A power series with decreasing positive coefficients has no zeroes in the disk












8












$begingroup$


Let $sum_{n=0}^infty a_n z^n$ be a formal complex power series, with $a_n$ strictly decreasing to 1 as $nto infty$. It is easy to see that the radius of convergence is 1, so that this power series represents a function $f$ holomorphic on the disk. The question is to



Show that f has no zeroes on the disk.



I've been looking at this a while and feel and tried a few different things.



$cdot$ One easy observation is the similarity of this series with $frac{1}{1-z}$, whose coefficients do not strictly decrease but which otherwise is an example of the thing to be proven.



$cdot$ Rouché's theorem doesn't seem to have an immediate application—I've tried comparing $f$ to $frac{1}{1-z}$ or to its partial sums by Rouché.



$cdot$ If the partial sums had no zeroes, then neither would $f$ by Hurwitz's theorem



$cdot$ If we could show the function had positive real part we'd obviously be done, which seems highly plausible—the condition gives that $f(-1) = sum a_n cospi n > 0$, but I don't know how to extend this argument to arguments other than $pi$. Besides, if this technique were to work, it seems it would rely most on algebraic manipulations of power series, and I would vastly prefer a classical complex analytic approach (argument principle, Rouché's theorem, etc)



Please advise!






complex-analysis power-series







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$endgroup$

















    8












    $begingroup$


    Let $sum_{n=0}^infty a_n z^n$ be a formal complex power series, with $a_n$ strictly decreasing to 1 as $nto infty$. It is easy to see that the radius of convergence is 1, so that this power series represents a function $f$ holomorphic on the disk. The question is to



    Show that f has no zeroes on the disk.



    I've been looking at this a while and feel and tried a few different things.



    $cdot$ One easy observation is the similarity of this series with $frac{1}{1-z}$, whose coefficients do not strictly decrease but which otherwise is an example of the thing to be proven.



    $cdot$ Rouché's theorem doesn't seem to have an immediate application—I've tried comparing $f$ to $frac{1}{1-z}$ or to its partial sums by Rouché.



    $cdot$ If the partial sums had no zeroes, then neither would $f$ by Hurwitz's theorem



    $cdot$ If we could show the function had positive real part we'd obviously be done, which seems highly plausible—the condition gives that $f(-1) = sum a_n cospi n > 0$, but I don't know how to extend this argument to arguments other than $pi$. Besides, if this technique were to work, it seems it would rely most on algebraic manipulations of power series, and I would vastly prefer a classical complex analytic approach (argument principle, Rouché's theorem, etc)



    Please advise!






    complex-analysis power-series







    share|cite|improve this question









    $endgroup$















      8












      8








      8


      2



      $begingroup$


      Let $sum_{n=0}^infty a_n z^n$ be a formal complex power series, with $a_n$ strictly decreasing to 1 as $nto infty$. It is easy to see that the radius of convergence is 1, so that this power series represents a function $f$ holomorphic on the disk. The question is to



      Show that f has no zeroes on the disk.



      I've been looking at this a while and feel and tried a few different things.



      $cdot$ One easy observation is the similarity of this series with $frac{1}{1-z}$, whose coefficients do not strictly decrease but which otherwise is an example of the thing to be proven.



      $cdot$ Rouché's theorem doesn't seem to have an immediate application—I've tried comparing $f$ to $frac{1}{1-z}$ or to its partial sums by Rouché.



      $cdot$ If the partial sums had no zeroes, then neither would $f$ by Hurwitz's theorem



      $cdot$ If we could show the function had positive real part we'd obviously be done, which seems highly plausible—the condition gives that $f(-1) = sum a_n cospi n > 0$, but I don't know how to extend this argument to arguments other than $pi$. Besides, if this technique were to work, it seems it would rely most on algebraic manipulations of power series, and I would vastly prefer a classical complex analytic approach (argument principle, Rouché's theorem, etc)



      Please advise!






      complex-analysis power-series







      share|cite|improve this question









      $endgroup$




      Let $sum_{n=0}^infty a_n z^n$ be a formal complex power series, with $a_n$ strictly decreasing to 1 as $nto infty$. It is easy to see that the radius of convergence is 1, so that this power series represents a function $f$ holomorphic on the disk. The question is to



      Show that f has no zeroes on the disk.



      I've been looking at this a while and feel and tried a few different things.



      $cdot$ One easy observation is the similarity of this series with $frac{1}{1-z}$, whose coefficients do not strictly decrease but which otherwise is an example of the thing to be proven.



      $cdot$ Rouché's theorem doesn't seem to have an immediate application—I've tried comparing $f$ to $frac{1}{1-z}$ or to its partial sums by Rouché.



      $cdot$ If the partial sums had no zeroes, then neither would $f$ by Hurwitz's theorem



      $cdot$ If we could show the function had positive real part we'd obviously be done, which seems highly plausible—the condition gives that $f(-1) = sum a_n cospi n > 0$, but I don't know how to extend this argument to arguments other than $pi$. Besides, if this technique were to work, it seems it would rely most on algebraic manipulations of power series, and I would vastly prefer a classical complex analytic approach (argument principle, Rouché's theorem, etc)



      Please advise!






      complex-analysis power-series




      complex-analysis power-series






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      share|cite|improve this question











