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Variation of residue theorem?
Clash Royale CLAN TAG#URR8PPP
up vote
6
down vote
favorite
Residue theorem can be stated informally as
$$oint_C f(z)dz=2pi isum a_-1$$
A contour integral sums up all the $-1$ coefficients inside.
Then, one would naturally ask:
Is there something like
$$textsomething of f(z)=sum a_-2$$ where $textsomething$ is a sort of operator?
I encounter difficulties as the residue theorem makes use of the fact that $frac1z-c$ has no antiderivative, but $frac1(z-c)^2$ does have an antiderivative, so the same trick can’t be used.
Any idea?
EDIT:
@David C. Ullrich's answer does rule out some possibilities. However, I am not desperate: I still have the hope that there might exist some operators $operatornameP$ that:
- only requires information of $f(z)$ along a contour $gamma$
- $$operatornameP_gamma[f(z)]=sum_textall poles included a_-2$$ where $a_-2$ is the coefficient of $z^-2$ of the Laurent expansion of $f(z)$ around the pole.
($f(z)$ can be assumed to be meromorphic on $mathbb C$.)
Therefore, I started a bounty to draw more attention.
residue-calculus
asked Jul 27 at 13:35
Szeto
3,9081421
 |Â
show 4 more comments
up vote
6
down vote
favorite
Residue theorem can be stated informally as
$$oint_C f(z)dz=2pi isum a_-1$$
A contour integral sums up all the $-1$ coefficients inside.
Then, one would naturally ask:
Is there something like
$$textsomething of f(z)=sum a_-2$$ where $textsomething$ is a sort of operator?
I encounter difficulties as the residue theorem makes use of the fact that $frac1z-c$ has no antiderivative, but $frac1(z-c)^2$ does have an antiderivative, so the same trick can’t be used.
Any idea?
EDIT:
@David C. Ullrich's answer does rule out some possibilities. However, I am not desperate: I still have the hope that there might exist some operators $operatornameP$ that:
- only requires information of $f(z)$ along a contour $gamma$
- $$operatornameP_gamma[f(z)]=sum_textall poles included a_-2$$ where $a_-2$ is the coefficient of $z^-2$ of the Laurent expansion of $f(z)$ around the pole.
($f(z)$ can be assumed to be meromorphic on $mathbb C$.)
Therefore, I started a bounty to draw more attention.
residue-calculus
asked Jul 27 at 13:35
Szeto
3,9081421
Doesn't $oint_C z f(z)dz$ work?
– pregunton
Jul 27 at 13:38
1
@pregunton You can try it out. It equals $$sum(a_-2+ca_-1)$$ where $c$ is location of pole.
– Szeto
Jul 27 at 13:41
1
@pregunton Yes, that is what I mean. Of course you will need locations of all singularities and all residues to do this calculation. If this works, the calculation will also similarly work if $oint f(z) dz=0$, you will just need to consider $tilde f(z)=f(z)+frac1z-A$ for some point $A$ inside $C$.
– Kusma
Jul 27 at 14:12
1
If you have a formula for $a_-2$, $a_-1$ and $a_-2+c a_-1$, then you have one for the pole $c$. That is, you would have a formula to compute poles (or zeroes) of meromorphic functions. That seems too much to be true. If you ever find such an operator, I suggest you try if you can prove the Riemann Hypothesis with it.
– barto
Aug 4 at 17:04
1
Well of course with no restrictions, not even linearity, on that operator there exists such an operator $P$. For example, define $P_gamma[f]$ to be the sum of $a_-2$ over poles inside $gamma$. (That requires only information about $f$ on $gamma$ by uniqueness: If $f=g$ on $gamma$ then $f=g$ inside $gamma$.)
– David C. Ullrich
Aug 6 at 21:56
 |Â
show 4 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Residue theorem can be stated informally as
$$oint_C f(z)dz=2pi isum a_-1$$
A contour integral sums up all the $-1$ coefficients inside.
Then, one would naturally ask:
Is there something like
$$textsomething of f(z)=sum a_-2$$ where $textsomething$ is a sort of operator?
I encounter difficulties as the residue theorem makes use of the fact that $frac1z-c$ has no antiderivative, but $frac1(z-c)^2$ does have an antiderivative, so the same trick can’t be used.
Any idea?
