Cubic interpolation coefficients and basis matrix

Question

Alessandro Maria Marco 2024-1-29

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https://ww2.mathworks.cn/matlabcentral/answers/2075831-cubic-interpolation-coefficients-and-basis-matrix

评论： Alessandro Maria Marco 2024-2-6

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 2024-1-30

编辑：Alessandro Maria Marco 2024-1-30

The basis functions are defined recursively. This stuff is known as B-splines, I think Matlab should have it

Torsten 2024-1-30

编辑：Torsten 2024-1-30

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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Answer 1

Matt J 2024-1-30

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2075831-cubic-interpolation-coefficients-and-basis-matrix#answer_1399991

编辑：Matt J 2024-1-30

在 MATLAB Online 中打开

Using this FEX downloadable,

https://www.mathworks.com/matlabcentral/fileexchange/44669-convert-linear-functions-to-matrix-form

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 2024-2-6

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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