How to solve $cos(2^n x) = cos(x)$












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I am given the function $F:mathbb{R}tomathbb{R}$, $xmapsto 4x(1-x)$ and I have to find all $n$-cycles of the function. I have already reduced this to the problem of solving $cos(2^n x)=cos(x)$ for $xin[0,pi]$. Using Mathematica and $nin{1,2,3,4}$, I concluded that the solutions must be ${frac{2kpi}{2^n-1},frac{2kpi}{2^n+1}}cap[0,pi]$. Since the equation obviously has $2^n$ solutions and the solution set has $2^n$ elements, it only remains to verify that they are indeed solutions to the equation. I tried using $cos(2x)=2cos^2(x)-1$ and proving this by induction on $n$, but I was not able to. Thank you for any help with this problem!






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    I am given the function $F:mathbb{R}tomathbb{R}$, $xmapsto 4x(1-x)$ and I have to find all $n$-cycles of the function. I have already reduced this to the problem of solving $cos(2^n x)=cos(x)$ for $xin[0,pi]$. Using Mathematica and $nin{1,2,3,4}$, I concluded that the solutions must be ${frac{2kpi}{2^n-1},frac{2kpi}{2^n+1}}cap[0,pi]$. Since the equation obviously has $2^n$ solutions and the solution set has $2^n$ elements, it only remains to verify that they are indeed solutions to the equation. I tried using $cos(2x)=2cos^2(x)-1$ and proving this by induction on $n$, but I was not able to. Thank you for any help with this problem!






    trigonometry







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      I am given the function $F:mathbb{R}tomathbb{R}$, $xmapsto 4x(1-x)$ and I have to find all $n$-cycles of the function. I have already reduced this to the problem of solving $cos(2^n x)=cos(x)$ for $xin[0,pi]$. Using Mathematica and $nin{1,2,3,4}$, I concluded that the solutions must be ${frac{2kpi}{2^n-1},frac{2kpi}{2^n+1}}cap[0,pi]$. Since the equation obviously has $2^n$ solutions and the solution set has $2^n$ elements, it only remains to verify that they are indeed solutions to the equation. I tried using $cos(2x)=2cos^2(x)-1$ and proving this by induction on $n$, but I was not able to. Thank you for any help with this problem!






      trigonometry







      share|cite|improve this question











      $endgroup$




      I am given the function $F:mathbb{R}tomathbb{R}$, $xmapsto 4x(1-x)$ and I have to find all $n$-cycles of the function. I have already reduced this to the problem of solving $cos(2^n x)=cos(x)$ for $xin[0,pi]$. Using Mathematica and $nin{1,2,3,4}$, I concluded that the solutions must be ${frac{2kpi}{2^n-1},frac{2kpi}{2^n+1}}cap[0,pi]$. Since the equation obviously has $2^n$ solutions and the solution set has $2^n$ elements, it only remains to verify that they are indeed solutions to the equation. I tried using $cos(2x)=2cos^2(x)-1$ and proving this by induction on $n$, but I was not able to. Thank you for any help with this problem!






      trigonometry




      trigonometry






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      edited Jan 17 at 7:09







      Analysis801

















      asked Jan 17 at 6:51









      Analysis801Analysis801

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          With only basic results on trigonometric equations:
          begin{align}cos(2^nx)=cos x &iff 2^nxequiv pm xmod 2piiffbegin{cases}(2^n-1)xequiv 0mod 2pi\(2^n+1)xequiv 0mod 2piend{cases}\[1ex]
          &iff xequiv 0mod frac{2pi}{2^n-1},bmod frac{2pi}{2^n+1}.
          end{align}






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            Using the identity $$cos A - cos B = 2sinleft(dfrac{B-A}{2}right)sinleft(dfrac{B+A}{2}right),$$ we have $$0 = cos(x)-cos(2^nx) = 2sinleft(dfrac{2^n-1}{2}xright)sinleft(dfrac{2^n+1}{2}xright),$$ which holds iff $$sinleft(dfrac{2^n-1}{2}xright) = 0 text{OR} sinleft(dfrac{2^n+1}{2}xright) = 0$$ $$dfrac{2^n-1}{2}x = kpi text{OR} dfrac{2^n+1}{2}x = kpi$$ $$x = dfrac{2kpi}{2^n-1} text{OR} x = dfrac{2kpi}{2^n+1}$$ for some $k in mathbb{Z}$.






