Roots of a polynomial with indeterminates $cos(x), sin(x)$












2














Suppose $P(cos(x), sin(x))$ is a polynomial in $mathbb R[X, Y]$ with degree $n$. I want to determine the number of roots of $P(cos(x), sin(x))$. Here I mean the possible values of the pair $(cos(x), sin(x))$, not roots in $x$. We exclude the case $P = cos^2(x) + sin^2(x) - 1$ and the ideal generated by $P$. Not sure whether this is enough but I meant to ask when we don't have this trivial case.



If we introduce the change of variable
$$ cos (x) = frac{ 1-t^2}{1+t^2}, sin(x) = frac{2t}{1+t^2}.$$
Then $P(cos(x), sin(x))$ can be considered as a rational function with numerator $n(t) in mathbb R[t]$ having degree at most $2n$. But this would imply we could have at most $2n$ solutions for $t$. Thus at most $2n$ solutions for the pair $(cos(x), sin(x))$. Is this correct? However, I had a feeling that the number should be $n$.






linear-algebra polynomials







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    2














    Suppose $P(cos(x), sin(x))$ is a polynomial in $mathbb R[X, Y]$ with degree $n$. I want to determine the number of roots of $P(cos(x), sin(x))$. Here I mean the possible values of the pair $(cos(x), sin(x))$, not roots in $x$. We exclude the case $P = cos^2(x) + sin^2(x) - 1$ and the ideal generated by $P$. Not sure whether this is enough but I meant to ask when we don't have this trivial case.



    If we introduce the change of variable
    $$ cos (x) = frac{ 1-t^2}{1+t^2}, sin(x) = frac{2t}{1+t^2}.$$
    Then $P(cos(x), sin(x))$ can be considered as a rational function with numerator $n(t) in mathbb R[t]$ having degree at most $2n$. But this would imply we could have at most $2n$ solutions for $t$. Thus at most $2n$ solutions for the pair $(cos(x), sin(x))$. Is this correct? However, I had a feeling that the number should be $n$.






    linear-algebra polynomials







    share|cite|improve this question



























      2












      2








      2







      Suppose $P(cos(x), sin(x))$ is a polynomial in $mathbb R[X, Y]$ with degree $n$. I want to determine the number of roots of $P(cos(x), sin(x))$. Here I mean the possible values of the pair $(cos(x), sin(x))$, not roots in $x$. We exclude the case $P = cos^2(x) + sin^2(x) - 1$ and the ideal generated by $P$. Not sure whether this is enough but I meant to ask when we don't have this trivial case.



      If we introduce the change of variable
      $$ cos (x) = frac{ 1-t^2}{1+t^2}, sin(x) = frac{2t}{1+t^2}.$$
      Then $P(cos(x), sin(x))$ can be considered as a rational function with numerator $n(t) in mathbb R[t]$ having degree at most $2n$. But this would imply we could have at most $2n$ solutions for $t$. Thus at most $2n$ solutions for the pair $(cos(x), sin(x))$. Is this correct? However, I had a feeling that the number should be $n$.






      linear-algebra polynomials







      share|cite|improve this question















      Suppose $P(cos(x), sin(x))$ is a polynomial in $mathbb R[X, Y]$ with degree $n$. I want to determine the number of roots of $P(cos(x), sin(x))$. Here I mean the possible values of the pair $(cos(x), sin(x))$, not roots in $x$. We exclude the case $P = cos^2(x) + sin^2(x) - 1$ and the ideal generated by $P$. Not sure whether this is enough but I meant to ask when we don't have this trivial case.



