Euclidean division exercise












3














I faced the following problem in a previous abstract algebra session in my university:
Let $omega$ be a non-zero real number and n be a non-zero natural integer both supposed to be fixed. Calculate the remainder of the Euclidean division of the polynomial $(cos{omega}+Xsin{omega}) ^n$ by $X^2 +1$.



I tried to expand $(cos{omega}+Xsin{omega}) ^n$ but it didn't look helpful for me. Can anyone help?






abstract-algebra polynomials







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    3














    I faced the following problem in a previous abstract algebra session in my university:
    Let $omega$ be a non-zero real number and n be a non-zero natural integer both supposed to be fixed. Calculate the remainder of the Euclidean division of the polynomial $(cos{omega}+Xsin{omega}) ^n$ by $X^2 +1$.



    I tried to expand $(cos{omega}+Xsin{omega}) ^n$ but it didn't look helpful for me. Can anyone help?






    abstract-algebra polynomials







    share|cite|improve this question



























      3












      3








      3







      I faced the following problem in a previous abstract algebra session in my university:
      Let $omega$ be a non-zero real number and n be a non-zero natural integer both supposed to be fixed. Calculate the remainder of the Euclidean division of the polynomial $(cos{omega}+Xsin{omega}) ^n$ by $X^2 +1$.



      I tried to expand $(cos{omega}+Xsin{omega}) ^n$ but it didn't look helpful for me. Can anyone help?






      abstract-algebra polynomials







      share|cite|improve this question















      I faced the following problem in a previous abstract algebra session in my university:
      Let $omega$ be a non-zero real number and n be a non-zero natural integer both supposed to be fixed. Calculate the remainder of the Euclidean division of the polynomial $(cos{omega}+Xsin{omega}) ^n$ by $X^2 +1$.



      I tried to expand $(cos{omega}+Xsin{omega}) ^n$ but it didn't look helpful for me. Can anyone help?






      abstract-algebra polynomials




      abstract-algebra polynomials






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      share|cite|improve this question













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      edited 3 hours ago









      José Carlos Santos

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      150k22120221










      asked 4 hours ago









      user7857462

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          2 Answers
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          Write the reminder $aX+b$. You have



          $$P(X) = (cos omega + sinomega X)^n =Q(X)(X^2+1) +aX+b.$$



          Substitute in this equation $X$ by $i$. You get
          $$e^{in omega} = ai+b.$$ Substitute now $X$ by $-i$. You get



          $$e^{-in omega} = -ai+b$$



          Solving in $a,b$ you finally get $b =cos nomega$ and $a = sin nomega$.






          share|cite|improve this answer





























            2














            This is essentially the same answer as mathcounterexamples.net's answer, but it's a different explanation, since frankly I don't follow several aspects of mathcounterexamples.net's thought process (though I did upvote it). Hence I'll demonstrate my thought process on seeing this question.



            We can calculate the remainder on division by a polynomial $fin k[x]$ by working in the ring $k[x]/(f)$, since the remainder of a polynomial $a$ on division by $f$ is the unique representative of $a$ in $k[x]/(f)$ of degree less than the degree of $f$.



            Then recognize that $Bbb{R}[X]/(X^2+1)cong Bbb{C}$ via $Xmapsto i$.
            Thus $(cos omega +Xsin omega)^nmapsto (e^{iomega})^n=e^{iomega n} = cos(nomega) + i sin(nomega)$. Thus the remainder on division by $X^2+1$ must be $cos(nomega)+Xsin(nomega)$.






            share|cite|improve this answer





















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              2 Answers
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              2 Answers
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              active

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              active

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              active

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              6














              Write the reminder $aX+b$. You have



              $$P(X) = (cos omega + sinomega X)^n =Q(X)(X^2+1) +aX+b.$$



              Substitute in this equation $X$ by $i$. You get
              $$e^{in omega} = ai+b.$$ Substitute now $X$ by $-i$. You get



              $$e^{-in omega} = -ai+b$$



              Solving in $a,b$ you finally get $b =cos nomega$ and $a = sin nomega$.






              share|cite|improve this answer


























                6














                Write the reminder $aX+b$. You have



                $$P(X) = (cos omega + sinomega X)^n =Q(X)(X^2+1) +aX+b.$$



                Substitute in this equation $X$ by $i$. You get
                $$e^{in omega} = ai+b.$$ Substitute now $X$ by $-i$. You get



                $$e^{-in omega} = -ai+b$$



                Solving in $a,b$ you finally get $b =cos nomega$ and $a = sin nomega$.






