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](https://mathoverflow.net/questions/38813/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!



Are Fourier series of length 2 'asymmetric enough' to generate all crossing patterns? - A reformulation of the Fourier-(1,1,2) knot question
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 $d=23$:


[3,13,2,12,16,18,8,9,10,19;1,17,14,7,4,15,20,1,6,5]/20


[7,9,5,4,14,16,10,20,6,15;17,19,8,12,3,1,18,2,11,13]/20


[9,8,12,19,1,16,18,10,5,17;7,13,4,20,6,21,2,14,11,15]/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: 


[10,16,11,3,9,20,15,6;12,4,19,5,1,2,13,8]/20,


[12,5,10,14,4,11,7;8,1,3,6,2,9,13]/14,


[10,11,3,2,1,7;8,9,4,5,6,12]/12.


