The infimum of the radii of convergence of a real analytic function over a compact
Suppose $f$ is real analytic on a domain $D$ of $mathbb{R}^n$ with $ngeq1$. By definition at each point $x$ of $D$ the Taylor series expansion of $f$ converges; let $r(x)$ be the radius of convergence of this series. Again, by definition, $r(x)$ is never zero as $x$ runs over $D$. Now, let $K$ be a compact of $D$. Prove that the infimum of these $r(x)$, as $x$ runs over $K$ is strictly positive.
real-analysis complex-analysis
asked Nov 29 at 5:59
M. Rahmat
309211
add a comment |
Suppose $f$ is real analytic on a domain $D$ of $mathbb{R}^n$ with $ngeq1$. By definition at each point $x$ of $D$ the Taylor series expansion of $f$ converges; let $r(x)$ be the radius of convergence of this series. Again, by definition, $r(x)$ is never zero as $x$ runs over $D$. Now, let $K$ be a compact of $D$. Prove that the infimum of these $r(x)$, as $x$ runs over $K$ is strictly positive.
real-analysis complex-analysis
asked Nov 29 at 5:59
M. Rahmat
309211
1
It's a compact set. Can you show $r(x)$ is continuous? Continuous functions achieve their extrema on compact sets, and the minimum is greater than zero so the infimum = minimum is as well.
– Brevan Ellefsen
Nov 29 at 8:24
Thanks for your reply. How would you do that?
– M. Rahmat
Nov 30 at 4:47
1
Depends on what tools you have. Given you have a complex analysis tag: The radius of convergence is simply the distance to the nearest singularity. The distance function is necessarily continuous (e.g., since $mathbb C$ has norm $|z|$)
– Brevan Ellefsen
Nov 30 at 7:16
Thanks. I will try...
– M. Rahmat
Nov 30 at 19:37
Success I hope?
– Brevan Ellefsen
Dec 3 at 16:45
add a comment |
Suppose $f$ is real analytic on a domain $D$ of $mathbb{R}^n$ with $ngeq1$. By definition at each point $x$ of $D$ the Taylor series expansion of $f$ converges; let $r(x)$ be the radius of convergence of this series. Again, by definition, $r(x)$ is never zero as $x$ runs over $D$. Now, let $K$ be a compact of $D$. Prove that the infimum of these $r(x)$, as $x$ runs over $K$ is strictly positive.
real-analysis complex-analysis
asked Nov 29 at 5:59
M. Rahmat
309211
Suppose $f$ is real analytic on a domain $D$ of $mathbb{R}^n$ with $ngeq1$. By definition at each point $x$ of $D$ the Taylor series expansion of $f$ converges; let $r(x)$ be the radius of convergence of this series. Again, by definition, $r(x)$ is never zero as $x$ runs over $D$. Now, let $K$ be a compact of $D$. Prove that the infimum of these $r(x)$, as $x$ runs over $K$ is strictly positive.
real-analysis complex-analysis
real-analysis complex-analysis
asked Nov 29 at 5:59
M. Rahmat
309211
asked Nov 29 at 5:59
M. Rahmat
309211
asked Nov 29 at 5:59
M. Rahmat
309211
asked Nov 29 at 5:59
M. Rahmat
309211
asked Nov 29 at 5:59
M. Rahmat
309211
309211
1
It's a compact set. Can you show $r(x)$ is continuous? Continuous functions achieve their extrema on compact sets, and the minimum is greater than zero so the infimum = minimum is as well.
– Brevan Ellefsen
Nov 29 at 8:24
Thanks for your reply. How would you do that?
– M. Rahmat
Nov 30 at 4:47
1
Depends on what tools you have. Given you have a complex analysis tag: The radius of convergence is simply the distance to the nearest singularity. The distance function is necessarily continuous (e.g., since $mathbb C$ has norm $|z|$)
– Brevan Ellefsen
Nov 30 at 7:16
Thanks. I will try...
– M. Rahmat
Nov 30 at 19:37
Success I hope?
– Brevan Ellefsen
Dec 3 at 16:45
add a comment |
1
It's a compact set. Can you show $r(x)$ is continuous? Continuous functions achieve their extrema on compact sets, and the minimum is greater than zero so the infimum = minimum is as well.
– Brevan Ellefsen
Nov 29 at 8:24
Thanks for your reply. How would you do that?
– M. Rahmat
Nov 30 at 4:47
1
Depends on what tools you have. Given you have a complex analysis tag: The radius of convergence is simply the distance to the nearest singularity. The distance function is necessarily continuous (e.g., since $mathbb C$ has norm $|z|$)
– Brevan Ellefsen
Nov 30 at 7:16
Thanks. I will try...
– M. Rahmat
Nov 30 at 19:37
Success I hope?
– Brevan Ellefsen
Dec 3 at 16:45
1
1
It's a compact set. Can you show $r(x)$ is continuous? Continuous functions achieve their extrema on compact sets, and the minimum is greater than zero so the infimum = minimum is as well.
– Brevan Ellefsen
Nov 29 at 8:24
It's a compact set. Can you show $r(x)$ is continuous? Continuous functions achieve their extrema on compact sets, and the minimum is greater than zero so the infimum = minimum is as well.
– Brevan Ellefsen
Nov 29 at 8:24
Thanks for your reply. How would you do that?
– M. Rahmat
Nov 30 at 4:47
Thanks for your reply. How would you do that?
– M. Rahmat
Nov 30 at 4:47
1
1
Depends on what tools you have. Given you have a complex analysis tag: The radius of convergence is simply the distance to the nearest singularity. The distance function is necessarily continuous (e.g., since $mathbb C$ has norm $|z|$)
– Brevan Ellefsen
Nov 30 at 7:16
Depends on what tools you have. Given you have a complex analysis tag: The radius of convergence is simply the distance to the nearest singularity. The distance function is necessarily continuous (e.g., since $mathbb C$ has norm $|z|$)
– Brevan Ellefsen
Nov 30 at 7:16
Thanks. I will try...
– M. Rahmat
Nov 30 at 19:37
Thanks. I will try...
– M. Rahmat
Nov 30 at 19:37
Success I hope?
– Brevan Ellefsen
Dec 3 at 16:45
Success I hope?
– Brevan Ellefsen
Dec 3 at 16:45
add a comment |
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1
It's a compact set. Can you show $r(x)$ is continuous? Continuous functions achieve their extrema on compact sets, and the minimum is greater than zero so the infimum = minimum is as well.
– Brevan Ellefsen
Nov 29 at 8:24
Thanks for your reply. How would you do that?
– M. Rahmat
Nov 30 at 4:47
1
Depends on what tools you have. Given you have a complex analysis tag: The radius of convergence is simply the distance to the nearest singularity. The distance function is necessarily continuous (e.g., since $mathbb C$ has norm $|z|$)
– Brevan Ellefsen
Nov 30 at 7:16
Thanks. I will try...
– M. Rahmat
Nov 30 at 19:37
Success I hope?
– Brevan Ellefsen
Dec 3 at 16:45