# [SciPy-User] Gradient of spline interpolation at a point wrt changing the knot points

Daπid davidmenhur@gmail....
Fri Jun 14 18:03:09 CDT 2013

```On 14 June 2013 23:47, Pauli Virtanen <pav@iki.fi> wrote:
> Sorry, wrote it too fast. Computing the spline coefficients
> is IIRC is a global process, and it is affected by knot locations,
> so I doubt there is a simple formula for the derivative.

Actually, it only depends on the surrounding points, and with linear
properties. See, for example, its matrix form [1]: a multi-diagonal
fixed matrix A multiplying a vector of coefficients b.

A b = y

So, one could solve the vector of coeficients:

b = A^-1 y

Now, the definition of derivative with respect to y_i (the value of y
at point i):

d b/dy_i = limit h->0 ((A^-1 (y + h)-A^-1 y)/h) = limit h->0 A^-1 h /h

where h = [0, 0 .... h ....0] in the i-th position. And this is the
sum of the i-th row of the matrix. In the case of cubic interpolation,
that would be 0.507 for any point not in the borders or next to one.

David.

[1] Formula 9: http://www.mechanicaldust.com/UCB/math128a/cubic.pdf
```