What's the meaning of being expressible as a convergent power series in a neighborhood of each point?
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The following pictures are from Lee's "Introduction to Smooth Manifolds".
What's the meaning of being expressible as a convergent power series in a neighborhood of each point? However, I only know convergent power series in real and complex fields.
manifolds smooth-manifolds manifolds-with-boundary
asked Nov 27 at 0:54
Born to be proud
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The following pictures are from Lee's "Introduction to Smooth Manifolds".
What's the meaning of being expressible as a convergent power series in a neighborhood of each point? However, I only know convergent power series in real and complex fields.
manifolds smooth-manifolds manifolds-with-boundary
asked Nov 27 at 0:54
Born to be proud
782510
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The following pictures are from Lee's "Introduction to Smooth Manifolds".
What's the meaning of being expressible as a convergent power series in a neighborhood of each point? However, I only know convergent power series in real and complex fields.
manifolds smooth-manifolds manifolds-with-boundary
asked Nov 27 at 0:54
Born to be proud
782510
The following pictures are from Lee's "Introduction to Smooth Manifolds".
What's the meaning of being expressible as a convergent power series in a neighborhood of each point? However, I only know convergent power series in real and complex fields.
manifolds smooth-manifolds manifolds-with-boundary
manifolds smooth-manifolds manifolds-with-boundary
asked Nov 27 at 0:54
Born to be proud
782510
asked Nov 27 at 0:54
Born to be proud
782510
asked Nov 27 at 0:54
Born to be proud
782510
asked Nov 27 at 0:54
Born to be proud
782510
asked Nov 27 at 0:54
Born to be proud
782510
782510
add a comment |
add a comment |
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I hope you have understood enough of the definition so far to understand that the geometry of the manifold is determined by overlapping patches each of which looks like a piece of real $n$-space. So where the patches overlap you have a transition map $T$ that is a function from $mathbb{R}^n$ to itself. So you can ask how smooth $T$ is. Perhaps it has derivatives up to order $k$. In the best case, it is infinitely differentiable, and, moreover, it has a Taylor series expansion (in $n$ variables) that converges to the function. That's like $T(x) = 1/(1-x)$ in one variable: the Taylor series for that function converges to the function in a neighborhood of each point in the interval $(=1,1)$.
answered Nov 27 at 1:21
Ethan Bolker
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1 Answer
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1 Answer
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I hope you have understood enough of the definition so far to understand that the geometry of the manifold is determined by overlapping patches each of which looks like a piece of real $n$-space. So where the patches overlap you have a transition map $T$ that is a function from $mathbb{R}^n$ to itself. So you can ask how smooth $T$ is. Perhaps it has derivatives up to order $k$. In the best case, it is infinitely differentiable, and, moreover, it has a Taylor series expansion (in $n$ variables) that converges to the function. That's like $T(x) = 1/(1-x)$ in one variable: the Taylor series for that function converges to the function in a neighborhood of each point in the interval $(=1,1)$.
answered Nov 27 at 1:21
Ethan Bolker
40.7k546108
add a comment |
up vote
1
down vote
I hope you have understood enough of the definition so far to understand that the geometry of the manifold is determined by overlapping patches each of which looks like a piece of real $n$-space. So where the patches overlap you have a transition map $T$ that is a function from $mathbb{R}^n$ to itself. So you can ask how smooth $T$ is. Perhaps it has derivatives up to order $k$. In the best case, it is infinitely differentiable, and, moreover, it has a Taylor series expansion (in $n$ variables) that converges to the function. That's like $T(x) = 1/(1-x)$ in one variable: the Taylor series for that function converges to the function in a neighborhood of each point in the interval $(=1,1)$.
answered Nov 27 at 1:21
Ethan Bolker
40.7k546108
add a comment |
up vote
1
down vote
up vote
1
down vote
I hope you have understood enough of the definition so far to understand that the geometry of the manifold is determined by overlapping patches each of which looks like a piece of real $n$-space. So where the patches overlap you have a transition map $T$ that is a function from $mathbb{R}^n$ to itself. So you can ask how smooth $T$ is. Perhaps it has derivatives up to order $k$. In the best case, it is infinitely differentiable, and, moreover, it has a Taylor series expansion (in $n$ variables) that converges to the function. That's like $T(x) = 1/(1-x)$ in one variable: the Taylor series for that function converges to the function in a neighborhood of each point in the interval $(=1,1)$.
answered Nov 27 at 1:21
Ethan Bolker
40.7k546108
I hope you have understood enough of the definition so far to understand that the geometry of the manifold is determined by overlapping patches each of which looks like a piece of real $n$-space. So where the patches overlap you have a transition map $T$ that is a function from $mathbb{R}^n$ to itself. So you can ask how smooth $T$ is. Perhaps it has derivatives up to order $k$. In the best case, it is infinitely differentiable, and, moreover, it has a Taylor series expansion (in $n$ variables) that converges to the function. That's like $T(x) = 1/(1-x)$ in one variable: the Taylor series for that function converges to the function in a neighborhood of each point in the interval $(=1,1)$.
answered Nov 27 at 1:21
Ethan Bolker
40.7k546108
edited Nov 27 at 2:01
answered Nov 27 at 1:21
Ethan Bolker
40.7k546108
answered Nov 27 at 1:21
Ethan Bolker
40.7k546108
answered Nov 27 at 1:21
Ethan Bolker
40.7k546108
40.7k546108
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