EDIT:
@David C. Ullrich's answer does rule out some possibilities. However, I am not desperate: I still have the hope that there might exist some operators $operatornameP$ that:
- only requires information of $f(z)$ along a contour $gamma$
- $$operatornameP_gamma[f(z)]=sum_textall poles included a_-2$$ where $a_-2$ is the coefficient of $z^-2$ of the Laurent expansion of $f(z)$ around the pole.
($f(z)$ can be assumed to be meromorphic on $mathbb C$.)
Therefore, I started a bounty to draw more attention.
residue-calculus
asked Jul 27 at 13:35
Szeto
3,9081421
Residue theorem can be stated informally as
$$oint_C f(z)dz=2pi isum a_-1$$
A contour integral sums up all the $-1$ coefficients inside.
Then, one would naturally ask:
Is there something like
$$textsomething of f(z)=sum a_-2$$ where $textsomething$ is a sort of operator?
I encounter difficulties as the residue theorem makes use of the fact that $frac1z-c$ has no antiderivative, but $frac1(z-c)^2$ does have an antiderivative, so the same trick can’t be used.
Any idea?
EDIT:
@David C. Ullrich's answer does rule out some possibilities. However, I am not desperate: I still have the hope that there might exist some operators $operatornameP$ that:
- only requires information of $f(z)$ along a contour $gamma$
- $$operatornameP_gamma[f(z)]=sum_textall poles included a_-2$$ where $a_-2$ is the coefficient of $z^-2$ of the Laurent expansion of $f(z)$ around the pole.
($f(z)$ can be assumed to be meromorphic on $mathbb C$.)
Therefore, I started a bounty to draw more attention.
residue-calculus
asked Jul 27 at 13:35
Szeto
3,9081421
edited Jul 31 at 0:50
asked Jul 27 at 13:35
Szeto
3,9081421
asked Jul 27 at 13:35
Szeto
3,9081421
asked Jul 27 at 13:35
Szeto
3,9081421
3,9081421
Doesn't $oint_C z f(z)dz$ work?
– pregunton
Jul 27 at 13:38
1
@pregunton You can try it out. It equals $$sum(a_-2+ca_-1)$$ where $c$ is location of pole.
– Szeto
Jul 27 at 13:41
1
@pregunton Yes, that is what I mean. Of course you will need locations of all singularities and all residues to do this calculation. If this works, the calculation will also similarly work if $oint f(z) dz=0$, you will just need to consider $tilde f(z)=f(z)+frac1z-A$ for some point $A$ inside $C$.
– Kusma
Jul 27 at 14:12
1
If you have a formula for $a_-2$, $a_-1$ and $a_-2+c a_-1$, then you have one for the pole $c$. That is, you would have a formula to compute poles (or zeroes) of meromorphic functions. That seems too much to be true. If you ever find such an operator, I suggest you try if you can prove the Riemann Hypothesis with it.
– barto
Aug 4 at 17:04
1
Well of course with no restrictions, not even linearity, on that operator there exists such an operator $P$. For example, define $P_gamma[f]$ to be the sum of $a_-2$ over poles inside $gamma$. (That requires only information about $f$ on $gamma$ by uniqueness: If $f=g$ on $gamma$ then $f=g$ inside $gamma$.)
– David C. Ullrich
Aug 6 at 21:56
 |Â
show 4 more comments
Doesn't $oint_C z f(z)dz$ work?
– pregunton
Jul 27 at 13:38
1
@pregunton You can try it out. It equals $$sum(a_-2+ca_-1)$$ where $c$ is location of pole.
– Szeto
Jul 27 at 13:41
1
@pregunton Yes, that is what I mean. Of course you will need locations of all singularities and all residues to do this calculation. If this works, the calculation will also similarly work if $oint f(z) dz=0$, you will just need to consider $tilde f(z)=f(z)+frac1z-A$ for some point $A$ inside $C$.
– Kusma
Jul 27 at 14:12
1
If you have a formula for $a_-2$, $a_-1$ and $a_-2+c a_-1$, then you have one for the pole $c$. That is, you would have a formula to compute poles (or zeroes) of meromorphic functions. That seems too much to be true. If you ever find such an operator, I suggest you try if you can prove the Riemann Hypothesis with it.
– barto
Aug 4 at 17:04
1
Well of course with no restrictions, not even linearity, on that operator there exists such an operator $P$. For example, define $P_gamma[f]$ to be the sum of $a_-2$ over poles inside $gamma$. (That requires only information about $f$ on $gamma$ by uniqueness: If $f=g$ on $gamma$ then $f=g$ inside $gamma$.)
