In my physics research I came across a mathematical proposition (translated into the mathematical language from the physical problem) that I feel to be true, and would like to prove it:

Proposition: Consider the function $f(m)=\cos(\frac{2m+1}{N}\pi)$ where the integer $m$ satisfies $0\leq m<N$ and the integer $N>2$. The situation $f(m_1)-f(m_2)=f(m_2)-f(m_3)$ while $f(m_1)>f(m_2)>f(m_3)$ can happen if and only if $\frac{2m_2+1}{N}=\frac{1}{2}$ or $\frac{2m_2+1}{N}=\frac{3}{2}$.

The if part is very simple. However I have not been able to prove the only if part. Without loss of generality it should be sufficient to consider only the case where $\frac{2m_i+1}{N}\pi\leq \pi$ because the cosine function is symmetric w.r.t $\pi$. If I restrict $m_1-m_2=m_2-m_3$ then I can show the proposition is true only if $\frac{2m_2+1}{N}=\frac{1}{2}$. Without the restriction however I'm having no luck.

The proposition probably relies on the fact that all the $m$'s and $N$ are integers. One idea I have tried is to notice that if $\theta=\frac{2m+1}{N}$ then $\cos(N\theta)=-1$ and there is a formula to expand $\cos(N\theta)$ into a polynomial $p_N(\cos(\theta))$ of $\cos(\theta)$, hence $f(m_1)$, $f(m_2)$ and $f(m_3)$ are all roots of the equation $p_N(\cos(\theta))$+1=0. But I haven't been able to proceed from there.

Of course the proposition may not be true after all despite my intuition, but if then it'd be nice to find out for which $N$ it fails. It certainly is true for $N\leq 8$ since I have calculated all solutions. Either way the proposition would have real physical consequences.

  • 1
    $\begingroup$ @MarkSapir so what? $\endgroup$ – Fedor Petrov Apr 14 '17 at 19:05
  • 1
    $\begingroup$ Yes, but OP asks about the case when $m_1-m_2\ne m_2-m_3$ $\endgroup$ – Fedor Petrov Apr 14 '17 at 19:11
  • $\begingroup$ This is certainly true if $N$ is prime, or more generally if $\phi(N)>\frac 56N$, where $\phi$ is the Euler phi-function, since a relation of the type you're interested in gives rise to a polynomial with integer coefficients satisfied by $e^{2\pi i/N}$ of degree at most $5N/6$. $\endgroup$ – Anthony Quas Apr 14 '17 at 23:26
  • $\begingroup$ @FedorPetrov: You are right! $\endgroup$ – user6976 Apr 15 '17 at 1:39
  • $\begingroup$ @AnthonyQuas After some learning in the Euler phi-function, I cannot see how it is related to the current problem in the way Anthony proposed, could you please elaborate? Thank you very much! $\endgroup$ – Andrew Apr 15 '17 at 13:52

Let $\zeta=e^{2\pi i/N}$ and $\alpha=e^{2\pi i/(2N)}$. Your assumption was $\Re(\alpha(\zeta^{m_3}-2\zeta^{m_2}+\zeta^{m_1}))=0$. This is equivalent to $\alpha(\zeta^{m_3}-2\zeta^{m_2}+\zeta^{m_1}))+\bar\alpha(\zeta^{-m_3}-2\zeta^{-m_2}+\zeta^{-m_1})=0$. Multiplying by $\bar\alpha$ and using the fact that $\alpha^2=\zeta$, you get $\zeta^{m_3}-2\zeta^{m_2}+\zeta^{m_1}+\zeta^{N-m_3-1}-2\zeta^{N-m_2-1}+\zeta^{N-m_1-1}=0$. You can multiply this equation by any power of $\zeta$ that you want (again using $\zeta^N=1$) to ensure that any term you want becomes the constant term. Since there are 6 powers of $\zeta$ arranged around the circle, if you set the power that immediately follows the longest gap to be the constant term, the polynomial you arrive at has degree at most $5N/6$.

On the other hand, the minimal polynomial of $\zeta$ is $\Phi_N$, the $N$th cyclotomic polynomial, which has degree $\phi(N)$. The polynomial that we arrived at above must be a multiple of $\Phi_N$. This is impossible if $\phi(N)>5N/6$ (which is true if $N$ is prime, and for infinitely many other numbers).

  • $\begingroup$ Do you mean "which is true if $N$ is prime and $N>5\,$"? I think that $\phi(5)=4<\frac{5}{6}\cdot 5$. $\endgroup$ – Mikhail Borovoi Apr 15 '17 at 20:27
  • $\begingroup$ I suppose so! Since the OP already dealt with $N\le 8$, I wasn't too worried about that case. $\endgroup$ – Anthony Quas Apr 15 '17 at 20:45
  • $\begingroup$ I am stuck at the last step: " if you set the power that immediately follows the longest gap to be the constant term, the polynomial you arrive at has degree at most 5N/6." Otherwise this is a very nice proof for the numbers that satisfy $\phi(N)>5N/6$. I will think more along this line to see if there can be a proof or counter proof for non-primes. $\endgroup$ – Andrew Apr 15 '17 at 22:15
  • 1
    $\begingroup$ Now that's been reduced to a sum of eight roots of unity, it suffices to apply theorem 6 of Conway and Jones, Acta Arith. XXX (1976) 229-240 to find all solutions. $\endgroup$ – Felipe Voloch Apr 15 '17 at 23:19
  • $\begingroup$ I'll illustrate my answer by an example. Let $N=67$ and imagine that $\zeta^6 - 2\zeta^{13}+\zeta^{18}+\zeta^{60}-2\zeta^{53}+\zeta^{48}=0$. There are non-zero coefficients in the 6th, 13th, 18th, 48th, 53rd and 60th powers of $\zeta$. The gap between 60th and 6th powers is 13 (going around the circle), which is smaller than the gap between the 18th and 48th. , so I'll multiply the entire equation by $\zeta^{-48}=\zeta^{19}$ to obtain $1-2\zeta^5+\zeta^{12}+\zeta^{25}-2\zeta^{32}+\zeta^{37}=0$. The degree of this polynomial is less than $5N/6$. $\endgroup$ – Anthony Quas Apr 16 '17 at 3:11

