Can we have a simple root of real polynomials with coefficients as linear functions such that the curve of...












2












$begingroup$


Let us consider a parametrized polynomial
begin{align*}
f(x, r) = x^n + ra_{n-1} x^{n-1} + dots + ra_0,
end{align*}

where $r in mathbb R$ is the parameter and $a = (a_{n-1}, dots, a_0)$ is fixed. Alternatively, we can consider $f(x) in mathbb R[x]$ with coefficients as linear functions over $mathbb R$
begin{align*}
f(x, r) = x^n + a_{n-1}(r) x^{n-1} + dots + a_0(r).
end{align*}



Since $r in mathbb R$, there exists $n$ continuous functions $b_1(r), dots, b_n(r)$ such that for each $r$, ${b_1(r), dots, b_n(r)}$ is the set of roots for $f(x,r)$ counting multiplicity. Now suppose for $r_0 in mathbb R$ and some $b in mathbb R$, $f(bi, r_0) = 0$ where $i$ denotes imaginary symbol. Further assume at $r_0$, $bi$ is a simple root. So at least locally around $r_0$ (if I am not mistaken), we should have a smooth function that tracks this root. Let us call this function $gamma(r)$.



My question is: is it possible that the curve $gamma(r)$ is tangent to the imaginary axis at $r_0$? Here is a plot to demonstrate what I have in mind. Note the plot is a parabola, but the shape is really not important. I only need the very local behavior about $r_0$.



enter image description here





This is related to a question I asked here Can roots of a polynomial stay on one side of the complex plane as the coefficients vary?










share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Let us consider a parametrized polynomial
    begin{align*}
    f(x, r) = x^n + ra_{n-1} x^{n-1} + dots + ra_0,
    end{align*}

    where $r in mathbb R$ is the parameter and $a = (a_{n-1}, dots, a_0)$ is fixed. Alternatively, we can consider $f(x) in mathbb R[x]$ with coefficients as linear functions over $mathbb R$
    begin{align*}
    f(x, r) = x^n + a_{n-1}(r) x^{n-1} + dots + a_0(r).
    end{align*}



    Since $r in mathbb R$, there exists $n$ continuous functions $b_1(r), dots, b_n(r)$ such that for each $r$, ${b_1(r), dots, b_n(r)}$ is the set of roots for $f(x,r)$ counting multiplicity. Now suppose for $r_0 in mathbb R$ and some $b in mathbb R$, $f(bi, r_0) = 0$ where $i$ denotes imaginary symbol. Further assume at $r_0$, $bi$ is a simple root. So at least locally around $r_0$ (if I am not mistaken), we should have a smooth function that tracks this root. Let us call this function $gamma(r)$.



    My question is: is it possible that the curve $gamma(r)$ is tangent to the imaginary axis at $r_0$? Here is a plot to demonstrate what I have in mind. Note the plot is a parabola, but the shape is really not important. I only need the very local behavior about $r_0$.



    enter image description here





    This is related to a question I asked here Can roots of a polynomial stay on one side of the complex plane as the coefficients vary?










    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Let us consider a parametrized polynomial
      begin{align*}
      f(x, r) = x^n + ra_{n-1} x^{n-1} + dots + ra_0,
      end{align*}

      where $r in mathbb R$ is the parameter and $a = (a_{n-1}, dots, a_0)$ is fixed. Alternatively, we can consider $f(x) in mathbb R[x]$ with coefficients as linear functions over $mathbb R$
      begin{align*}
      f(x, r) = x^n + a_{n-1}(r) x^{n-1} + dots + a_0(r).
      end{align*}



      Since $r in mathbb R$, there exists $n$ continuous functions $b_1(r), dots, b_n(r)$ such that for each $r$, ${b_1(r), dots, b_n(r)}$ is the set of roots for $f(x,r)$ counting multiplicity. Now suppose for $r_0 in mathbb R$ and some $b in mathbb R$, $f(bi, r_0) = 0$ where $i$ denotes imaginary symbol. Further assume at $r_0$, $bi$ is a simple root. So at least locally around $r_0$ (if I am not mistaken), we should have a smooth function that tracks this root. Let us call this function $gamma(r)$.



