Construct an analytic function with zeros at intersection of zeros of other analytic functions
$begingroup$
Given two complex analytic functions f and g on some compact domain $D$, I want to construct a new analytic function h on $D$ such that
$$h(z) = 0 iff f(z) = 0 text{ and } g(z) = 0$$
I want to do this without finding the zeros of $f$ and $g$.
If I wanted to find $h$ such that
$$
h(z) = 0 iff f(z) = 0 text{ or } g(z) = 0,
$$
this can be easily done by letting
$$
h(z) = f(z)g(z).
$$
In a more specific version, we can assume that zeros of $f$ and $g$ lie on the real line.
Thanks!
complex-analysis roots
edited Jan 7 at 7:53
Daniele Tampieri
1,9841619
asked Jan 7 at 5:43
Jaeyoon KimJaeyoon Kim
84
$endgroup$
add a comment |
$begingroup$
Given two complex analytic functions f and g on some compact domain $D$, I want to construct a new analytic function h on $D$ such that
$$h(z) = 0 iff f(z) = 0 text{ and } g(z) = 0$$
I want to do this without finding the zeros of $f$ and $g$.
If I wanted to find $h$ such that
$$
h(z) = 0 iff f(z) = 0 text{ or } g(z) = 0,
$$
this can be easily done by letting
$$
h(z) = f(z)g(z).
$$
In a more specific version, we can assume that zeros of $f$ and $g$ lie on the real line.
Thanks!
complex-analysis roots
edited Jan 7 at 7:53
Daniele Tampieri
1,9841619
asked Jan 7 at 5:43
Jaeyoon KimJaeyoon Kim
84
$endgroup$
2
$begingroup$
If you are asking for a simple formula for $h$ I doubt if such a thing exists.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 6:07
$begingroup$
For polynomials it's just the greatest common denominator. For general analytic functions I have no idea.
$endgroup$
– Arthur
Jan 7 at 6:41
$begingroup$
@KaviRamaMurthy such $h$ does exists. If we assume that $f$ and $g$ are not identically zero, then by isolated zeros, we know that there are finitely many zeros. Let $z_1,ldots,z_n$ be all the values such that $f(z_i)$ and $g(z_i)$ are zero. Let $h(z) = prod_{i}(z-z_i)$. But yes, I don't see how to easily compute for $h$.
$endgroup$
– Jaeyoon Kim
Jan 7 at 7:29
$begingroup$
@JaeyoonKim Surely, $h$ exists with the desired property. But you appear to seek a simple formula for $h$ and I suspect that no such formula exists.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 7:37
$begingroup$
If the zeros are simple you may try to find $u,v$ analytic such that $w(z)=frac{f'(z)}{f(z)}+e^{u(z)} frac{g'(z)}{g(z)}+v(z)$ has no zeros, so $f(z)g(z)w(z)$ is what you want. Asking if $u,v$ always exist is an interesting problem.
$endgroup$
– reuns
Jan 7 at 8:08
add a comment |
$begingroup$
Given two complex analytic functions f and g on some compact domain $D$, I want to construct a new analytic function h on $D$ such that
$$h(z) = 0 iff f(z) = 0 text{ and } g(z) = 0$$
I want to do this without finding the zeros of $f$ and $g$.
If I wanted to find $h$ such that
$$
h(z) = 0 iff f(z) = 0 text{ or } g(z) = 0,
$$
this can be easily done by letting
$$
h(z) = f(z)g(z).
$$
In a more specific version, we can assume that zeros of $f$ and $g$ lie on the real line.
Thanks!
complex-analysis roots
edited Jan 7 at 7:53
Daniele Tampieri
1,9841619
asked Jan 7 at 5:43
Jaeyoon KimJaeyoon Kim
84
$endgroup$
Given two complex analytic functions f and g on some compact domain $D$, I want to construct a new analytic function h on $D$ such that
$$h(z) = 0 iff f(z) = 0 text{ and } g(z) = 0$$
I want to do this without finding the zeros of $f$ and $g$.
If I wanted to find $h$ such that
$$
h(z) = 0 iff f(z) = 0 text{ or } g(z) = 0,
$$
this can be easily done by letting
$$
h(z) = f(z)g(z).
$$
In a more specific version, we can assume that zeros of $f$ and $g$ lie on the real line.