– David C. Ullrich
Aug 6 at 21:56
Doesn't $oint_C z f(z)dz$ work?
– pregunton
Jul 27 at 13:38
Doesn't $oint_C z f(z)dz$ work?
– pregunton
Jul 27 at 13:38
1
1
@pregunton You can try it out. It equals $$sum(a_-2+ca_-1)$$ where $c$ is location of pole.
– Szeto
Jul 27 at 13:41
@pregunton You can try it out. It equals $$sum(a_-2+ca_-1)$$ where $c$ is location of pole.
– Szeto
Jul 27 at 13:41
1
1
@pregunton Yes, that is what I mean. Of course you will need locations of all singularities and all residues to do this calculation. If this works, the calculation will also similarly work if $oint f(z) dz=0$, you will just need to consider $tilde f(z)=f(z)+frac1z-A$ for some point $A$ inside $C$.
– Kusma
Jul 27 at 14:12
@pregunton Yes, that is what I mean. Of course you will need locations of all singularities and all residues to do this calculation. If this works, the calculation will also similarly work if $oint f(z) dz=0$, you will just need to consider $tilde f(z)=f(z)+frac1z-A$ for some point $A$ inside $C$.
– Kusma
Jul 27 at 14:12
1
1
If you have a formula for $a_-2$, $a_-1$ and $a_-2+c a_-1$, then you have one for the pole $c$. That is, you would have a formula to compute poles (or zeroes) of meromorphic functions. That seems too much to be true. If you ever find such an operator, I suggest you try if you can prove the Riemann Hypothesis with it.
– barto
Aug 4 at 17:04
If you have a formula for $a_-2$, $a_-1$ and $a_-2+c a_-1$, then you have one for the pole $c$. That is, you would have a formula to compute poles (or zeroes) of meromorphic functions. That seems too much to be true. If you ever find such an operator, I suggest you try if you can prove the Riemann Hypothesis with it.
– barto
Aug 4 at 17:04
1
1
Well of course with no restrictions, not even linearity, on that operator there exists such an operator $P$. For example, define $P_gamma[f]$ to be the sum of $a_-2$ over poles inside $gamma$. (That requires only information about $f$ on $gamma$ by uniqueness: If $f=g$ on $gamma$ then $f=g$ inside $gamma$.)
– David C. Ullrich
Aug 6 at 21:56
Well of course with no restrictions, not even linearity, on that operator there exists such an operator $P$. For example, define $P_gamma[f]$ to be the sum of $a_-2$ over poles inside $gamma$. (That requires only information about $f$ on $gamma$ by uniqueness: If $f=g$ on $gamma$ then $f=g$ inside $gamma$.)
– David C. Ullrich
Aug 6 at 21:56
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
up vote
6
down vote
Various positive suggestions in comments depend on for example knowing where all the poles are. This is not a residue-theoremish thing; an analog of RT would give, for the disk, a complex measure $mu$ on $|z|=1$ such that $int_f,dmu$ equals the sum of the $a_-2$ at all the poles. With no information except the values of $f$ on the boundary, is the point.
There is no such measure. Taking $f(z)=z^n$ for $ninBbb Z$ tells you what the Fourier coefficients of $mu$ would be, and it follows that the only $mu$ that could possibly work is the one defined by $$int_f(z),dmu=frac12pi iint_zf(z),dz.$$
But that gives the wrong answer for other $f$, for example $f(z)=1/(z-1/2)$.
In more detail: First, it's clear that $$frac12pi iint_zcdot z^n,dz=begincases1,&(n=-2),
\0,&(nne-2).endcases$$
Now assume that $mu$ "works". Then we have $int_f(z),dmu=frac12pi iint_zf(z),dz$ if $f(z)=z^n$. Since trigonometric polynomials are uniformly dense in the continuous functions on the circle it follows that $$int_f(z),dmu=frac12pi iint_zf(z),dz$$for every $f$ continuous on the unit circle.
But now let $f(z)=1/(z-1/2)$. It follows that $$int_f(z),dmu=frac12pi iint_zf(z),dzne0,$$by the actual Residue Theorem. But $sum a_-2=0$, so $mu$ doesn't work for this $f$.
answered Jul 27 at 15:25
David C. Ullrich
54k33481
Thank you for your answer. Can you please elaborate on the proof of the non-existence of the measure?
– Szeto
Jul 27 at 15:33
@Szeto Added a detailed version.
