# Lower bounding sum of cosines

Let $$N$$ be an odd integer $$\geq 3$$.

I am interested in finding a lower bound for $$\min \vert{\cos(\frac{2\pi k}{N})}+\cos(\frac{2\pi l}{N})-\cos(\frac{2\pi m}{N})-\cos(\frac{2\pi n}{N})\vert,$$ where $$k,l,m,n \in \{0,\dots, N-1\}$$ and $$\{k,l\}\not\subset \{m,n,N-m, N-n\}$$ (without a comparable assumption the above quantity is trivially zero).

Numerically, one sees (at least for reasonably large $$N\sim 100$$) that this quantity is strictly greater than zero, and that it decreases with increasing $$N$$ (although not in a monotone fashion). Does anybody have an idea of how one could go about lower bounding this quantity? I am also interested in finding an argument for why it is strictly greater than zero.

Thanks!

• Empirically the minima seem to be bounded above by a constant times $N^{-4}$, which is consistent with there being nothing "interesting" going on, given that there are on the order of $N^4$ possible choices of $k, l, m, n$. Jun 7 at 18:00
• @MichaelLugo. Do you have an estimate of the constant? I'd be interested to see whether it corresponds with my hypothesis below.
– mcd
Jun 7 at 18:25
• @mcd The constant seems to be around 1168 or so... which matches your $12\pi^4$. Jun 7 at 18:38
• @MichaelLugo. Thank you.
– mcd
Jun 7 at 18:45

You could think about it geometrically: $$\cos\big(\tfrac{2\pi k}{N}\big)+\cos\big(\tfrac{2\pi l}{N}\big)-\cos\big(\tfrac{2\pi m}{N}\big)-\cos\big(\tfrac{2\pi n}{N}\big)=\Re\big(e^{\frac{2\pi i k}{N}}+e^{\frac{2\pi i l}{N}}-e^{\frac{2\pi i m}{N}}-e^{\frac{2\pi i n}{N}}\big)$$ so you are trying to choose four vertices of a regular $$N$$-gon inscribed in the unit circle and minimise the sum of two $$x$$-coordinates minus the sum of two others (the absolute value is irrelevant as you can always swap your choices of $$\{k, l\}$$ and $$\{m, n\}$$). When $$N$$ is odd, as you require, say $$N=2r+1$$, you can choose, for example $$k=0$$, $$l=r-1$$, $$m=2$$, $$n=r-2$$. Then a lower bound is $$1+\cos\big(\tfrac{2(r-1)\pi}{2r+1}\big)-\cos\big(\tfrac{4\pi}{2r+1}\big)-\cos\big(\tfrac{2(r-2)\pi}{2r+1}\big).$$ Using trig identities, this comes to $$2\sin^2\big(\tfrac{2\pi}{2r+1}\big)-2\sin\big(\tfrac{\pi}{2r+1}\big)\sin\big(\tfrac{4\pi}{2r+1}\big).$$ As $$N \to \infty$$, using the first two terms of the Maclaurin series for $$\sin$$, this is approximately $$\frac{12\pi^4}{N^4}$$.
I think this is the lower bound, but I don't (yet) have a proof of this, though this does seem to be the combination of $$\{k, l, m, n\}$$ where the quadratic Maclaurin terms vanish and the quartic ones have minimum coefficient, so gives the minimum asymptotically at least.