      My question is: is it possible that the curve $gamma(r)$ is tangent to the imaginary axis at $r_0$? Here is a plot to demonstrate what I have in mind. Note the plot is a parabola, but the shape is really not important. I only need the very local behavior about $r_0$.



      enter image description here





      This is related to a question I asked here Can roots of a polynomial stay on one side of the complex plane as the coefficients vary?










      share|cite|improve this question











      $endgroup$




      Let us consider a parametrized polynomial
      begin{align*}
      f(x, r) = x^n + ra_{n-1} x^{n-1} + dots + ra_0,
      end{align*}

      where $r in mathbb R$ is the parameter and $a = (a_{n-1}, dots, a_0)$ is fixed. Alternatively, we can consider $f(x) in mathbb R[x]$ with coefficients as linear functions over $mathbb R$
      begin{align*}
      f(x, r) = x^n + a_{n-1}(r) x^{n-1} + dots + a_0(r).
      end{align*}



      Since $r in mathbb R$, there exists $n$ continuous functions $b_1(r), dots, b_n(r)$ such that for each $r$, ${b_1(r), dots, b_n(r)}$ is the set of roots for $f(x,r)$ counting multiplicity. Now suppose for $r_0 in mathbb R$ and some $b in mathbb R$, $f(bi, r_0) = 0$ where $i$ denotes imaginary symbol. Further assume at $r_0$, $bi$ is a simple root. So at least locally around $r_0$ (if I am not mistaken), we should have a smooth function that tracks this root. Let us call this function $gamma(r)$.



      My question is: is it possible that the curve $gamma(r)$ is tangent to the imaginary axis at $r_0$? Here is a plot to demonstrate what I have in mind. Note the plot is a parabola, but the shape is really not important. I only need the very local behavior about $r_0$.



      enter image description here





      This is related to a question I asked here Can roots of a polynomial stay on one side of the complex plane as the coefficients vary?







      linear-algebra abstract-algebra algebraic-geometry polynomials






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 6 '18 at 6:29







      user1101010

















      asked Dec 6 '18 at 5:05









      user1101010user1101010

      7751730




      7751730






















          1 Answer
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          $begingroup$

          Yes, it is possible. Try
          $$
          p_r (x) = x^5+r(2x^4+5x^3+4x^2+4x+2).
          $$
          Let $epsilon = r-1$ and write $p_r(x) = 0$ as
          $$
          x^5+2x^4+5x^3+4x^2+4x+2 = -epsilon (2x^4+5x^3+4x^2+4x+2)quadcdots (*).
          $$
          At $epsilon = 0$, we see that $x= i$ is a root of $(*)$. And a numerical plot shows that $gamma(1+epsilon)$ has a positive real part for all $epsilon$ with sufficiently small absolute value. You can check this easily on the web using https://www.wolframalpha.com/.



          $textbf{EDIT:}$ Here's how I got this example. First, we can without loss of generality, assume that $r_0 = 1$ (through rescaling) and thus any such example is of the form of $(*)$:
          $$
          x^n + b_{n-1}x^{n-1} + cdots + b_1x + b_0 = -epsilon (b_{n-1}x^{n-1} + cdots + b_1x + b_0)=-epsilon Q(x)quadcdots(**).
          $$
          We can further assume that $x=i $ is a root of $(**)$ when $epsilon=0$. That is, $i^n + Q(i)=0$. Secondly, we have
          $$
          gamma(1+epsilon) = i -epsilon frac{Q(i)}{ni^{n-1}+ Q'(i)}+mathcal{o}(epsilon),
          $$
          and hence $gamma'(1) = -frac{Q(i)}{ni^{n-1}+ Q'(i)} =ibeta$ for some real $beta$. It is recommended to choose a high $n$ to produce such a plot in your figure. On the other hand, choosing a high $n$ produces too many coefficients to be determined. My choice was $n=5$. So, let $Q(x) = ax^4 + bx^3 + cx^2 +dx +e$ and do the calculation using $i + Q(i)=0$ and $-frac{Q(i)}{5+ Q'(i)} = frac{i}{5+ Q'(i)}=ibeta.$ This yields:
          $$
          (a-c+e) +i(-b+d+1) =0,
          $$

          $$
          (-3b +d+5)+i(-4a+2c) =frac{1}{beta}=beta'.
          $$
          This gives restrictions on the coefficients:
          $$
          c=2a, e=a, d=b-1,
          $$
          and $a,b$ are free variables. Finally, choose arbitrary $(a,b)$ and execute the test whether it produces such a plot. You may be able to find many other examples using this method.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you. May I ask how you come up with the example? What's your thought process?
            $endgroup$
            – user1101010
            Dec 6 '18 at 22:09










