Entire function either constant or has a zero
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Suppose that a Taylor series of an entire function $f$ converges to $f$ uniformly in $mathbb{C}$. How do I show that either $f$ is a non-zero constant or $f$ has a zero?
I was thinking about either: if f is everywhere nonzero then form $g = 1/f $ and then use $ |fg| = 1 $ to show that $ f $ is bounded, hence constant, or suppose that $ f $ is not constant, then show that $ f $ has a zero somehow. But I don't know how to do either. How do I prove this?
complex-analysis entire-functions
edited Dec 30 '18 at 15:05
José Carlos Santos
170k23132238
asked Dec 30 '18 at 14:31
calmcalm
1387
$endgroup$
add a comment |
$begingroup$
Suppose that a Taylor series of an entire function $f$ converges to $f$ uniformly in $mathbb{C}$. How do I show that either $f$ is a non-zero constant or $f$ has a zero?
I was thinking about either: if f is everywhere nonzero then form $g = 1/f $ and then use $ |fg| = 1 $ to show that $ f $ is bounded, hence constant, or suppose that $ f $ is not constant, then show that $ f $ has a zero somehow. But I don't know how to do either. How do I prove this?
complex-analysis entire-functions
edited Dec 30 '18 at 15:05
José Carlos Santos
170k23132238
asked Dec 30 '18 at 14:31
calmcalm
1387
$endgroup$
$begingroup$
You might want to drop the condition that the Taylor series converges uniformly on $mathbb{C}$, I think this is only satisfied by Polynomials.
$endgroup$
– 0x539
Dec 30 '18 at 14:37
$begingroup$
@0x539 You can't drop it:for example the exponential function doesn't have a zero.
$endgroup$
– ploosu2
Dec 30 '18 at 15:04
$begingroup$
@ploosu2 You're right, my bad.
$endgroup$
– 0x539
Dec 30 '18 at 15:09
add a comment |
$begingroup$
Suppose that a Taylor series of an entire function $f$ converges to $f$ uniformly in $mathbb{C}$. How do I show that either $f$ is a non-zero constant or $f$ has a zero?
I was thinking about either: if f is everywhere nonzero then form $g = 1/f $ and then use $ |fg| = 1 $ to show that $ f $ is bounded, hence constant, or suppose that $ f $ is not constant, then show that $ f $ has a zero somehow. But I don't know how to do either. How do I prove this?
complex-analysis entire-functions
edited Dec 30 '18 at 15:05
José Carlos Santos
170k23132238
asked Dec 30 '18 at 14:31
calmcalm
1387
$endgroup$
Suppose that a Taylor series of an entire function $f$ converges to $f$ uniformly in $mathbb{C}$. How do I show that either $f$ is a non-zero constant or $f$ has a zero?
I was thinking about either: if f is everywhere nonzero then form $g = 1/f $ and then use $ |fg| = 1 $ to show that $ f $ is bounded, hence constant, or suppose that $ f $ is not constant, then show that $ f $ has a zero somehow. But I don't know how to do either. How do I prove this?
complex-analysis entire-functions
complex-analysis entire-functions
edited Dec 30 '18 at 15:05
José Carlos Santos
170k23132238
asked Dec 30 '18 at 14:31
calmcalm
1387
edited Dec 30 '18 at 15:05
José Carlos Santos
170k23132238
asked Dec 30 '18 at 14:31
calmcalm
1387
edited Dec 30 '18 at 15:05
José Carlos Santos
170k23132238
edited Dec 30 '18 at 15:05
José Carlos Santos
170k23132238
edited Dec 30 '18 at 15:05
José Carlos Santos
170k23132238
170k23132238
asked Dec 30 '18 at 14:31
calmcalm
1387
asked Dec 30 '18 at 14:31
calmcalm
1387
asked Dec 30 '18 at 14:31
calmcalm
1387
1387
$begingroup$
You might want to drop the condition that the Taylor series converges uniformly on $mathbb{C}$, I think this is only satisfied by Polynomials.
