在Matlab / Octave中构造样条方程

Question

I am aware that in matlab/octave, the polyfit can extrapolate constants used to build the polynomial equation that fits the given set of data x and y. Polyout outputs the polynomial equation with its constants. see below the link for the example:

https://en.wikibooks.org/wiki/Octave_Programming_Tutorial/Polynomials

How do I 'polyout' the splinefit constants to construct the equation? How do spline equation looks like? Is it possible to polyfit an almost perfect line to 6 decimal precision? Thanks.

n=5;
m=3;   x=0:0.000001:1;   y=asin(x);
p = polyfit(x,y,n);
f = polyval(p,x);
pp = splinefit(x,y,m);
g = ppval(pp,x);
plot(x,y,'o',x,f,'-',x,g,'o');
format long;
disp ("The value of p is:"), disp (p);
polyout(p, 'x');
disp ("The value of pp is:"), disp (pp);  

Answer 1

I figure it out. Unfortunately polyout(pp,'x') does not work for spline. However, the results can be interpreted manually. For the following code:

x=0:0.1:1;   
y=sin(x);
m=2;
format long;
disp ("The value of pp is:"), disp (pp)

Mathcad/Octave gives the answer it the following format:

The value of pp is:

scalar structure containing the fields:

form = pp
breaks =

   0.000000000000000   0.500000000000000   1.000000000000000

coefs =

 Columns 1 through 4:

   7.91957255741094e-03   3.68579287957638e-04  -1.66790993823105e-01   1.71448144544173e-05
   6.55744943740794e-03   2.01675106814849e-02  -1.46254903853663e-01  -2.39717011291503e-01

 Columns 5 and 6:

   9.99999254411779e-01  -1.23307748213508e-09
   8.77582309927074e-01   4.79425560796454e-01

pieces =  2
order =  6
dim =  1

To interpret the output, the output means
y1= -1.23307748213508e-09 + 9.99999254411779e-01*(x-1) + 1.71448144544173e-05*(x-1)^2 .....up to eight polynomial (set up by "order" in spline equation.)
y2 = the same way with the second row except instead (x-1) there is (x -1.5) defined by pieces. Hence y = y1 + y2