Do the zeroes of this polynomial lie inside, outside, or on the unit circle? $P_n(z)=1^3 z + 2^3 z^2 + 3^3...












7















For each positive integer $n$, let's define the polynomial $$P_n(z)=1^3 z + 2^3 z^2 + 3^3 z^3 + cdots + n^3 z^n$$
Do the zeroes of $P_n$ lie inside, outside, or on the unit circle $|z|=1$?




I tried to find a formula for $displaystyle sum_{k=1}^n k^3 z^k$ by repeatedly taking derivatives of $z^k$ but it was so tough. Initial investigation showed that the zeroes lie inside the unit circle but I couldn't generalize the result.



Any help would be appreciated!



Source : The Arts and Crafts of Problem Solving










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  • 2




    hint: have you heard of Rouché's theorem? mathworld.wolfram.com/RouchesTheorem.html this might help
    – Pink Panther
    yesterday


















7















For each positive integer $n$, let's define the polynomial $$P_n(z)=1^3 z + 2^3 z^2 + 3^3 z^3 + cdots + n^3 z^n$$
Do the zeroes of $P_n$ lie inside, outside, or on the unit circle $|z|=1$?




I tried to find a formula for $displaystyle sum_{k=1}^n k^3 z^k$ by repeatedly taking derivatives of $z^k$ but it was so tough. Initial investigation showed that the zeroes lie inside the unit circle but I couldn't generalize the result.



Any help would be appreciated!



Source : The Arts and Crafts of Problem Solving










share|cite|improve this question




















  • 2




    hint: have you heard of Rouché's theorem? mathworld.wolfram.com/RouchesTheorem.html this might help
    – Pink Panther
    yesterday
















7












7








7








For each positive integer $n$, let's define the polynomial $$P_n(z)=1^3 z + 2^3 z^2 + 3^3 z^3 + cdots + n^3 z^n$$
Do the zeroes of $P_n$ lie inside, outside, or on the unit circle $|z|=1$?




I tried to find a formula for $displaystyle sum_{k=1}^n k^3 z^k$ by repeatedly taking derivatives of $z^k$ but it was so tough. Initial investigation showed that the zeroes lie inside the unit circle but I couldn't generalize the result.



Any help would be appreciated!



Source : The Arts and Crafts of Problem Solving










share|cite|improve this question
















For each positive integer $n$, let's define the polynomial $$P_n(z)=1^3 z + 2^3 z^2 + 3^3 z^3 + cdots + n^3 z^n$$
Do the zeroes of $P_n$ lie inside, outside, or on the unit circle $|z|=1$?




I tried to find a formula for $displaystyle sum_{k=1}^n k^3 z^k$ by repeatedly taking derivatives of $z^k$ but it was so tough. Initial investigation showed that the zeroes lie inside the unit circle but I couldn't generalize the result.



Any help would be appreciated!



Source : The Arts and Crafts of Problem Solving







polynomials power-series roots roots-of-unity






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edited 18 hours ago









Blue

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asked yesterday









Atiq Rahman

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  • 2




    hint: have you heard of Rouché's theorem? mathworld.wolfram.com/RouchesTheorem.html this might help
    – Pink Panther
    yesterday
















  • 2




    hint: have you heard of Rouché's theorem? mathworld.wolfram.com/RouchesTheorem.html this might help
    – Pink Panther
    yesterday










2




2




hint: have you heard of Rouché's theorem? mathworld.wolfram.com/RouchesTheorem.html this might help
– Pink Panther
yesterday






hint: have you heard of Rouché's theorem? mathworld.wolfram.com/RouchesTheorem.html this might help
– Pink Panther
yesterday












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12














Let $$g(z) = sum_{k=0}^n z^k = frac{z^{n+1}-1}{z-1}$$



It is easy to see the roots of $g(z)$ lie on the unit circle $|z|= 1 $ and all of them are simple.



By Gauss-Lucas theorem, the roots of $g'(z)$ belong to the convex hull of the roots of $g$. Since the closed unit disk $|z| le 1$ is convex, this convex hull is a subset of the closed unit disk.
Notice the convex hull is a $n$-gon which intersect the unit circle $|z| = 1$ only at the roots of $g(z)$. Since the roots of $g(z)$ are simple, none of them can be root of $g'(z)$. As a result, the roots of $g'(z)$ belongs to the open unit disk $|z| < 1$.



Since $zg'(z)$ differs from $g'(z)$ by only a root at $0$, the roots of



$$zg'(z) = sum_{k=1}^n k z^k$$



belong to the open unit disk $|z| < 1$.



