Where does a Laurent Series converge?
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From the Cauchy integral formula, I get
$f(z)=int_{C} frac{f(zeta)}{zeta -z}dzeta$
manipulating, =$int_{C} f(zeta)(frac{1}{zeta -p -(z-p)})$=$int_{C} f(zeta)(frac{1}{(zeta -p)(1 -frac{(z-p)}{zeta -p}})$=$int_{C} f(zeta) sum_{n=0}^{infty} (frac{z-p}{zeta -p})^n$ which is valid for $|frac{z-p}{zeta -p}|<1$.
Geometrically, I interpret this as the points closer to $z$ than the boundary of C
Or, I can choose to do a different power series.
$f(z)=int_{C} frac{f(zeta)}{zeta -z}dzeta$ = manipulating differently, =$int_{C} f(zeta)(frac{1}{zeta -p -(z-p)})$=$int_{C} f(zeta)(frac{1}{-(z -p)(1 -frac{(zeta-p)}{z -p}})$=$-int_{C} f(zeta) sum_{n=0}^{infty} (frac{zeta-p}{z -p})^n$ which is valid for $|frac{zeta-p}{z -p}|<1$.
Which I am then interpreting as the outside region of the circle
My question is, as you may have guessed, Where does the Laurent series given by
$f(z)=sum_{n=1}^{infty} a_n(z-p)^n +sum_{n=1}^{infty} frac{b_n}{(z-p)^n}$ converge? I do not understand where the annulus of convergence can be.
Thanks in advance.
complex-analysis
asked Nov 27 at 4:07
Jungleshrimp
18410
add a comment |
up vote
1
down vote
favorite
From the Cauchy integral formula, I get
$f(z)=int_{C} frac{f(zeta)}{zeta -z}dzeta$
manipulating, =$int_{C} f(zeta)(frac{1}{zeta -p -(z-p)})$=$int_{C} f(zeta)(frac{1}{(zeta -p)(1 -frac{(z-p)}{zeta -p}})$=$int_{C} f(zeta) sum_{n=0}^{infty} (frac{z-p}{zeta -p})^n$ which is valid for $|frac{z-p}{zeta -p}|<1$.
Geometrically, I interpret this as the points closer to $z$ than the boundary of C
Or, I can choose to do a different power series.
$f(z)=int_{C} frac{f(zeta)}{zeta -z}dzeta$ = manipulating differently, =$int_{C} f(zeta)(frac{1}{zeta -p -(z-p)})$=$int_{C} f(zeta)(frac{1}{-(z -p)(1 -frac{(zeta-p)}{z -p}})$=$-int_{C} f(zeta) sum_{n=0}^{infty} (frac{zeta-p}{z -p})^n$ which is valid for $|frac{zeta-p}{z -p}|<1$.
Which I am then interpreting as the outside region of the circle
My question is, as you may have guessed, Where does the Laurent series given by
$f(z)=sum_{n=1}^{infty} a_n(z-p)^n +sum_{n=1}^{infty} frac{b_n}{(z-p)^n}$ converge? I do not understand where the annulus of convergence can be.
Thanks in advance.
complex-analysis
asked Nov 27 at 4:07
Jungleshrimp
18410
a Laurent series converges between the point where it is constructed around (in your case $p$) and up to the closer singularity to the function that it represent. If the function doesn't have any singularity then the Laurent series is just a Taylor series with radius of convergence infinity
– Masacroso
Nov 27 at 4:17
@Masacroso Here yes but in general the domain of convergence can be any anulus, eg. $sum_n (1/2)^{|n|} z^n$ converges for $|z| < 1/2, |1/z| < 1/2 = |z| in (1/2,2)$. The (open) annulus where $sum_n a_n z^n$ converges is $|z|in (1/r,R)$ where $R,r$ are the radius of convergence of $sum_{n ge 0} a_n z^n,sum_{n > 0} a_{-n} z^n$. There is a singularity on $|z| = R, |z| =1/r$
– reuns
Nov 27 at 4:25
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
From the Cauchy integral formula, I get
$f(z)=int_{C} frac{f(zeta)}{zeta -z}dzeta$
manipulating, =$int_{C} f(zeta)(frac{1}{zeta -p -(z-p)})$=$int_{C} f(zeta)(frac{1}{(zeta -p)(1 -frac{(z-p)}{zeta -p}})$=$int_{C} f(zeta) sum_{n=0}^{infty} (frac{z-p}{zeta -p})^n$ which is valid for $|frac{z-p}{zeta -p}|<1$.
Geometrically, I interpret this as the points closer to $z$ than the boundary of C
Or, I can choose to do a different power series.
$f(z)=int_{C} frac{f(zeta)}{zeta -z}dzeta$ = manipulating differently, =$int_{C} f(zeta)(frac{1}{zeta -p -(z-p)})$=$int_{C} f(zeta)(frac{1}{-(z -p)(1 -frac{(zeta-p)}{z -p}})$=$-int_{C} f(zeta) sum_{n=0}^{infty} (frac{zeta-p}{z -p})^n$ which is valid for $|frac{zeta-p}{z -p}|<1$.
Which I am then interpreting as the outside region of the circle
My question is, as you may have guessed, Where does the Laurent series given by
$f(z)=sum_{n=1}^{infty} a_n(z-p)^n +sum_{n=1}^{infty} frac{b_n}{(z-p)^n}$ converge? I do not understand where the annulus of convergence can be.
