Mixing
Show that $sum_{r=1}^infty frac{1}{(r-z)^2}$ is holomorphic on $mathbb{C}backslashmathbb{N}$
I'm asked to show that $displaystylesum_{r=1}^infty dfrac{1}{(r-z)^2}$ is holomorphic on $mathbb{C}backslashmathbb{N}$, and am given the hint that for any $z in mathbb{C}backslashmathbb{N}$ the series is uniform on some neighbourhood of $z$.
Someone has asked the same question here:
Showing $sum_{r=1}^infty frac{1}{(r-z)^2}$ is holomorphic on $mathbb{C}backslashmathbb{N}$
but I haven't been shown the result that the answer uses. Does anyone know a better way to do it?
I know I can use Morera's theorem to show that $displaystyleint_{gamma} sum_{r=1}^infty dfrac{1}{(r-z)^2} text{dz} = 0 $ for any closed path $gamma$, but I'm not sure how to justify exchanging the integral and the sum for any arbitrary path, when the convergence is only uniform on some neighbourhood of each point - not necessarily along the whole arbitrary closed path.
Thanks in advance.
sequences-and-series complex-analysis analysis derivatives holomorphic-functions
asked Nov 20 '18 at 14:13
fobena
11
add a comment |
I'm asked to show that $displaystylesum_{r=1}^infty dfrac{1}{(r-z)^2}$ is holomorphic on $mathbb{C}backslashmathbb{N}$, and am given the hint that for any $z in mathbb{C}backslashmathbb{N}$ the series is uniform on some neighbourhood of $z$.
Someone has asked the same question here:
Showing $sum_{r=1}^infty frac{1}{(r-z)^2}$ is holomorphic on $mathbb{C}backslashmathbb{N}$
but I haven't been shown the result that the answer uses. Does anyone know a better way to do it?
I know I can use Morera's theorem to show that $displaystyleint_{gamma} sum_{r=1}^infty dfrac{1}{(r-z)^2} text{dz} = 0 $ for any closed path $gamma$, but I'm not sure how to justify exchanging the integral and the sum for any arbitrary path, when the convergence is only uniform on some neighbourhood of each point - not necessarily along the whole arbitrary closed path.
Thanks in advance.
sequences-and-series complex-analysis analysis derivatives holomorphic-functions
asked Nov 20 '18 at 14:13
fobena
11
If there already is an answer, it is probably easiest to understand the result used there.
– James
Nov 20 '18 at 15:09
HINT: Write $z=x+iy$ and verify Wirtinger's equations (equivalent to $overlinepartial f=0$).
– James
Nov 20 '18 at 15:11
add a comment |
I'm asked to show that $displaystylesum_{r=1}^infty dfrac{1}{(r-z)^2}$ is holomorphic on $mathbb{C}backslashmathbb{N}$, and am given the hint that for any $z in mathbb{C}backslashmathbb{N}$ the series is uniform on some neighbourhood of $z$.
Someone has asked the same question here:
Showing $sum_{r=1}^infty frac{1}{(r-z)^2}$ is holomorphic on $mathbb{C}backslashmathbb{N}$
but I haven't been shown the result that the answer uses. Does anyone know a better way to do it?
I know I can use Morera's theorem to show that $displaystyleint_{gamma} sum_{r=1}^infty dfrac{1}{(r-z)^2} text{dz} = 0 $ for any closed path $gamma$, but I'm not sure how to justify exchanging the integral and the sum for any arbitrary path, when the convergence is only uniform on some neighbourhood of each point - not necessarily along the whole arbitrary closed path.
Thanks in advance.
sequences-and-series complex-analysis analysis derivatives holomorphic-functions
asked Nov 20 '18 at 14:13
fobena
11
I'm asked to show that $displaystylesum_{r=1}^infty dfrac{1}{(r-z)^2}$ is holomorphic on $mathbb{C}backslashmathbb{N}$, and am given the hint that for any $z in mathbb{C}backslashmathbb{N}$ the series is uniform on some neighbourhood of $z$.
Someone has asked the same question here:
Showing $sum_{r=1}^infty frac{1}{(r-z)^2}$ is holomorphic on $mathbb{C}backslashmathbb{N}$
but I haven't been shown the result that the answer uses. Does anyone know a better way to do it?
I know I can use Morera's theorem to show that $displaystyleint_{gamma} sum_{r=1}^infty dfrac{1}{(r-z)^2} text{dz} = 0 $ for any closed path $gamma$, but I'm not sure how to justify exchanging the integral and the sum for any arbitrary path, when the convergence is only uniform on some neighbourhood of each point - not necessarily along the whole arbitrary closed path.
