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Antiderivative of a Laurent series: how to choose branch cut?
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In the annulus $0<|z-c|<R$, $$f'(z)=sum^infty_k=-inftya_k(z-c)^k$$ where $a_-1ne0$.
Because the series is uniformly convergent within the annulus of convergence, after performing term-by-term indefinite integration, one obtains (for $0<|z-c|<R$):
$$f(z)=K+a_-1log(z-c)+sum^infty_k=-infty,,kne0fraca_k-1k(z-c)^k$$ where $K$ is the constant of integration.
But, how can I determine the branch cut of the logarithm? Is it arbitrary?
My attempt:
In complex analysis, the antiderivative of $f'(z)$ should be written formally as $$f(z)=int f'(z)dz=int_0^1f(gamma(t))gamma'(t)dt$$
where $gamma(t):[0,1]tomathbb Csetminusc$, $gamma(0)=textarbitrary constant L$, $0<|L-c|<R$, $gamma(1)=z$.
If $gamma$ goes around $z=c$ positively, doing the integration would give us one of the possibilities of $f(z)$, call it $f_1(z)$. If $gamma$ goes around $z=c$ negatively, we obtain another function $f_2(z)$, with $f_2(z)=f_1(z)-2pi ia_-1$.
It appears to me that if we walk from $L$, go around $c$ positively once, and go back to $L$, we would get an extra $2pi ia_-1$, thus the branch cut must connect $c$ and $L$.
This argument is quite vague. Moreover, I've seen a theorem that guarantees the region of convergence of an uniformly convergent series would remain unchanged after term-by-term integration. However, in this case, any branch cut would remove continuity of a line of points, and $f(z)$ no longer has the original size of annulus of convergence.
Can you help me to determine the relation between $L$ and the choice of branch cut, and help me to solve the above contradiction?
Thanks in advance.
complex-analysis complex-integration
asked Jul 24 at 3:51
Szeto
4,1281521
add a comment |Â
up vote
0
down vote
favorite
In the annulus $0<|z-c|<R$, $$f'(z)=sum^infty_k=-inftya_k(z-c)^k$$ where $a_-1ne0$.
Because the series is uniformly convergent within the annulus of convergence, after performing term-by-term indefinite integration, one obtains (for $0<|z-c|<R$):
$$f(z)=K+a_-1log(z-c)+sum^infty_k=-infty,,kne0fraca_k-1k(z-c)^k$$ where $K$ is the constant of integration.
But, how can I determine the branch cut of the logarithm? Is it arbitrary?
My attempt:
In complex analysis, the antiderivative of $f'(z)$ should be written formally as $$f(z)=int f'(z)dz=int_0^1f(gamma(t))gamma'(t)dt$$
where $gamma(t):[0,1]tomathbb Csetminusc$, $gamma(0)=textarbitrary constant L$, $0<|L-c|<R$, $gamma(1)=z$.
If $gamma$ goes around $z=c$ positively, doing the integration would give us one of the possibilities of $f(z)$, call it $f_1(z)$. If $gamma$ goes around $z=c$ negatively, we obtain another function $f_2(z)$, with $f_2(z)=f_1(z)-2pi ia_-1$.
It appears to me that if we walk from $L$, go around $c$ positively once, and go back to $L$, we would get an extra $2pi ia_-1$, thus the branch cut must connect $c$ and $L$.
This argument is quite vague. Moreover, I've seen a theorem that guarantees the region of convergence of an uniformly convergent series would remain unchanged after term-by-term integration. However, in this case, any branch cut would remove continuity of a line of points, and $f(z)$ no longer has the original size of annulus of convergence.
Can you help me to determine the relation between $L$ and the choice of branch cut, and help me to solve the above contradiction?
Thanks in advance.
complex-analysis complex-integration
asked Jul 24 at 3:51
Szeto
4,1281521
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In the annulus $0<|z-c|<R$, $$f'(z)=sum^infty_k=-inftya_k(z-c)^k$$ where $a_-1ne0$.
Because the series is uniformly convergent within the annulus of convergence, after performing term-by-term indefinite integration, one obtains (for $0<|z-c|<R$):
$$f(z)=K+a_-1log(z-c)+sum^infty_k=-infty,,kne0fraca_k-1k(z-c)^k$$ where $K$ is the constant of integration.
But, how can I determine the branch cut of the logarithm? Is it arbitrary?