This proof uses the reference suggested by Felipe Voloch in a comment below the answer provided by Anthony Quas:

In Conway and Jones, Acta Arith. XXX (1976) 229-240, Theorem 7 (abbreviated to CJ7 in the following discussions) states that:

Suppose we have at most four distinct rational multiples of $\pi$ lying strictly between 0 and $\pi/2$ for which some rational linear combination of their cosines is rational but no proper subset has this property. Then the appropriate linear combination is proportional to one from the following list: $\cos \pi/3 =1/2$, $-\cos \phi+\cos(\pi/3-\phi)+\cos(\pi/3+\phi)=0$, $\cos \pi/5-\cos 2\pi/5=1/2$, $\cos \pi/7-\cos 2\pi/7+\cos 3\pi/7=1/2$, $\cos \pi/5-\cos \pi/15+\cos 4\pi/15=1/2$, $-\cos 2\pi/5+\cos 2\pi/15-\cos 7\pi/15=1/2$, $\cos \pi/7+\cos 3\pi/7-\cos \pi/21+\cos 8\pi/21=1/2$, $\cos \pi/7-\cos 2\pi/7+\cos 2\pi/21-\cos 5\pi/21=1/2$, $-\cos 2\pi/7+\cos 3\pi/7+\cos 4\pi/21+\cos 10\pi/21=1/2$, $-\cos \pi/15+\cos 2\pi/15+\cos 4\pi/15-\cos 7\pi/15=1/2$,

In the current problem we want to find the conditions for $\cos(\frac{2m_1+1}{N}\pi)+\cos(\frac{2m_3+1}{N}\pi)-2\cos(\frac{2m_2+1}{N}\pi)=0$ which for the moment can be written as $\cos\theta_1+\cos\theta_3-2\cos\theta_2=0$, where $0<\theta_i\leq \pi$.

Before we can use CJ7, we need to first consider the case when $\theta_1=\pi$ or $\theta_2=\pi$. If $\theta_1=\pi$, $\cos\theta_3-2\cos\theta_2=1$, which is not possible by CJ7 if none of $\theta_2$ and $\theta_3$ is $\pi/2$. If indeed we allow $\theta_3=\pi/2$ and $\theta_2=2\pi/3$ the equation is satisfied. However in our original problem $\theta_i=\frac{2m_i+1}{N}\pi$, and there is no integers of $N$ and $m_i$ that can make both $\theta_3=\pi/2$ and $\theta_1=\pi$, so this is not an option. On the other hand if If $\theta_2=\pi$, $\cos\theta_1+\cos\theta_3=-2$, simply impossible.

So now we consider the case none of the three angles is $\pi$. Again we cannot use CJ7 unless we first exclude the case when one of them is $\pi/2$. Suppose $\theta_1=\pi/2$ then $\cos\theta_3-2\cos\theta_2=0$ which is not possible by CJ7. On the other hand if $\theta_2=\pi/2$ then $\cos\theta_1+\cos\theta_3=0$, which is impossible by CJ7 if $\theta_1$ and $\theta_3$ are distinct as defined in CJ7. If $\theta_1$ and $\theta_3$ are allowed to be antisymmetric pairs w.r.t. $\pi/2$ CJ7 doesn't apply and the sum is 0. Note that this case is just the one we originally proposed.

Now suppose none of the three angles is either $\pi$ or $\pi/2$, then if they are all distinct as defined by CJ7, we know $\cos\theta_1+\cos\theta_3-2\cos\theta_2$ cannot be rational, and hence cannot be 0. If two of them are antisymmetric pairs w.r.t. $\pi/2$ then either $-2\cos\theta_2=0$ or $\cos\theta_1+3\cos\theta_3=0$, which we know are impossible by CJ7.

To summary, for all the cases where CJ7 does not apply, we have shown that the equality to zero is satisfied if and only if $\theta_2=\pi/2$ and $\theta_1$ and $\theta_3$ are antisymmetric pairs w.r.t. $\pi/2$. For the cases where CJ7 applies, we have shown that the equality to zero is never possible. Hence our original proposition is true.

  • $\begingroup$ Does accepting an answer prevent people from commenting further? I will un-accept it then because I'd like it to be discussed and criticized. $\endgroup$ – Andrew Apr 19 '17 at 18:16
  • $\begingroup$ Accepting an answer doesn't stop anyone from doing anything. It may discourage some people from looking at the question, since they figure there's nothing they can add – but it may encourage some people to look at the question, since they figure they can learn something. Anyway, accepting an answer is generally considered a good thing to do, once OP finds an answer acceptable. $\endgroup$ – Gerry Myerson Apr 19 '17 at 23:36
  • 1
    $\begingroup$ @GerryMyerson ok I see! Yea since it is my own answer so I obviously think it is a good answer, but I figure maybe I should wait for people to criticize it a little bit before I accept it. If by the end of tomorrow nobody finds anything wrong about it I'll accept it. $\endgroup$ – Andrew Apr 20 '17 at 1:10

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.