k+1) = 6*(d(k) - d(k-1))
The m(k) are unknowns we wish to solve for. The h(k) and d(k) are defined by the input data. This three-term recurrence relation defines a tridiagonal linear system. Once the m(k) are determined the coefficients for Sk are given by
sk0 = y(k) sk1 = d(k) - h(k)*(2*m(k) + m(k-1))/6 sk2 = m(k)/2 sk3 = m(k+1) - m(k)/(6*h(k))
Okay that is all the math you need to know to completely define the algorithm to compute a cubic spline. Here it is in Matlab:
function [s0,s1,s2,s3]=cubic_spline(x,y) if any(size(x) ~= size(y)) || size(x,2) ~= 1 error('inputs x and y must be column vectors of equal length'); end n = length(x) h = x(2:n) - x(1:n-1); d = (y(2:n) - y(1:n-1))./h; lower = h(1:end-1); main = 2*(h(1:end-1) + h(2:end)); upper = h(2:end); T = spdiags([lower main upper], [-1 0 1], n-2, n-2); rhs = 6*(d(2:end)-d(1:end-1)); m = T\rhs; % Use natural boundary conditions where second derivative % is zero at the endpoints m = [ 0; m; 0]; s0 = y; s1 = d - h.*(2*m(1:end-1) + m(2:end))/6; s2 = m/2; s3 =(m(2:end)-m(1:end-1))./(6*h);
Here is some code to plot a cubic spline:
function plot_cubic_spline(x,s0,s1,s2,s3) n = length(x); inner_points = 20; for i=1:n-1 xx = linspace(x(i),x(i+1),inner_points); xi = repmat(x(i),1,inner_points); yy = s0(i) + s1(i)*(xx-xi) + ... s2(i)*(xx-xi).^2 + s3(i)*(xx - xi).^3; plot(xx,yy,'b') plot(x(i),0,'r'); end
Here is a function that constructs a cubic spline and plots in on the famous Runge function:
function cubic_driver(num_points) runge = @(x) 1./(1+ 25*x.^2); x = linspace(-1,1,num_points); y = runge(x); [s0,s1,s2,s3] = cubic_spline(x',y'); plot_points = 1000; xx = linspace(-1,1,plot_points); yy = runge(xx); plot(xx,yy,'g'); hold on; plot_cubic_spline(x,s0,s1,s2,s3);
You can see it in action by running the following at the Matlab prompt
>> cubic_driver(5) >> clf >> cubic_driver(10) >> clf >> cubic_driver(20)
By the time you have twenty nodes your interpolant is visually indistinguishable from the Runge function.
Some comments on the Matlab code: I don't use any for or while loops. I am able to vectorize all operations. I quickly form the sparse tridiagonal matrix with spdiags. I solve it using the backslash operator. I counting on Tim Davis's UMFPACK to handle the decomposition and forward and backward solves.
Hope that helps. The code is available as a gist on github https://gist.github.com/1269709