My attempt:
In complex analysis, the antiderivative of $f'(z)$ should be written formally as $$f(z)=int f'(z)dz=int_0^1f(gamma(t))gamma'(t)dt$$
where $gamma(t):[0,1]tomathbb Csetminusc$, $gamma(0)=textarbitrary constant L$, $0<|L-c|<R$, $gamma(1)=z$.
If $gamma$ goes around $z=c$ positively, doing the integration would give us one of the possibilities of $f(z)$, call it $f_1(z)$. If $gamma$ goes around $z=c$ negatively, we obtain another function $f_2(z)$, with $f_2(z)=f_1(z)-2pi ia_-1$.
It appears to me that if we walk from $L$, go around $c$ positively once, and go back to $L$, we would get an extra $2pi ia_-1$, thus the branch cut must connect $c$ and $L$.
This argument is quite vague. Moreover, I've seen a theorem that guarantees the region of convergence of an uniformly convergent series would remain unchanged after term-by-term integration. However, in this case, any branch cut would remove continuity of a line of points, and $f(z)$ no longer has the original size of annulus of convergence.
Can you help me to determine the relation between $L$ and the choice of branch cut, and help me to solve the above contradiction?
Thanks in advance.
complex-analysis complex-integration
asked Jul 24 at 3:51
Szeto
4,1281521
In the annulus $0<|z-c|<R$, $$f'(z)=sum^infty_k=-inftya_k(z-c)^k$$ where $a_-1ne0$.
Because the series is uniformly convergent within the annulus of convergence, after performing term-by-term indefinite integration, one obtains (for $0<|z-c|<R$):
$$f(z)=K+a_-1log(z-c)+sum^infty_k=-infty,,kne0fraca_k-1k(z-c)^k$$ where $K$ is the constant of integration.
But, how can I determine the branch cut of the logarithm? Is it arbitrary?
My attempt:
In complex analysis, the antiderivative of $f'(z)$ should be written formally as $$f(z)=int f'(z)dz=int_0^1f(gamma(t))gamma'(t)dt$$
where $gamma(t):[0,1]tomathbb Csetminusc$, $gamma(0)=textarbitrary constant L$, $0<|L-c|<R$, $gamma(1)=z$.
If $gamma$ goes around $z=c$ positively, doing the integration would give us one of the possibilities of $f(z)$, call it $f_1(z)$. If $gamma$ goes around $z=c$ negatively, we obtain another function $f_2(z)$, with $f_2(z)=f_1(z)-2pi ia_-1$.
It appears to me that if we walk from $L$, go around $c$ positively once, and go back to $L$, we would get an extra $2pi ia_-1$, thus the branch cut must connect $c$ and $L$.
This argument is quite vague. Moreover, I've seen a theorem that guarantees the region of convergence of an uniformly convergent series would remain unchanged after term-by-term integration. However, in this case, any branch cut would remove continuity of a line of points, and $f(z)$ no longer has the original size of annulus of convergence.
Can you help me to determine the relation between $L$ and the choice of branch cut, and help me to solve the above contradiction?
Thanks in advance.
complex-analysis complex-integration
asked Jul 24 at 3:51
Szeto
4,1281521
asked Jul 24 at 3:51
Szeto
4,1281521
asked Jul 24 at 3:51
Szeto
4,1281521
asked Jul 24 at 3:51
Szeto
4,1281521
4,1281521
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
You can write the derivative of the intended $f$ in the form
$$f'(z)=a_-1over z-c+sum_kne-1a_kz^k=a_-1over z+g'(z) .$$
Here $g'$ is the derivative of a bona fide analytic function $g:>Omegatomathbb C$, where $Omega$ is the given annulus. It follows that
$$f(z)=a_-1 rm ''log(z-c)>rm ''+g(z)+K ,$$
whereby we now have to make some sense of the $ rm ''log(z-c)>rm ''$ appearing here. After fixing an initial (admissible) value at some point $z_0inOmega$ the $log(z-c)$ picks up an additive constant $2pi i$ with every complete loop around the point $c$. In order to obtain a true analytic function $f$ we have to shrink $Omega$ to a subdomain $Omega'$ that does not contain such loops. How you do this is arbitrary (you could even draw a logarithmic spiral with center $c$). Make the cut that is most practical in the case of your concrete situation. You should be forced to "look over the cut" as little as possible.
answered Jul 24 at 13:01
Christian Blatter
163k7107306
Sorry...but what is ‘bona fide’?
– Szeto
Jul 24 at 14:01
It means: "as advertised", in other words: no additional questions necessary, as with $sin xover x$ at $x=0$.
