Why is the magnitude of the sum of two adjacent nth roots always an 'interesting' number, and what do these...












4












$begingroup$


While doing something completely unrelated, I discovered an interesting function:



$$f(x)=2leftvertcos{frac{pi}{x}}rightvert$$



Which gives the absolute value of the sum of any two adjacent $x^text{th}$-roots of unity for $xinmathbb{R}$.



This function seemed interesting to me because, for whatever reason, integer (and certain rational) values of $x$ seem only to yield 'interesting' numbers.



For instance:



$f(5/2)=Phi$ (golden ratio conjugate)
, $f(4)=sqrt{2}$, $f(5)=phi$ (golden ratio),



$f(7)=sqrt{mathcal{S}}$ (squareroot of the silver constant),



$f(9/7)=frac{1}{mathcal{P}_c(6^3)}$ ($mathcal{P}_cleft(6^3right)$ is the bond percolation threshold for a honeycomb lattice),



$f(9)=frac{1}{mathcal{P}_c(3^6)}-1$ ($mathcal{P}_cleft(3^6right)$ is the bond percolation threshold for a triangular lattice)



...and so on.



I found out shortly before writing this that these are the square roots of Beraha numbers, but I have no idea what that means or what it has to do with roots of unity, so the question hasn't changed.



Why are these numbers showing up? And what do Beraha numbers, roots of unity, lattices, and algebraic roots have to do with each other?




Note: if $x$ is rational, then $f(x)$ is algebraic. I can neither confirm nor deny that $f(x)$ is algebraic if $x$ is algebraic. In any case, transcendental numbers don't show up here. It is also easy to show that the sum of any two $x^text{th}$ roots of unity are algebraic if $x$ is rational.



Note: two $x^text{th}$-roots of unity are 'adjacent' if the distance between them is minimal. i.e. $z_1,z_2=sqrt[x]{1}$ are adjacent if there is no $z_3=sqrt[x]{1}$ such that $d(z_1,z_2)>d(z_1,z_3)$ or $d(z_1,z_2)>d(z_2,z_3)$



If you count the roots of unity by 'stepping' clockwise or counterclockwise around the unit circle, then the magnitude of the sum of two roots of unity separated by $n$ steps is given by the function:
$$f(x)=2leftvertcos{frac{pi n}{x}}rightvert$$






number-theory complex-numbers soft-question roots-of-unity







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    While doing something completely unrelated, I discovered an interesting function:



    $$f(x)=2leftvertcos{frac{pi}{x}}rightvert$$



    Which gives the absolute value of the sum of any two adjacent $x^text{th}$-roots of unity for $xinmathbb{R}$.



    This function seemed interesting to me because, for whatever reason, integer (and certain rational) values of $x$ seem only to yield 'interesting' numbers.



    For instance:



    $f(5/2)=Phi$ (golden ratio conjugate)
    , $f(4)=sqrt{2}$, $f(5)=phi$ (golden ratio),



    $f(7)=sqrt{mathcal{S}}$ (squareroot of the silver constant),



    $f(9/7)=frac{1}{mathcal{P}_c(6^3)}$ ($mathcal{P}_cleft(6^3right)$ is the bond percolation threshold for a honeycomb lattice),



    $f(9)=frac{1}{mathcal{P}_c(3^6)}-1$ ($mathcal{P}_cleft(3^6right)$ is the bond percolation threshold for a triangular lattice)



    ...and so on.



    I found out shortly before writing this that these are the square roots of Beraha numbers, but I have no idea what that means or what it has to do with roots of unity, so the question hasn't changed.



    Why are these numbers showing up? And what do Beraha numbers, roots of unity, lattices, and algebraic roots have to do with each other?




    Note: if $x$ is rational, then $f(x)$ is algebraic. I can neither confirm nor deny that $f(x)$ is algebraic if $x$ is algebraic. In any case, transcendental numbers don't show up here. It is also easy to show that the sum of any two $x^text{th}$ roots of unity are algebraic if $x$ is rational.