                share|cite|improve this answer
























                  6












                  6








                  6






                  Write the reminder $aX+b$. You have



                  $$P(X) = (cos omega + sinomega X)^n =Q(X)(X^2+1) +aX+b.$$



                  Substitute in this equation $X$ by $i$. You get
                  $$e^{in omega} = ai+b.$$ Substitute now $X$ by $-i$. You get



                  $$e^{-in omega} = -ai+b$$



                  Solving in $a,b$ you finally get $b =cos nomega$ and $a = sin nomega$.






                  share|cite|improve this answer












                  Write the reminder $aX+b$. You have



                  $$P(X) = (cos omega + sinomega X)^n =Q(X)(X^2+1) +aX+b.$$



                  Substitute in this equation $X$ by $i$. You get
                  $$e^{in omega} = ai+b.$$ Substitute now $X$ by $-i$. You get



                  $$e^{-in omega} = -ai+b$$



                  Solving in $a,b$ you finally get $b =cos nomega$ and $a = sin nomega$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  mathcounterexamples.net

                  24.2k21753




                  24.2k21753























                      2














                      This is essentially the same answer as mathcounterexamples.net's answer, but it's a different explanation, since frankly I don't follow several aspects of mathcounterexamples.net's thought process (though I did upvote it). Hence I'll demonstrate my thought process on seeing this question.



                      We can calculate the remainder on division by a polynomial $fin k[x]$ by working in the ring $k[x]/(f)$, since the remainder of a polynomial $a$ on division by $f$ is the unique representative of $a$ in $k[x]/(f)$ of degree less than the degree of $f$.



                      Then recognize that $Bbb{R}[X]/(X^2+1)cong Bbb{C}$ via $Xmapsto i$.
                      Thus $(cos omega +Xsin omega)^nmapsto (e^{iomega})^n=e^{iomega n} = cos(nomega) + i sin(nomega)$. Thus the remainder on division by $X^2+1$ must be $cos(nomega)+Xsin(nomega)$.






                      share|cite|improve this answer


























                        2














                        This is essentially the same answer as mathcounterexamples.net's answer, but it's a different explanation, since frankly I don't follow several aspects of mathcounterexamples.net's thought process (though I did upvote it). Hence I'll demonstrate my thought process on seeing this question.



                        We can calculate the remainder on division by a polynomial $fin k[x]$ by working in the ring $k[x]/(f)$, since the remainder of a polynomial $a$ on division by $f$ is the unique representative of $a$ in $k[x]/(f)$ of degree less than the degree of $f$.



                        Then recognize that $Bbb{R}[X]/(X^2+1)cong Bbb{C}$ via $Xmapsto i$.
                        Thus $(cos omega +Xsin omega)^nmapsto (e^{iomega})^n=e^{iomega n} = cos(nomega) + i sin(nomega)$. Thus the remainder on division by $X^2+1$ must be $cos(nomega)+Xsin(nomega)$.






                        share|cite|improve this answer
























                          2












                          2








                          2






                          This is essentially the same answer as mathcounterexamples.net's answer, but it's a different explanation, since frankly I don't follow several aspects of mathcounterexamples.net's thought process (though I did upvote it). Hence I'll demonstrate my thought process on seeing this question.



                          We can calculate the remainder on division by a polynomial $fin k[x]$ by working in the ring $k[x]/(f)$, since the remainder of a polynomial $a$ on division by $f$ is the unique representative of $a$ in $k[x]/(f)$ of degree less than the degree of $f$.



                          Then recognize that $Bbb{R}[X]/(X^2+1)cong Bbb{C}$ via $Xmapsto i$.
                          Thus $(cos omega +Xsin omega)^nmapsto (e^{iomega})^n=e^{iomega n} = cos(nomega) + i sin(nomega)$. Thus the remainder on division by $X^2+1$ must be $cos(nomega)+Xsin(nomega)$.






                          share|cite|improve this answer












                          This is essentially the same answer as mathcounterexamples.net's answer, but it's a different explanation, since frankly I don't follow several aspects of mathcounterexamples.net's thought process (though I did upvote it). Hence I'll demonstrate my thought process on seeing this question.



                          We can calculate the remainder on division by a polynomial $fin k[x]$ by working in the ring $k[x]/(f)$, since the remainder of a polynomial $a$ on division by $f$ is the unique representative of $a$ in $k[x]/(f)$ of degree less than the degree of $f$.



                          Then recognize that $Bbb{R}[X]/(X^2+1)cong Bbb{C}$ via $Xmapsto i$.
                          Thus $(cos omega +Xsin omega)^nmapsto (e^{iomega})^n=e^{iomega n} = cos(nomega) + i sin(nomega)$. Thus the remainder on division by $X^2+1$ must be $cos(nomega)+Xsin(nomega)$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 2 hours ago









                          jgon

                          12.7k21940




                          12.7k21940






























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