– Christian Blatter
Jul 24 at 18:02
add a comment |Â
up vote
0
down vote
The integration $f'(z)$ into $f(z)$ requires a constant. If $F(z)$ is a antiderivative of $f'(z)$ then $f(z) = F(z) + c$, $c in mathbbC$. When integrating complex function this constant is being determined by the branch of the $log$ so if given a $log$ branch (usually given by the value of $f(w)$ for a given $w$) this will determine the value of $c$.
edited Jul 25 at 12:31
Daniel Buck
2,3041623
answered Jul 25 at 12:19
José Carlos
101
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You can write the derivative of the intended $f$ in the form
$$f'(z)=a_-1over z-c+sum_kne-1a_kz^k=a_-1over z+g'(z) .$$
Here $g'$ is the derivative of a bona fide analytic function $g:>Omegatomathbb C$, where $Omega$ is the given annulus. It follows that
$$f(z)=a_-1 rm ''log(z-c)>rm ''+g(z)+K ,$$
whereby we now have to make some sense of the $ rm ''log(z-c)>rm ''$ appearing here. After fixing an initial (admissible) value at some point $z_0inOmega$ the $log(z-c)$ picks up an additive constant $2pi i$ with every complete loop around the point $c$. In order to obtain a true analytic function $f$ we have to shrink $Omega$ to a subdomain $Omega'$ that does not contain such loops. How you do this is arbitrary (you could even draw a logarithmic spiral with center $c$). Make the cut that is most practical in the case of your concrete situation. You should be forced to "look over the cut" as little as possible.
answered Jul 24 at 13:01
Christian Blatter
163k7107306
Sorry...but what is ‘bona fide’?
– Szeto
Jul 24 at 14:01
It means: "as advertised", in other words: no additional questions necessary, as with $sin xover x$ at $x=0$.
– Christian Blatter
Jul 24 at 18:02
add a comment |Â
up vote
1
down vote
You can write the derivative of the intended $f$ in the form
$$f'(z)=a_-1over z-c+sum_kne-1a_kz^k=a_-1over z+g'(z) .$$
Here $g'$ is the derivative of a bona fide analytic function $g:>Omegatomathbb C$, where $Omega$ is the given annulus. It follows that
$$f(z)=a_-1 rm ''log(z-c)>rm ''+g(z)+K ,$$
whereby we now have to make some sense of the $ rm ''log(z-c)>rm ''$ appearing here. After fixing an initial (admissible) value at some point $z_0inOmega$ the $log(z-c)$ picks up an additive constant $2pi i$ with every complete loop around the point $c$. In order to obtain a true analytic function $f$ we have to shrink $Omega$ to a subdomain $Omega'$ that does not contain such loops. How you do this is arbitrary (you could even draw a logarithmic spiral with center $c$). Make the cut that is most practical in the case of your concrete situation. You should be forced to "look over the cut" as little as possible.
answered Jul 24 at 13:01
Christian Blatter
163k7107306
Sorry...but what is ‘bona fide’?
– Szeto
Jul 24 at 14:01
It means: "as advertised", in other words: no additional questions necessary, as with $sin xover x$ at $x=0$.
– Christian Blatter
Jul 24 at 18:02
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You can write the derivative of the intended $f$ in the form
$$f'(z)=a_-1over z-c+sum_kne-1a_kz^k=a_-1over z+g'(z) .$$
Here $g'$ is the derivative of a bona fide analytic function $g:>Omegatomathbb C$, where $Omega$ is the given annulus. It follows that
$$f(z)=a_-1 rm ''log(z-c)>rm ''+g(z)+K ,$$
whereby we now have to make some sense of the $ rm ''log(z-c)>rm ''$ appearing here. After fixing an initial (admissible) value at some point $z_0inOmega$ the $log(z-c)$ picks up an additive constant $2pi i$ with every complete loop around the point $c$. In order to obtain a true analytic function $f$ we have to shrink $Omega$ to a subdomain $Omega'$ that does not contain such loops. How you do this is arbitrary (you could even draw a logarithmic spiral with center $c$). Make the cut that is most practical in the case of your concrete situation. You should be forced to "look over the cut" as little as possible.
