Mixing
Prove $frac12 pi iint_gamma fracgf'f = sum_m=1^j g(z_m) - sum_n=1^k g(p_n)$.
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka
Pf:
By Thm 8.14 and Cor 9.6, $exists$ holomorphic $h: G to mathbb C$ s.t. $h(z) ne 0$ for $z=p_1,...,p_k,z_1,...,z_k$ and $$f = frach prod_m=1^j (z-z_m)^textorder(z_m)prod_n=1^k (z-p_n)^textmult(p_n)$$
$$implies f'/f = frach'h + sum_m=1^j fractextorder(z_m)z-z_m - sum_n=1^k fractextmult(p_n)z-p_n$$
$$implies gf'/f = g[frach'h + sum_m=1^j fractextorder(z_m)z-z_m - sum_n=1^k fractextmult(p_n)z-p_n]$$
$$implies int_gamma gf'/f dz = int_gamma g[frach'h + sum_m=1^j fractextorder(z_m)z-z_m - sum_n=1^k fractextmult(p_n)z-p_n] dz$$
$$ = int_gamma g[frach'h] dz + int_gamma gsum_m=1^j fractextorder(z_m)z-z_m dz - int_gamma gsum_n=1^k fractextmult(p_n)z-p_n dz$$
$$ = int_gamma g[frach'h] dz + int_gamma sum_m=1^j fracg textorder(z_m)z-z_m dz - int_gamma sum_n=1^k fracg textmult(p_n)z-p_n dz$$
$$ = int_gamma g[frach'h] dz + sum_m=1^j int_gamma fracg textorder(z_m)z-z_m dz - sum_n=1^k int_gamma fracg textmult(p_n)z-p_n dz$$
$$ = int_gamma g[frach'h] dz + sum_m=1^j textorder(z_m) int_gamma fracgz-z_m dz - sum_n=1^k textmult(p_n) int_gamma fracgz-p_n dz$$
Now, $g[frach'h]$ is holomorphic in G because it is the product of functions holomorphic in $G$: $g$ and $frach'h$, the former by assumption and the latter for the same reason the '$fracg'g$' in Argument Principle 9.17 is holomorphic. Thus,
$$ = 0 + sum_m=1^j textorder(z_m) int_gamma fracgz-z_m dz - sum_n=1^k textmult(p_n) int_gamma fracgz-p_n dz$$
Now we apply Residue Thm 9.10, Prop 9.11 or Cauchy's Integral Formula 4.27 to get
$$ = 0 + sum_m=1^j textorder(z_m) [2 pi i g(z_m)] - sum_n=1^k textmult(p_n) [2 pi i g(p_n)]$$
QED
If all the zeroes and multiplicities are somehow 1, then why are they 1 here, and why were they not 1 in Argument Principle 9.17? Also, where have I gone wrong?
Otherwise, why is it that $$frac12 pi iint_gamma fracgf'f = sum_m=1^j g(z_m) - sum_n=1^k g(p_n)$$ instead of
$$frac12 pi iint_gamma fracgf'f = sum_m=1^j textorder(z_m) g(z_m) - sum_n=1^k textmult(p_n) g(p_n)$$
?
