11
$\begingroup$

[This is a follow-up question to this one: Figures and Numbers: Relating properties of geometric shapes and their Fourier series.]


Drawing shapes by some predefined Fourier series I found this square with rounded corners which is given by the Fourier series

$$a_k = \begin{cases} +\pi^{-k} & \text{ for } k \equiv 1 \pmod 4 \\ +\pi^{-k} & \text{ for } k \equiv 3 \pmod 4\\ 0 & \text{ otherwise } \end{cases}$$

$$b_k = \begin{cases} +\pi^{-k} & \text{ for } k \equiv 1 \pmod 4\\ -\pi^{-k} & \text{ for } k \equiv 3 \pmod 4\\ 0 & \text{ otherwise } \end{cases}$$

enter image description here

Compare this to the perfect square given by

$$a_k = \begin{cases} +k^{-2} & \text{ for } k \equiv 1 \pmod 4 \\ +k^{-2} & \text{ for } k \equiv 3 \pmod 4\\ 0 & \text{ otherwise } \end{cases}$$

$$b_k = \begin{cases} +k^{-2} & \text{ for } k \equiv 1 \pmod 4\\ -k^{-2} & \text{ for } k \equiv 3 \pmod 4\\ 0 & \text{ otherwise } \end{cases}$$

enter image description here

Note that for bases different from $\pi$ in the formulas for $a_k$ and $b_k$, the rounded square hasn't got so "straight" sides anymore, e.g. for base $2.5$ and $4.5$:

enter image description here

My question is:

Might it be true, that for base $\pi$ in the formulas for $a_k$, $b_k$ above, the resulting shape has "maximally straight" sides? And if so: How to prove it?

First, of course, one has to define what "maximally straight" means.


For the sake of comparison here is the unrounded square with exponents $1.5$ and $2.5$ instead of $2$ in the corresponding formulas. As in the case of the rounded square, $2$ (instead of $\pi$) yields the "maximally straight" sides.

enter image description here

$\endgroup$
  • 4
    $\begingroup$ I guess maximally straight could mean that $\pi$ is the largest base such that the resulting shape is convex. Whether that's true I don't know $\endgroup$ – Bananach Apr 6 at 8:20
  • 2
    $\begingroup$ @Banach: I guess you mean the smallest base such that the resulting shape is convex. $\endgroup$ – Hans-Peter Stricker Apr 6 at 8:22
  • 1
    $\begingroup$ Yes, I do. And since we are only interested in the sign of the curvature, which is equal to that of $x'y''-x''y'$, this might actually be feasible $\endgroup$ – Bananach Apr 6 at 8:48
  • 1
    $\begingroup$ @Banach: Do you plan to give it a try? $\endgroup$ – Hans-Peter Stricker Apr 6 at 8:50
  • 3
    $\begingroup$ It looks like $\pi$ doesn't give a convex shape, one can visually check that the line $y=x+0.40065$ cuts the shape in four points, so the shape's "sides" bend inwards. Following Bananach's approach and setting the curvature $K(a,t)$ to zero at $t=\pi/4$, and if WolframAlpha is not mistaken, it seems that the minimum such number is actually $a=\sqrt{2+2\sqrt{3}+\sqrt{15+8\sqrt{3}}} = 3.291795...$, the greatest positive solution to $a^{10}-9a^8-22a^6+22a^4+9a^2-1=0$. $\endgroup$ – pregunton Apr 6 at 14:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.