Talk:Spline interpolation

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[edit] Definition

i have some questions about this phrase:

Given n+1 distinct knots xi such that
x_0 < x_1 < ... < x_{n-1} < x_n, \,\!
with n+1 knot values yi we are trying to find a spline function of degree n
S(x) := \left\{\begin{matrix}      S_0(x) & x \in [x_0, x_1] \\     S_1(x) & x \in [x_1, x_2] \\     \vdots & \vdots \\ S_{k-1}(x) & x \in [x_{k-1}, x_k]  \end{matrix}\right.
with each Si(x) a polynomial of degree n.

Is this not confusing, using n both for the degree of the polynomial, and for the number of points? --Anonymus, wiki nl

There's a k for that. Very well explained then ;) --217.136.81.22, 11:16, 13 Jun 2005 (UTC)

[edit] Natural cubic spline oscillation

Clamped and natural cubic splines yield the least oscillation about f than any other twice continuously differentiable function.

In the above sentence from the article, just what is f? Perhaps the article can be updated to clarify this. --Abelani, 19 November 2005

This is to confirm that someone has posted a clarification. --Abelani, 2:16, 27 November 2005 (UTC)


Amongst all twice continuously differentiable functions, clamped and natural cubic splines yield the least oscillation about the function f which is interpolated.

In the above sentence from the article, surely f itself is the function with the least oscillation about f. What is the restricted set of interpolation functions for which the statement is true and interesting? Harold f 03:27, 20 August 2006 (UTC)

[edit] Interpolation using natural cubic spline

In the formulas for interpolation using natural cubic spline, it seems that one could replace each zi by 6zi, enabling one to cancel 6s and obtain

S_i(x) = \frac{z_{i+1} (x-x_i)^3 + z_i (x_{i+1}-x)^3}{h_i}        + \left(\frac{y_{i+1}}{h_i} - h_i z_{i+1}\right)(x-x_i)        + \left(\frac{y_{i}}{h_i} - h_i z_i\right) (x_{i+1}-x)

and

\begin{matrix}         h_{i-1}            z_{i-1}        + 2(h_{i-1} + h_i) z_i        + h_i              z_{i+1}         =             \frac{y_{i+1}-y_i}{h_i} -             \frac{y_i-y_{i-1}}{h_{i-1}} \end{matrix}

Is there any reason not to do that? --Jwwalker 01:51, 26 July 2006 (UTC)