complex-analysis proof-verification contour-integration residue-calculus cauchy-integral-formula
asked Jul 23 at 10:09
BCLC
6,89921973
add a comment |Â
up vote
2
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka
Pf:
By Thm 8.14 and Cor 9.6, $exists$ holomorphic $h: G to mathbb C$ s.t. $h(z) ne 0$ for $z=p_1,...,p_k,z_1,...,z_k$ and $$f = frach prod_m=1^j (z-z_m)^textorder(z_m)prod_n=1^k (z-p_n)^textmult(p_n)$$
$$implies f'/f = frach'h + sum_m=1^j fractextorder(z_m)z-z_m - sum_n=1^k fractextmult(p_n)z-p_n$$
$$implies gf'/f = g[frach'h + sum_m=1^j fractextorder(z_m)z-z_m - sum_n=1^k fractextmult(p_n)z-p_n]$$
$$implies int_gamma gf'/f dz = int_gamma g[frach'h + sum_m=1^j fractextorder(z_m)z-z_m - sum_n=1^k fractextmult(p_n)z-p_n] dz$$
$$ = int_gamma g[frach'h] dz + int_gamma gsum_m=1^j fractextorder(z_m)z-z_m dz - int_gamma gsum_n=1^k fractextmult(p_n)z-p_n dz$$
$$ = int_gamma g[frach'h] dz + int_gamma sum_m=1^j fracg textorder(z_m)z-z_m dz - int_gamma sum_n=1^k fracg textmult(p_n)z-p_n dz$$
$$ = int_gamma g[frach'h] dz + sum_m=1^j int_gamma fracg textorder(z_m)z-z_m dz - sum_n=1^k int_gamma fracg textmult(p_n)z-p_n dz$$
$$ = int_gamma g[frach'h] dz + sum_m=1^j textorder(z_m) int_gamma fracgz-z_m dz - sum_n=1^k textmult(p_n) int_gamma fracgz-p_n dz$$
Now, $g[frach'h]$ is holomorphic in G because it is the product of functions holomorphic in $G$: $g$ and $frach'h$, the former by assumption and the latter for the same reason the '$fracg'g$' in Argument Principle 9.17 is holomorphic. Thus,
$$ = 0 + sum_m=1^j textorder(z_m) int_gamma fracgz-z_m dz - sum_n=1^k textmult(p_n) int_gamma fracgz-p_n dz$$
Now we apply Residue Thm 9.10, Prop 9.11 or Cauchy's Integral Formula 4.27 to get
$$ = 0 + sum_m=1^j textorder(z_m) [2 pi i g(z_m)] - sum_n=1^k textmult(p_n) [2 pi i g(p_n)]$$
QED
If all the zeroes and multiplicities are somehow 1, then why are they 1 here, and why were they not 1 in Argument Principle 9.17? Also, where have I gone wrong?
Otherwise, why is it that $$frac12 pi iint_gamma fracgf'f = sum_m=1^j g(z_m) - sum_n=1^k g(p_n)$$ instead of
$$frac12 pi iint_gamma fracgf'f = sum_m=1^j textorder(z_m) g(z_m) - sum_n=1^k textmult(p_n) g(p_n)$$
?
complex-analysis proof-verification contour-integration residue-calculus cauchy-integral-formula
asked Jul 23 at 10:09
BCLC
6,89921973
2
"counted according to multiplicity" means a zero/pole of higher is listed multiple times.
– Daniel Fischer♦
Jul 23 at 10:52
@DanielFischer Thanks! Good to know you're still around maths SE, all the more that that you're a mod. Anyhoo, so what does it mean for this? What exactly happened to the $textorder(z_m)$ and $textmult(p_n)$ please?
