Sum infinite sum for a complex variable not in the integers The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Convergence of the infinite product $prod_n = 1^infty fracz - alpha_nz - beta_n$Suppose $sum_k=-infty^inftya_kz^k$ and $sum_-infty^inftyb_kz^k$ converge to $1/sin(pi z)$. Find $b_k-a_k$.Laurent series of $ 1over (z - i) $Laurent series for $z^2 e^1/z$ at $z = infty$Write $sumlimits_n=0^infty e^-xn^3$ in the form $sumlimits_n=-infty^infty a_nx^n$Help needed on laurent series for a complex functionShow that $sum_-infty^infty (-1)^nexp(nz-frac12(n+frac12)^2omega)$ converges and is entireΑn entire function as an infinite sum of entire functionsClassify singularities in the extended complex planeFinding the laurent series around z = 0
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Sum infinite sum for a complex variable not in the integers
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Convergence of the infinite product $prod_n = 1^infty fracz - alpha_nz - beta_n$Suppose $sum_k=-infty^inftya_kz^k$ and $sum_-infty^inftyb_kz^k$ converge to $1/sin(pi z)$. Find $b_k-a_k$.Laurent series of $ 1over (z - i) $Laurent series for $z^2 e^1/z$ at $z = infty$Write $sumlimits_n=0^infty e^-xn^3$ in the form $sumlimits_n=-infty^infty a_nx^n$Help needed on laurent series for a complex functionShow that $sum_-infty^infty (-1)^nexp(nz-frac12(n+frac12)^2omega)$ converges and is entireΑn entire function as an infinite sum of entire functionsClassify singularities in the extended complex planeFinding the laurent series around z = 0
$begingroup$
i'm trying to sum the series
$sum_-infty^infty(n + a)^-2$ for a complex number $a notin mathbbZ$
without any luck. I am not sure how to approach this question, other than it should be tricks using laurent series and residuals i think.
Would love some help and tips to attack the problem.
complex-analysis
asked Mar 24 at 18:20
Pernk DernetsPernk Dernets
386
$endgroup$
add a comment |
$begingroup$
i'm trying to sum the series
$sum_-infty^infty(n + a)^-2$ for a complex number $a notin mathbbZ$
without any luck. I am not sure how to approach this question, other than it should be tricks using laurent series and residuals i think.
Would love some help and tips to attack the problem.
complex-analysis
asked Mar 24 at 18:20
Pernk DernetsPernk Dernets
386
$endgroup$
add a comment |
$begingroup$
i'm trying to sum the series
$sum_-infty^infty(n + a)^-2$ for a complex number $a notin mathbbZ$
without any luck. I am not sure how to approach this question, other than it should be tricks using laurent series and residuals i think.
Would love some help and tips to attack the problem.
complex-analysis
asked Mar 24 at 18:20
Pernk DernetsPernk Dernets
386
$endgroup$
i'm trying to sum the series
$sum_-infty^infty(n + a)^-2$ for a complex number $a notin mathbbZ$
without any luck. I am not sure how to approach this question, other than it should be tricks using laurent series and residuals i think.
Would love some help and tips to attack the problem.
complex-analysis
complex-analysis
asked Mar 24 at 18:20
Pernk DernetsPernk Dernets
386
asked Mar 24 at 18:20
Pernk DernetsPernk Dernets
386
asked Mar 24 at 18:20
Pernk DernetsPernk Dernets
386
asked Mar 24 at 18:20
Pernk DernetsPernk Dernets
386
asked Mar 24 at 18:20
Pernk DernetsPernk Dernets
386
386
add a comment |
add a comment |
1 Answer
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$begingroup$
It's well-known that $$sin pi x=pi xprod_nne 0left(1-fracx^2n^2right).$$Taking the log-derivative, $$picot pi x=frac1x+sum_nne 0frac-2xn^2-x^2=frac1x+sum_nne 0left(frac1n+x-frac1n-xright).$$Splitting the infinite sum, $$picotpi x=frac1x+sum_nge 1frac1x+n+sum_nge 1frac1x-n=sum_ninBbb Zfrac1x+n.$$Finally, take $-fracddx$ at $x=a$: $$pi^2csc^2pi a=sum_ninBbb Zfrac1(n+a)^2.$$This result makes obvious why we had the restriction $anotinBbb Z$ to begin with.
answered Mar 24 at 18:33
J.G.J.G.