      If we introduce the change of variable
      $$ cos (x) = frac{ 1-t^2}{1+t^2}, sin(x) = frac{2t}{1+t^2}.$$
      Then $P(cos(x), sin(x))$ can be considered as a rational function with numerator $n(t) in mathbb R[t]$ having degree at most $2n$. But this would imply we could have at most $2n$ solutions for $t$. Thus at most $2n$ solutions for the pair $(cos(x), sin(x))$. Is this correct? However, I had a feeling that the number should be $n$.






      linear-algebra polynomials




      linear-algebra polynomials






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      share|cite|improve this question













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      edited Nov 30 at 6:33

























      asked Nov 30 at 6:17









      user1101010

      7551630




      7551630






















          1 Answer
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          4














          That's basically right. You have to be a little careful in that $(cos x,sin x)
          =(-1,0)$           isn't included in your parametrisation, but the conclusion still holds.



          You do need $2n$, not $n$. You can see this even when $n=1$. For instance
          $cos x=sin x$ has two solutions $(cos x,sin x)=(1/sqrt2,1/sqrt2)$
          and $(-1/sqrt2,-1/sqrt2)$.






          share|cite|improve this answer





















          • Thanks. You said "the conclusion still holds" without parametrize the point $(-1, 0)$. I did not notice this part. Could you please elaborate why?
            – user1101010
            Nov 30 at 6:27










          • @nicomezi: I noticed this. I should exclude this case or any ideal generated by this one. Not sure whether this is enough. I meant to ask when we don't have this particular case.
            – user1101010
            Nov 30 at 6:28












          • OK. Is it because $(-1,0) = lim_{t toinfty} (frac{1-t^2}{1+t^2}, frac{2t}{1+t^2})$? This means if $(-1,0)$ is a solution to $P(t)$, then $lim_{t to infty} P(t) = 0$ by continuity and this would imply $P(t)$ is a zero polynomial.
            – user1101010
            Nov 30 at 7:04













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          1 Answer
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          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4














          That's basically right. You have to be a little careful in that $(cos x,sin x)
          =(-1,0)$           isn't included in your parametrisation, but the conclusion still holds.



          You do need $2n$, not $n$. You can see this even when $n=1$. For instance
          $cos x=sin x$ has two solutions $(cos x,sin x)=(1/sqrt2,1/sqrt2)$
          and $(-1/sqrt2,-1/sqrt2)$.






          share|cite|improve this answer





















          • Thanks. You said "the conclusion still holds" without parametrize the point $(-1, 0)$. I did not notice this part. Could you please elaborate why?
            – user1101010
            Nov 30 at 6:27










          • @nicomezi: I noticed this. I should exclude this case or any ideal generated by this one. Not sure whether this is enough. I meant to ask when we don't have this particular case.
            – user1101010
            Nov 30 at 6:28












          • OK. Is it because $(-1,0) = lim_{t toinfty} (frac{1-t^2}{1+t^2}, frac{2t}{1+t^2})$? This means if $(-1,0)$ is a solution to $P(t)$, then $lim_{t to infty} P(t) = 0$ by continuity and this would imply $P(t)$ is a zero polynomial.
            – user1101010
            Nov 30 at 7:04


















          4














          That's basically right. You have to be a little careful in that $(cos x,sin x)
          =(-1,0)$           isn't included in your parametrisation, but the conclusion still holds.



          You do need $2n$, not $n$. You can see this even when $n=1$. For instance
          $cos x=sin x$ has two solutions $(cos x,sin x)=(1/sqrt2,1/sqrt2)$
          and $(-1/sqrt2,-1/sqrt2)$.






          share|cite|improve this answer





















          • Thanks. You said "the conclusion still holds" without parametrize the point $(-1, 0)$. I did not notice this part. Could you please elaborate why?
            – user1101010
            Nov 30 at 6:27










          • @nicomezi: I noticed this. I should exclude this case or any ideal generated by this one. Not sure whether this is enough. I meant to ask when we don't have this particular case.
            – user1101010
            Nov 30 at 6:28












          • OK. Is it because $(-1,0) = lim_{t toinfty} (frac{1-t^2}{1+t^2}, frac{2t}{1+t^2})$? This means if $(-1,0)$ is a solution to $P(t)$, then $lim_{t to infty} P(t) = 0$ by continuity and this would imply $P(t)$ is a zero polynomial.
            – user1101010
            Nov 30 at 7:04
















          4












          4








          4






          That's basically right. You have to be a little careful in that $(cos x,sin x)
          =(-1,0)$           isn't included in your parametrisation, but the conclusion still holds.