– BCLC
Jul 23 at 11:22
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka
Pf:
By Thm 8.14 and Cor 9.6, $exists$ holomorphic $h: G to mathbb C$ s.t. $h(z) ne 0$ for $z=p_1,...,p_k,z_1,...,z_k$ and $$f = frach prod_m=1^j (z-z_m)^textorder(z_m)prod_n=1^k (z-p_n)^textmult(p_n)$$
$$implies f'/f = frach'h + sum_m=1^j fractextorder(z_m)z-z_m - sum_n=1^k fractextmult(p_n)z-p_n$$
$$implies gf'/f = g[frach'h + sum_m=1^j fractextorder(z_m)z-z_m - sum_n=1^k fractextmult(p_n)z-p_n]$$
$$implies int_gamma gf'/f dz = int_gamma g[frach'h + sum_m=1^j fractextorder(z_m)z-z_m - sum_n=1^k fractextmult(p_n)z-p_n] dz$$
$$ = int_gamma g[frach'h] dz + int_gamma gsum_m=1^j fractextorder(z_m)z-z_m dz - int_gamma gsum_n=1^k fractextmult(p_n)z-p_n dz$$
$$ = int_gamma g[frach'h] dz + int_gamma sum_m=1^j fracg textorder(z_m)z-z_m dz - int_gamma sum_n=1^k fracg textmult(p_n)z-p_n dz$$
$$ = int_gamma g[frach'h] dz + sum_m=1^j int_gamma fracg textorder(z_m)z-z_m dz - sum_n=1^k int_gamma fracg textmult(p_n)z-p_n dz$$
$$ = int_gamma g[frach'h] dz + sum_m=1^j textorder(z_m) int_gamma fracgz-z_m dz - sum_n=1^k textmult(p_n) int_gamma fracgz-p_n dz$$
Now, $g[frach'h]$ is holomorphic in G because it is the product of functions holomorphic in $G$: $g$ and $frach'h$, the former by assumption and the latter for the same reason the '$fracg'g$' in Argument Principle 9.17 is holomorphic. Thus,
$$ = 0 + sum_m=1^j textorder(z_m) int_gamma fracgz-z_m dz - sum_n=1^k textmult(p_n) int_gamma fracgz-p_n dz$$
Now we apply Residue Thm 9.10, Prop 9.11 or Cauchy's Integral Formula 4.27 to get
$$ = 0 + sum_m=1^j textorder(z_m) [2 pi i g(z_m)] - sum_n=1^k textmult(p_n) [2 pi i g(p_n)]$$
QED
If all the zeroes and multiplicities are somehow 1, then why are they 1 here, and why were they not 1 in Argument Principle 9.17? Also, where have I gone wrong?
Otherwise, why is it that $$frac12 pi iint_gamma fracgf'f = sum_m=1^j g(z_m) - sum_n=1^k g(p_n)$$ instead of
$$frac12 pi iint_gamma fracgf'f = sum_m=1^j textorder(z_m) g(z_m) - sum_n=1^k textmult(p_n) g(p_n)$$
?
complex-analysis proof-verification contour-integration residue-calculus cauchy-integral-formula
asked Jul 23 at 10:09
BCLC
6,89921973
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka
Pf:
By Thm 8.14 and Cor 9.6, $exists$ holomorphic $h: G to mathbb C$ s.t. $h(z) ne 0$ for $z=p_1,...,p_k,z_1,...,z_k$ and $$f = frach prod_m=1^j (z-z_m)^textorder(z_m)prod_n=1^k (z-p_n)^textmult(p_n)$$
$$implies f'/f = frach'h + sum_m=1^j fractextorder(z_m)z-z_m - sum_n=1^k fractextmult(p_n)z-p_n$$
$$implies gf'/f = g[frach'h + sum_m=1^j fractextorder(z_m)z-z_m - sum_n=1^k fractextmult(p_n)z-p_n]$$
$$implies int_gamma gf'/f dz = int_gamma g[frach'h + sum_m=1^j fractextorder(z_m)z-z_m - sum_n=1^k fractextmult(p_n)z-p_n] dz$$
$$ = int_gamma g[frach'h] dz + int_gamma gsum_m=1^j fractextorder(z_m)z-z_m dz - int_gamma gsum_n=1^k fractextmult(p_n)z-p_n dz$$
$$ = int_gamma g[frach'h] dz + int_gamma sum_m=1^j fracg textorder(z_m)z-z_m dz - int_gamma sum_n=1^k fracg textmult(p_n)z-p_n dz$$
$$ = int_gamma g[frach'h] dz + sum_m=1^j int_gamma fracg textorder(z_m)z-z_m dz - sum_n=1^k int_gamma fracg textmult(p_n)z-p_n dz$$
$$ = int_gamma g[frach'h] dz + sum_m=1^j textorder(z_m) int_gamma fracgz-z_m dz - sum_n=1^k textmult(p_n) int_gamma fracgz-p_n dz$$
Now, $g[frach'h]$ is holomorphic in G because it is the product of functions holomorphic in $G$: $g$ and $frach'h$, the former by assumption and the latter for the same reason the '$fracg'g$' in Argument Principle 9.17 is holomorphic. Thus,
$$ = 0 + sum_m=1^j textorder(z_m) int_gamma fracgz-z_m dz - sum_n=1^k textmult(p_n) int_gamma fracgz-p_n dz$$
Now we apply Residue Thm 9.10, Prop 9.11 or Cauchy's Integral Formula 4.27 to get
$$ = 0 + sum_m=1^j textorder(z_m) [2 pi i g(z_m)] - sum_n=1^k textmult(p_n) [2 pi i g(p_n)]$$
QED
If all the zeroes and multiplicities are somehow 1, then why are they 1 here, and why were they not 1 in Argument Principle 9.17? Also, where have I gone wrong?