Thanks in advance.
sequences-and-series complex-analysis analysis derivatives holomorphic-functions
sequences-and-series complex-analysis analysis derivatives holomorphic-functions
asked Nov 20 '18 at 14:13
fobena
11
asked Nov 20 '18 at 14:13
fobena
11
edited Nov 20 '18 at 14:50
asked Nov 20 '18 at 14:13
fobena
11
asked Nov 20 '18 at 14:13
fobena
11
asked Nov 20 '18 at 14:13
fobena
11
11
If there already is an answer, it is probably easiest to understand the result used there.
– James
Nov 20 '18 at 15:09
HINT: Write $z=x+iy$ and verify Wirtinger's equations (equivalent to $overlinepartial f=0$).
– James
Nov 20 '18 at 15:11
add a comment |
If there already is an answer, it is probably easiest to understand the result used there.
– James
Nov 20 '18 at 15:09
HINT: Write $z=x+iy$ and verify Wirtinger's equations (equivalent to $overlinepartial f=0$).
– James
Nov 20 '18 at 15:11
If there already is an answer, it is probably easiest to understand the result used there.
– James
Nov 20 '18 at 15:09
If there already is an answer, it is probably easiest to understand the result used there.
– James
Nov 20 '18 at 15:09
HINT: Write $z=x+iy$ and verify Wirtinger's equations (equivalent to $overlinepartial f=0$).
– James
Nov 20 '18 at 15:11
HINT: Write $z=x+iy$ and verify Wirtinger's equations (equivalent to $overlinepartial f=0$).
– James
Nov 20 '18 at 15:11
add a comment |
1 Answer
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You can use the following simpler fact. If
$f_n$ are holomorphic in $Omega$,
$f_nto f$ uniformly in $Omega$,
$f_n'to g$ uniformly in $Omega$,
then $f$ is holomorphic with $f'=g$.
Then check that with $f_n$ equal to the partial sum and $OmegaSubsetmathbb Csetminusmathbb N$ the assumptions are verified.
answered Nov 20 '18 at 15:18
Federico
4,679514
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can use the following simpler fact. If
$f_n$ are holomorphic in $Omega$,
$f_nto f$ uniformly in $Omega$,
$f_n'to g$ uniformly in $Omega$,
then $f$ is holomorphic with $f'=g$.
Then check that with $f_n$ equal to the partial sum and $OmegaSubsetmathbb Csetminusmathbb N$ the assumptions are verified.
answered Nov 20 '18 at 15:18
Federico
4,679514
add a comment |
You can use the following simpler fact. If
$f_n$ are holomorphic in $Omega$,
$f_nto f$ uniformly in $Omega$,
$f_n'to g$ uniformly in $Omega$,
then $f$ is holomorphic with $f'=g$.
Then check that with $f_n$ equal to the partial sum and $OmegaSubsetmathbb Csetminusmathbb N$ the assumptions are verified.
answered Nov 20 '18 at 15:18
Federico
4,679514
add a comment |
You can use the following simpler fact. If
$f_n$ are holomorphic in $Omega$,
$f_nto f$ uniformly in $Omega$,
$f_n'to g$ uniformly in $Omega$,
then $f$ is holomorphic with $f'=g$.
Then check that with $f_n$ equal to the partial sum and $OmegaSubsetmathbb Csetminusmathbb N$ the assumptions are verified.
answered Nov 20 '18 at 15:18
Federico
4,679514
You can use the following simpler fact. If
$f_n$ are holomorphic in $Omega$,
$f_nto f$ uniformly in $Omega$,
$f_n'to g$ uniformly in $Omega$,
then $f$ is holomorphic with $f'=g$.
Then check that with $f_n$ equal to the partial sum and $OmegaSubsetmathbb Csetminusmathbb N$ the assumptions are verified.
answered Nov 20 '18 at 15:18
Federico
4,679514
answered Nov 20 '18 at 15:18
Federico
4,679514
answered Nov 20 '18 at 15:18
Federico
4,679514
answered Nov 20 '18 at 15:18
Federico
4,679514
4,679514
add a comment |
add a comment |
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If there already is an answer, it is probably easiest to understand the result used there.
– James
Nov 20 '18 at 15:09
HINT: Write $z=x+iy$ and verify Wirtinger's equations (equivalent to $overlinepartial f=0$).
– James
Nov 20 '18 at 15:11