          You do need $2n$, not $n$. You can see this even when $n=1$. For instance
          $cos x=sin x$ has two solutions $(cos x,sin x)=(1/sqrt2,1/sqrt2)$
          and $(-1/sqrt2,-1/sqrt2)$.






          share|cite|improve this answer












          That's basically right. You have to be a little careful in that $(cos x,sin x)
          =(-1,0)$           isn't included in your parametrisation, but the conclusion still holds.



          You do need $2n$, not $n$. You can see this even when $n=1$. For instance
          $cos x=sin x$ has two solutions $(cos x,sin x)=(1/sqrt2,1/sqrt2)$
          and $(-1/sqrt2,-1/sqrt2)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 at 6:24









          Lord Shark the Unknown

          101k958131




          101k958131












          • Thanks. You said "the conclusion still holds" without parametrize the point $(-1, 0)$. I did not notice this part. Could you please elaborate why?
            – user1101010
            Nov 30 at 6:27










          • @nicomezi: I noticed this. I should exclude this case or any ideal generated by this one. Not sure whether this is enough. I meant to ask when we don't have this particular case.
            – user1101010
            Nov 30 at 6:28












          • OK. Is it because $(-1,0) = lim_{t toinfty} (frac{1-t^2}{1+t^2}, frac{2t}{1+t^2})$? This means if $(-1,0)$ is a solution to $P(t)$, then $lim_{t to infty} P(t) = 0$ by continuity and this would imply $P(t)$ is a zero polynomial.
            – user1101010
            Nov 30 at 7:04




















          • Thanks. You said "the conclusion still holds" without parametrize the point $(-1, 0)$. I did not notice this part. Could you please elaborate why?
            – user1101010
            Nov 30 at 6:27










          • @nicomezi: I noticed this. I should exclude this case or any ideal generated by this one. Not sure whether this is enough. I meant to ask when we don't have this particular case.
            – user1101010
            Nov 30 at 6:28












          • OK. Is it because $(-1,0) = lim_{t toinfty} (frac{1-t^2}{1+t^2}, frac{2t}{1+t^2})$? This means if $(-1,0)$ is a solution to $P(t)$, then $lim_{t to infty} P(t) = 0$ by continuity and this would imply $P(t)$ is a zero polynomial.
            – user1101010
            Nov 30 at 7:04


















          Thanks. You said "the conclusion still holds" without parametrize the point $(-1, 0)$. I did not notice this part. Could you please elaborate why?
          – user1101010
          Nov 30 at 6:27




          Thanks. You said "the conclusion still holds" without parametrize the point $(-1, 0)$. I did not notice this part. Could you please elaborate why?
          – user1101010
          Nov 30 at 6:27












          @nicomezi: I noticed this. I should exclude this case or any ideal generated by this one. Not sure whether this is enough. I meant to ask when we don't have this particular case.
          – user1101010
          Nov 30 at 6:28






          @nicomezi: I noticed this. I should exclude this case or any ideal generated by this one. Not sure whether this is enough. I meant to ask when we don't have this particular case.
          – user1101010
          Nov 30 at 6:28














          OK. Is it because $(-1,0) = lim_{t toinfty} (frac{1-t^2}{1+t^2}, frac{2t}{1+t^2})$? This means if $(-1,0)$ is a solution to $P(t)$, then $lim_{t to infty} P(t) = 0$ by continuity and this would imply $P(t)$ is a zero polynomial.
          – user1101010
          Nov 30 at 7:04






          OK. Is it because $(-1,0) = lim_{t toinfty} (frac{1-t^2}{1+t^2}, frac{2t}{1+t^2})$? This means if $(-1,0)$ is a solution to $P(t)$, then $lim_{t to infty} P(t) = 0$ by continuity and this would imply $P(t)$ is a zero polynomial.
          – user1101010
          Nov 30 at 7:04




















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