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Gerald Edgar
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!!Without regard for computability!!
One classic criterion is this. $(U,f)$ analytically continues to $(V,g)$ iff there is a finite sequence $(U_k,f_k), k=0,\dots,n$ of analytic elements, with $(U_0,f_0) = (U,f)$ and $(U_n,g_n) = (V,g)$ where, for $k=1,\dots,n$, we have $U_{k-1}\cap U_k \ne \varnothing$ and $f_{k-1}(z) = f_k(z)$ for all $z \in U_{k-1}\cap U_k$.


A simple example.
Take $(U,f)$ where $f(z) = \log z$ where $f(1) = 0$ and $U$ is the disk $|z-1|<1$.
Take $(V,g)$ where $g(z) = \log(-z)+C$ where $g(-1) = C$, $C$ a fixed constant, and $V$ is the disk $|z+1|<1$.
Question: Is $(U,f)$ coninuablecontinuable to $(V,g)$?
Answer: If and only if $C$ is an odd integer multiple of $i\pi$.

!!Without regard for computability!!
One classic criterion is this. $(U,f)$ analytically continues to $(V,g)$ iff there is a finite sequence $(U_k,f_k), k=0,\dots,n$ of analytic elements, with $(U_0,f_0) = (U,f)$ and $(U_n,g_n) = (V,g)$ where, for $k=1,\dots,n$, we have $U_{k-1}\cap U_k \ne \varnothing$ and $f_{k-1}(z) = f_k(z)$ for all $z \in U_{k-1}\cap U_k$.


A simple example.
Take $(U,f)$ where $f(z) = \log z$ where $f(1) = 0$ and $U$ is the disk $|z-1|<1$.
Take $(V,g)$ where $g(z) = \log(-z)+C$ where $g(-1) = C$, $C$ a fixed constant, and $V$ is the disk $|z+1|<1$.
Question: Is $(U,f)$ coninuable to $(V,g)$?
Answer: If and only if $C$ is an odd integer multiple of $i\pi$.

!!Without regard for computability!!
One classic criterion is this. $(U,f)$ analytically continues to $(V,g)$ iff there is a finite sequence $(U_k,f_k), k=0,\dots,n$ of analytic elements, with $(U_0,f_0) = (U,f)$ and $(U_n,g_n) = (V,g)$ where, for $k=1,\dots,n$, we have $U_{k-1}\cap U_k \ne \varnothing$ and $f_{k-1}(z) = f_k(z)$ for all $z \in U_{k-1}\cap U_k$.


A simple example.
Take $(U,f)$ where $f(z) = \log z$ where $f(1) = 0$ and $U$ is the disk $|z-1|<1$.
Take $(V,g)$ where $g(z) = \log(-z)+C$ where $g(-1) = C$, $C$ a fixed constant, and $V$ is the disk $|z+1|<1$.
Question: Is $(U,f)$ continuable to $(V,g)$?
Answer: If and only if $C$ is an odd integer multiple of $i\pi$.

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Source Link
Gerald Edgar
  • 41.1k
  • 5
  • 125
  • 219

!!Without regard for computability!!
One classic criterion is this. $(U,f)$ analytically continues to $(V,g)$ iff there is a finite sequence $(U_k,f_k), k=0,\dots,n$ of analytic elements, with $(U_0,f_0) = (U,f)$ and $(U_n,g_n) = (V,g)$ where, for $k=1,\dots,n$, we have $U_{k-1}\cap U_k \ne \varnothing$ and $f_{k-1}(z) = f_k(z)$ for all $z \in U_{k-1}\cap U_k$.


A simple example.
Take $(U,f)$ where $f(z) = \log z$ where $f(1) = 0$ and $U$ is the disk $|z-1|<1$.
Take $(V,g)$ where $g(z) = \log(-z)+C$ where $g(-1) = C$, $C$ a fixed constant, and $V$ is the disk $|z+1|<1$.
Question: Is $(U,f)$ coninuable to $(V,g)$?
Answer: If and only if $C$ is an odd integer multiple of $i\pi$.

!!Without regard for computability!!
One classic criterion is this. $(U,f)$ analytically continues to $(V,g)$ iff there is a finite sequence $(U_k,f_k), k=0,\dots,n$ of analytic elements, with $(U_0,f_0) = (U,f)$ and $(U_n,g_n) = (V,g)$ where, for $k=1,\dots,n$, we have $U_{k-1}\cap U_k \ne \varnothing$ and $f_{k-1}(z) = f_k(z)$ for all $z \in U_{k-1}\cap U_k$.

!!Without regard for computability!!
One classic criterion is this. $(U,f)$ analytically continues to $(V,g)$ iff there is a finite sequence $(U_k,f_k), k=0,\dots,n$ of analytic elements, with $(U_0,f_0) = (U,f)$ and $(U_n,g_n) = (V,g)$ where, for $k=1,\dots,n$, we have $U_{k-1}\cap U_k \ne \varnothing$ and $f_{k-1}(z) = f_k(z)$ for all $z \in U_{k-1}\cap U_k$.


A simple example.
Take $(U,f)$ where $f(z) = \log z$ where $f(1) = 0$ and $U$ is the disk $|z-1|<1$.
Take $(V,g)$ where $g(z) = \log(-z)+C$ where $g(-1) = C$, $C$ a fixed constant, and $V$ is the disk $|z+1|<1$.
Question: Is $(U,f)$ coninuable to $(V,g)$?
Answer: If and only if $C$ is an odd integer multiple of $i\pi$.

Source Link
Gerald Edgar
  • 41.1k
  • 5
  • 125
  • 219

!!Without regard for computability!!
One classic criterion is this. $(U,f)$ analytically continues to $(V,g)$ iff there is a finite sequence $(U_k,f_k), k=0,\dots,n$ of analytic elements, with $(U_0,f_0) = (U,f)$ and $(U_n,g_n) = (V,g)$ where, for $k=1,\dots,n$, we have $U_{k-1}\cap U_k \ne \varnothing$ and $f_{k-1}(z) = f_k(z)$ for all $z \in U_{k-1}\cap U_k$.