Once upon a time, I was taught how to play connect the dots.

Some years later, I was given pseudo-code for an algorithm which would compute a polynomial of minimum degree passing through some points.

However, the local minima and maxima of the interpolation are located far away from where the discrete points are.

Suppose that we are interpolating the points $x_1, x_2, x_3, \cdots, x_n$ where $n$ is some whole number $n$, such as $761$ or $81$.

$\text{id est,} \text{ } n \in \mathbb{N} \text{ and } \forall k \in \mathbb{N}, \thinspace x_{k} \in \mathbb{R}$.

In the two diagrams shown below, the same set of points is interpolated, but the wiggly line is worse than the smooth curve.

a picture of both a smooth curve and a wiggly curve

The fact that the wiggly curve is not the curve we want is known as the Runge Phenomenon and you can read all about it in the encyclopedia named $`` \thinspace Wikipedia"$.

Suppose that $\mathcal{DATA} = \begin{Bmatrix} (x_{k}, y_{k}): x_{k}, y_{k} \in \mathbb{R} \end{Bmatrix}$ is a finite set of ordered pairs of real numbers.

Also, suppose that:

  • $\mathcal{DATA} \subseteq \mathcal{PS}$
  • $\mathcal{PS} \subseteq \mathbb{R}^{2}$
  • $\mathcal{PS}$ is a non-finite set of points such that for all $x \in \mathbb{R}$, there exists unique $y \in \mathbb{R}$ such that $(x, y) \in \mathcal{PS}$.

I hope that you will imagine a very small circle such that the circle is centered on the curve described by $\mathcal{PS}$.

There are two points on the outside of the circle (periphery) which intersect curve $\mathcal{PS}$.


The purple curve represents the subset of $\mathbb{R}^{2}$ I named $\mathcal{PS}$.

enter image description here

There are three triangles based on five points. We name the five points $p_1, p_2, p_3, p_4, p_5$.

$p_1, p_2, p_3, p_4, p_5 \in \mathcal{PS}$.

That is points $p_1, p_2, p_3, p_4, p_5$ are all points in the set we wish to interpolate.


So, we have three triangles:

  1. $△p_1, p_2, p_3$
  2. $△p_2, p_3, p_4$
  3. $△p_3, p_4, p_5$

Also, there are three angles of interest:

  1. The angle at vertex $p_2$ in triangle $△p_1, p_2, p_3$
  2. The angle at vertex $p_3$ in triangle $△p_2, p_3, p_4$
  3. The angle at vertex $p_4$ in triangle $△p_3, p_4, p_5$

How would you define a smooth curve in terms of the three triangles?

In the limit, the angles associated with all three triangles go to zero.

However, the angles go to zero at different rates.

I would say that a curve is smooth if for any $x, y \in \mathbb{R}$ if $(x, y)$ is an element of the curve we wish to interpolate, and we form a circle centered at point $(x, y)$, then the angles of the three associated triangles are almost the same as the radius of the circle gets increasingly small.

Can we describe this in terms of formal limits and a lot of logic using the universal quantifier, existential quantifier, and set theory.

For the absolute value function at $x = 0$, the angle at the middle of the triangle is $90° = \frac{\pi}{2}$, and the two outer triangles have angles of $0°$


Clearly, the absolute value function is not smooth at exactly $x = 0$ because the angles of the three triangles are different.

  • $\begingroup$ Thanks for the high quality question $\endgroup$
    – Klangen
    Jan 23 at 8:50
  • $\begingroup$ One aspect you seem to be looking at, given your pictures, is to avoid having a turning point for each interpolation point - your "wiggly curve" has at least one local extremum between each interpolation point but your smooth curve and circle-based diagram does not. Is this a behaviour you'd want to be part of a 'smooth' definition? $\endgroup$ Jan 23 at 9:05

1 Answer 1


I think you may be looking for an interpolation using cubic splines: piecewise cubic curves whose derivatives match at the interpolation points.

That is not a direct answer to your question asking for a better definition of "smooth" but may answer the question behind your question.


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