Thanks!
complex-analysis roots
complex-analysis roots
edited Jan 7 at 7:53
Daniele Tampieri
1,9841619
asked Jan 7 at 5:43
Jaeyoon KimJaeyoon Kim
84
edited Jan 7 at 7:53
Daniele Tampieri
1,9841619
asked Jan 7 at 5:43
Jaeyoon KimJaeyoon Kim
84
edited Jan 7 at 7:53
Daniele Tampieri
1,9841619
edited Jan 7 at 7:53
Daniele Tampieri
1,9841619
edited Jan 7 at 7:53
Daniele Tampieri
1,9841619
1,9841619
asked Jan 7 at 5:43
Jaeyoon KimJaeyoon Kim
84
asked Jan 7 at 5:43
Jaeyoon KimJaeyoon Kim
84
asked Jan 7 at 5:43
Jaeyoon KimJaeyoon Kim
84
84
2
$begingroup$
If you are asking for a simple formula for $h$ I doubt if such a thing exists.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 6:07
$begingroup$
For polynomials it's just the greatest common denominator. For general analytic functions I have no idea.
$endgroup$
– Arthur
Jan 7 at 6:41
$begingroup$
@KaviRamaMurthy such $h$ does exists. If we assume that $f$ and $g$ are not identically zero, then by isolated zeros, we know that there are finitely many zeros. Let $z_1,ldots,z_n$ be all the values such that $f(z_i)$ and $g(z_i)$ are zero. Let $h(z) = prod_{i}(z-z_i)$. But yes, I don't see how to easily compute for $h$.
$endgroup$
– Jaeyoon Kim
Jan 7 at 7:29
$begingroup$
@JaeyoonKim Surely, $h$ exists with the desired property. But you appear to seek a simple formula for $h$ and I suspect that no such formula exists.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 7:37
$begingroup$
If the zeros are simple you may try to find $u,v$ analytic such that $w(z)=frac{f'(z)}{f(z)}+e^{u(z)} frac{g'(z)}{g(z)}+v(z)$ has no zeros, so $f(z)g(z)w(z)$ is what you want. Asking if $u,v$ always exist is an interesting problem.
$endgroup$
– reuns
Jan 7 at 8:08
add a comment |
2
$begingroup$
If you are asking for a simple formula for $h$ I doubt if such a thing exists.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 6:07
$begingroup$
For polynomials it's just the greatest common denominator. For general analytic functions I have no idea.
$endgroup$
– Arthur
Jan 7 at 6:41
$begingroup$
@KaviRamaMurthy such $h$ does exists. If we assume that $f$ and $g$ are not identically zero, then by isolated zeros, we know that there are finitely many zeros. Let $z_1,ldots,z_n$ be all the values such that $f(z_i)$ and $g(z_i)$ are zero. Let $h(z) = prod_{i}(z-z_i)$. But yes, I don't see how to easily compute for $h$.
$endgroup$
– Jaeyoon Kim
Jan 7 at 7:29
$begingroup$
@JaeyoonKim Surely, $h$ exists with the desired property. But you appear to seek a simple formula for $h$ and I suspect that no such formula exists.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 7:37
$begingroup$
If the zeros are simple you may try to find $u,v$ analytic such that $w(z)=frac{f'(z)}{f(z)}+e^{u(z)} frac{g'(z)}{g(z)}+v(z)$ has no zeros, so $f(z)g(z)w(z)$ is what you want. Asking if $u,v$ always exist is an interesting problem.
$endgroup$
– reuns
Jan 7 at 8:08
2
2
$begingroup$
If you are asking for a simple formula for $h$ I doubt if such a thing exists.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 6:07
$begingroup$
If you are asking for a simple formula for $h$ I doubt if such a thing exists.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 6:07
$begingroup$
For polynomials it's just the greatest common denominator. For general analytic functions I have no idea.
$endgroup$
– Arthur
Jan 7 at 6:41
$begingroup$
For polynomials it's just the greatest common denominator. For general analytic functions I have no idea.
$endgroup$
– Arthur
Jan 7 at 6:41
$begingroup$
@KaviRamaMurthy such $h$ does exists. If we assume that $f$ and $g$ are not identically zero, then by isolated zeros, we know that there are finitely many zeros. Let $z_1,ldots,z_n$ be all the values such that $f(z_i)$ and $g(z_i)$ are zero. Let $h(z) = prod_{i}(z-z_i)$. But yes, I don't see how to easily compute for $h$.
$endgroup$
– Jaeyoon Kim
Jan 7 at 7:29
$begingroup$
@KaviRamaMurthy such $h$ does exists. If we assume that $f$ and $g$ are not identically zero, then by isolated zeros, we know that there are finitely many zeros. Let $z_1,ldots,z_n$ be all the values such that $f(z_i)$ and $g(z_i)$ are zero. Let $h(z) = prod_{i}(z-z_i)$. But yes, I don't see how to easily compute for $h$.
$endgroup$
– Jaeyoon Kim
Jan 7 at 7:29
$begingroup$
@JaeyoonKim Surely, $h$ exists with the desired property. But you appear to seek a simple formula for $h$ and I suspect that no such formula exists.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 7:37
$begingroup$
@JaeyoonKim Surely, $h$ exists with the desired property. But you appear to seek a simple formula for $h$ and I suspect that no such formula exists.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 7:37
$begingroup$
If the zeros are simple you may try to find $u,v$ analytic such that $w(z)=frac{f'(z)}{f(z)}+e^{u(z)} frac{g'(z)}{g(z)}+v(z)$ has no zeros, so $f(z)g(z)w(z)$ is what you want. Asking if $u,v$ always exist is an interesting problem.
