# express basis spline derivative in terms of interpolation values

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SA-W on 28 Mar 2024
Commented: Bruno Luong on 4 Apr 2024
x = [1 2 3 4 5];
y = randn(size(x));
f = spapi(5,x,y);
fdd = fnder(f, 2);
If we evaluate the second derivative at a given point x* as
xeval = 3.5;
val = fnval(fdd, xeval)
is there a way to express "val" as a linear combination of the interpolation values as
val = y(1)*c1 + ... + y(5)*c5
Then, the task is to find the coefficients ci.
Can this be done with symbolic differentiation or other techniques? Or is there an easy analytical representation for that?
Thank you!

Bruno Luong on 28 Mar 2024
x = [1 2 3 4 5];
y = randn(size(x));
k = 5;
f = spapi(k,x,y);
B = spapi(k,x,eye(length(x)));
fdd = fnder(f, 2);
Bdd = fnder(B,2);
xeval = 3.5;
val = fnval(fdd, xeval)
val = 1.5349
c = fnval(Bdd, xeval);
vcomb = y * c
vcomb = 1.5349
err = val-vcomb
err = 8.8818e-16
SA-W on 4 Apr 2024

Honestly I dont understand what you want to achieve your code. Why do you put positive constraint on fdddd(x(end)) and not fddd(x(end))? I don't know what is your goal

Oh, you are right. To have s''(x)>0 for x>x(end), I thought the fourth derivative must be positive at x(end). But s'''(x(end)) is sufficient, right?

(I will not ask further after this and say thanks a lot for helping!! :-))

Bruno Luong on 4 Apr 2024
You might need to impose both conditions of third and fourth derivative on x(end).