– David C. Ullrich
Jul 27 at 15:55
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
Various positive suggestions in comments depend on for example knowing where all the poles are. This is not a residue-theoremish thing; an analog of RT would give, for the disk, a complex measure $mu$ on $|z|=1$ such that $int_f,dmu$ equals the sum of the $a_-2$ at all the poles. With no information except the values of $f$ on the boundary, is the point.
There is no such measure. Taking $f(z)=z^n$ for $ninBbb Z$ tells you what the Fourier coefficients of $mu$ would be, and it follows that the only $mu$ that could possibly work is the one defined by $$int_f(z),dmu=frac12pi iint_zf(z),dz.$$
But that gives the wrong answer for other $f$, for example $f(z)=1/(z-1/2)$.
In more detail: First, it's clear that $$frac12pi iint_zcdot z^n,dz=begincases1,&(n=-2),
\0,&(nne-2).endcases$$
Now assume that $mu$ "works". Then we have $int_f(z),dmu=frac12pi iint_zf(z),dz$ if $f(z)=z^n$. Since trigonometric polynomials are uniformly dense in the continuous functions on the circle it follows that $$int_f(z),dmu=frac12pi iint_zf(z),dz$$for every $f$ continuous on the unit circle.
But now let $f(z)=1/(z-1/2)$. It follows that $$int_f(z),dmu=frac12pi iint_zf(z),dzne0,$$by the actual Residue Theorem. But $sum a_-2=0$, so $mu$ doesn't work for this $f$.
answered Jul 27 at 15:25
David C. Ullrich
54k33481
Thank you for your answer. Can you please elaborate on the proof of the non-existence of the measure?
– Szeto
Jul 27 at 15:33
@Szeto Added a detailed version.
– David C. Ullrich
Jul 27 at 15:55
add a comment |Â
up vote
6
down vote
Various positive suggestions in comments depend on for example knowing where all the poles are. This is not a residue-theoremish thing; an analog of RT would give, for the disk, a complex measure $mu$ on $|z|=1$ such that $int_f,dmu$ equals the sum of the $a_-2$ at all the poles. With no information except the values of $f$ on the boundary, is the point.
There is no such measure. Taking $f(z)=z^n$ for $ninBbb Z$ tells you what the Fourier coefficients of $mu$ would be, and it follows that the only $mu$ that could possibly work is the one defined by $$int_f(z),dmu=frac12pi iint_zf(z),dz.$$
But that gives the wrong answer for other $f$, for example $f(z)=1/(z-1/2)$.
In more detail: First, it's clear that $$frac12pi iint_zcdot z^n,dz=begincases1,&(n=-2),
\0,&(nne-2).endcases$$
Now assume that $mu$ "works". Then we have $int_f(z),dmu=frac12pi iint_zf(z),dz$ if $f(z)=z^n$. Since trigonometric polynomials are uniformly dense in the continuous functions on the circle it follows that $$int_f(z),dmu=frac12pi iint_zf(z),dz$$for every $f$ continuous on the unit circle.
But now let $f(z)=1/(z-1/2)$. It follows that $$int_f(z),dmu=frac12pi iint_zf(z),dzne0,$$by the actual Residue Theorem. But $sum a_-2=0$, so $mu$ doesn't work for this $f$.
answered Jul 27 at 15:25
David C. Ullrich
54k33481
Thank you for your answer. Can you please elaborate on the proof of the non-existence of the measure?
– Szeto
Jul 27 at 15:33
@Szeto Added a detailed version.
– David C. Ullrich
Jul 27 at 15:55
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Various positive suggestions in comments depend on for example knowing where all the poles are. This is not a residue-theoremish thing; an analog of RT would give, for the disk, a complex measure $mu$ on $|z|=1$ such that $int_f,dmu$ equals the sum of the $a_-2$ at all the poles. With no information except the values of $f$ on the boundary, is the point.
There is no such measure. Taking $f(z)=z^n$ for $ninBbb Z$ tells you what the Fourier coefficients of $mu$ would be, and it follows that the only $mu$ that could possibly work is the one defined by $$int_f(z),dmu=frac12pi iint_zf(z),dz.$$
But that gives the wrong answer for other $f$, for example $f(z)=1/(z-1/2)$.
In more detail: First, it's clear that $$frac12pi iint_zcdot z^n,dz=begincases1,&(n=-2),
\0,&(nne-2).endcases$$
Now assume that $mu$ "works". Then we have $int_f(z),dmu=frac12pi iint_zf(z),dz$ if $f(z)=z^n$. Since trigonometric polynomials are uniformly dense in the continuous functions on the circle it follows that $$int_f(z),dmu=frac12pi iint_zf(z),dz$$for every $f$ continuous on the unit circle.