$endgroup$
– reuns
Jan 7 at 8:08
$begingroup$
If the zeros are simple you may try to find $u,v$ analytic such that $w(z)=frac{f'(z)}{f(z)}+e^{u(z)} frac{g'(z)}{g(z)}+v(z)$ has no zeros, so $f(z)g(z)w(z)$ is what you want. Asking if $u,v$ always exist is an interesting problem.
$endgroup$
– reuns
Jan 7 at 8:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is no formula of the form $$h(z)=F(f(z),g(z)).$$
Proof: Letting $f(z)=z$, $g(z)=c$ shows that $zmapsto F(z,c)$ is holomorphic. Similarly for $F(c,w)$, so a theorem of Hartogs shows that $F$ is a holomorphic function of two variables. Now $F(0,0)=0$, but the zeroes of a holomorphic function in $Bbb C^2$ cannot be isolated, by another theorem of Hartogs. So it cannot happen that $F(f(z),g(z))=0$ if and only if $f(z)=g(z)=0$.
answered Jan 7 at 16:10
David C. UllrichDavid C. Ullrich
59.9k43994
$endgroup$
$begingroup$
Thank you for your help! I am actually asking for a weaker condition. I don't require $F$ to be holomorphic.
$endgroup$
– Jaeyoon Kim
Jan 7 at 18:14
$begingroup$
@JaeyoonKim ??? I didn't require $F$ to be holomorphic! If $F(f,g)$ is a holomorphic function with zero set the intersection of the two zero sets, for eery $f$ and $g$, it follows that $F$ is holomorphic - I gave a proof of that.
$endgroup$
– David C. Ullrich
Jan 7 at 18:22
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
There is no formula of the form $$h(z)=F(f(z),g(z)).$$
Proof: Letting $f(z)=z$, $g(z)=c$ shows that $zmapsto F(z,c)$ is holomorphic. Similarly for $F(c,w)$, so a theorem of Hartogs shows that $F$ is a holomorphic function of two variables. Now $F(0,0)=0$, but the zeroes of a holomorphic function in $Bbb C^2$ cannot be isolated, by another theorem of Hartogs. So it cannot happen that $F(f(z),g(z))=0$ if and only if $f(z)=g(z)=0$.
answered Jan 7 at 16:10
David C. UllrichDavid C. Ullrich
59.9k43994
$endgroup$
$begingroup$
Thank you for your help! I am actually asking for a weaker condition. I don't require $F$ to be holomorphic.
$endgroup$
– Jaeyoon Kim
Jan 7 at 18:14
$begingroup$
@JaeyoonKim ??? I didn't require $F$ to be holomorphic! If $F(f,g)$ is a holomorphic function with zero set the intersection of the two zero sets, for eery $f$ and $g$, it follows that $F$ is holomorphic - I gave a proof of that.
$endgroup$
– David C. Ullrich
Jan 7 at 18:22
add a comment |
$begingroup$
There is no formula of the form $$h(z)=F(f(z),g(z)).$$
Proof: Letting $f(z)=z$, $g(z)=c$ shows that $zmapsto F(z,c)$ is holomorphic. Similarly for $F(c,w)$, so a theorem of Hartogs shows that $F$ is a holomorphic function of two variables. Now $F(0,0)=0$, but the zeroes of a holomorphic function in $Bbb C^2$ cannot be isolated, by another theorem of Hartogs. So it cannot happen that $F(f(z),g(z))=0$ if and only if $f(z)=g(z)=0$.
answered Jan 7 at 16:10
David C. UllrichDavid C. Ullrich
59.9k43994
$endgroup$
$begingroup$
Thank you for your help! I am actually asking for a weaker condition. I don't require $F$ to be holomorphic.
$endgroup$
– Jaeyoon Kim
Jan 7 at 18:14
$begingroup$
@JaeyoonKim ??? I didn't require $F$ to be holomorphic! If $F(f,g)$ is a holomorphic function with zero set the intersection of the two zero sets, for eery $f$ and $g$, it follows that $F$ is holomorphic - I gave a proof of that.
