A power series with decreasing positive coefficients has no zeroes in the disk
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$begingroup$
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 $frac11-z$, 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 $frac11-z$ 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
asked Feb 18 at 16:04
Van LatimerVan Latimer
363110
$endgroup$
add a comment |
$begingroup$
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 $frac11-z$, 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 $frac11-z$ 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
asked Feb 18 at 16:04
Van LatimerVan Latimer
363110
$endgroup$
1
$begingroup$
Possible duplicate of Let $(a_n)_n geq 0$ be a strictly decreasing sequence of positive real numbers , and let $z in mathbb C$ , $|z| < 1$.
$endgroup$
– Martin R
Feb 19 at 8:02
add a comment |
$begingroup$
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 $frac11-z$, 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 $frac11-z$ 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
asked Feb 18 at 16:04
Van LatimerVan Latimer
363110
$endgroup$
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 $frac11-z$, 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 $frac11-z$ 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
asked Feb 18 at 16:04
Van LatimerVan Latimer
363110
asked Feb 18 at 16:04
Van LatimerVan Latimer
363110
asked Feb 18 at 16:04
Van LatimerVan Latimer
363110
asked Feb 18 at 16:04
Van LatimerVan Latimer
363110
asked Feb 18 at 16:04
Van LatimerVan Latimer
363110
363110
1
$begingroup$
Possible duplicate of Let $(a_n)_n geq 0$ be a strictly decreasing sequence of positive real numbers , and let $z in mathbb C$ , $|z| < 1$.
$endgroup$
– Martin R
Feb 19 at 8:02
add a comment |
1
$begingroup$
Possible duplicate of Let $(a_n)_n geq 0$ be a strictly decreasing sequence of positive real numbers , and let $z in mathbb C$ , $|z| < 1$.
$endgroup$
– Martin R
Feb 19 at 8:02
1
1
$begingroup$
Possible duplicate of Let $(a_n)_n geq 0$ be a strictly decreasing sequence of positive real numbers , and let $z in mathbb C$ , $|z| < 1$.
$endgroup$
– Martin R
Feb 19 at 8:02
$begingroup$
Possible duplicate of Let $(a_n)_n geq 0$ be a strictly decreasing sequence of positive real numbers , and let $z in mathbb C$ , $|z| < 1$.
$endgroup$
– Martin R
Feb 19 at 8:02
add a comment |
1 Answer
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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
$$
a_0=left|sum_n=1^infty (a_n-1-a_n)r^ne^inthetaright|le sum_n=1^infty (a_n-1-a_n)r^nlesum_n=1^infty(a_n-1-a_n)=a_0-1,
$$ which leads to a contradiction. Since $f$ has no zeros on the unit disk, neither does $fracf(z)1-z=sumlimits_n=0^infty a_n z^n$.
answered Feb 18 at 16:31
SongSong
17.5k21346
$endgroup$
$begingroup$
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
Feb 18 at 16:50
add a comment |
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1 Answer
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$begingroup$
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
$$
a_0=left|sum_n=1^infty (a_n-1-a_n)r^ne^inthetaright|le sum_n=1^infty (a_n-1-a_n)r^nlesum_n=1^infty(a_n-1-a_n)=a_0-1,
$$ which leads to a contradiction. Since $f$ has no zeros on the unit disk, neither does $fracf(z)1-z=sumlimits_n=0^infty a_n z^n$.
answered Feb 18 at 16:31
SongSong
17.5k21346
$endgroup$
$begingroup$
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
Feb 18 at 16:50
add a comment |
$begingroup$
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
$$
a_0=left|sum_n=1^infty (a_n-1-a_n)r^ne^inthetaright|le sum_n=1^infty (a_n-1-a_n)r^nlesum_n=1^infty(a_n-1-a_n)=a_0-1,
$$ which leads to a contradiction. Since $f$ has no zeros on the unit disk, neither does $fracf(z)1-z=sumlimits_n=0^infty a_n z^n$.
answered Feb 18 at 16:31
SongSong
17.5k21346
$endgroup$
$begingroup$
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
Feb 18 at 16:50
add a comment |
$begingroup$
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
$$
a_0=left|sum_n=1^infty (a_n-1-a_n)r^ne^inthetaright|le sum_n=1^infty (a_n-1-a_n)r^nlesum_n=1^infty(a_n-1-a_n)=a_0-1,
$$ which leads to a contradiction. Since $f$ has no zeros on the unit disk, neither does $fracf(z)1-z=sumlimits_n=0^infty a_n z^n$.
answered Feb 18 at 16:31
SongSong
17.5k21346
$endgroup$
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
$$
a_0=left|sum_n=1^infty (a_n-1-a_n)r^ne^inthetaright|le sum_n=1^infty (a_n-1-a_n)r^nlesum_n=1^infty(a_n-1-a_n)=a_0-1,
$$ which leads to a contradiction. Since $f$ has no zeros on the unit disk, neither does $fracf(z)1-z=sumlimits_n=0^infty a_n z^n$.
answered Feb 18 at 16:31
SongSong
17.5k21346
edited Feb 19 at 2:06
answered Feb 18 at 16:31
SongSong
17.5k21346
answered Feb 18 at 16:31
SongSong
17.5k21346
answered Feb 18 at 16:31
SongSong
17.5k21346
17.5k21346
$begingroup$
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
Feb 18 at 16:50
add a comment |
$begingroup$
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
Feb 18 at 16:50
$begingroup$
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
Feb 18 at 16:50
$begingroup$
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
Feb 18 at 16:50
add a comment |
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$begingroup$
Possible duplicate of Let $(a_n)_n geq 0$ be a strictly decreasing sequence of positive real numbers , and let $z in mathbb C$ , $|z| < 1$.
$endgroup$
– Martin R
Feb 19 at 8:02