33.3k23252
$endgroup$
add a comment |
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$begingroup$
It's well-known that $$sin pi x=pi xprod_nne 0left(1-fracx^2n^2right).$$Taking the log-derivative, $$picot pi x=frac1x+sum_nne 0frac-2xn^2-x^2=frac1x+sum_nne 0left(frac1n+x-frac1n-xright).$$Splitting the infinite sum, $$picotpi x=frac1x+sum_nge 1frac1x+n+sum_nge 1frac1x-n=sum_ninBbb Zfrac1x+n.$$Finally, take $-fracddx$ at $x=a$: $$pi^2csc^2pi a=sum_ninBbb Zfrac1(n+a)^2.$$This result makes obvious why we had the restriction $anotinBbb Z$ to begin with.
answered Mar 24 at 18:33
J.G.J.G.
33.3k23252
$endgroup$
add a comment |
$begingroup$
It's well-known that $$sin pi x=pi xprod_nne 0left(1-fracx^2n^2right).$$Taking the log-derivative, $$picot pi x=frac1x+sum_nne 0frac-2xn^2-x^2=frac1x+sum_nne 0left(frac1n+x-frac1n-xright).$$Splitting the infinite sum, $$picotpi x=frac1x+sum_nge 1frac1x+n+sum_nge 1frac1x-n=sum_ninBbb Zfrac1x+n.$$Finally, take $-fracddx$ at $x=a$: $$pi^2csc^2pi a=sum_ninBbb Zfrac1(n+a)^2.$$This result makes obvious why we had the restriction $anotinBbb Z$ to begin with.
answered Mar 24 at 18:33
J.G.J.G.
33.3k23252
$endgroup$
add a comment |
$begingroup$
It's well-known that $$sin pi x=pi xprod_nne 0left(1-fracx^2n^2right).$$Taking the log-derivative, $$picot pi x=frac1x+sum_nne 0frac-2xn^2-x^2=frac1x+sum_nne 0left(frac1n+x-frac1n-xright).$$Splitting the infinite sum, $$picotpi x=frac1x+sum_nge 1frac1x+n+sum_nge 1frac1x-n=sum_ninBbb Zfrac1x+n.$$Finally, take $-fracddx$ at $x=a$: $$pi^2csc^2pi a=sum_ninBbb Zfrac1(n+a)^2.$$This result makes obvious why we had the restriction $anotinBbb Z$ to begin with.
answered Mar 24 at 18:33
J.G.J.G.
33.3k23252
$endgroup$
It's well-known that $$sin pi x=pi xprod_nne 0left(1-fracx^2n^2right).$$Taking the log-derivative, $$picot pi x=frac1x+sum_nne 0frac-2xn^2-x^2=frac1x+sum_nne 0left(frac1n+x-frac1n-xright).$$Splitting the infinite sum, $$picotpi x=frac1x+sum_nge 1frac1x+n+sum_nge 1frac1x-n=sum_ninBbb Zfrac1x+n.$$Finally, take $-fracddx$ at $x=a$: $$pi^2csc^2pi a=sum_ninBbb Zfrac1(n+a)^2.$$This result makes obvious why we had the restriction $anotinBbb Z$ to begin with.
answered Mar 24 at 18:33
J.G.J.G.
33.3k23252
answered Mar 24 at 18:33
J.G.J.G.
33.3k23252
answered Mar 24 at 18:33
J.G.J.G.
33.3k23252
answered Mar 24 at 18:33
J.G.J.G.
33.3k23252
33.3k23252
add a comment |
add a comment |
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