How to check the smoothness of the following function?
up vote
0
down vote
favorite
I was solving a problem which boiled down to checking the smoothness of the following function.
$$
tilde{p} (z) = begin{cases} frac{1}{p(1/z)} &quad zneq 0 \
0 &quad z = 0.end{cases}
$$
where $p$ is a polynomial not identically zero and the domain of the function is chosen so that $p(1/z) neq 0$ in all of the domain.
Continuity of the function is easy to see as $z to 0$ implies $p(1/z) to infty$ hence $tilde{p}(z) to 0$. But I do not see how to imply smoothness of $tilde{p}(z)$. Here I do not want it to be holomorphic, it is enough if I can see that $tilde{p}$ is smooth as function of two variables.
complex-analysis smooth-manifolds smooth-functions
asked Nov 14 at 11:12
Prakhar Gupta
366112
add a comment |
up vote
0
down vote
favorite
I was solving a problem which boiled down to checking the smoothness of the following function.
$$
tilde{p} (z) = begin{cases} frac{1}{p(1/z)} &quad zneq 0 \
0 &quad z = 0.end{cases}
$$
where $p$ is a polynomial not identically zero and the domain of the function is chosen so that $p(1/z) neq 0$ in all of the domain.
Continuity of the function is easy to see as $z to 0$ implies $p(1/z) to infty$ hence $tilde{p}(z) to 0$. But I do not see how to imply smoothness of $tilde{p}(z)$. Here I do not want it to be holomorphic, it is enough if I can see that $tilde{p}$ is smooth as function of two variables.
complex-analysis smooth-manifolds smooth-functions
asked Nov 14 at 11:12
Prakhar Gupta
366112
Do you also want to show analyticity?
– edm
Nov 14 at 11:16
@edm. No, I just need to show that $tilde{p}$ is smooth as a map from $mathbb{R}^{2}$ to itself. But looking at the expression it seems that it could be holomorphic as well, but I don't see how I can prove or disprove such questions.
– Prakhar Gupta
Nov 14 at 11:18
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was solving a problem which boiled down to checking the smoothness of the following function.
$$
tilde{p} (z) = begin{cases} frac{1}{p(1/z)} &quad zneq 0 \
0 &quad z = 0.end{cases}
$$
where $p$ is a polynomial not identically zero and the domain of the function is chosen so that $p(1/z) neq 0$ in all of the domain.
Continuity of the function is easy to see as $z to 0$ implies $p(1/z) to infty$ hence $tilde{p}(z) to 0$. But I do not see how to imply smoothness of $tilde{p}(z)$. Here I do not want it to be holomorphic, it is enough if I can see that $tilde{p}$ is smooth as function of two variables.
complex-analysis smooth-manifolds smooth-functions
asked Nov 14 at 11:12
Prakhar Gupta
366112
I was solving a problem which boiled down to checking the smoothness of the following function.
$$
tilde{p} (z) = begin{cases} frac{1}{p(1/z)} &quad zneq 0 \
0 &quad z = 0.end{cases}
$$
where $p$ is a polynomial not identically zero and the domain of the function is chosen so that $p(1/z) neq 0$ in all of the domain.
Continuity of the function is easy to see as $z to 0$ implies $p(1/z) to infty$ hence $tilde{p}(z) to 0$. But I do not see how to imply smoothness of $tilde{p}(z)$. Here I do not want it to be holomorphic, it is enough if I can see that $tilde{p}$ is smooth as function of two variables.
complex-analysis smooth-manifolds smooth-functions
complex-analysis smooth-manifolds smooth-functions
asked Nov 14 at 11:12
Prakhar Gupta
366112
asked Nov 14 at 11:12
Prakhar Gupta
366112
asked Nov 14 at 11:12
Prakhar Gupta
366112
asked Nov 14 at 11:12
Prakhar Gupta
366112
asked Nov 14 at 11:12
Prakhar Gupta
366112
366112
Do you also want to show analyticity?
– edm
Nov 14 at 11:16
@edm. No, I just need to show that $tilde{p}$ is smooth as a map from $mathbb{R}^{2}$ to itself. But looking at the expression it seems that it could be holomorphic as well, but I don't see how I can prove or disprove such questions.