Otherwise, why is it that $$frac12 pi iint_gamma fracgf'f = sum_m=1^j g(z_m) - sum_n=1^k g(p_n)$$ instead of
$$frac12 pi iint_gamma fracgf'f = sum_m=1^j textorder(z_m) g(z_m) - sum_n=1^k textmult(p_n) g(p_n)$$
?
complex-analysis proof-verification contour-integration residue-calculus cauchy-integral-formula
asked Jul 23 at 10:09
BCLC
6,89921973
edited yesterday
asked Jul 23 at 10:09
BCLC
6,89921973
asked Jul 23 at 10:09
BCLC
6,89921973
asked Jul 23 at 10:09
BCLC
6,89921973
6,89921973
2
"counted according to multiplicity" means a zero/pole of higher is listed multiple times.
– Daniel Fischer♦
Jul 23 at 10:52
@DanielFischer Thanks! Good to know you're still around maths SE, all the more that that you're a mod. Anyhoo, so what does it mean for this? What exactly happened to the $textorder(z_m)$ and $textmult(p_n)$ please?
– BCLC
Jul 23 at 11:22
add a comment |Â
2
"counted according to multiplicity" means a zero/pole of higher is listed multiple times.
– Daniel Fischer♦
Jul 23 at 10:52
@DanielFischer Thanks! Good to know you're still around maths SE, all the more that that you're a mod. Anyhoo, so what does it mean for this? What exactly happened to the $textorder(z_m)$ and $textmult(p_n)$ please?
– BCLC
Jul 23 at 11:22
2
2
"counted according to multiplicity" means a zero/pole of higher is listed multiple times.
– Daniel Fischer♦
Jul 23 at 10:52
"counted according to multiplicity" means a zero/pole of higher is listed multiple times.
– Daniel Fischer♦
Jul 23 at 10:52
@DanielFischer Thanks! Good to know you're still around maths SE, all the more that that you're a mod. Anyhoo, so what does it mean for this? What exactly happened to the $textorder(z_m)$ and $textmult(p_n)$ please?
– BCLC
Jul 23 at 11:22
@DanielFischer Thanks! Good to know you're still around maths SE, all the more that that you're a mod. Anyhoo, so what does it mean for this? What exactly happened to the $textorder(z_m)$ and $textmult(p_n)$ please?
– BCLC
Jul 23 at 11:22
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
The "counted according to multiplicity" in the statement means that a zero of higher order appears several times among the $z_i$, as many times as the order says.
If $zeta_1,dotsc, zeta_r$ are the distinct zeros of $f$ inside $gamma$, and $nu_rho$ is the order of the zero $zeta_rho$, then $zeta_rho$ occurs $nu_rho$ times among the $z_i$. If ordered appropriately, we would have
beginalign
z_1 = z_2 = dotsc = z_nu_1 &= zeta_1,, \
z_nu_1 + 1 = dotsc = z_nu_1 + nu_2 &= zeta_2,, \
hfill vdots hspace3em &quadvdots \
z_nu_1 + dotsc + nu_rho-1 + 1 = dotsc = z_j &=zeta_rho,,
endalign
and similarly for the poles. So in the sums, we have
$$g(z_1) + dotsc + g(z_nu_1) = nu_1cdot g(zeta_1) = operatornameorder(zeta_1)g(zeta_1)$$
and so on.
