Please let me know in case the question is not clear or is not well-posed and if further clarifications are needed
Cubic interpolation coefficients and basis matrix
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Alessandro Maria Marco
on 29 Jan 2024
Commented: Alessandro Maria Marco
on 6 Feb 2024
Suppose I have n data points (x(i),y(i)) with i=1,..,n. I want to compute a cubic interpolant that fits exactly these points (interpolation, not least squares fit). I can do the following in Matlab:
x_min = -1;
x_max = 1;
nx = 10;
x = linspace(x_min,x_max,nx)';
y = exp(-x);
pp = csapi(x,y); %Can also use spline
disp(pp)
This returns the coefficients in the matrix pp.coefs: each row l (for l=1,..,n-1) of this matrix gives the 4 coefficients of the cubic polynomial for the specific subinterval l. However, I would like the cubic in another form. A cubic spline can be written as
(1)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1604136/image.png)
where phi_j(x) are the basis functions and c is a vector of n+2 coefficients. How can I get these n+2 coefficients (and, optionally the basis matrix Phi)? The Matlab functions csapi and spline give this (n-1)*4 matrix of coefficients which is not what I want.
Reference: The source for (1) is Fehr and Kindermann, "Computational Economics", Oxford University Press, page 93.
Any help is greatly appreciated, thanks!
5 Comments
Torsten
on 30 Jan 2024
Edited: Torsten
on 30 Jan 2024
Then you will have to sum the B_(j,3) functions that have non-zero support in [t_i,t_i+1] and equate this sum to the usual cubic spline representation coming from "csape" in [t_i,t_i+1]. Comparison of coefficients of the two cubic polynomials in all subintervals should give you a linear system of equations to determine the alpha_i coeffcients from the "csape" coefficients.
Accepted Answer
Matt J
on 30 Jan 2024
Edited: Matt J
on 30 Jan 2024
Using this FEX downloadable,
x_min = -1;
x_max = 1;
nx = 10;
x = linspace(x_min,x_max,nx)';
xx=linspace(x(1),x(end),1000)';
fun=@(in)csapi(x,in,xx);
Basis=func2mat(fun,x); %Columns of this matrix are basis functions
y=exp(-x);
yy=exp(-xx);
c=Basis\yy; %Coefficients
subplot(1,2,1)
h=plot(x,y,'o',xx,Basis*c,'.r'); axis square
title({'Interp by Basis'; 'Matrix Multiplication'})
subplot(1,2,2)
plot(Basis); axis square
title 'Basis functions'
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