– Prakhar Gupta
Nov 14 at 11:18
add a comment |
Do you also want to show analyticity?
– edm
Nov 14 at 11:16
@edm. No, I just need to show that $tilde{p}$ is smooth as a map from $mathbb{R}^{2}$ to itself. But looking at the expression it seems that it could be holomorphic as well, but I don't see how I can prove or disprove such questions.
– Prakhar Gupta
Nov 14 at 11:18
Do you also want to show analyticity?
– edm
Nov 14 at 11:16
Do you also want to show analyticity?
– edm
Nov 14 at 11:16
@edm. No, I just need to show that $tilde{p}$ is smooth as a map from $mathbb{R}^{2}$ to itself. But looking at the expression it seems that it could be holomorphic as well, but I don't see how I can prove or disprove such questions.
– Prakhar Gupta
Nov 14 at 11:18
@edm. No, I just need to show that $tilde{p}$ is smooth as a map from $mathbb{R}^{2}$ to itself. But looking at the expression it seems that it could be holomorphic as well, but I don't see how I can prove or disprove such questions.
– Prakhar Gupta
Nov 14 at 11:18
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
The reciprocal map $rho: zmapsto{1over z}$ is continuous througout its domain of definition $dot{mathbb C}:={mathbb C}setminus{0}$. Let $Omega$ be the envisaged domain of $tilde p$. Since it is assumed that $p(1/z)ne0$ for all points $zindot Omega$ the composition
$$tilde p(z)=(rhocirc pcircrho)(z)qquad(zindotOmega)$$
is continuous throughout $dotOmega$.
The point $z=0$ needs special treatment. If ${rm deg}(p)geq1$ then $$lim_{zto0}p(1/ z)=lim_{wtoinfty}p(w)=infty ,$$
and therefore
$$lim_{zto0}tilde p(z)=lim_{zto0}{1over p(1/z)}=0=tilde p(0) .$$
It follows that $tilde p$ is continuous at $z=0$ in this case. If, however, ${rm deg}(p)=0$ then $p(z)equiv cne0$, and we have $$lim_{zto0}p(1/ z)=lim_{wtoinfty}p(w)=c ,$$ and therefore
$$lim_{zto0}tilde p(z)=lim_{zto0}{1over p(1/z)}={1over c}netilde p(0) ,$$
so that $tilde p$ is not continuous at $0$ in this case.
answered Nov 14 at 13:37
Christian Blatter
170k7111324
This means that if $p$ is a constant polynomial, then we have a trouble. Thanks for pointing this out, and I think the following will resolve this problem. We can redefine $tilde{p}(z)$ by defining it to be $1/c$ everywhere. Then $tilde{p}$ will be smooth. Also, I realized later that in the other case we can use the fact that on $dot{Omega}$, $tilde{p}$ will be holomorphic and $0$ will be removable singularity making $tilde{p}$ holomorphic on all of $Omega$ hence smooth.
– Prakhar Gupta
Nov 14 at 14:44
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The reciprocal map $rho: zmapsto{1over z}$ is continuous througout its domain of definition $dot{mathbb C}:={mathbb C}setminus{0}$. Let $Omega$ be the envisaged domain of $tilde p$. Since it is assumed that $p(1/z)ne0$ for all points $zindot Omega$ the composition
$$tilde p(z)=(rhocirc pcircrho)(z)qquad(zindotOmega)$$
is continuous throughout $dotOmega$.
