# natural cubic spline interpolation of y-values: how to get derivative of the spline wrt the y-values?

17 vues (au cours des 30 derniers jours)
SA-W le 8 Mar 2023
Modifié(e) : Torsten le 9 Mar 2023
Given a data set with support points x_1,...,x_n and corresponding y-values y_1,...,y_n.
My objective is to create a cubic spline f (with natural boundary conditions) that passes through the y_values. There are, of course, plenty of functions for doing this.
However, for a parameter identification procedure, I have to compute the derivative of the spline f with respect to the y-values -- at arbitrary points within [x1, x_n].
Is there an easy way using built-in functions of Matlab to compute the sensitivities?
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Torsten le 8 Mar 2023
I can understand that you have to compute the derivative of the spline with respect to the parameters, but why with respect to the y-values ?
SA-W le 8 Mar 2023
Because the values y_1,...,y_n are the parameters of my optimization. My objective is to optimize the values y_1,...,y_n and interpolate between them with cubic splines. Is that clear?

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### Réponse acceptée

Bruno Luong le 8 Mar 2023
Modifié(e) : Bruno Luong le 8 Mar 2023
The derivative f wrt to y_i is the spline interpolate b_i := (0,0,...,1,0...) where 1 is at ith position, since the spline is linear to y values.
Replace spline command with your function that computes natural spline pp form
x=cumsum(rand(1,10));
y=rand(size(x));
xi=linspace(min(x),max(x),500);
f=ppval(spline(x,y),xi)
f = 1×500
0.8578 0.8215 0.7861 0.7516 0.7180 0.6852 0.6533 0.6223 0.5920 0.5627 0.5341 0.5063 0.4794 0.4532 0.4278 0.4032 0.3794 0.3563 0.3340 0.3124 0.2915 0.2714 0.2519 0.2332 0.2152 0.1978 0.1812 0.1652 0.1498 0.1351
plot(x,y,'or',xi,f,'b') %
b=eye(length(x));
yd=spline(x,b);
dfdy=ppval(yd,xi); % dfdy(i,j) is the derivative of f(xi(j)) with respect to y(i))
figure
plot(xi,dfdy')
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SA-W le 9 Mar 2023
That said, if I wanted to calculate
dspline/dy = (spline(y + e_i * h) - spline(y)) / h
by finite differencing,
I could choose an arbitrary value for h as it cancels out anyway, right?
Bruno Luong le 9 Mar 2023
Right

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### Plus de réponses (2)

Bruno Luong le 8 Mar 2023
Modifié(e) : Bruno Luong le 8 Mar 2023
you can find my function that compute the derivative of a piecewise polynomiall function (pp), inclusing pp form of the spline functions. This function returns the pp form of the derivative, so you can evaluate using MATLAB ppval.
function ppd = ppder(pp)
ppd = pp;
coefs = ppd.coefs;
n = size(coefs,2);
ppd.coefs = coefs(:,1:n-1).*(n-1:-1:1);
ppd.order = ppd.order-1;
end
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SA-W le 8 Mar 2023
The derivative f wrt to y_i is the spline interpolate b_i := (0,0,...,1,0...) where 1 is at ith position, since the spline is linear to y values.
This is not clear to me. If I take the derivative of f (a scalar function) with respect to y_i (a scalar), the result should be a single number and not a vector b_i := (0,0,...,1,0...)...
Also, why is a cubic spline linear to y values? If I change yn to yn+h, f will be more affected at yn than at y1, for instance.
Bruno Luong le 8 Mar 2023
Modifié(e) : Bruno Luong le 8 Mar 2023
"the result should be a single number"
No the result is a scalar function. If we take f at a given point x then it is a scalar.
"...and not a vector b_i := (0,0,...,1,0...)."
I did not tell the derivative is b, the derivative is the spline interpolating b
"why is a cubic spline linear to y values?"
You clearly missunderstand and male confusion betwen being linear and being a lilnear function

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Torsten le 8 Mar 2023
Déplacé(e) : Torsten le 8 Mar 2023
Why do you want to compute the sensitivities manually ?
Usually, the fitting software computes them using a finite-difference approximation, i.e. by calling your function with
y_i
and
y_i+h
getting back
f_j(y_1,...,y_i,...,y_n) and f_j(y_1,...,y_i+h,...,y_n)
and approximating
df_j/dy_i = (f_j(y_1,...,y_i+h,...,y_n)-f_j(y_1,...,y_i,...,y_n))/h
And this would also be my suggestion on how to do it manually if it is really needed.
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