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      asked 14 hours ago









      Van LatimerVan Latimer

      363110




      363110






















          1 Answer
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          9












          $begingroup$

          Let us consider $f(z) =(1-z)sumlimits_{n=0}^infty a_n z^n =a_0+sumlimits_{n=1}^infty (a_n-a_{n-1})z^n$. Suppose $f(re^{itheta})=0$ for $rle 1$. Then it follows
          $$
          a_0 =sum_{n=1}^infty (a_{n-1}-a_n)r^ne^{intheta},
          $$           hence
          $$
          a_0=left|sum_{n=1}^infty (a_{n-1}-a_n)r^ne^{intheta}right|le sum_{n=1}^infty (a_{n-1}-a_n)r^nlesum_{n=1}^infty(a_{n-1}-a_n)=a_0-1,
          $$           which leads to a contradiction. Since $f$ has no zeros on the unit disk, neither does $frac{f(z)}{1-z}=sumlimits_{n=0}^infty a_n z^n$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That's beautiful, thank you. I looked at the series (1-z) sum a_n z^n but didn't see how to get this contradiction.
            $endgroup$
            – Van Latimer
            14 hours ago











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          1 Answer
          1






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          active

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          active

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          9












          $begingroup$

          Let us consider $f(z) =(1-z)sumlimits_{n=0}^infty a_n z^n =a_0+sumlimits_{n=1}^infty (a_n-a_{n-1})z^n$. Suppose $f(re^{itheta})=0$ for $rle 1$. Then it follows
          $$
          a_0 =sum_{n=1}^infty (a_{n-1}-a_n)r^ne^{intheta},
          $$           hence
          $$
          a_0=left|sum_{n=1}^infty (a_{n-1}-a_n)r^ne^{intheta}right|le sum_{n=1}^infty (a_{n-1}-a_n)r^nlesum_{n=1}^infty(a_{n-1}-a_n)=a_0-1,
          $$           which leads to a contradiction. Since $f$ has no zeros on the unit disk, neither does $frac{f(z)}{1-z}=sumlimits_{n=0}^infty a_n z^n$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That's beautiful, thank you. I looked at the series (1-z) sum a_n z^n but didn't see how to get this contradiction.
            $endgroup$
            – Van Latimer
            14 hours ago
















          9












          $begingroup$

          Let us consider $f(z) =(1-z)sumlimits_{n=0}^infty a_n z^n =a_0+sumlimits_{n=1}^infty (a_n-a_{n-1})z^n$. Suppose $f(re^{itheta})=0$ for $rle 1$. Then it follows
          $$
          a_0 =sum_{n=1}^infty (a_{n-1}-a_n)r^ne^{intheta},
          $$           hence
          $$
          a_0=left|sum_{n=1}^infty (a_{n-1}-a_n)r^ne^{intheta}right|le sum_{n=1}^infty (a_{n-1}-a_n)r^nlesum_{n=1}^infty(a_{n-1}-a_n)=a_0-1,
          $$           which leads to a contradiction. Since $f$ has no zeros on the unit disk, neither does $frac{f(z)}{1-z}=sumlimits_{n=0}^infty a_n z^n$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That's beautiful, thank you. I looked at the series (1-z) sum a_n z^n but didn't see how to get this contradiction.
            $endgroup$
            – Van Latimer
            14 hours ago














          9












          9








          9





          $begingroup$

          Let us consider $f(z) =(1-z)sumlimits_{n=0}^infty a_n z^n =a_0+sumlimits_{n=1}^infty (a_n-a_{n-1})z^n$. Suppose $f(re^{itheta})=0$ for $rle 1$. Then it follows
          $$
          a_0 =sum_{n=1}^infty (a_{n-1}-a_n)r^ne^{intheta},
          $$           hence
          $$
          a_0=left|sum_{n=1}^infty (a_{n-1}-a_n)r^ne^{intheta}right|le sum_{n=1}^infty (a_{n-1}-a_n)r^nlesum_{n=1}^infty(a_{n-1}-a_n)=a_0-1,
          $$           which leads to a contradiction. Since $f$ has no zeros on the unit disk, neither does $frac{f(z)}{1-z}=sumlimits_{n=0}^infty a_n z^n$.






          share|cite|improve this answer











          $endgroup$



          Let us consider $f(z) =(1-z)sumlimits_{n=0}^infty a_n z^n =a_0+sumlimits_{n=1}^infty (a_n-a_{n-1})z^n$. Suppose $f(re^{itheta})=0$ for $rle 1$. Then it follows
          $$
          a_0 =sum_{n=1}^infty (a_{n-1}-a_n)r^ne^{intheta},
          $$           hence
          $$
          a_0=left|sum_{n=1}^infty (a_{n-1}-a_n)r^ne^{intheta}right|le sum_{n=1}^infty (a_{n-1}-a_n)r^nlesum_{n=1}^infty(a_{n-1}-a_n)=a_0-1,
          $$           which leads to a contradiction. Since $f$ has no zeros on the unit disk, neither does $frac{f(z)}{1-z}=sumlimits_{n=0}^infty a_n z^n$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 4 hours ago

























          answered 14 hours ago









          SongSong

          14.1k1633




          14.1k1633












          • $begingroup$
            That's beautiful, thank you. I looked at the series (1-z) sum a_n z^n but didn't see how to get this contradiction.
            $endgroup$
            – Van Latimer
            14 hours ago


















          • $begingroup$
            That's beautiful, thank you. I looked at the series (1-z) sum a_n z^n but didn't see how to get this contradiction.
            $endgroup$
            – Van Latimer
            14 hours ago
















          $begingroup$
          That's beautiful, thank you. I looked at the series (1-z) sum a_n z^n but didn't see how to get this contradiction.
          $endgroup$
          – Van Latimer
          14 hours ago




          $begingroup$
          That's beautiful, thank you. I looked at the series (1-z) sum a_n z^n but didn't see how to get this contradiction.
          $endgroup$
          – Van Latimer
          14 hours ago


















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