connection between cubic interpolation and curvature of a function

Question

I am reading "Numerical Optimization", second edition, written by Joerge Nocedal and Stephen Wright. On page 59 the authors claim:

Cubic interpolation provides a good model for functions with significant changes of curvature.

I don't understand the connection between the curvature and the possibility of a cubic interpolation.

Answer 1

Suppose the author wants to express the advantage of 3rd order w.r.t. linear or 2nd order. Cubic interpolation deals with curve trend and fitting with analytical polynomials while curvature studies the curve property at local points.

Hope can be helpful...

Answer 2

Here is a connection between curvature and splines function.

Let us consider a set of $n$ points $(x_1,y_1), \ (x_2,y_2), \ \cdots (x_n,y_n)$ with $a=x_1<x_2<\cdots<x_n=b$. An interpolating function of these points is a function $f$ such that

$$\forall k, 1 \leq k \leq n, \ \ f(x_k)=y_k$$

Among all such interpolating functions $y=f(x)$, the cubic spline interpolation minimizes the integral :

$$\int_a^b (f''(t))^2 dt$$

(see for example (https://arxiv.org/pdf/1302.5248))

In fact, curvature, given by

$$K(t)=\dfrac{(f''(t))^2}{(1+f'(t)^2)^{3/2}}$$

is close to the integrand $(f''(t))^2$ if the variations of $f$ (measured by $f'(t)$) are not rapid.

See (https://www.math24.net/curvature-radius/).