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              2 Answers
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              2 Answers
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              $begingroup$

              With only basic results on trigonometric equations:
              begin{align}cos(2^nx)=cos x &iff 2^nxequiv pm xmod 2piiffbegin{cases}(2^n-1)xequiv 0mod 2pi\(2^n+1)xequiv 0mod 2piend{cases}\[1ex]
              &iff xequiv 0mod frac{2pi}{2^n-1},bmod frac{2pi}{2^n+1}.
              end{align}






              share|cite|improve this answer









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                1












                $begingroup$

                With only basic results on trigonometric equations:
                begin{align}cos(2^nx)=cos x &iff 2^nxequiv pm xmod 2piiffbegin{cases}(2^n-1)xequiv 0mod 2pi\(2^n+1)xequiv 0mod 2piend{cases}\[1ex]
                &iff xequiv 0mod frac{2pi}{2^n-1},bmod frac{2pi}{2^n+1}.
                end{align}






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  With only basic results on trigonometric equations:
                  begin{align}cos(2^nx)=cos x &iff 2^nxequiv pm xmod 2piiffbegin{cases}(2^n-1)xequiv 0mod 2pi\(2^n+1)xequiv 0mod 2piend{cases}\[1ex]
                  &iff xequiv 0mod frac{2pi}{2^n-1},bmod frac{2pi}{2^n+1}.
                  end{align}






                  share|cite|improve this answer









                  $endgroup$



                  With only basic results on trigonometric equations:
                  begin{align}cos(2^nx)=cos x &iff 2^nxequiv pm xmod 2piiffbegin{cases}(2^n-1)xequiv 0mod 2pi\(2^n+1)xequiv 0mod 2piend{cases}\[1ex]
                  &iff xequiv 0mod frac{2pi}{2^n-1},bmod frac{2pi}{2^n+1}.
                  end{align}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 17 at 9:23









                  BernardBernard

                  121k740116




                  121k740116























                      5












                      $begingroup$

                      Using the identity $$cos A - cos B = 2sinleft(dfrac{B-A}{2}right)sinleft(dfrac{B+A}{2}right),$$ we have $$0 = cos(x)-cos(2^nx) = 2sinleft(dfrac{2^n-1}{2}xright)sinleft(dfrac{2^n+1}{2}xright),$$ which holds iff $$sinleft(dfrac{2^n-1}{2}xright) = 0 text{OR} sinleft(dfrac{2^n+1}{2}xright) = 0$$ $$dfrac{2^n-1}{2}x = kpi text{OR} dfrac{2^n+1}{2}x = kpi$$ $$x = dfrac{2kpi}{2^n-1} text{OR} x = dfrac{2kpi}{2^n+1}$$ for some $k in mathbb{Z}$.






                      share|cite|improve this answer









                      $endgroup$


















                        5












                        $begingroup$

                        Using the identity $$cos A - cos B = 2sinleft(dfrac{B-A}{2}right)sinleft(dfrac{B+A}{2}right),$$ we have $$0 = cos(x)-cos(2^nx) = 2sinleft(dfrac{2^n-1}{2}xright)sinleft(dfrac{2^n+1}{2}xright),$$ which holds iff $$sinleft(dfrac{2^n-1}{2}xright) = 0 text{OR} sinleft(dfrac{2^n+1}{2}xright) = 0$$ $$dfrac{2^n-1}{2}x = kpi text{OR} dfrac{2^n+1}{2}x = kpi$$ $$x = dfrac{2kpi}{2^n-1} text{OR} x = dfrac{2kpi}{2^n+1}$$ for some $k in mathbb{Z}$.






                        share|cite|improve this answer









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                          5












                          5








                          5





                          $begingroup$

                          Using the identity $$cos A - cos B = 2sinleft(dfrac{B-A}{2}right)sinleft(dfrac{B+A}{2}right),$$ we have $$0 = cos(x)-cos(2^nx) = 2sinleft(dfrac{2^n-1}{2}xright)sinleft(dfrac{2^n+1}{2}xright),$$ which holds iff $$sinleft(dfrac{2^n-1}{2}xright) = 0 text{OR} sinleft(dfrac{2^n+1}{2}xright) = 0$$ $$dfrac{2^n-1}{2}x = kpi text{OR} dfrac{2^n+1}{2}x = kpi$$ $$x = dfrac{2kpi}{2^n-1} text{OR} x = dfrac{2kpi}{2^n+1}$$ for some $k in mathbb{Z}$.






                          share|cite|improve this answer









                          $endgroup$



                          Using the identity $$cos A - cos B = 2sinleft(dfrac{B-A}{2}right)sinleft(dfrac{B+A}{2}right),$$ we have $$0 = cos(x)-cos(2^nx) = 2sinleft(dfrac{2^n-1}{2}xright)sinleft(dfrac{2^n+1}{2}xright),$$ which holds iff $$sinleft(dfrac{2^n-1}{2}xright) = 0 text{OR} sinleft(dfrac{2^n+1}{2}xright) = 0$$ $$dfrac{2^n-1}{2}x = kpi text{OR} dfrac{2^n+1}{2}x = kpi$$ $$x = dfrac{2kpi}{2^n-1} text{OR} x = dfrac{2kpi}{2^n+1}$$ for some $k in mathbb{Z}$.







                          share|cite|improve this answer












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                          answered Jan 17 at 7:16









                          JimmyK4542JimmyK4542

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                          41.1k245107






























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