In the identity $sumlimits_{n=-∞}^∞g(n)=frac1{2i}oint_{|x|=∞}frac{cos πx}{sin πx}g(x),mathrm dx$, why...
3
$begingroup$
I am trying to prove this sum formula:$$sum_{n=-∞}^∞g(n)=frac1{2i}oint_{|x|=∞}frac{cos πx}{sin πx}g(x),mathrm dx.$$I do not understand why $cos πx$ is needed here. This is because poles of $sin πx$ are at $x=0,±1,±2,cdots$
Any help will be highly appreciated.
complex-analysis
edited Dec 6 '18 at 10:48
Saad
19.7k92352
asked Dec 6 '18 at 2:30
Hazoor ImranHazoor Imran
161
$endgroup$
|
show 7 more comments
3
$begingroup$
I am trying to prove this sum formula:$$sum_{n=-∞}^∞g(n)=frac1{2i}oint_{|x|=∞}frac{cos πx}{sin πx}g(x),mathrm dx.$$I do not understand why $cos πx$ is needed here. This is because poles of $sin πx$ are at $x=0,±1,±2,cdots$
Any help will be highly appreciated.
complex-analysis
edited Dec 6 '18 at 10:48
Saad
19.7k92352
asked Dec 6 '18 at 2:30
Hazoor ImranHazoor Imran
161
$endgroup$
$begingroup$
What are you integrating over?
$endgroup$
– greelious
Dec 6 '18 at 2:38
$begingroup$
Hint: $(sin pi x)'=picospi x$.
$endgroup$
– user10354138
Dec 6 '18 at 2:46
3
$begingroup$
Up to issue how should one take the limit, one has $$frac{1}{sinpi x} = sumlimits_{k=-infty}^{infty} frac{1}{(x-k)(sin(pi x)')|_{x=k}} = sumlimits_{k=-infty}^infty frac{(-1)^k}{pi(x-k)}$$ The $cos(pi x)$ factor is there to get rid of the $(-1)^k$ factor in the residues.
$endgroup$
– achille hui
Dec 6 '18 at 3:01
$begingroup$
@user302797, You changed the notationspiinftyto∞andπ. In my personal opinion, it is more suitable to use the former. Can you please explain why you use the latter?
$endgroup$
– Kemono Chen
Dec 6 '18 at 3:02
1
$begingroup$
@user302797 Using unicode to replace correct mathJAX markup is not acceptable. As the system won't know how to replace the unicode on different rendering engine and expression may not scale/style properly.
$endgroup$
– achille hui
Dec 6 '18 at 3:15
|
show 7 more comments
3
3
3
$begingroup$
I am trying to prove this sum formula:$$sum_{n=-∞}^∞g(n)=frac1{2i}oint_{|x|=∞}frac{cos πx}{sin πx}g(x),mathrm dx.$$I do not understand why $cos πx$ is needed here. This is because poles of $sin πx$ are at $x=0,±1,±2,cdots$
Any help will be highly appreciated.
complex-analysis
edited Dec 6 '18 at 10:48
Saad
19.7k92352
asked Dec 6 '18 at 2:30
Hazoor ImranHazoor Imran
161
$endgroup$
I am trying to prove this sum formula:$$sum_{n=-∞}^∞g(n)=frac1{2i}oint_{|x|=∞}frac{cos πx}{sin πx}g(x),mathrm dx.$$I do not understand why $cos πx$ is needed here. This is because poles of $sin πx$ are at $x=0,±1,±2,cdots$
Any help will be highly appreciated.
complex-analysis
complex-analysis
edited Dec 6 '18 at 10:48
Saad
19.7k92352
asked Dec 6 '18 at 2:30
Hazoor ImranHazoor Imran
161
edited Dec 6 '18 at 10:48
Saad
19.7k92352
asked Dec 6 '18 at 2:30
Hazoor ImranHazoor Imran
161
edited Dec 6 '18 at 10:48
Saad
19.7k92352
edited Dec 6 '18 at 10:48
Saad
19.7k92352
edited Dec 6 '18 at 10:48
Saad
19.7k92352
19.7k92352
asked Dec 6 '18 at 2:30
Hazoor ImranHazoor Imran
161
asked Dec 6 '18 at 2:30
Hazoor ImranHazoor Imran
161
asked Dec 6 '18 at 2:30
Hazoor ImranHazoor Imran
161
161
$begingroup$
What are you integrating over?
$endgroup$
– greelious
Dec 6 '18 at 2:38
$begingroup$
Hint: $(sin pi x)'=picospi x$.
$endgroup$
– user10354138
Dec 6 '18 at 2:46
3
$begingroup$
Up to issue how should one take the limit, one has $$frac{1}{sinpi x} = sumlimits_{k=-infty}^{infty} frac{1}{(x-k)(sin(pi x)')|_{x=k}} = sumlimits_{k=-infty}^infty frac{(-1)^k}{pi(x-k)}$$ The $cos(pi x)$ factor is there to get rid of the $(-1)^k$ factor in the residues.
$endgroup$
– achille hui
Dec 6 '18 at 3:01
$begingroup$
@user302797, You changed the notationspiinftyto∞andπ. In my personal opinion, it is more suitable to use the former. Can you please explain why you use the latter?
