data interpolation using cubic splines and transform them to B-form

I have some data points given in space describing a closed periodic curve that i want to interpolate. I know that the curve fitting toolbox offers functions that do that.

For instance csape(x,fx,'periodic') is working perfect, but i'd like to have a B- form representation to have control over the basis. To do so i can use fn2fm(fnk,'B-') to transform the pp-form to B-form but then i gets confusing.

Usually when you read mathematical textbooks that explain how to interpolate periodc function, recommend using following b-spline basis:

In this case we look at the b-splines , , and , since they are local polynomials of degree 3. In matlab they are printed out as b-splines of order 4 since you need four coefficients to describe a polynomial of degree 3. (I have chosen order 4 to have global interpolant)

However to interpolate 8 given data points you need 7 basis functions. To get the right coefficients you have to determine the values of your choosen basis in each node and solve the linear equation. ( For closed periodic curves you have to solve the linear equations for each component)

%% generate data
n = 8; 
a = 0;
b = 2*pi;
t = linspace(a,b,n);
int = linspace(a,b,300);

% data points to interpolate in one component
fx = cos(t);
%fy = sin(t);
%fz = zeros(1,length(t));

N = n-1;                               % # elements and of my basis functions !
x_node = linspace(0,2*pi,N+1);         % nodes
element = [1:N;2:N+1]'                 % elements
A = my_matrix(N,x_node)                % load values of all my basis functions 

% solve linear equation for one component 
ux = A\fx';
%uy = A\fy';
%uz = A\fz';

A =

         0         0         0         0    0.1667    0.6667    0.1667
    0.1667         0         0         0         0    0.1667    0.6667
    0.6667    0.1667         0         0         0         0    0.1667
    0.1667    0.6667    0.1667         0         0         0         0
         0    0.1667    0.6667    0.1667         0         0         0
         0         0    0.1667    0.6667    0.1667         0         0
         0         0         0    0.1667    0.6667    0.1667         0
         0         0         0         0    0.1667    0.6667    0.1667
         

Compering this to matlab we get the following:

% cubic spline interpolation for the first component
X = [fx]
     %fy;
     %fz];
F = csape(t,X,'periodic');
% transformation to B-form 
B = (fn2fm(F,'B-'))

B = 

  struct with fields:

      form: 'B-'
     knots: [0 0 0 0 0.8976 1.7952 2.6928 3.5904 4.4880 5.3856 6.2832 6.2832 6.2832 6.2832]
     coefs: [3×10 double]
    number: 10
     order: 4
       dim: 3 
       
 B.knots      
   ans =

         0         0         0         0    3.1416    3.1416    3.1416    6.2832    6.2832    6.2832    6.2832    

As it can be seen matlab is using way more knots then i do.

Is it is not like that, if i choose m identical knots that the b-spline at this node / knot is ? This would mean that the b-spline is at the endpoints, since it is actually of order 4 (in matlab) (i.e ) ?

Thats the point where my confusion starts, since we have data that we want to interpolate by a function that is in each node (global ).

Looking at the B-spline basis matlab chooses for a periodic interpolation the first Basis (blue) and the last one (yellow) are those that shoud be , but matlab prits them out to have order 4, and the produced interpolant is .

basis = (spmak(B.knots(1:1+4),1))
basis = 

  struct with fields:

      form: 'B-'
     knots: [0 0 0 0 0.8976]
     coefs: 1
    number: 1
     order: 4
       dim: 1

Looking at the structure B above it can be seen that the number of basis functions is 10 (for 8 datat points). It seems that if you do interpolation for space curves with csape using periodic condition, matlab after transforming the sturcture to B-form, always takes n+2 b-spline basis functions. But if you doing the interpolation for each coordinate separatly the number of basis functions can differ from coordinate to coordinate you are interpolating.

I wanted to check if i figured out the right basis. Useing the B.coefs and multiply them with the basis lead to the right/same interpolant that i get with my chosen basis.

%plot the first component of the space curve generated by matlab 
fig2 = subplot(2,1,1);
    p1 = plot(B.knots,B.knots*0,'b-o','MarkerSize',10,'DisplayName','matlab knots');
    hold on 
    p2 = plot(t,fx,'-o','DisplayName','data');
    hold on 

val = 0;

for i = 1:length(B.coefs)
    % generate basis of order 4
    B_basis = (spmak(B.knots(i:i+4),1));
    fnplt(B_basis)
    val = val + fnval(B_basis,int)*B.coefs(1,i);
    hold on 
end
    hold on 
    p3 = plot(int,val,'-','LineWidth',2,'DisplayName','interpoland by matlab');
    legend([p1,p2,p3],'Location','southeast')
    title('first component fx by matlab')
    
    
% plot first component of the space curve with my basis
f2 = subplot(2,2,1);
    plot(int,fval_x,'-','LineWidth',2)
    hold on 
    plot(x_node,0*x_node,'b-o','MarkerFaceColor','r')
    hold on 
    plot(t,fx,'-o')
    hold on 
    plot(int,my_spline(int,element,x_node,'basis'))
    legend({'interpolant','xnode','data'},'Location','southeast')
    title('first component fx')
    hold off    

After that i tryed to calculate the coefs matlab is using for his basis. For this i looked up witch values the basis of matlab gives at each node to get a matrix that contains all information. If you use the usual approach you will end up here:

    colmat = spcol(knott,4,brk2knt(tau,1)) 
colmat =

  Columns 1 through 6

    1.0000         0         0         0         0         0
         0    0.2500    0.5833    0.1667         0         0
         0         0    0.1667    0.6667    0.1667         0
         0         0         0    0.1667    0.6667    0.1667
         0         0         0         0    0.1667    0.6667
         0         0         0         0         0    0.1667
         0         0         0         0         0         0
         0         0         0         0         0         0
         
     Columns 7 through 10

         0         0         0         0
         0         0         0         0
         0         0         0         0
         0         0         0         0
    0.1667         0         0         0
    0.6667    0.1667         0         0
    0.1667    0.5833    0.2500         0
         0         0         0    1.0000      
         
% solve linear equation for the first component        
cofcol = (colmat\fx');
    
cofcol =

ans =

  Columns 1 through 6

    1.0000         0    1.1764   -0.3764   -1.0059   -1.0059

  Columns 7 through 10

   -0.3764    1.1764         0    1.0000
   
   
% matlab coefficients    
B.coefs(1,:)

ans =

  Columns 1 through 6

    1.0000    1.0000    0.7130   -0.2545   -1.0303   -1.0303

  Columns 7 through 10

   -0.2545    0.7130    1.0000    1.0000

Obviously wirth wrong coefficients. This leads to the question whitch matrix is used by matlab ? and how to build it by my own?

----------------------------------------------

Questions:

1) does someone know why matlab is using that specific b-spline basis for periodic interpolation ?

2) how is it possible to get a global interpolant using this basis, since at the endpoints the basis shoud be ?

3) how to get the right matrix that contains also the periodic information to determine the coefficients

i.e how to determine the B-form structure interpolating with periodic condtion without using csape

4) is it possible to transform my interpolation to a B-structure ?

5) why do we get n+2 basis functions when we have n data points ?

Thanks in advance.

%%%%%%%%%% ---- %%%%%%%%%

look at the file jump.m, there is a example for a basis by matlab that is of order 4 but its first derivative in the choosen node has a jump. (this happens if n=3)

Derivative of the 4th basis function

The file my_spline.m gives you the basis i have chosen.

Interpolation.m contains all plots that i used above and also some more.

(i.e interpolation for curves in )