The order/multiplicity appears as an explicit factor if one lists the distinct zeros and poles, it appears implicitly by repeating a summand the appropriate number of times when one lists the zeros and poles according to multiplicity.
answered Jul 23 at 12:00
Daniel Fischer♦
171k16154274
thanks. 1. ummmmmmm soooooo order and mult are in the equation implicitly? 2. is it consistent that the order and mult are not implicit in the proof of the argument principle 9.17?
– BCLC
Jul 23 at 12:32
1
1. Yes, 2. consistent in which sense? Of course it's logically consistent to use different styles at different points. If it's a work with multiple authors, that's a very common occurrence, and even in single-author works it's not exactly rare.
– Daniel Fischer♦
Jul 23 at 12:50
Daniel Fischer, here, this whole page shows all the order and mult.
– BCLC
Jul 23 at 13:09
Anyhoo, is my amendment of statement to be explicit and proof of explicit version of statement right?
– BCLC
Jul 23 at 13:10
1
Apart from a notational headache caused by writing e.g. $fracgz-z_m$ - which ought to be $fracg(z)z-z_m$, or if you want to write your integrals without arguments for the integrand, then you'd have to define a function, say $varphi_m colon z mapsto frac1z-z_m$ and write $gvarphi_m$; mixing the with and the without arguments style in one integral is un-pretty - your proof is good, provided that one takes the $z_i$ and $p_i$ to denote the distinct zeros and poles. It does not match the statement of the exercise, however, which says they are listed according to multiplicity.
– Daniel Fischer♦
Jul 23 at 13:24
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The "counted according to multiplicity" in the statement means that a zero of higher order appears several times among the $z_i$, as many times as the order says.
If $zeta_1,dotsc, zeta_r$ are the distinct zeros of $f$ inside $gamma$, and $nu_rho$ is the order of the zero $zeta_rho$, then $zeta_rho$ occurs $nu_rho$ times among the $z_i$. If ordered appropriately, we would have
beginalign
z_1 = z_2 = dotsc = z_nu_1 &= zeta_1,, \
z_nu_1 + 1 = dotsc = z_nu_1 + nu_2 &= zeta_2,, \
hfill vdots hspace3em &quadvdots \
z_nu_1 + dotsc + nu_rho-1 + 1 = dotsc = z_j &=zeta_rho,,
endalign
and similarly for the poles. So in the sums, we have
$$g(z_1) + dotsc + g(z_nu_1) = nu_1cdot g(zeta_1) = operatornameorder(zeta_1)g(zeta_1)$$
and so on.
The order/multiplicity appears as an explicit factor if one lists the distinct zeros and poles, it appears implicitly by repeating a summand the appropriate number of times when one lists the zeros and poles according to multiplicity.
answered Jul 23 at 12:00
Daniel Fischer♦
171k16154274
thanks. 1. ummmmmmm soooooo order and mult are in the equation implicitly? 2. is it consistent that the order and mult are not implicit in the proof of the argument principle 9.17?
– BCLC
Jul 23 at 12:32
1
1. Yes, 2. consistent in which sense? Of course it's logically consistent to use different styles at different points. If it's a work with multiple authors, that's a very common occurrence, and even in single-author works it's not exactly rare.
– Daniel Fischer♦
Jul 23 at 12:50
Daniel Fischer, here, this whole page shows all the order and mult.
– BCLC
Jul 23 at 13:09
Anyhoo, is my amendment of statement to be explicit and proof of explicit version of statement right?