answered Jul 24 at 13:01
Christian Blatter
163k7107306
You can write the derivative of the intended $f$ in the form
$$f'(z)=a_-1over z-c+sum_kne-1a_kz^k=a_-1over z+g'(z) .$$
Here $g'$ is the derivative of a bona fide analytic function $g:>Omegatomathbb C$, where $Omega$ is the given annulus. It follows that
$$f(z)=a_-1 rm ''log(z-c)>rm ''+g(z)+K ,$$
whereby we now have to make some sense of the $ rm ''log(z-c)>rm ''$ appearing here. After fixing an initial (admissible) value at some point $z_0inOmega$ the $log(z-c)$ picks up an additive constant $2pi i$ with every complete loop around the point $c$. In order to obtain a true analytic function $f$ we have to shrink $Omega$ to a subdomain $Omega'$ that does not contain such loops. How you do this is arbitrary (you could even draw a logarithmic spiral with center $c$). Make the cut that is most practical in the case of your concrete situation. You should be forced to "look over the cut" as little as possible.
answered Jul 24 at 13:01
Christian Blatter
163k7107306
answered Jul 24 at 13:01
Christian Blatter
163k7107306
answered Jul 24 at 13:01
Christian Blatter
163k7107306
answered Jul 24 at 13:01
Christian Blatter
163k7107306
163k7107306
Sorry...but what is ‘bona fide’?
– Szeto
Jul 24 at 14:01
It means: "as advertised", in other words: no additional questions necessary, as with $sin xover x$ at $x=0$.
– Christian Blatter
Jul 24 at 18:02
add a comment |Â
Sorry...but what is ‘bona fide’?
– Szeto
Jul 24 at 14:01
It means: "as advertised", in other words: no additional questions necessary, as with $sin xover x$ at $x=0$.
– Christian Blatter
Jul 24 at 18:02
Sorry...but what is ‘bona fide’?
– Szeto
Jul 24 at 14:01
Sorry...but what is ‘bona fide’?
– Szeto
Jul 24 at 14:01
It means: "as advertised", in other words: no additional questions necessary, as with $sin xover x$ at $x=0$.
– Christian Blatter
Jul 24 at 18:02
It means: "as advertised", in other words: no additional questions necessary, as with $sin xover x$ at $x=0$.
– Christian Blatter
Jul 24 at 18:02
add a comment |Â
up vote
0
down vote
The integration $f'(z)$ into $f(z)$ requires a constant. If $F(z)$ is a antiderivative of $f'(z)$ then $f(z) = F(z) + c$, $c in mathbbC$. When integrating complex function this constant is being determined by the branch of the $log$ so if given a $log$ branch (usually given by the value of $f(w)$ for a given $w$) this will determine the value of $c$.
edited Jul 25 at 12:31
Daniel Buck
2,3041623
answered Jul 25 at 12:19
José Carlos
101
add a comment |Â
up vote
0
down vote
The integration $f'(z)$ into $f(z)$ requires a constant. If $F(z)$ is a antiderivative of $f'(z)$ then $f(z) = F(z) + c$, $c in mathbbC$. When integrating complex function this constant is being determined by the branch of the $log$ so if given a $log$ branch (usually given by the value of $f(w)$ for a given $w$) this will determine the value of $c$.
edited Jul 25 at 12:31
Daniel Buck
2,3041623
answered Jul 25 at 12:19
José Carlos
101
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The integration $f'(z)$ into $f(z)$ requires a constant. If $F(z)$ is a antiderivative of $f'(z)$ then $f(z) = F(z) + c$, $c in mathbbC$. When integrating complex function this constant is being determined by the branch of the $log$ so if given a $log$ branch (usually given by the value of $f(w)$ for a given $w$) this will determine the value of $c$.
edited Jul 25 at 12:31
Daniel Buck
2,3041623
answered Jul 25 at 12:19
José Carlos
101
The integration $f'(z)$ into $f(z)$ requires a constant. If $F(z)$ is a antiderivative of $f'(z)$ then $f(z) = F(z) + c$, $c in mathbbC$. When integrating complex function this constant is being determined by the branch of the $log$ so if given a $log$ branch (usually given by the value of $f(w)$ for a given $w$) this will determine the value of $c$.
edited Jul 25 at 12:31
Daniel Buck
2,3041623
answered Jul 25 at 12:19
José Carlos
101
edited Jul 25 at 12:31
Daniel Buck
2,3041623
edited Jul 25 at 12:31
Daniel Buck
2,3041623
edited Jul 25 at 12:31
Daniel Buck
2,3041623
2,3041623
answered Jul 25 at 12:19
José Carlos
101
answered Jul 25 at 12:19
José Carlos
101
answered Jul 25 at 12:19
José Carlos
101
101
add a comment |Â
add a comment |Â
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