$endgroup$
– David C. Ullrich
Jan 7 at 18:22
add a comment |
$begingroup$
There is no formula of the form $$h(z)=F(f(z),g(z)).$$
Proof: Letting $f(z)=z$, $g(z)=c$ shows that $zmapsto F(z,c)$ is holomorphic. Similarly for $F(c,w)$, so a theorem of Hartogs shows that $F$ is a holomorphic function of two variables. Now $F(0,0)=0$, but the zeroes of a holomorphic function in $Bbb C^2$ cannot be isolated, by another theorem of Hartogs. So it cannot happen that $F(f(z),g(z))=0$ if and only if $f(z)=g(z)=0$.
answered Jan 7 at 16:10
David C. UllrichDavid C. Ullrich
59.9k43994
$endgroup$
There is no formula of the form $$h(z)=F(f(z),g(z)).$$
Proof: Letting $f(z)=z$, $g(z)=c$ shows that $zmapsto F(z,c)$ is holomorphic. Similarly for $F(c,w)$, so a theorem of Hartogs shows that $F$ is a holomorphic function of two variables. Now $F(0,0)=0$, but the zeroes of a holomorphic function in $Bbb C^2$ cannot be isolated, by another theorem of Hartogs. So it cannot happen that $F(f(z),g(z))=0$ if and only if $f(z)=g(z)=0$.
answered Jan 7 at 16:10
David C. UllrichDavid C. Ullrich
59.9k43994
answered Jan 7 at 16:10
David C. UllrichDavid C. Ullrich
59.9k43994
answered Jan 7 at 16:10
David C. UllrichDavid C. Ullrich
59.9k43994
answered Jan 7 at 16:10
David C. UllrichDavid C. Ullrich
59.9k43994
59.9k43994
$begingroup$
Thank you for your help! I am actually asking for a weaker condition. I don't require $F$ to be holomorphic.
$endgroup$
– Jaeyoon Kim
Jan 7 at 18:14
$begingroup$
@JaeyoonKim ??? I didn't require $F$ to be holomorphic! If $F(f,g)$ is a holomorphic function with zero set the intersection of the two zero sets, for eery $f$ and $g$, it follows that $F$ is holomorphic - I gave a proof of that.
$endgroup$
– David C. Ullrich
Jan 7 at 18:22
add a comment |
$begingroup$
Thank you for your help! I am actually asking for a weaker condition. I don't require $F$ to be holomorphic.
$endgroup$
– Jaeyoon Kim
Jan 7 at 18:14
$begingroup$
@JaeyoonKim ??? I didn't require $F$ to be holomorphic! If $F(f,g)$ is a holomorphic function with zero set the intersection of the two zero sets, for eery $f$ and $g$, it follows that $F$ is holomorphic - I gave a proof of that.
$endgroup$
– David C. Ullrich
Jan 7 at 18:22
$begingroup$
Thank you for your help! I am actually asking for a weaker condition. I don't require $F$ to be holomorphic.
$endgroup$
– Jaeyoon Kim
Jan 7 at 18:14
$begingroup$
Thank you for your help! I am actually asking for a weaker condition. I don't require $F$ to be holomorphic.
$endgroup$
– Jaeyoon Kim
Jan 7 at 18:14
$begingroup$
@JaeyoonKim ??? I didn't require $F$ to be holomorphic! If $F(f,g)$ is a holomorphic function with zero set the intersection of the two zero sets, for eery $f$ and $g$, it follows that $F$ is holomorphic - I gave a proof of that.
$endgroup$
– David C. Ullrich
Jan 7 at 18:22
$begingroup$
@JaeyoonKim ??? I didn't require $F$ to be holomorphic! If $F(f,g)$ is a holomorphic function with zero set the intersection of the two zero sets, for eery $f$ and $g$, it follows that $F$ is holomorphic - I gave a proof of that.
$endgroup$
– David C. Ullrich
Jan 7 at 18:22
add a comment |
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2
$begingroup$
If you are asking for a simple formula for $h$ I doubt if such a thing exists.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 6:07
$begingroup$
For polynomials it's just the greatest common denominator. For general analytic functions I have no idea.
$endgroup$
– Arthur
Jan 7 at 6:41
$begingroup$
@KaviRamaMurthy such $h$ does exists. If we assume that $f$ and $g$ are not identically zero, then by isolated zeros, we know that there are finitely many zeros. Let $z_1,ldots,z_n$ be all the values such that $f(z_i)$ and $g(z_i)$ are zero. Let $h(z) = prod_{i}(z-z_i)$. But yes, I don't see how to easily compute for $h$.
$endgroup$
– Jaeyoon Kim
Jan 7 at 7:29
$begingroup$
@JaeyoonKim Surely, $h$ exists with the desired property. But you appear to seek a simple formula for $h$ and I suspect that no such formula exists.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 7:37
$begingroup$
If the zeros are simple you may try to find $u,v$ analytic such that $w(z)=frac{f'(z)}{f(z)}+e^{u(z)} frac{g'(z)}{g(z)}+v(z)$ has no zeros, so $f(z)g(z)w(z)$ is what you want. Asking if $u,v$ always exist is an interesting problem.
$endgroup$
– reuns
Jan 7 at 8:08