Is the principal root defined at the right half-plane continuous at $0$?
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I'm reading a proof of a theorem. In the proof the author says that the principal root $z^{1- delta/2}$ (that is, the function $z^{1- delta/2}:=exp(mbox{Log}(z)(1- delta/2))$, where $mbox{Log}$ is the principal branch of logarithm function) for $ delta in (0,1]$ is continuous at the boundary of the right half-plane. But I think that maybe there is problem at $0$. Is the principal root $z^{1- delta/2}$ continuous at $0$? Can we say that $exp(z^{1- delta/2})$ is continuous at $0$?
complex-analysis complex-numbers
asked Dec 2 '18 at 20:24
KSM1743KSM1743
164
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add a comment |
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I'm reading a proof of a theorem. In the proof the author says that the principal root $z^{1- delta/2}$ (that is, the function $z^{1- delta/2}:=exp(mbox{Log}(z)(1- delta/2))$, where $mbox{Log}$ is the principal branch of logarithm function) for $ delta in (0,1]$ is continuous at the boundary of the right half-plane. But I think that maybe there is problem at $0$. Is the principal root $z^{1- delta/2}$ continuous at $0$? Can we say that $exp(z^{1- delta/2})$ is continuous at $0$?
complex-analysis complex-numbers
asked Dec 2 '18 at 20:24
KSM1743KSM1743
164
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What is your doubt? Can you explain why you think that the function you have described might fail to be continuous at zero?
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– Xander Henderson
Dec 2 '18 at 20:29
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I think that the function might fail to be continuous at zero because the principal branch of $z^{1-delta/2}$ is defined using the principal branch $ mbox{Log}$ of logarithm function.
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– KSM1743
Dec 2 '18 at 20:32
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I think you can define that function at $0$ as $0$.
$endgroup$
– Minysh
Dec 3 '18 at 0:33
add a comment |
$begingroup$
I'm reading a proof of a theorem. In the proof the author says that the principal root $z^{1- delta/2}$ (that is, the function $z^{1- delta/2}:=exp(mbox{Log}(z)(1- delta/2))$, where $mbox{Log}$ is the principal branch of logarithm function) for $ delta in (0,1]$ is continuous at the boundary of the right half-plane. But I think that maybe there is problem at $0$. Is the principal root $z^{1- delta/2}$ continuous at $0$? Can we say that $exp(z^{1- delta/2})$ is continuous at $0$?
complex-analysis complex-numbers
asked Dec 2 '18 at 20:24
KSM1743KSM1743
164
$endgroup$
I'm reading a proof of a theorem. In the proof the author says that the principal root $z^{1- delta/2}$ (that is, the function $z^{1- delta/2}:=exp(mbox{Log}(z)(1- delta/2))$, where $mbox{Log}$ is the principal branch of logarithm function) for $ delta in (0,1]$ is continuous at the boundary of the right half-plane. But I think that maybe there is problem at $0$. Is the principal root $z^{1- delta/2}$ continuous at $0$? Can we say that $exp(z^{1- delta/2})$ is continuous at $0$?
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Dec 2 '18 at 20:24
KSM1743KSM1743
164
asked Dec 2 '18 at 20:24
KSM1743KSM1743
164
edited Dec 2 '18 at 20:41
KSM1743
asked Dec 2 '18 at 20:24
KSM1743KSM1743
164
asked Dec 2 '18 at 20:24
KSM1743KSM1743
164
asked Dec 2 '18 at 20:24
KSM1743KSM1743
164
164
$begingroup$
What is your doubt? Can you explain why you think that the function you have described might fail to be continuous at zero?
$endgroup$
– Xander Henderson
Dec 2 '18 at 20:29
$begingroup$
I think that the function might fail to be continuous at zero because the principal branch of $z^{1-delta/2}$ is defined using the principal branch $ mbox{Log}$ of logarithm function.
$endgroup$
– KSM1743
Dec 2 '18 at 20:32
$begingroup$
I think you can define that function at $0$ as $0$.
$endgroup$
– Minysh
Dec 3 '18 at 0:33
add a comment |
$begingroup$
What is your doubt? Can you explain why you think that the function you have described might fail to be continuous at zero?
$endgroup$
– Xander Henderson
Dec 2 '18 at 20:29
$begingroup$
I think that the function might fail to be continuous at zero because the principal branch of $z^{1-delta/2}$ is defined using the principal branch $ mbox{Log}$ of logarithm function.
$endgroup$
– KSM1743
Dec 2 '18 at 20:32
$begingroup$
I think you can define that function at $0$ as $0$.
$endgroup$
– Minysh
Dec 3 '18 at 0:33
$begingroup$
What is your doubt? Can you explain why you think that the function you have described might fail to be continuous at zero?
$endgroup$
– Xander Henderson
Dec 2 '18 at 20:29
$begingroup$
What is your doubt? Can you explain why you think that the function you have described might fail to be continuous at zero?
