# Cubic interpolation coefficients and basis matrix

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Alessandro Maria Marco el 29 de En. de 2024
Comentada: Alessandro Maria Marco el 6 de Feb. de 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)
form: 'pp' breaks: [-1 -0.7778 -0.5556 -0.3333 -0.1111 0.1111 0.3333 0.5556 0.7778 1] coefs: [9×4 double] pieces: 9 order: 4 dim: 1
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)
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!
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Alessandro Maria Marco el 30 de En. de 2024
Editada: Alessandro Maria Marco el 30 de En. de 2024
The basis functions are defined recursively. This stuff is known as B-splines, I think Matlab should have it
Torsten el 30 de En. de 2024
Editada: Torsten el 30 de En. de 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.

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Matt J el 30 de En. de 2024
Editada: Matt J el 30 de En. de 2024
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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Alessandro Maria Marco el 6 de Feb. de 2024
Thanks for your answer! I wanted to note that there is another way to do this using the routine "spcol" from the curve fitting toolbox. Might be useful for those users who have the curve fitting toolbox and prefer not to download extra functions from FEX.

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