    Note: two $x^text{th}$-roots of unity are 'adjacent' if the distance between them is minimal. i.e. $z_1,z_2=sqrt[x]{1}$ are adjacent if there is no $z_3=sqrt[x]{1}$ such that $d(z_1,z_2)>d(z_1,z_3)$ or $d(z_1,z_2)>d(z_2,z_3)$



    If you count the roots of unity by 'stepping' clockwise or counterclockwise around the unit circle, then the magnitude of the sum of two roots of unity separated by $n$ steps is given by the function:
    $$f(x)=2leftvertcos{frac{pi n}{x}}rightvert$$






    number-theory complex-numbers soft-question roots-of-unity







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      2



      $begingroup$


      While doing something completely unrelated, I discovered an interesting function:



      $$f(x)=2leftvertcos{frac{pi}{x}}rightvert$$



      Which gives the absolute value of the sum of any two adjacent $x^text{th}$-roots of unity for $xinmathbb{R}$.



      This function seemed interesting to me because, for whatever reason, integer (and certain rational) values of $x$ seem only to yield 'interesting' numbers.



      For instance:



      $f(5/2)=Phi$ (golden ratio conjugate)
      , $f(4)=sqrt{2}$, $f(5)=phi$ (golden ratio),



      $f(7)=sqrt{mathcal{S}}$ (squareroot of the silver constant),



      $f(9/7)=frac{1}{mathcal{P}_c(6^3)}$ ($mathcal{P}_cleft(6^3right)$ is the bond percolation threshold for a honeycomb lattice),



      $f(9)=frac{1}{mathcal{P}_c(3^6)}-1$ ($mathcal{P}_cleft(3^6right)$ is the bond percolation threshold for a triangular lattice)



      ...and so on.



      I found out shortly before writing this that these are the square roots of Beraha numbers, but I have no idea what that means or what it has to do with roots of unity, so the question hasn't changed.



      Why are these numbers showing up? And what do Beraha numbers, roots of unity, lattices, and algebraic roots have to do with each other?




      Note: if $x$ is rational, then $f(x)$ is algebraic. I can neither confirm nor deny that $f(x)$ is algebraic if $x$ is algebraic. In any case, transcendental numbers don't show up here. It is also easy to show that the sum of any two $x^text{th}$ roots of unity are algebraic if $x$ is rational.



      Note: two $x^text{th}$-roots of unity are 'adjacent' if the distance between them is minimal. i.e. $z_1,z_2=sqrt[x]{1}$ are adjacent if there is no $z_3=sqrt[x]{1}$ such that $d(z_1,z_2)>d(z_1,z_3)$ or $d(z_1,z_2)>d(z_2,z_3)$



      If you count the roots of unity by 'stepping' clockwise or counterclockwise around the unit circle, then the magnitude of the sum of two roots of unity separated by $n$ steps is given by the function:
      $$f(x)=2leftvertcos{frac{pi n}{x}}rightvert$$






      number-theory complex-numbers soft-question roots-of-unity







      share|cite|improve this question











      $endgroup$




      While doing something completely unrelated, I discovered an interesting function:



      $$f(x)=2leftvertcos{frac{pi}{x}}rightvert$$



      Which gives the absolute value of the sum of any two adjacent $x^text{th}$-roots of unity for $xinmathbb{R}$.



      This function seemed interesting to me because, for whatever reason, integer (and certain rational) values of $x$ seem only to yield 'interesting' numbers.



      For instance:



      $f(5/2)=Phi$ (golden ratio conjugate)
      , $f(4)=sqrt{2}$, $f(5)=phi$ (golden ratio),



      $f(7)=sqrt{mathcal{S}}$ (squareroot of the silver constant),



      $f(9/7)=frac{1}{mathcal{P}_c(6^3)}$ ($mathcal{P}_cleft(6^3right)$ is the bond percolation threshold for a honeycomb lattice),



      $f(9)=frac{1}{mathcal{P}_c(3^6)}-1$ ($mathcal{P}_cleft(3^6right)$ is the bond percolation threshold for a triangular lattice)



      ...and so on.