Apply Gauss-Lucas theorem again and then add a root at $z = 0$, we find the roots of



$$left(zfrac{d}{dz}right)^2 g(z) = sum_{k=1}^n k^2 z^k$$
belong to the open unit disk $|z| < 1$.
Repeat this process one more time, we find all the zeros of



$$left(zfrac{d}{dz}right)^3 g(z) = sum_{k=1}^n k^3 z^k = P_n(z)$$
lie inside the unit circle.



Notes



There are other ways to arrive at same conclusion. In particular, we can use following results:




Let $displaystyle;f(z) = sum_{k=0}^m a_k z^k$ be any polynomial with real and positive coefficients.




  1. If the coefficients $a_k$ are non-descending,
    $$0 < a_0 le a_2 le cdots le a_m$$
    then roots of $f(z)$ belong to the closed unit disk $|z| le 1$.


  2. If the coefficients $a_k$ are non-ascending,
    $$ a_0 ge a_1 ge cdots ge a_m > 0$$
    then roots of $f(z)$ lie outside the open unit disk (i.e. $|z| ge 1$ for all the roots)


  3. In general, the roots of $f(z)$ belong to the closed annulus
    $$min_{1le k le m} frac{a_{k-1}}{a_k} le |z| le max_{1 le k le m}frac{a_{k-1}}{a_k}$$





Since it is easy to derive any one of these results from the other two,
these results are typically treated as a single theorem known as the Eneström-Kakeya Theorem.



For a proof of the first result, see answers of a related question. In particular, the answer by Ayman Hourieh which uses Rouché's theorem.



Back to the problem at hand. It is easy to see we can rewrite $P_n(z)$ as $z f(z)$ for some polynomial $f(z)$ with real and positive coefficients. Apply the third result, we immediately find aside from a root at $z = 0$, the remaining $n-1$ roots of $P_n(z)$ lies within the closed annulus $$frac18 le |z| le left(frac{n-1
}{n}right)^3 < 1$$






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    12














    Let $$g(z) = sum_{k=0}^n z^k = frac{z^{n+1}-1}{z-1}$$



    It is easy to see the roots of $g(z)$ lie on the unit circle $|z|= 1 $ and all of them are simple.



    By Gauss-Lucas theorem, the roots of $g'(z)$ belong to the convex hull of the roots of $g$. Since the closed unit disk $|z| le 1$ is convex, this convex hull is a subset of the closed unit disk.
    Notice the convex hull is a $n$-gon which intersect the unit circle $|z| = 1$ only at the roots of $g(z)$. Since the roots of $g(z)$ are simple, none of them can be root of $g'(z)$. As a result, the roots of $g'(z)$ belongs to the open unit disk $|z| < 1$.



    Since $zg'(z)$ differs from $g'(z)$ by only a root at $0$, the roots of



    $$zg'(z) = sum_{k=1}^n k z^k$$



    belong to the open unit disk $|z| < 1$.



    Apply Gauss-Lucas theorem again and then add a root at $z = 0$, we find the roots of



    $$left(zfrac{d}{dz}right)^2 g(z) = sum_{k=1}^n k^2 z^k$$
    belong to the open unit disk $|z| < 1$.
    Repeat this process one more time, we find all the zeros of



    $$left(zfrac{d}{dz}right)^3 g(z) = sum_{k=1}^n k^3 z^k = P_n(z)$$
    lie inside the unit circle.



    Notes



    There are other ways to arrive at same conclusion. In particular, we can use following results:




    Let $displaystyle;f(z) = sum_{k=0}^m a_k z^k$ be any polynomial with real and positive coefficients.




    1. If the coefficients $a_k$ are non-descending,
      $$0 < a_0 le a_2 le cdots le a_m$$
      then roots of $f(z)$ belong to the closed unit disk $|z| le 1$.


    2. If the coefficients $a_k$ are non-ascending,
      $$ a_0 ge a_1 ge cdots ge a_m > 0$$
      then roots of $f(z)$ lie outside the open unit disk (i.e. $|z| ge 1$ for all the roots)


    3. In general, the roots of $f(z)$ belong to the closed annulus
      $$min_{1le k le m} frac{a_{k-1}}{a_k} le |z| le max_{1 le k le m}frac{a_{k-1}}{a_k}$$





    Since it is easy to derive any one of these results from the other two,
    these results are typically treated as a single theorem known as the Eneström-Kakeya Theorem.



    For a proof of the first result, see answers of a related question. In particular, the answer by Ayman Hourieh which uses Rouché's theorem.