$endgroup$
– 0x539
Dec 30 '18 at 14:37
$begingroup$
@0x539 You can't drop it:for example the exponential function doesn't have a zero.
$endgroup$
– ploosu2
Dec 30 '18 at 15:04
$begingroup$
@ploosu2 You're right, my bad.
$endgroup$
– 0x539
Dec 30 '18 at 15:09
add a comment |
$begingroup$
You might want to drop the condition that the Taylor series converges uniformly on $mathbb{C}$, I think this is only satisfied by Polynomials.
$endgroup$
– 0x539
Dec 30 '18 at 14:37
$begingroup$
@0x539 You can't drop it:for example the exponential function doesn't have a zero.
$endgroup$
– ploosu2
Dec 30 '18 at 15:04
$begingroup$
@ploosu2 You're right, my bad.
$endgroup$
– 0x539
Dec 30 '18 at 15:09
$begingroup$
You might want to drop the condition that the Taylor series converges uniformly on $mathbb{C}$, I think this is only satisfied by Polynomials.
$endgroup$
– 0x539
Dec 30 '18 at 14:37
$begingroup$
You might want to drop the condition that the Taylor series converges uniformly on $mathbb{C}$, I think this is only satisfied by Polynomials.
$endgroup$
– 0x539
Dec 30 '18 at 14:37
$begingroup$
@0x539 You can't drop it:for example the exponential function doesn't have a zero.
$endgroup$
– ploosu2
Dec 30 '18 at 15:04
$begingroup$
@0x539 You can't drop it:for example the exponential function doesn't have a zero.
$endgroup$
– ploosu2
Dec 30 '18 at 15:04
$begingroup$
@ploosu2 You're right, my bad.
$endgroup$
– 0x539
Dec 30 '18 at 15:09
$begingroup$
@ploosu2 You're right, my bad.
$endgroup$
– 0x539
Dec 30 '18 at 15:09
add a comment |
1 Answer
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Since the Taylor series of $f$ converges uniformly to $f$, $f$ is a polynomial function. Therefore, if $f$ is not constant, then it has at least a zero, by the Fundamental Theorem of Algebra.
answered Dec 30 '18 at 14:37
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Since the Taylor series of $f$ converges uniformly to $f$, $f$ is a polynomial function. Therefore, if $f$ is not constant, then it has at least a zero, by the Fundamental Theorem of Algebra.
answered Dec 30 '18 at 14:37
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
add a comment |
$begingroup$
Since the Taylor series of $f$ converges uniformly to $f$, $f$ is a polynomial function. Therefore, if $f$ is not constant, then it has at least a zero, by the Fundamental Theorem of Algebra.
answered Dec 30 '18 at 14:37
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
add a comment |
$begingroup$
Since the Taylor series of $f$ converges uniformly to $f$, $f$ is a polynomial function. Therefore, if $f$ is not constant, then it has at least a zero, by the Fundamental Theorem of Algebra.
answered Dec 30 '18 at 14:37
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
Since the Taylor series of $f$ converges uniformly to $f$, $f$ is a polynomial function. Therefore, if $f$ is not constant, then it has at least a zero, by the Fundamental Theorem of Algebra.
answered Dec 30 '18 at 14:37
José Carlos SantosJosé Carlos Santos
170k23132238
answered Dec 30 '18 at 14:37
José Carlos SantosJosé Carlos Santos
170k23132238
answered Dec 30 '18 at 14:37
José Carlos SantosJosé Carlos Santos
170k23132238
answered Dec 30 '18 at 14:37
José Carlos SantosJosé Carlos Santos
170k23132238
170k23132238
add a comment |
add a comment |
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$begingroup$
You might want to drop the condition that the Taylor series converges uniformly on $mathbb{C}$, I think this is only satisfied by Polynomials.
$endgroup$
– 0x539
Dec 30 '18 at 14:37
$begingroup$
@0x539 You can't drop it:for example the exponential function doesn't have a zero.
$endgroup$
– ploosu2
Dec 30 '18 at 15:04
$begingroup$
@ploosu2 You're right, my bad.
$endgroup$
– 0x539
Dec 30 '18 at 15:09