How to prove that $f(x)=fracxsqrtx$ is smooth?Show that an open subset of a smooth manifold is again smooth.Smooth chart in what sense?Proof that this is a smooth manifoldSmooth coverings are open maps proof verificationWhat does it take for a smooth homeomorphism to be a diffeomorphism?Construct a diffeomorphism $psi: B_1 to epsilontext-neighborhood of K$, where $K$ is a subset of a smooth manifold.Diffeomorphism, from class $C^1$ to $C^k$Abstract Smooth Manifolds vs Embedded Smooth ManifoldsSmooth curves that are submanifoldsMaximal Smooth Atlas For Stereographic Projection
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How to prove that $f(x)=fracxsqrtx$ is smooth?
Show that an open subset of a smooth manifold is again smooth.Smooth chart in what sense?Proof that this is a smooth manifoldSmooth coverings are open maps proof verificationWhat does it take for a smooth homeomorphism to be a diffeomorphism?Construct a diffeomorphism $psi: B_1 to epsilontext-neighborhood of K$, where $K$ is a subset of a smooth manifold.Diffeomorphism, from class $C^1$ to $C^k$Abstract Smooth Manifolds vs Embedded Smooth ManifoldsSmooth curves that are submanifoldsMaximal Smooth Atlas For Stereographic Projection
$begingroup$
I have a function $f(x)=fracxsqrtx$ on the open ball $B_1(0)$ in $mathbbR^n$. I have to show that this function is a diffeomorphism between $mathbbR^n$ and $B_1(0)$. I have already proved that the function is bijective with inverse $fracxsqrtx$ and it is continuous. How to prove that the function is smooth?
Thanks!
differential-geometry smooth-manifolds
edited May 7 at 13:33
Asaf Karagila♦
310k33443776
asked May 7 at 8:56
sakasaka
504
$endgroup$
add a comment |
$begingroup$
I have a function $f(x)=fracxsqrtx$ on the open ball $B_1(0)$ in $mathbbR^n$. I have to show that this function is a diffeomorphism between $mathbbR^n$ and $B_1(0)$. I have already proved that the function is bijective with inverse $fracxsqrtx$ and it is continuous. How to prove that the function is smooth?
Thanks!
differential-geometry smooth-manifolds
edited May 7 at 13:33
Asaf Karagila♦
310k33443776
asked May 7 at 8:56
sakasaka
504
$endgroup$
add a comment |
$begingroup$
I have a function $f(x)=fracxsqrtx$ on the open ball $B_1(0)$ in $mathbbR^n$. I have to show that this function is a diffeomorphism between $mathbbR^n$ and $B_1(0)$. I have already proved that the function is bijective with inverse $fracxsqrtx$ and it is continuous. How to prove that the function is smooth?
Thanks!
differential-geometry smooth-manifolds
edited May 7 at 13:33
Asaf Karagila♦
310k33443776
asked May 7 at 8:56
sakasaka
504
$endgroup$
I have a function $f(x)=fracxsqrtx$ on the open ball $B_1(0)$ in $mathbbR^n$. I have to show that this function is a diffeomorphism between $mathbbR^n$ and $B_1(0)$. I have already proved that the function is bijective with inverse $fracxsqrtx$ and it is continuous. How to prove that the function is smooth?
Thanks!
differential-geometry smooth-manifolds
differential-geometry smooth-manifolds
edited May 7 at 13:33
Asaf Karagila♦
310k33443776
asked May 7 at 8:56
sakasaka
504
edited May 7 at 13:33
Asaf Karagila♦
310k33443776
asked May 7 at 8:56
sakasaka
504
edited May 7 at 13:33
Asaf Karagila♦
310k33443776
edited May 7 at 13:33
Asaf Karagila♦
310k33443776
edited May 7 at 13:33
Asaf Karagila♦
310k33443776
310k33443776
asked May 7 at 8:56
sakasaka
504
asked May 7 at 8:56
sakasaka
504
asked May 7 at 8:56
sakasaka
504
504
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can apply the theorem that the sum, product, quotient (whenever defined), square root (of a positive smooth function), and composition of smooth functions are smooth. In your examples, all the component functions $x$, $|x|^2$, $1 pm |x|^2$, and so on are all smooth.
answered May 7 at 9:06
Loring TuLoring Tu
1912
$endgroup$
add a comment |
$begingroup$
Just use the fact that compositions of smooth functions is smooth and ratio of two smooth functions is smooth as long as the denominator never vanishes.
answered May 7 at 9:04
Kavi Rama MurthyKavi Rama Murthy
82k53673
$endgroup$
add a comment |
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2 Answers
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$begingroup$
You can apply the theorem that the sum, product, quotient (whenever defined), square root (of a positive smooth function), and composition of smooth functions are smooth. In your examples, all the component functions $x$, $|x|^2$, $1 pm |x|^2$, and so on are all smooth.
answered May 7 at 9:06
Loring TuLoring Tu
1912
$endgroup$
add a comment |
$begingroup$
You can apply the theorem that the sum, product, quotient (whenever defined), square root (of a positive smooth function), and composition of smooth functions are smooth. In your examples, all the component functions $x$, $|x|^2$, $1 pm |x|^2$, and so on are all smooth.
answered May 7 at 9:06
Loring TuLoring Tu
1912
$endgroup$
add a comment |
$begingroup$
You can apply the theorem that the sum, product, quotient (whenever defined), square root (of a positive smooth function), and composition of smooth functions are smooth. In your examples, all the component functions $x$, $|x|^2$, $1 pm |x|^2$, and so on are all smooth.
answered May 7 at 9:06
Loring TuLoring Tu
1912
$endgroup$
You can apply the theorem that the sum, product, quotient (whenever defined), square root (of a positive smooth function), and composition of smooth functions are smooth. In your examples, all the component functions $x$, $|x|^2$, $1 pm |x|^2$, and so on are all smooth.
answered May 7 at 9:06
Loring TuLoring Tu
1912
answered May 7 at 9:06
Loring TuLoring Tu
1912
answered May 7 at 9:06
Loring TuLoring Tu
1912
answered May 7 at 9:06
Loring TuLoring Tu
1912
1912
add a comment |
add a comment |
$begingroup$
Just use the fact that compositions of smooth functions is smooth and ratio of two smooth functions is smooth as long as the denominator never vanishes.
answered May 7 at 9:04
Kavi Rama MurthyKavi Rama Murthy
82k53673
$endgroup$
add a comment |
$begingroup$
Just use the fact that compositions of smooth functions is smooth and ratio of two smooth functions is smooth as long as the denominator never vanishes.
answered May 7 at 9:04
Kavi Rama MurthyKavi Rama Murthy
82k53673
$endgroup$
add a comment |
$begingroup$
Just use the fact that compositions of smooth functions is smooth and ratio of two smooth functions is smooth as long as the denominator never vanishes.
answered May 7 at 9:04
Kavi Rama MurthyKavi Rama Murthy
82k53673
$endgroup$
Just use the fact that compositions of smooth functions is smooth and ratio of two smooth functions is smooth as long as the denominator never vanishes.
answered May 7 at 9:04
Kavi Rama MurthyKavi Rama Murthy
82k53673
answered May 7 at 9:04
Kavi Rama MurthyKavi Rama Murthy
82k53673
answered May 7 at 9:04
Kavi Rama MurthyKavi Rama Murthy
82k53673
answered May 7 at 9:04
Kavi Rama MurthyKavi Rama Murthy
82k53673
82k53673
add a comment |
add a comment |
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