Given $N$ pairs of distinct real numbers $t_i, t'_i \in [0,1]$, $i = 1,\ldots,N$, we ask if there is a function $f(x) = \cos(2\pi mx+\alpha) + \gamma\cdot \cos(2\pi nx+\beta)$, with $m, n \in \mathbb{N}$, $\alpha, \beta, \gamma \in \mathbb{R}$, so that for all $i$: $f(t_i) > f(t'_i)$?

If yes, this proves that all knots are Fourier-(1,1,2) knots, that is, possess a parametrization with Fourier series of length 1,1,2 in coordinates $x,y,z$.

Because: a) the parametrization in the x-y plane is rich enough to generate all knots (see Lamm, 1998) and b) by interchanging $t_i$ and $t'_i$ every crossing pattern can be achieved.

As references see the articles

http://arxiv.org/abs/q-alg/9711013 (Kauffman, 'Fourier knots', 1997)

http://arxiv.org/abs/1210.4543 (Lamm, 'Fourier knots', 1998)

http://arxiv.org/abs/0707.4210 (Boocher et. al, 'Sampling Lissajous and Fourier knots', 2007)

http://arxiv.org/abs/0708.3590 (Hoste, 'Torus knots are Fourier-(1,1,2) knots', 2007)

If the answer is no, we ask more generally if functions with a bounded number of cosine terms suffice (e.g. with bound 3).

We remark that Fourier-(1,1,1) knots are Lissajous knots. These are too symmetric to yield all knots (see also Tying knots with reflecting lightrays).

Edit1 (12Feb14). We give an example with 4 pairs of distinct numbers for which a single cosine function $\cos(2\pi mx+\alpha)$ does not suffice:

Choose $t_1, t'_1$ and $t_2, t'_2$ arbitrarily (but distinct) in $[0,1]$.

Let $t_3 = t_1 + 0.5$, $t'_3 = t'_1 + 0.5$ and $t_4 = t'_2 + 0.5$, $t'_4 = t_2 + 0.5$ (note the interchanged ' in $t_2$).

We then have:

if $f(t_1) > f(t'_1)$ then for odd $m$: $f(t_3) < f(t'_3)$ and for even $m$: $f(t_4) < f(t'_4)$.

This shows how the symmetries $\cos(x+2\pi) = \cos(x)$ and $\cos(x+\pi) = -\cos(x)$ prevent a solution in a Fourier series of length 1. The same argument applies for a series of length two (or more) if the frequencies are all odd or all even.

Edit2 (09July15).

The article

http://arxiv.org/abs/1507.00880 (Marc Soret, Marina Ville)

solves the Fourier-(1,1,2) knot problem using Kronecker's theorem!


For almost all $(t_1, t_2, \ldots, t_N, t'_1, t'_2, \ldots, t'_N)$, just one Fourier term suffices. As I will explain below, I think this good enough for your goal of showing that all knots are Fourier-$(1,1,2)$.

Claim Suppose that $t_1$, $t_2$, ..., $t_N$, $t'_1$, $t'_2$, ..., $t'_N$ and $\pi$ are linearly independent over $\mathbb{Q}$. Then there is an integer $n$ so that $\sin(n t_i) < \sin(n t'_i)$ for all $i$.

Proof We write $\{ x \}$ for the fractional part $x - \lfloor x \rfloor$.