– BCLC
Jul 23 at 13:10
1
Apart from a notational headache caused by writing e.g. $fracgz-z_m$ - which ought to be $fracg(z)z-z_m$, or if you want to write your integrals without arguments for the integrand, then you'd have to define a function, say $varphi_m colon z mapsto frac1z-z_m$ and write $gvarphi_m$; mixing the with and the without arguments style in one integral is un-pretty - your proof is good, provided that one takes the $z_i$ and $p_i$ to denote the distinct zeros and poles. It does not match the statement of the exercise, however, which says they are listed according to multiplicity.
– Daniel Fischer♦
Jul 23 at 13:24
 |Â
show 1 more comment
up vote
1
down vote
The "counted according to multiplicity" in the statement means that a zero of higher order appears several times among the $z_i$, as many times as the order says.
If $zeta_1,dotsc, zeta_r$ are the distinct zeros of $f$ inside $gamma$, and $nu_rho$ is the order of the zero $zeta_rho$, then $zeta_rho$ occurs $nu_rho$ times among the $z_i$. If ordered appropriately, we would have
beginalign
z_1 = z_2 = dotsc = z_nu_1 &= zeta_1,, \
z_nu_1 + 1 = dotsc = z_nu_1 + nu_2 &= zeta_2,, \
hfill vdots hspace3em &quadvdots \
z_nu_1 + dotsc + nu_rho-1 + 1 = dotsc = z_j &=zeta_rho,,
endalign
and similarly for the poles. So in the sums, we have
$$g(z_1) + dotsc + g(z_nu_1) = nu_1cdot g(zeta_1) = operatornameorder(zeta_1)g(zeta_1)$$
and so on.
The order/multiplicity appears as an explicit factor if one lists the distinct zeros and poles, it appears implicitly by repeating a summand the appropriate number of times when one lists the zeros and poles according to multiplicity.
answered Jul 23 at 12:00
Daniel Fischer♦
171k16154274
thanks. 1. ummmmmmm soooooo order and mult are in the equation implicitly? 2. is it consistent that the order and mult are not implicit in the proof of the argument principle 9.17?
– BCLC
Jul 23 at 12:32
1
1. Yes, 2. consistent in which sense? Of course it's logically consistent to use different styles at different points. If it's a work with multiple authors, that's a very common occurrence, and even in single-author works it's not exactly rare.
– Daniel Fischer♦
Jul 23 at 12:50
Daniel Fischer, here, this whole page shows all the order and mult.
– BCLC
Jul 23 at 13:09
Anyhoo, is my amendment of statement to be explicit and proof of explicit version of statement right?
– BCLC
Jul 23 at 13:10
1
Apart from a notational headache caused by writing e.g. $fracgz-z_m$ - which ought to be $fracg(z)z-z_m$, or if you want to write your integrals without arguments for the integrand, then you'd have to define a function, say $varphi_m colon z mapsto frac1z-z_m$ and write $gvarphi_m$; mixing the with and the without arguments style in one integral is un-pretty - your proof is good, provided that one takes the $z_i$ and $p_i$ to denote the distinct zeros and poles. It does not match the statement of the exercise, however, which says they are listed according to multiplicity.
– Daniel Fischer♦
Jul 23 at 13:24
 |Â
show 1 more comment
up vote
1
down vote
up vote
1
down vote
The "counted according to multiplicity" in the statement means that a zero of higher order appears several times among the $z_i$, as many times as the order says.
If $zeta_1,dotsc, zeta_r$ are the distinct zeros of $f$ inside $gamma$, and $nu_rho$ is the order of the zero $zeta_rho$, then $zeta_rho$ occurs $nu_rho$ times among the $z_i$. If ordered appropriately, we would have
beginalign
z_1 = z_2 = dotsc = z_nu_1 &= zeta_1,, \
z_nu_1 + 1 = dotsc = z_nu_1 + nu_2 &= zeta_2,, \
hfill vdots hspace3em &quadvdots \
z_nu_1 + dotsc + nu_rho-1 + 1 = dotsc = z_j &=zeta_rho,,
endalign
and similarly for the poles. So in the sums, we have
$$g(z_1) + dotsc + g(z_nu_1) = nu_1cdot g(zeta_1) = operatornameorder(zeta_1)g(zeta_1)$$
and so on.
