how to interpolate the non uniform resolution in cubic spline?

Hi all, if I have 6 segments with resolution of 8,8,8,8,8,12, why can not cubic spline interpolate correctly for the last segment? How to solve this problem? Thank you very much

[John D'Errico]

Why can't you? Because you did not write the correct code. Sorry, but the computation is trivial with a cubic spline, written correctly to handle unequal spacing between the breaks. (Triviality is sometimes in the eye of the beholder.)

What did you do wrong? That is impossible to know, because we cannot magically see inside your computer, and you did not provide your code.

Most likely, you assumed that the data spacing was uniform, and wrote the equations and code to follow that assumption. So don't do it that way.

If this is for homework, then the issue is you need to write code using the correct equations. If this is for your work or whatever task you are trying to perform, then DON'T WRITE YOUR OWN SPLINE CODE. Use existing tools like spline, or pchip, or interp1, etc.

[AniCnx]

Thank you for your nice comments, John D'Errico. actually, i did code it on my own as following and i attached the result as pic. in pic, that can be seen the last segment has error of interpolation.

%if true
seg=24;
NL=[10;10;10;10;10;10;10;10;10;10;10;10;10;10;10;10;10;10;10;10;10;10;10;17];
Ck=[-266.479532163743,-266.479532163743,-236.267393053600,-162.405313757590,-92.1402364342393,-27.3010252757420,10.5878240764911,37.7567786138143,50.6856443813328,48.3565527346632,41.6192148682732,38.9397293502820,12.0864493612835,-8.65639937685077,12.7171882144171,51.7106955326762,42.4350761397639,112.616420850404,119.405044382449,81.1969741122724,26.0127908204208,-20.8259537029191,-43.0898340533435,-6.51334559323291,-8.89813958239892,-11.2829335715649,-11.2829335715649,-11.2829335715649];
 x=1;
 for s=1:seg;   
     for k=1:NL(s,1);
     %cubic spline      
         y=x/NL(s,1); fx=floor(y)+1;      
         for p=-2:1; BV(p+3,1)=B3spline(y-(fx+p)); end  %compute BV 
         Gcs(x,1)=Ck(1,fx:fx+3)*BV;
         x=x+1;
     end
 end
 %for Ck plot
 A=zeros(1,seg);
 for i=2:seg;
     temp=NL(i,1);
     A(1,i)=A(1,i-1)+temp;
 end

   plot(Gcs(:,1))
   hold on
   plot(A,Ck(1,2:25,1),['*','r']);

function B=B3spline(x) x=abs(x); if x<1 B=(2/3) - (x*x) + (x*x*x)/2; elseif x>=1 && x<2 B=(x-2)*(x-2)*(x-2)/(-6); else B=0; end %end

Answer 1

[Image Analyst]

See my attached spline demo. Of course, you can put in any x value you want to get the estimated/spline value at that location.

[AniCnx]

Thank you for your suggestion, Image analysis. in case of your demo, i quite pretty understand that. but i face to the problem as each segment has non-uniform resolution as my result is wrong interpolation as shown in picture.