          • $begingroup$
            Actually, it was a guess but a 'logical' guess partially based on my argument in your previous post math.stackexchange.com/questions/3006711/…. I'll edit my answer when I have time.
            $endgroup$
            – Song
            Dec 6 '18 at 22:16











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          $begingroup$

          Yes, it is possible. Try
          $$
          p_r (x) = x^5+r(2x^4+5x^3+4x^2+4x+2).
          $$
          Let $epsilon = r-1$ and write $p_r(x) = 0$ as
          $$
          x^5+2x^4+5x^3+4x^2+4x+2 = -epsilon (2x^4+5x^3+4x^2+4x+2)quadcdots (*).
          $$
          At $epsilon = 0$, we see that $x= i$ is a root of $(*)$. And a numerical plot shows that $gamma(1+epsilon)$ has a positive real part for all $epsilon$ with sufficiently small absolute value. You can check this easily on the web using https://www.wolframalpha.com/.



          $textbf{EDIT:}$ Here's how I got this example. First, we can without loss of generality, assume that $r_0 = 1$ (through rescaling) and thus any such example is of the form of $(*)$:
          $$
          x^n + b_{n-1}x^{n-1} + cdots + b_1x + b_0 = -epsilon (b_{n-1}x^{n-1} + cdots + b_1x + b_0)=-epsilon Q(x)quadcdots(**).
          $$
          We can further assume that $x=i $ is a root of $(**)$ when $epsilon=0$. That is, $i^n + Q(i)=0$. Secondly, we have
          $$
          gamma(1+epsilon) = i -epsilon frac{Q(i)}{ni^{n-1}+ Q'(i)}+mathcal{o}(epsilon),
          $$
          and hence $gamma'(1) = -frac{Q(i)}{ni^{n-1}+ Q'(i)} =ibeta$ for some real $beta$. It is recommended to choose a high $n$ to produce such a plot in your figure. On the other hand, choosing a high $n$ produces too many coefficients to be determined. My choice was $n=5$. So, let $Q(x) = ax^4 + bx^3 + cx^2 +dx +e$ and do the calculation using $i + Q(i)=0$ and $-frac{Q(i)}{5+ Q'(i)} = frac{i}{5+ Q'(i)}=ibeta.$ This yields:
          $$
          (a-c+e) +i(-b+d+1) =0,
          $$

          $$
          (-3b +d+5)+i(-4a+2c) =frac{1}{beta}=beta'.
          $$
          This gives restrictions on the coefficients:
          $$
          c=2a, e=a, d=b-1,
          $$
          and $a,b$ are free variables. Finally, choose arbitrary $(a,b)$ and execute the test whether it produces such a plot. You may be able to find many other examples using this method.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you. May I ask how you come up with the example? What's your thought process?
            $endgroup$
            – user1101010
            Dec 6 '18 at 22:09










          • $begingroup$
            Actually, it was a guess but a 'logical' guess partially based on my argument in your previous post math.stackexchange.com/questions/3006711/…. I'll edit my answer when I have time.
            $endgroup$
            – Song
            Dec 6 '18 at 22:16
















          1












          $begingroup$

          Yes, it is possible. Try
          $$
          p_r (x) = x^5+r(2x^4+5x^3+4x^2+4x+2).
          $$
          Let $epsilon = r-1$ and write $p_r(x) = 0$ as
          $$
          x^5+2x^4+5x^3+4x^2+4x+2 = -epsilon (2x^4+5x^3+4x^2+4x+2)quadcdots (*).
          $$
          At $epsilon = 0$, we see that $x= i$ is a root of $(*)$. And a numerical plot shows that $gamma(1+epsilon)$ has a positive real part for all $epsilon$ with sufficiently small absolute value. You can check this easily on the web using https://www.wolframalpha.com/.