The order/multiplicity appears as an explicit factor if one lists the distinct zeros and poles, it appears implicitly by repeating a summand the appropriate number of times when one lists the zeros and poles according to multiplicity.
answered Jul 23 at 12:00
Daniel Fischer♦
171k16154274
The "counted according to multiplicity" in the statement means that a zero of higher order appears several times among the $z_i$, as many times as the order says.
If $zeta_1,dotsc, zeta_r$ are the distinct zeros of $f$ inside $gamma$, and $nu_rho$ is the order of the zero $zeta_rho$, then $zeta_rho$ occurs $nu_rho$ times among the $z_i$. If ordered appropriately, we would have
beginalign
z_1 = z_2 = dotsc = z_nu_1 &= zeta_1,, \
z_nu_1 + 1 = dotsc = z_nu_1 + nu_2 &= zeta_2,, \
hfill vdots hspace3em &quadvdots \
z_nu_1 + dotsc + nu_rho-1 + 1 = dotsc = z_j &=zeta_rho,,
endalign
and similarly for the poles. So in the sums, we have
$$g(z_1) + dotsc + g(z_nu_1) = nu_1cdot g(zeta_1) = operatornameorder(zeta_1)g(zeta_1)$$
and so on.
The order/multiplicity appears as an explicit factor if one lists the distinct zeros and poles, it appears implicitly by repeating a summand the appropriate number of times when one lists the zeros and poles according to multiplicity.
answered Jul 23 at 12:00
Daniel Fischer♦
171k16154274
answered Jul 23 at 12:00
Daniel Fischer♦
171k16154274
answered Jul 23 at 12:00
Daniel Fischer♦
171k16154274
answered Jul 23 at 12:00
Daniel Fischer♦
171k16154274
171k16154274
thanks. 1. ummmmmmm soooooo order and mult are in the equation implicitly? 2. is it consistent that the order and mult are not implicit in the proof of the argument principle 9.17?
– BCLC
Jul 23 at 12:32
1
1. Yes, 2. consistent in which sense? Of course it's logically consistent to use different styles at different points. If it's a work with multiple authors, that's a very common occurrence, and even in single-author works it's not exactly rare.
– Daniel Fischer♦
Jul 23 at 12:50
Daniel Fischer, here, this whole page shows all the order and mult.
– BCLC
Jul 23 at 13:09
Anyhoo, is my amendment of statement to be explicit and proof of explicit version of statement right?
– BCLC
Jul 23 at 13:10
1
Apart from a notational headache caused by writing e.g. $fracgz-z_m$ - which ought to be $fracg(z)z-z_m$, or if you want to write your integrals without arguments for the integrand, then you'd have to define a function, say $varphi_m colon z mapsto frac1z-z_m$ and write $gvarphi_m$; mixing the with and the without arguments style in one integral is un-pretty - your proof is good, provided that one takes the $z_i$ and $p_i$ to denote the distinct zeros and poles. It does not match the statement of the exercise, however, which says they are listed according to multiplicity.
– Daniel Fischer♦
Jul 23 at 13:24
 |Â
show 1 more comment
thanks. 1. ummmmmmm soooooo order and mult are in the equation implicitly? 2. is it consistent that the order and mult are not implicit in the proof of the argument principle 9.17?
– BCLC
Jul 23 at 12:32
1
1. Yes, 2. consistent in which sense? Of course it's logically consistent to use different styles at different points. If it's a work with multiple authors, that's a very common occurrence, and even in single-author works it's not exactly rare.
– Daniel Fischer♦
Jul 23 at 12:50
Daniel Fischer, here, this whole page shows all the order and mult.