The point $z=0$ needs special treatment. If ${rm deg}(p)geq1$ then $$lim_{zto0}p(1/ z)=lim_{wtoinfty}p(w)=infty ,$$
and therefore
$$lim_{zto0}tilde p(z)=lim_{zto0}{1over p(1/z)}=0=tilde p(0) .$$
It follows that $tilde p$ is continuous at $z=0$ in this case. If, however, ${rm deg}(p)=0$ then $p(z)equiv cne0$, and we have $$lim_{zto0}p(1/ z)=lim_{wtoinfty}p(w)=c ,$$ and therefore
$$lim_{zto0}tilde p(z)=lim_{zto0}{1over p(1/z)}={1over c}netilde p(0) ,$$
so that $tilde p$ is not continuous at $0$ in this case.
answered Nov 14 at 13:37
Christian Blatter
170k7111324
This means that if $p$ is a constant polynomial, then we have a trouble. Thanks for pointing this out, and I think the following will resolve this problem. We can redefine $tilde{p}(z)$ by defining it to be $1/c$ everywhere. Then $tilde{p}$ will be smooth. Also, I realized later that in the other case we can use the fact that on $dot{Omega}$, $tilde{p}$ will be holomorphic and $0$ will be removable singularity making $tilde{p}$ holomorphic on all of $Omega$ hence smooth.
– Prakhar Gupta
Nov 14 at 14:44
add a comment |
up vote
0
down vote
The reciprocal map $rho: zmapsto{1over z}$ is continuous througout its domain of definition $dot{mathbb C}:={mathbb C}setminus{0}$. Let $Omega$ be the envisaged domain of $tilde p$. Since it is assumed that $p(1/z)ne0$ for all points $zindot Omega$ the composition
$$tilde p(z)=(rhocirc pcircrho)(z)qquad(zindotOmega)$$
is continuous throughout $dotOmega$.
The point $z=0$ needs special treatment. If ${rm deg}(p)geq1$ then $$lim_{zto0}p(1/ z)=lim_{wtoinfty}p(w)=infty ,$$
and therefore
$$lim_{zto0}tilde p(z)=lim_{zto0}{1over p(1/z)}=0=tilde p(0) .$$
It follows that $tilde p$ is continuous at $z=0$ in this case. If, however, ${rm deg}(p)=0$ then $p(z)equiv cne0$, and we have $$lim_{zto0}p(1/ z)=lim_{wtoinfty}p(w)=c ,$$ and therefore
$$lim_{zto0}tilde p(z)=lim_{zto0}{1over p(1/z)}={1over c}netilde p(0) ,$$
so that $tilde p$ is not continuous at $0$ in this case.
answered Nov 14 at 13:37
Christian Blatter
170k7111324
This means that if $p$ is a constant polynomial, then we have a trouble. Thanks for pointing this out, and I think the following will resolve this problem. We can redefine $tilde{p}(z)$ by defining it to be $1/c$ everywhere. Then $tilde{p}$ will be smooth. Also, I realized later that in the other case we can use the fact that on $dot{Omega}$, $tilde{p}$ will be holomorphic and $0$ will be removable singularity making $tilde{p}$ holomorphic on all of $Omega$ hence smooth.
– Prakhar Gupta
Nov 14 at 14:44
add a comment |
up vote
0
down vote
up vote
0
down vote
The reciprocal map $rho: zmapsto{1over z}$ is continuous througout its domain of definition $dot{mathbb C}:={mathbb C}setminus{0}$. Let $Omega$ be the envisaged domain of $tilde p$. Since it is assumed that $p(1/z)ne0$ for all points $zindot Omega$ the composition
$$tilde p(z)=(rhocirc pcircrho)(z)qquad(zindotOmega)$$
is continuous throughout $dotOmega$.
The point $z=0$ needs special treatment. If ${rm deg}(p)geq1$ then $$lim_{zto0}p(1/ z)=lim_{wtoinfty}p(w)=infty ,$$
and therefore
$$lim_{zto0}tilde p(z)=lim_{zto0}{1over p(1/z)}=0=tilde p(0) .$$
It follows that $tilde p$ is continuous at $z=0$ in this case. If, however, ${rm deg}(p)=0$ then $p(z)equiv cne0$, and we have $$lim_{zto0}p(1/ z)=lim_{wtoinfty}p(w)=c ,$$ and therefore
$$lim_{zto0}tilde p(z)=lim_{zto0}{1over p(1/z)}={1over c}netilde p(0) ,$$
so that $tilde p$ is not continuous at $0$ in this case.
answered Nov 14 at 13:37
Christian Blatter
170k7111324
The reciprocal map $rho: zmapsto{1over z}$ is continuous througout its domain of definition $dot{mathbb C}:={mathbb C}setminus{0}$. Let $Omega$ be the envisaged domain of $tilde p$. Since it is assumed that $p(1/z)ne0$ for all points $zindot Omega$ the composition
$$tilde p(z)=(rhocirc pcircrho)(z)qquad(zindotOmega)$$
is continuous throughout $dotOmega$.