But now let $f(z)=1/(z-1/2)$. It follows that $$int_f(z),dmu=frac12pi iint_zf(z),dzne0,$$by the actual Residue Theorem. But $sum a_-2=0$, so $mu$ doesn't work for this $f$.
answered Jul 27 at 15:25
David C. Ullrich
54k33481
Various positive suggestions in comments depend on for example knowing where all the poles are. This is not a residue-theoremish thing; an analog of RT would give, for the disk, a complex measure $mu$ on $|z|=1$ such that $int_f,dmu$ equals the sum of the $a_-2$ at all the poles. With no information except the values of $f$ on the boundary, is the point.
There is no such measure. Taking $f(z)=z^n$ for $ninBbb Z$ tells you what the Fourier coefficients of $mu$ would be, and it follows that the only $mu$ that could possibly work is the one defined by $$int_f(z),dmu=frac12pi iint_zf(z),dz.$$
But that gives the wrong answer for other $f$, for example $f(z)=1/(z-1/2)$.
In more detail: First, it's clear that $$frac12pi iint_zcdot z^n,dz=begincases1,&(n=-2),
\0,&(nne-2).endcases$$
Now assume that $mu$ "works". Then we have $int_f(z),dmu=frac12pi iint_zf(z),dz$ if $f(z)=z^n$. Since trigonometric polynomials are uniformly dense in the continuous functions on the circle it follows that $$int_f(z),dmu=frac12pi iint_zf(z),dz$$for every $f$ continuous on the unit circle.
But now let $f(z)=1/(z-1/2)$. It follows that $$int_f(z),dmu=frac12pi iint_zf(z),dzne0,$$by the actual Residue Theorem. But $sum a_-2=0$, so $mu$ doesn't work for this $f$.
answered Jul 27 at 15:25
David C. Ullrich
54k33481
edited Jul 27 at 15:54
answered Jul 27 at 15:25
David C. Ullrich
54k33481
answered Jul 27 at 15:25
David C. Ullrich
54k33481
answered Jul 27 at 15:25
David C. Ullrich
54k33481
54k33481
Thank you for your answer. Can you please elaborate on the proof of the non-existence of the measure?
– Szeto
Jul 27 at 15:33
@Szeto Added a detailed version.
– David C. Ullrich
Jul 27 at 15:55
add a comment |Â
Thank you for your answer. Can you please elaborate on the proof of the non-existence of the measure?
– Szeto
Jul 27 at 15:33
@Szeto Added a detailed version.
– David C. Ullrich
Jul 27 at 15:55
Thank you for your answer. Can you please elaborate on the proof of the non-existence of the measure?
– Szeto
Jul 27 at 15:33
Thank you for your answer. Can you please elaborate on the proof of the non-existence of the measure?
– Szeto
Jul 27 at 15:33
@Szeto Added a detailed version.
– David C. Ullrich
Jul 27 at 15:55
@Szeto Added a detailed version.
– David C. Ullrich
Jul 27 at 15:55
add a comment |Â
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Doesn't $oint_C z f(z)dz$ work?
– pregunton
Jul 27 at 13:38
1
@pregunton You can try it out. It equals $$sum(a_-2+ca_-1)$$ where $c$ is location of pole.
– Szeto
Jul 27 at 13:41
1
@pregunton Yes, that is what I mean. Of course you will need locations of all singularities and all residues to do this calculation. If this works, the calculation will also similarly work if $oint f(z) dz=0$, you will just need to consider $tilde f(z)=f(z)+frac1z-A$ for some point $A$ inside $C$.
– Kusma
Jul 27 at 14:12
1
If you have a formula for $a_-2$, $a_-1$ and $a_-2+c a_-1$, then you have one for the pole $c$. That is, you would have a formula to compute poles (or zeroes) of meromorphic functions. That seems too much to be true. If you ever find such an operator, I suggest you try if you can prove the Riemann Hypothesis with it.
– barto
Aug 4 at 17:04
1
Well of course with no restrictions, not even linearity, on that operator there exists such an operator $P$. For example, define $P_gamma[f]$ to be the sum of $a_-2$ over poles inside $gamma$. (That requires only information about $f$ on $gamma$ by uniqueness: If $f=g$ on $gamma$ then $f=g$ inside $gamma$.)
– David C. Ullrich
Aug 6 at 21:56