Thanks in advance.
complex-analysis
asked Nov 27 at 4:07
Jungleshrimp
18410
From the Cauchy integral formula, I get
$f(z)=int_{C} frac{f(zeta)}{zeta -z}dzeta$
manipulating, =$int_{C} f(zeta)(frac{1}{zeta -p -(z-p)})$=$int_{C} f(zeta)(frac{1}{(zeta -p)(1 -frac{(z-p)}{zeta -p}})$=$int_{C} f(zeta) sum_{n=0}^{infty} (frac{z-p}{zeta -p})^n$ which is valid for $|frac{z-p}{zeta -p}|<1$.
Geometrically, I interpret this as the points closer to $z$ than the boundary of C
Or, I can choose to do a different power series.
$f(z)=int_{C} frac{f(zeta)}{zeta -z}dzeta$ = manipulating differently, =$int_{C} f(zeta)(frac{1}{zeta -p -(z-p)})$=$int_{C} f(zeta)(frac{1}{-(z -p)(1 -frac{(zeta-p)}{z -p}})$=$-int_{C} f(zeta) sum_{n=0}^{infty} (frac{zeta-p}{z -p})^n$ which is valid for $|frac{zeta-p}{z -p}|<1$.
Which I am then interpreting as the outside region of the circle
My question is, as you may have guessed, Where does the Laurent series given by
$f(z)=sum_{n=1}^{infty} a_n(z-p)^n +sum_{n=1}^{infty} frac{b_n}{(z-p)^n}$ converge? I do not understand where the annulus of convergence can be.
Thanks in advance.
complex-analysis
complex-analysis
asked Nov 27 at 4:07
Jungleshrimp
18410
asked Nov 27 at 4:07
Jungleshrimp
18410
asked Nov 27 at 4:07
Jungleshrimp
18410
asked Nov 27 at 4:07
Jungleshrimp
18410
asked Nov 27 at 4:07
Jungleshrimp
18410
18410
a Laurent series converges between the point where it is constructed around (in your case $p$) and up to the closer singularity to the function that it represent. If the function doesn't have any singularity then the Laurent series is just a Taylor series with radius of convergence infinity
– Masacroso
Nov 27 at 4:17
@Masacroso Here yes but in general the domain of convergence can be any anulus, eg. $sum_n (1/2)^{|n|} z^n$ converges for $|z| < 1/2, |1/z| < 1/2 = |z| in (1/2,2)$. The (open) annulus where $sum_n a_n z^n$ converges is $|z|in (1/r,R)$ where $R,r$ are the radius of convergence of $sum_{n ge 0} a_n z^n,sum_{n > 0} a_{-n} z^n$. There is a singularity on $|z| = R, |z| =1/r$
– reuns
Nov 27 at 4:25
add a comment |
a Laurent series converges between the point where it is constructed around (in your case $p$) and up to the closer singularity to the function that it represent. If the function doesn't have any singularity then the Laurent series is just a Taylor series with radius of convergence infinity
– Masacroso
Nov 27 at 4:17
@Masacroso Here yes but in general the domain of convergence can be any anulus, eg. $sum_n (1/2)^{|n|} z^n$ converges for $|z| < 1/2, |1/z| < 1/2 = |z| in (1/2,2)$. The (open) annulus where $sum_n a_n z^n$ converges is $|z|in (1/r,R)$ where $R,r$ are the radius of convergence of $sum_{n ge 0} a_n z^n,sum_{n > 0} a_{-n} z^n$. There is a singularity on $|z| = R, |z| =1/r$
– reuns
Nov 27 at 4:25
a Laurent series converges between the point where it is constructed around (in your case $p$) and up to the closer singularity to the function that it represent. If the function doesn't have any singularity then the Laurent series is just a Taylor series with radius of convergence infinity
– Masacroso
Nov 27 at 4:17
a Laurent series converges between the point where it is constructed around (in your case $p$) and up to the closer singularity to the function that it represent. If the function doesn't have any singularity then the Laurent series is just a Taylor series with radius of convergence infinity
– Masacroso
Nov 27 at 4:17
@Masacroso Here yes but in general the domain of convergence can be any anulus, eg. $sum_n (1/2)^{|n|} z^n$ converges for $|z| < 1/2, |1/z| < 1/2 = |z| in (1/2,2)$. The (open) annulus where $sum_n a_n z^n$ converges is $|z|in (1/r,R)$ where $R,r$ are the radius of convergence of $sum_{n ge 0} a_n z^n,sum_{n > 0} a_{-n} z^n$. There is a singularity on $|z| = R, |z| =1/r$
– reuns
Nov 27 at 4:25
@Masacroso Here yes but in general the domain of convergence can be any anulus, eg. $sum_n (1/2)^{|n|} z^n$ converges for $|z| < 1/2, |1/z| < 1/2 = |z| in (1/2,2)$. The (open) annulus where $sum_n a_n z^n$ converges is $|z|in (1/r,R)$ where $R,r$ are the radius of convergence of $sum_{n ge 0} a_n z^n,sum_{n > 0} a_{-n} z^n$. There is a singularity on $|z| = R, |z| =1/r$
– reuns
Nov 27 at 4:25
add a comment |
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a Laurent series converges between the point where it is constructed around (in your case $p$) and up to the closer singularity to the function that it represent. If the function doesn't have any singularity then the Laurent series is just a Taylor series with radius of convergence infinity
– Masacroso
Nov 27 at 4:17
@Masacroso Here yes but in general the domain of convergence can be any anulus, eg. $sum_n (1/2)^{|n|} z^n$ converges for $|z| < 1/2, |1/z| < 1/2 = |z| in (1/2,2)$. The (open) annulus where $sum_n a_n z^n$ converges is $|z|in (1/r,R)$ where $R,r$ are the radius of convergence of $sum_{n ge 0} a_n z^n,sum_{n > 0} a_{-n} z^n$. There is a singularity on $|z| = R, |z| =1/r$
– reuns
Nov 27 at 4:25