i need to calculate the value of the cubic spline not a knot in the point x=1.97, the problem gives me coordinates of x=[0.0 0.5 1.0 1.5 2.0] and the function y=(sin(x)-(x+1).^2)/(x.^2+3). i don't understand why sometimes when i do y=f(x) it gives me a vector and sometimes like now it gives me only a value, and as a result it gives me an error when calculating the spline.

 Accepted Answer

KSSV
KSSV on 20 Jun 2018
x=[0.0 0.5 1.0 1.5 2.0] ;
y = @(x) (sin(x)-(x+1).^2)./(x.^2+3) ;
y = y(x) ;
xi = 1.97 ;
yi = interp1(x,y,xi,'spline')
plot(x,y,'b')
hold on
plot(xi,yi,'*r')

2 Comments

thanks a lot, i didn't put the point before the "/" that's why i wasn't able of solving the exercise.
Sorry I dont understand, if you have the function why not just calculate the function with x = 1.97?

Sign in to comment.

More Answers (0)

Categories

Find more on Mathematics in Help Center and File Exchange

Products

Release

R2018a

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!