$endgroup$
– Kemono Chen
Dec 6 '18 at 3:02
1
$begingroup$
@user302797 Using unicode to replace correct mathJAX markup is not acceptable. As the system won't know how to replace the unicode on different rendering engine and expression may not scale/style properly.
$endgroup$
– achille hui
Dec 6 '18 at 3:15
|
show 7 more comments
$begingroup$
What are you integrating over?
$endgroup$
– greelious
Dec 6 '18 at 2:38
$begingroup$
Hint: $(sin pi x)'=picospi x$.
$endgroup$
– user10354138
Dec 6 '18 at 2:46
3
$begingroup$
Up to issue how should one take the limit, one has $$frac{1}{sinpi x} = sumlimits_{k=-infty}^{infty} frac{1}{(x-k)(sin(pi x)')|_{x=k}} = sumlimits_{k=-infty}^infty frac{(-1)^k}{pi(x-k)}$$ The $cos(pi x)$ factor is there to get rid of the $(-1)^k$ factor in the residues.
$endgroup$
– achille hui
Dec 6 '18 at 3:01
$begingroup$
@user302797, You changed the notationspiinftyto∞andπ. In my personal opinion, it is more suitable to use the former. Can you please explain why you use the latter?
$endgroup$
– Kemono Chen
Dec 6 '18 at 3:02
1
$begingroup$
@user302797 Using unicode to replace correct mathJAX markup is not acceptable. As the system won't know how to replace the unicode on different rendering engine and expression may not scale/style properly.
$endgroup$
– achille hui
Dec 6 '18 at 3:15
$begingroup$
What are you integrating over?
$endgroup$
– greelious
Dec 6 '18 at 2:38
$begingroup$
What are you integrating over?
$endgroup$
– greelious
Dec 6 '18 at 2:38
$begingroup$
Hint: $(sin pi x)'=picospi x$.
$endgroup$
– user10354138
Dec 6 '18 at 2:46
$begingroup$
Hint: $(sin pi x)'=picospi x$.
$endgroup$
– user10354138
Dec 6 '18 at 2:46
3
3
$begingroup$
Up to issue how should one take the limit, one has $$frac{1}{sinpi x} = sumlimits_{k=-infty}^{infty} frac{1}{(x-k)(sin(pi x)')|_{x=k}} = sumlimits_{k=-infty}^infty frac{(-1)^k}{pi(x-k)}$$ The $cos(pi x)$ factor is there to get rid of the $(-1)^k$ factor in the residues.
$endgroup$
– achille hui
Dec 6 '18 at 3:01
$begingroup$
Up to issue how should one take the limit, one has $$frac{1}{sinpi x} = sumlimits_{k=-infty}^{infty} frac{1}{(x-k)(sin(pi x)')|_{x=k}} = sumlimits_{k=-infty}^infty frac{(-1)^k}{pi(x-k)}$$ The $cos(pi x)$ factor is there to get rid of the $(-1)^k$ factor in the residues.
$endgroup$
– achille hui
Dec 6 '18 at 3:01
$begingroup$
@user302797, You changed the notations
pi infty to ∞ and π. In my personal opinion, it is more suitable to use the former. Can you please explain why you use the latter?$endgroup$
– Kemono Chen
Dec 6 '18 at 3:02
$begingroup$
@user302797, You changed the notations
pi infty to ∞ and π. In my personal opinion, it is more suitable to use the former. Can you please explain why you use the latter?$endgroup$
– Kemono Chen
Dec 6 '18 at 3:02
1
1
$begingroup$
@user302797 Using unicode to replace correct mathJAX markup is not acceptable. As the system won't know how to replace the unicode on different rendering engine and expression may not scale/style properly.
$endgroup$
– achille hui
Dec 6 '18 at 3:15
$begingroup$
@user302797 Using unicode to replace correct mathJAX markup is not acceptable. As the system won't know how to replace the unicode on different rendering engine and expression may not scale/style properly.
$endgroup$
– achille hui
Dec 6 '18 at 3:15
|
show 7 more comments
0
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$begingroup$
What are you integrating over?
$endgroup$
– greelious
Dec 6 '18 at 2:38
$begingroup$
Hint: $(sin pi x)'=picospi x$.
$endgroup$
– user10354138
Dec 6 '18 at 2:46
3
$begingroup$
Up to issue how should one take the limit, one has $$frac{1}{sinpi x} = sumlimits_{k=-infty}^{infty} frac{1}{(x-k)(sin(pi x)')|_{x=k}} = sumlimits_{k=-infty}^infty frac{(-1)^k}{pi(x-k)}$$ The $cos(pi x)$ factor is there to get rid of the $(-1)^k$ factor in the residues.
$endgroup$
– achille hui
Dec 6 '18 at 3:01
$begingroup$
@user302797, You changed the notations
piinftyto∞andπ. In my personal opinion, it is more suitable to use the former. Can you please explain why you use the latter?$endgroup$
– Kemono Chen
Dec 6 '18 at 3:02
1
$begingroup$
@user302797 Using unicode to replace correct mathJAX markup is not acceptable. As the system won't know how to replace the unicode on different rendering engine and expression may not scale/style properly.
$endgroup$
– achille hui
Dec 6 '18 at 3:15