$endgroup$
– Xander Henderson
Dec 2 '18 at 20:29
$begingroup$
I think that the function might fail to be continuous at zero because the principal branch of $z^{1-delta/2}$ is defined using the principal branch $ mbox{Log}$ of logarithm function.
$endgroup$
– KSM1743
Dec 2 '18 at 20:32
$begingroup$
I think that the function might fail to be continuous at zero because the principal branch of $z^{1-delta/2}$ is defined using the principal branch $ mbox{Log}$ of logarithm function.
$endgroup$
– KSM1743
Dec 2 '18 at 20:32
$begingroup$
I think you can define that function at $0$ as $0$.
$endgroup$
– Minysh
Dec 3 '18 at 0:33
$begingroup$
I think you can define that function at $0$ as $0$.
$endgroup$
– Minysh
Dec 3 '18 at 0:33
add a comment |
1 Answer
1
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Let $z = re^{itheta}$ and let $epsilon = 1 - frac delta 2$. Note that $frac 12 le epsilon < 1$. Now $$|exp(operatorname{Log}(z)epsilon)| = r^epsilon$$
As $z to 0, r = |z| to 0$, and therefore $r^epsilon to 0$. That is, $$exp(operatorname{Log}(z)epsilon) to 0$$
While $0$ is not in the domain of $operatorname{Log}$, and therefore the full expression, this is a straight-forward continuous extension.
answered Dec 3 '18 at 4:09
Paul SinclairPaul Sinclair
20.2k21443
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1 Answer
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1 Answer
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$begingroup$
Let $z = re^{itheta}$ and let $epsilon = 1 - frac delta 2$. Note that $frac 12 le epsilon < 1$. Now $$|exp(operatorname{Log}(z)epsilon)| = r^epsilon$$
As $z to 0, r = |z| to 0$, and therefore $r^epsilon to 0$. That is, $$exp(operatorname{Log}(z)epsilon) to 0$$
While $0$ is not in the domain of $operatorname{Log}$, and therefore the full expression, this is a straight-forward continuous extension.
answered Dec 3 '18 at 4:09
Paul SinclairPaul Sinclair
20.2k21443
$endgroup$
add a comment |
$begingroup$
Let $z = re^{itheta}$ and let $epsilon = 1 - frac delta 2$. Note that $frac 12 le epsilon < 1$. Now $$|exp(operatorname{Log}(z)epsilon)| = r^epsilon$$
As $z to 0, r = |z| to 0$, and therefore $r^epsilon to 0$. That is, $$exp(operatorname{Log}(z)epsilon) to 0$$
While $0$ is not in the domain of $operatorname{Log}$, and therefore the full expression, this is a straight-forward continuous extension.
answered Dec 3 '18 at 4:09
Paul SinclairPaul Sinclair
20.2k21443
$endgroup$
add a comment |
$begingroup$
Let $z = re^{itheta}$ and let $epsilon = 1 - frac delta 2$. Note that $frac 12 le epsilon < 1$. Now $$|exp(operatorname{Log}(z)epsilon)| = r^epsilon$$
As $z to 0, r = |z| to 0$, and therefore $r^epsilon to 0$. That is, $$exp(operatorname{Log}(z)epsilon) to 0$$
While $0$ is not in the domain of $operatorname{Log}$, and therefore the full expression, this is a straight-forward continuous extension.
answered Dec 3 '18 at 4:09
Paul SinclairPaul Sinclair
20.2k21443
$endgroup$
Let $z = re^{itheta}$ and let $epsilon = 1 - frac delta 2$. Note that $frac 12 le epsilon < 1$. Now $$|exp(operatorname{Log}(z)epsilon)| = r^epsilon$$
As $z to 0, r = |z| to 0$, and therefore $r^epsilon to 0$. That is, $$exp(operatorname{Log}(z)epsilon) to 0$$
While $0$ is not in the domain of $operatorname{Log}$, and therefore the full expression, this is a straight-forward continuous extension.
answered Dec 3 '18 at 4:09
Paul SinclairPaul Sinclair
20.2k21443
answered Dec 3 '18 at 4:09
Paul SinclairPaul Sinclair
20.2k21443
answered Dec 3 '18 at 4:09
Paul SinclairPaul Sinclair
20.2k21443
answered Dec 3 '18 at 4:09
Paul SinclairPaul Sinclair
20.2k21443
20.2k21443
add a comment |
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$begingroup$
What is your doubt? Can you explain why you think that the function you have described might fail to be continuous at zero?
$endgroup$
– Xander Henderson
Dec 2 '18 at 20:29
$begingroup$
I think that the function might fail to be continuous at zero because the principal branch of $z^{1-delta/2}$ is defined using the principal branch $ mbox{Log}$ of logarithm function.
$endgroup$
– KSM1743
Dec 2 '18 at 20:32
$begingroup$
I think you can define that function at $0$ as $0$.
$endgroup$
– Minysh
Dec 3 '18 at 0:33