The point $z=0$ needs special treatment. If ${rm deg}(p)geq1$ then $$lim_{zto0}p(1/ z)=lim_{wtoinfty}p(w)=infty ,$$
and therefore
$$lim_{zto0}tilde p(z)=lim_{zto0}{1over p(1/z)}=0=tilde p(0) .$$
It follows that $tilde p$ is continuous at $z=0$ in this case. If, however, ${rm deg}(p)=0$ then $p(z)equiv cne0$, and we have $$lim_{zto0}p(1/ z)=lim_{wtoinfty}p(w)=c ,$$ and therefore
$$lim_{zto0}tilde p(z)=lim_{zto0}{1over p(1/z)}={1over c}netilde p(0) ,$$
so that $tilde p$ is not continuous at $0$ in this case.
answered Nov 14 at 13:37
Christian Blatter
170k7111324
answered Nov 14 at 13:37
Christian Blatter
170k7111324
answered Nov 14 at 13:37
Christian Blatter
170k7111324
answered Nov 14 at 13:37
Christian Blatter
170k7111324
170k7111324
This means that if $p$ is a constant polynomial, then we have a trouble. Thanks for pointing this out, and I think the following will resolve this problem. We can redefine $tilde{p}(z)$ by defining it to be $1/c$ everywhere. Then $tilde{p}$ will be smooth. Also, I realized later that in the other case we can use the fact that on $dot{Omega}$, $tilde{p}$ will be holomorphic and $0$ will be removable singularity making $tilde{p}$ holomorphic on all of $Omega$ hence smooth.
– Prakhar Gupta
Nov 14 at 14:44
add a comment |
This means that if $p$ is a constant polynomial, then we have a trouble. Thanks for pointing this out, and I think the following will resolve this problem. We can redefine $tilde{p}(z)$ by defining it to be $1/c$ everywhere. Then $tilde{p}$ will be smooth. Also, I realized later that in the other case we can use the fact that on $dot{Omega}$, $tilde{p}$ will be holomorphic and $0$ will be removable singularity making $tilde{p}$ holomorphic on all of $Omega$ hence smooth.
– Prakhar Gupta
Nov 14 at 14:44
This means that if $p$ is a constant polynomial, then we have a trouble. Thanks for pointing this out, and I think the following will resolve this problem. We can redefine $tilde{p}(z)$ by defining it to be $1/c$ everywhere. Then $tilde{p}$ will be smooth. Also, I realized later that in the other case we can use the fact that on $dot{Omega}$, $tilde{p}$ will be holomorphic and $0$ will be removable singularity making $tilde{p}$ holomorphic on all of $Omega$ hence smooth.
– Prakhar Gupta
Nov 14 at 14:44
This means that if $p$ is a constant polynomial, then we have a trouble. Thanks for pointing this out, and I think the following will resolve this problem. We can redefine $tilde{p}(z)$ by defining it to be $1/c$ everywhere. Then $tilde{p}$ will be smooth. Also, I realized later that in the other case we can use the fact that on $dot{Omega}$, $tilde{p}$ will be holomorphic and $0$ will be removable singularity making $tilde{p}$ holomorphic on all of $Omega$ hence smooth.
– Prakhar Gupta
Nov 14 at 14:44
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2998141%2fhow-to-check-the-smoothness-of-the-following-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Do you also want to show analyticity?
– edm
Nov 14 at 11:16
@edm. No, I just need to show that $tilde{p}$ is smooth as a map from $mathbb{R}^{2}$ to itself. But looking at the expression it seems that it could be holomorphic as well, but I don't see how I can prove or disprove such questions.
– Prakhar Gupta
Nov 14 at 11:18