    Back to the problem at hand. It is easy to see we can rewrite $P_n(z)$ as $z f(z)$ for some polynomial $f(z)$ with real and positive coefficients. Apply the third result, we immediately find aside from a root at $z = 0$, the remaining $n-1$ roots of $P_n(z)$ lies within the closed annulus $$frac18 le |z| le left(frac{n-1
    }{n}right)^3 < 1$$






    share|cite|improve this answer




























      12














      Let $$g(z) = sum_{k=0}^n z^k = frac{z^{n+1}-1}{z-1}$$



      It is easy to see the roots of $g(z)$ lie on the unit circle $|z|= 1 $ and all of them are simple.



      By Gauss-Lucas theorem, the roots of $g'(z)$ belong to the convex hull of the roots of $g$. Since the closed unit disk $|z| le 1$ is convex, this convex hull is a subset of the closed unit disk.
      Notice the convex hull is a $n$-gon which intersect the unit circle $|z| = 1$ only at the roots of $g(z)$. Since the roots of $g(z)$ are simple, none of them can be root of $g'(z)$. As a result, the roots of $g'(z)$ belongs to the open unit disk $|z| < 1$.



      Since $zg'(z)$ differs from $g'(z)$ by only a root at $0$, the roots of



      $$zg'(z) = sum_{k=1}^n k z^k$$



      belong to the open unit disk $|z| < 1$.



      Apply Gauss-Lucas theorem again and then add a root at $z = 0$, we find the roots of



      $$left(zfrac{d}{dz}right)^2 g(z) = sum_{k=1}^n k^2 z^k$$
      belong to the open unit disk $|z| < 1$.
      Repeat this process one more time, we find all the zeros of



      $$left(zfrac{d}{dz}right)^3 g(z) = sum_{k=1}^n k^3 z^k = P_n(z)$$
      lie inside the unit circle.



      Notes



      There are other ways to arrive at same conclusion. In particular, we can use following results:




      Let $displaystyle;f(z) = sum_{k=0}^m a_k z^k$ be any polynomial with real and positive coefficients.




      1. If the coefficients $a_k$ are non-descending,
        $$0 < a_0 le a_2 le cdots le a_m$$
        then roots of $f(z)$ belong to the closed unit disk $|z| le 1$.


      2. If the coefficients $a_k$ are non-ascending,
        $$ a_0 ge a_1 ge cdots ge a_m > 0$$
        then roots of $f(z)$ lie outside the open unit disk (i.e. $|z| ge 1$ for all the roots)


      3. In general, the roots of $f(z)$ belong to the closed annulus
        $$min_{1le k le m} frac{a_{k-1}}{a_k} le |z| le max_{1 le k le m}frac{a_{k-1}}{a_k}$$





      Since it is easy to derive any one of these results from the other two,
      these results are typically treated as a single theorem known as the Eneström-Kakeya Theorem.



      For a proof of the first result, see answers of a related question. In particular, the answer by Ayman Hourieh which uses Rouché's theorem.



      Back to the problem at hand. It is easy to see we can rewrite $P_n(z)$ as $z f(z)$ for some polynomial $f(z)$ with real and positive coefficients. Apply the third result, we immediately find aside from a root at $z = 0$, the remaining $n-1$ roots of $P_n(z)$ lies within the closed annulus $$frac18 le |z| le left(frac{n-1
      }{n}right)^3 < 1$$






      share|cite|improve this answer


























        12












        12








        12






        Let $$g(z) = sum_{k=0}^n z^k = frac{z^{n+1}-1}{z-1}$$



        It is easy to see the roots of $g(z)$ lie on the unit circle $|z|= 1 $ and all of them are simple.



        By Gauss-Lucas theorem, the roots of $g'(z)$ belong to the convex hull of the roots of $g$. Since the closed unit disk $|z| le 1$ is convex, this convex hull is a subset of the closed unit disk.
        Notice the convex hull is a $n$-gon which intersect the unit circle $|z| = 1$ only at the roots of $g(z)$. Since the roots of $g(z)$ are simple, none of them can be root of $g'(z)$. As a result, the roots of $g'(z)$ belongs to the open unit disk $|z| < 1$.



        Since $zg'(z)$ differs from $g'(z)$ by only a root at $0$, the roots of



        $$zg'(z) = sum_{k=1}^n k z^k$$



        belong to the open unit disk $|z| < 1$.



        Apply Gauss-Lucas theorem again and then add a root at $z = 0$, we find the roots of



        $$left(zfrac{d}{dz}right)^2 g(z) = sum_{k=1}^n k^2 z^k$$
        belong to the open unit disk $|z| < 1$.
        Repeat this process one more time, we find all the zeros of



        $$left(zfrac{d}{dz}right)^3 g(z) = sum_{k=1}^n k^3 z^k = P_n(z)$$
        lie inside the unit circle.