We use Kronecker's theorem on diophantine approximation to conclude that the fractional parts $(\{n t_1/2\pi \} , \{n t_2/2\pi \}, \ldots, \{n t_N/2\pi \}, \{ n t'_1/2 \pi \}, \ldots, \{n t'_N/2 \pi \})$ are dense in the torus $(\mathbb{R}/\mathbb{Z})^{2N}$. In particularly, there is some $n$ so that $$0 < \{n t_1/2\pi \} < \{n t_2/2\pi \} < \cdots < \{n t_N/2\pi \} < \{n t'_1/2\pi \} < \{n t'_2/2\pi \} < \cdots < \{n t'_N/2\pi \} < 1/4$$ and so $$0 < \sin(n t_1) < \sin(n t_2) < \cdots < \sin(n t_N) < \sin(n t'_1) < \cdots < \sin(n t'_N) < 1. \quad \square$$

I went to look at your first paper on Fourier knots to see whether this is good enough to prove that all knots are $(1,1,1)$. The answer is no, because the checkerboard diagrams you create do not give values of $t_i$ and $t'_i$ which are linearly independent over $\mathbb{Q}$. For example, you make a checker-board for the trefoil by $$x = \cos(2t + 6) \quad y = \cos(3t + 0.15),$$ and the nodes of this checkerboard are at $$(t,t') = (-0.05+\pi/2, -0.05-\pi/2), (-0.05+\pi/2+\pi/3, -0.05-\pi/2+\pi/3), (-0.05+\pi/2+2 \pi/3, -0.05-\pi/2+ 2 \pi/3), (-3+ \pi/3, -3 - \pi/3), (-3+ 2\pi/3, -3 - 2\pi/3), (-3+\pi/3+\pi/2, -3-\pi/3+\pi/2), (-3+2\pi/3+\pi/2, -3-2\pi/3+\pi/2)$$ which are very obviously dependent.

However, suppose you take one of your checkerboard parameterizations $(\cos(jt+\alpha), \cos(kt + \beta))$ and replace it by $(r \cos(jt+\alpha)+\epsilon, s \cos(kt + \beta))$ for some very small transcendental $\epsilon$. This won't change the topology of the checkerboard, and I bet (no proof) it will make the $t$ values of the nodes become linearly independent. So this would show that every knot can be parametrized as $(r \sin(jt+\alpha)+\epsilon, s \sin(kt + \beta), \sin(n t))$. UPDATE This doesn't work, see comments below. I continue to believe some small modification of this strategy could work, but I don't have a concrete proposal at the current time.

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    $\begingroup$ Thank you for your answer. It highlights that a deformation method can be useful because then Kronecker's theorem applies. This was used successfully in the proof of Daniel Pecker that every knot is a billiard knot in an elliptic prism (= deformation of the too symmetric situation of cylinders), see also the MO discussion on Tying knots with reflecting lightrays. Note, however, that the idea of scaling and adding a small constant seems not to work: if the t-values of intersections are calculated by setting equal two such terms, the small constant and the scaling will cancel. $\endgroup$ Feb 11 '14 at 8:02
  • 1
    $\begingroup$ You're right, the small constant isn't good enough. $\endgroup$ Feb 12 '14 at 14:37

If all $t_i$, $t'_i$ are rational numbers with common denominator $d$, then it is enough to examine frequencies $m, n \le d$. After David's answer I decided that it could be worthwhile to look for counterexamples for small $d$ and used an optimization routine to minimize the sum over $\max(0,f(t'_i)-f(t_i))$ for each combination of $m, n$.

Counterexamples seem to exist for 10 or more pairs. I found the following for $d=20$ and $d=23$:




This should be read as $t_1=3/20$, $t'_1=1/20, \ldots$ for the first example.

Open questions:

a) Show that the above examples are indeed counterexamples for Fourier series of length 2.

b) Can you find counterexamples with less than 10 pairs?

c) Study the question for Fourier series of length 3.

I would like to add that in a numerical study a small tolerance should be introduced: e.g. sum over $\max(0,f(t'_i)-f(t_i)+10^{-6})$ to avoid functions which are zero at $t_i$, $t'_i$. These occur because of the identity $\cos(x)-\cos(y)=-2 \sin(\frac{x+y}{2})\sin(\frac{x-y}{2})$ for frequencies with $m+n=d$ and $\gamma=-1$.

Update (6Mar14):

I found potential counterexamples with 8, 7 and 6 pairs:





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