      I found out shortly before writing this that these are the square roots of Beraha numbers, but I have no idea what that means or what it has to do with roots of unity, so the question hasn't changed.



      Why are these numbers showing up? And what do Beraha numbers, roots of unity, lattices, and algebraic roots have to do with each other?




      Note: if $x$ is rational, then $f(x)$ is algebraic. I can neither confirm nor deny that $f(x)$ is algebraic if $x$ is algebraic. In any case, transcendental numbers don't show up here. It is also easy to show that the sum of any two $x^text{th}$ roots of unity are algebraic if $x$ is rational.



      Note: two $x^text{th}$-roots of unity are 'adjacent' if the distance between them is minimal. i.e. $z_1,z_2=sqrt[x]{1}$ are adjacent if there is no $z_3=sqrt[x]{1}$ such that $d(z_1,z_2)>d(z_1,z_3)$ or $d(z_1,z_2)>d(z_2,z_3)$



      If you count the roots of unity by 'stepping' clockwise or counterclockwise around the unit circle, then the magnitude of the sum of two roots of unity separated by $n$ steps is given by the function:
      $$f(x)=2leftvertcos{frac{pi n}{x}}rightvert$$






      number-theory complex-numbers soft-question roots-of-unity




      number-theory complex-numbers soft-question roots-of-unity






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 18 at 18:19







      R. Burton

















      asked Jan 18 at 2:28









      R. BurtonR. Burton

      612110




      612110






















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          $begingroup$

          Before all these $$f(x)=sqrt{4cos^2frac{pi}{x}}=2left|cosfrac{pi}{x}right|.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. I've added this to the question, but I'm not sure how this connects to the aforementioned lattices and Beraha numbers.
            $endgroup$
            – R. Burton
            Jan 18 at 18:20











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          $begingroup$

          Before all these $$f(x)=sqrt{4cos^2frac{pi}{x}}=2left|cosfrac{pi}{x}right|.$$






          share|cite|improve this answer









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          • $begingroup$
            Thanks. I've added this to the question, but I'm not sure how this connects to the aforementioned lattices and Beraha numbers.
            $endgroup$
            – R. Burton
            Jan 18 at 18:20
















          3












          $begingroup$

          Before all these $$f(x)=sqrt{4cos^2frac{pi}{x}}=2left|cosfrac{pi}{x}right|.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. I've added this to the question, but I'm not sure how this connects to the aforementioned lattices and Beraha numbers.
            $endgroup$
            – R. Burton
            Jan 18 at 18:20














          3












          3








          3





          $begingroup$

          Before all these $$f(x)=sqrt{4cos^2frac{pi}{x}}=2left|cosfrac{pi}{x}right|.$$






          share|cite|improve this answer









          $endgroup$



          Before all these $$f(x)=sqrt{4cos^2frac{pi}{x}}=2left|cosfrac{pi}{x}right|.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 18 at 7:18









          Michael RozenbergMichael Rozenberg

          105k1892198




          105k1892198












          • $begingroup$
            Thanks. I've added this to the question, but I'm not sure how this connects to the aforementioned lattices and Beraha numbers.
            $endgroup$
            – R. Burton
            Jan 18 at 18:20


















          • $begingroup$
            Thanks. I've added this to the question, but I'm not sure how this connects to the aforementioned lattices and Beraha numbers.
            $endgroup$
            – R. Burton
            Jan 18 at 18:20
















          $begingroup$
          Thanks. I've added this to the question, but I'm not sure how this connects to the aforementioned lattices and Beraha numbers.
          $endgroup$
          – R. Burton
          Jan 18 at 18:20




          $begingroup$
          Thanks. I've added this to the question, but I'm not sure how this connects to the aforementioned lattices and Beraha numbers.
          $endgroup$
          – R. Burton
          Jan 18 at 18:20


















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