        Notes



        There are other ways to arrive at same conclusion. In particular, we can use following results:




        Let $displaystyle;f(z) = sum_{k=0}^m a_k z^k$ be any polynomial with real and positive coefficients.




        1. If the coefficients $a_k$ are non-descending,
          $$0 < a_0 le a_2 le cdots le a_m$$
          then roots of $f(z)$ belong to the closed unit disk $|z| le 1$.


        2. If the coefficients $a_k$ are non-ascending,
          $$ a_0 ge a_1 ge cdots ge a_m > 0$$
          then roots of $f(z)$ lie outside the open unit disk (i.e. $|z| ge 1$ for all the roots)


        3. In general, the roots of $f(z)$ belong to the closed annulus
          $$min_{1le k le m} frac{a_{k-1}}{a_k} le |z| le max_{1 le k le m}frac{a_{k-1}}{a_k}$$





        Since it is easy to derive any one of these results from the other two,
        these results are typically treated as a single theorem known as the Eneström-Kakeya Theorem.



        For a proof of the first result, see answers of a related question. In particular, the answer by Ayman Hourieh which uses Rouché's theorem.



        Back to the problem at hand. It is easy to see we can rewrite $P_n(z)$ as $z f(z)$ for some polynomial $f(z)$ with real and positive coefficients. Apply the third result, we immediately find aside from a root at $z = 0$, the remaining $n-1$ roots of $P_n(z)$ lies within the closed annulus $$frac18 le |z| le left(frac{n-1
        }{n}right)^3 < 1$$






        share|cite|improve this answer














        Let $$g(z) = sum_{k=0}^n z^k = frac{z^{n+1}-1}{z-1}$$



        It is easy to see the roots of $g(z)$ lie on the unit circle $|z|= 1 $ and all of them are simple.



        By Gauss-Lucas theorem, the roots of $g'(z)$ belong to the convex hull of the roots of $g$. Since the closed unit disk $|z| le 1$ is convex, this convex hull is a subset of the closed unit disk.
        Notice the convex hull is a $n$-gon which intersect the unit circle $|z| = 1$ only at the roots of $g(z)$. Since the roots of $g(z)$ are simple, none of them can be root of $g'(z)$. As a result, the roots of $g'(z)$ belongs to the open unit disk $|z| < 1$.



        Since $zg'(z)$ differs from $g'(z)$ by only a root at $0$, the roots of



        $$zg'(z) = sum_{k=1}^n k z^k$$



        belong to the open unit disk $|z| < 1$.



        Apply Gauss-Lucas theorem again and then add a root at $z = 0$, we find the roots of



        $$left(zfrac{d}{dz}right)^2 g(z) = sum_{k=1}^n k^2 z^k$$
        belong to the open unit disk $|z| < 1$.
        Repeat this process one more time, we find all the zeros of



        $$left(zfrac{d}{dz}right)^3 g(z) = sum_{k=1}^n k^3 z^k = P_n(z)$$
        lie inside the unit circle.



        Notes



        There are other ways to arrive at same conclusion. In particular, we can use following results:




        Let $displaystyle;f(z) = sum_{k=0}^m a_k z^k$ be any polynomial with real and positive coefficients.




        1. If the coefficients $a_k$ are non-descending,
          $$0 < a_0 le a_2 le cdots le a_m$$
          then roots of $f(z)$ belong to the closed unit disk $|z| le 1$.


        2. If the coefficients $a_k$ are non-ascending,
          $$ a_0 ge a_1 ge cdots ge a_m > 0$$
          then roots of $f(z)$ lie outside the open unit disk (i.e. $|z| ge 1$ for all the roots)


        3. In general, the roots of $f(z)$ belong to the closed annulus
          $$min_{1le k le m} frac{a_{k-1}}{a_k} le |z| le max_{1 le k le m}frac{a_{k-1}}{a_k}$$





        Since it is easy to derive any one of these results from the other two,
        these results are typically treated as a single theorem known as the Eneström-Kakeya Theorem.



        For a proof of the first result, see answers of a related question. In particular, the answer by Ayman Hourieh which uses Rouché's theorem.



        Back to the problem at hand. It is easy to see we can rewrite $P_n(z)$ as $z f(z)$ for some polynomial $f(z)$ with real and positive coefficients. Apply the third result, we immediately find aside from a root at $z = 0$, the remaining $n-1$ roots of $P_n(z)$ lies within the closed annulus $$frac18 le |z| le left(frac{n-1
        }{n}right)^3 < 1$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 18 hours ago

























        answered yesterday









        achille hui

        95.6k5130257




        95.6k5130257






























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