          $textbf{EDIT:}$ Here's how I got this example. First, we can without loss of generality, assume that $r_0 = 1$ (through rescaling) and thus any such example is of the form of $(*)$:
          $$
          x^n + b_{n-1}x^{n-1} + cdots + b_1x + b_0 = -epsilon (b_{n-1}x^{n-1} + cdots + b_1x + b_0)=-epsilon Q(x)quadcdots(**).
          $$
          We can further assume that $x=i $ is a root of $(**)$ when $epsilon=0$. That is, $i^n + Q(i)=0$. Secondly, we have
          $$
          gamma(1+epsilon) = i -epsilon frac{Q(i)}{ni^{n-1}+ Q'(i)}+mathcal{o}(epsilon),
          $$
          and hence $gamma'(1) = -frac{Q(i)}{ni^{n-1}+ Q'(i)} =ibeta$ for some real $beta$. It is recommended to choose a high $n$ to produce such a plot in your figure. On the other hand, choosing a high $n$ produces too many coefficients to be determined. My choice was $n=5$. So, let $Q(x) = ax^4 + bx^3 + cx^2 +dx +e$ and do the calculation using $i + Q(i)=0$ and $-frac{Q(i)}{5+ Q'(i)} = frac{i}{5+ Q'(i)}=ibeta.$ This yields:
          $$
          (a-c+e) +i(-b+d+1) =0,
          $$

          $$
          (-3b +d+5)+i(-4a+2c) =frac{1}{beta}=beta'.
          $$
          This gives restrictions on the coefficients:
          $$
          c=2a, e=a, d=b-1,
          $$
          and $a,b$ are free variables. Finally, choose arbitrary $(a,b)$ and execute the test whether it produces such a plot. You may be able to find many other examples using this method.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you. May I ask how you come up with the example? What's your thought process?
            $endgroup$
            – user1101010
            Dec 6 '18 at 22:09










          • $begingroup$
            Actually, it was a guess but a 'logical' guess partially based on my argument in your previous post math.stackexchange.com/questions/3006711/…. I'll edit my answer when I have time.
            $endgroup$
            – Song
            Dec 6 '18 at 22:16














          1












          1








          1





          $begingroup$

          Yes, it is possible. Try
          $$
          p_r (x) = x^5+r(2x^4+5x^3+4x^2+4x+2).
          $$
          Let $epsilon = r-1$ and write $p_r(x) = 0$ as
          $$
          x^5+2x^4+5x^3+4x^2+4x+2 = -epsilon (2x^4+5x^3+4x^2+4x+2)quadcdots (*).
          $$
          At $epsilon = 0$, we see that $x= i$ is a root of $(*)$. And a numerical plot shows that $gamma(1+epsilon)$ has a positive real part for all $epsilon$ with sufficiently small absolute value. You can check this easily on the web using https://www.wolframalpha.com/.



          $textbf{EDIT:}$ Here's how I got this example. First, we can without loss of generality, assume that $r_0 = 1$ (through rescaling) and thus any such example is of the form of $(*)$:
          $$
          x^n + b_{n-1}x^{n-1} + cdots + b_1x + b_0 = -epsilon (b_{n-1}x^{n-1} + cdots + b_1x + b_0)=-epsilon Q(x)quadcdots(**).
          $$
          We can further assume that $x=i $ is a root of $(**)$ when $epsilon=0$. That is, $i^n + Q(i)=0$. Secondly, we have
          $$
          gamma(1+epsilon) = i -epsilon frac{Q(i)}{ni^{n-1}+ Q'(i)}+mathcal{o}(epsilon),
          $$
          and hence $gamma'(1) = -frac{Q(i)}{ni^{n-1}+ Q'(i)} =ibeta$ for some real $beta$. It is recommended to choose a high $n$ to produce such a plot in your figure. On the other hand, choosing a high $n$ produces too many coefficients to be determined. My choice was $n=5$. So, let $Q(x) = ax^4 + bx^3 + cx^2 +dx +e$ and do the calculation using $i + Q(i)=0$ and $-frac{Q(i)}{5+ Q'(i)} = frac{i}{5+ Q'(i)}=ibeta.$ This yields:
          $$
          (a-c+e) +i(-b+d+1) =0,
          $$

          $$
          (-3b +d+5)+i(-4a+2c) =frac{1}{beta}=beta'.
          $$
          This gives restrictions on the coefficients:
          $$
          c=2a, e=a, d=b-1,
          $$
          and $a,b$ are free variables. Finally, choose arbitrary $(a,b)$ and execute the test whether it produces such a plot. You may be able to find many other examples using this method.