– BCLC
Jul 23 at 13:09
Anyhoo, is my amendment of statement to be explicit and proof of explicit version of statement right?
– BCLC
Jul 23 at 13:10
1
Apart from a notational headache caused by writing e.g. $fracgz-z_m$ - which ought to be $fracg(z)z-z_m$, or if you want to write your integrals without arguments for the integrand, then you'd have to define a function, say $varphi_m colon z mapsto frac1z-z_m$ and write $gvarphi_m$; mixing the with and the without arguments style in one integral is un-pretty - your proof is good, provided that one takes the $z_i$ and $p_i$ to denote the distinct zeros and poles. It does not match the statement of the exercise, however, which says they are listed according to multiplicity.
– Daniel Fischer♦
Jul 23 at 13:24
thanks. 1. ummmmmmm soooooo order and mult are in the equation implicitly? 2. is it consistent that the order and mult are not implicit in the proof of the argument principle 9.17?
– BCLC
Jul 23 at 12:32
thanks. 1. ummmmmmm soooooo order and mult are in the equation implicitly? 2. is it consistent that the order and mult are not implicit in the proof of the argument principle 9.17?
– BCLC
Jul 23 at 12:32
1
1
1. Yes, 2. consistent in which sense? Of course it's logically consistent to use different styles at different points. If it's a work with multiple authors, that's a very common occurrence, and even in single-author works it's not exactly rare.
– Daniel Fischer♦
Jul 23 at 12:50
1. Yes, 2. consistent in which sense? Of course it's logically consistent to use different styles at different points. If it's a work with multiple authors, that's a very common occurrence, and even in single-author works it's not exactly rare.
– Daniel Fischer♦
Jul 23 at 12:50
Daniel Fischer, here, this whole page shows all the order and mult.
– BCLC
Jul 23 at 13:09
Daniel Fischer, here, this whole page shows all the order and mult.
– BCLC
Jul 23 at 13:09
Anyhoo, is my amendment of statement to be explicit and proof of explicit version of statement right?
– BCLC
Jul 23 at 13:10
Anyhoo, is my amendment of statement to be explicit and proof of explicit version of statement right?
– BCLC
Jul 23 at 13:10
1
1
Apart from a notational headache caused by writing e.g. $fracgz-z_m$ - which ought to be $fracg(z)z-z_m$, or if you want to write your integrals without arguments for the integrand, then you'd have to define a function, say $varphi_m colon z mapsto frac1z-z_m$ and write $gvarphi_m$; mixing the with and the without arguments style in one integral is un-pretty - your proof is good, provided that one takes the $z_i$ and $p_i$ to denote the distinct zeros and poles. It does not match the statement of the exercise, however, which says they are listed according to multiplicity.
– Daniel Fischer♦
Jul 23 at 13:24
Apart from a notational headache caused by writing e.g. $fracgz-z_m$ - which ought to be $fracg(z)z-z_m$, or if you want to write your integrals without arguments for the integrand, then you'd have to define a function, say $varphi_m colon z mapsto frac1z-z_m$ and write $gvarphi_m$; mixing the with and the without arguments style in one integral is un-pretty - your proof is good, provided that one takes the $z_i$ and $p_i$ to denote the distinct zeros and poles. It does not match the statement of the exercise, however, which says they are listed according to multiplicity.
– Daniel Fischer♦
Jul 23 at 13:24
 |Â
show 1 more comment
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2860194%2fprove-frac12-pi-i-int-gamma-fracgff-sum-m-1j-gz-m-s%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
"counted according to multiplicity" means a zero/pole of higher is listed multiple times.
– Daniel Fischer♦
Jul 23 at 10:52
@DanielFischer Thanks! Good to know you're still around maths SE, all the more that that you're a mod. Anyhoo, so what does it mean for this? What exactly happened to the $textorder(z_m)$ and $textmult(p_n)$ please?
– BCLC
Jul 23 at 11:22