          share|cite|improve this answer











          $endgroup$



          Yes, it is possible. Try
          $$
          p_r (x) = x^5+r(2x^4+5x^3+4x^2+4x+2).
          $$
          Let $epsilon = r-1$ and write $p_r(x) = 0$ as
          $$
          x^5+2x^4+5x^3+4x^2+4x+2 = -epsilon (2x^4+5x^3+4x^2+4x+2)quadcdots (*).
          $$
          At $epsilon = 0$, we see that $x= i$ is a root of $(*)$. And a numerical plot shows that $gamma(1+epsilon)$ has a positive real part for all $epsilon$ with sufficiently small absolute value. You can check this easily on the web using https://www.wolframalpha.com/.



          $textbf{EDIT:}$ Here's how I got this example. First, we can without loss of generality, assume that $r_0 = 1$ (through rescaling) and thus any such example is of the form of $(*)$:
          $$
          x^n + b_{n-1}x^{n-1} + cdots + b_1x + b_0 = -epsilon (b_{n-1}x^{n-1} + cdots + b_1x + b_0)=-epsilon Q(x)quadcdots(**).
          $$
          We can further assume that $x=i $ is a root of $(**)$ when $epsilon=0$. That is, $i^n + Q(i)=0$. Secondly, we have
          $$
          gamma(1+epsilon) = i -epsilon frac{Q(i)}{ni^{n-1}+ Q'(i)}+mathcal{o}(epsilon),
          $$
          and hence $gamma'(1) = -frac{Q(i)}{ni^{n-1}+ Q'(i)} =ibeta$ for some real $beta$. It is recommended to choose a high $n$ to produce such a plot in your figure. On the other hand, choosing a high $n$ produces too many coefficients to be determined. My choice was $n=5$. So, let $Q(x) = ax^4 + bx^3 + cx^2 +dx +e$ and do the calculation using $i + Q(i)=0$ and $-frac{Q(i)}{5+ Q'(i)} = frac{i}{5+ Q'(i)}=ibeta.$ This yields:
          $$
          (a-c+e) +i(-b+d+1) =0,
          $$

          $$
          (-3b +d+5)+i(-4a+2c) =frac{1}{beta}=beta'.
          $$
          This gives restrictions on the coefficients:
          $$
          c=2a, e=a, d=b-1,
          $$
          and $a,b$ are free variables. Finally, choose arbitrary $(a,b)$ and execute the test whether it produces such a plot. You may be able to find many other examples using this method.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 6 '18 at 23:10

























          answered Dec 6 '18 at 9:40









          SongSong

          10.4k627




          10.4k627












          • $begingroup$
            Thank you. May I ask how you come up with the example? What's your thought process?
            $endgroup$
            – user1101010
            Dec 6 '18 at 22:09










          • $begingroup$
            Actually, it was a guess but a 'logical' guess partially based on my argument in your previous post math.stackexchange.com/questions/3006711/…. I'll edit my answer when I have time.
            $endgroup$
            – Song
            Dec 6 '18 at 22:16


















          • $begingroup$
            Thank you. May I ask how you come up with the example? What's your thought process?
            $endgroup$
            – user1101010
            Dec 6 '18 at 22:09










          • $begingroup$
            Actually, it was a guess but a 'logical' guess partially based on my argument in your previous post math.stackexchange.com/questions/3006711/…. I'll edit my answer when I have time.
            $endgroup$
            – Song
            Dec 6 '18 at 22:16
















          $begingroup$
          Thank you. May I ask how you come up with the example? What's your thought process?
          $endgroup$
          – user1101010
          Dec 6 '18 at 22:09




          $begingroup$
          Thank you. May I ask how you come up with the example? What's your thought process?
          $endgroup$
          – user1101010
          Dec 6 '18 at 22:09












          $begingroup$
          Actually, it was a guess but a 'logical' guess partially based on my argument in your previous post math.stackexchange.com/questions/3006711/…. I'll edit my answer when I have time.
          $endgroup$
          – Song
          Dec 6 '18 at 22:16




          $begingroup$
          Actually, it was a guess but a 'logical' guess partially based on my argument in your previous post math.stackexchange.com/questions/3006711/…. I'll edit my answer when I have time.
          $endgroup$
          – Song
          Dec 6 '18 at 22:16


















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