Finding $cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+…+cos(theta+nalpha)$ with complex variable analysis Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How do we know Taylor's Series works with complex numbers?Computing an Integral Using Complex AnalysisEvaluation the integral in complex analysisFourier series without Fourier analysis techniquescomplex analysis - differentiabiliitycomplex analysis exponential series evaluationQuadratic Polynomial with complex coefficientsUsing complex variables to find sums of Fourier seriesGeometric series and complex numbersRelationship between hyperbolic functions and complex analysis
The answer of a series with complex variable analysis
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Finding $cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+…+cos(theta+nalpha)$ with complex variable analysis
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How do we know Taylor's Series works with complex numbers?Computing an Integral Using Complex AnalysisEvaluation the integral in complex analysisFourier series without Fourier analysis techniquescomplex analysis - differentiabiliitycomplex analysis exponential series evaluationQuadratic Polynomial with complex coefficientsUsing complex variables to find sums of Fourier seriesGeometric series and complex numbersRelationship between hyperbolic functions and complex analysis
$begingroup$
We have a series as
$cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+...+cos(theta+nalpha)=U$
How can we make use of complex variable analysis to arrive at the term below which is equivalent to the above series?
$U=fracsin(fracn+12alpha)sin(frac12alpha)cos(theta+frac12nalpha)$
sequences-and-series complex-analysis complex-numbers
edited 5 mins ago
YuiTo Cheng
2,58641037
asked 1 hour ago
UnbelievableUnbelievable
1163
$endgroup$
add a comment |
$begingroup$
We have a series as
$cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+...+cos(theta+nalpha)=U$
How can we make use of complex variable analysis to arrive at the term below which is equivalent to the above series?
$U=fracsin(fracn+12alpha)sin(frac12alpha)cos(theta+frac12nalpha)$
sequences-and-series complex-analysis complex-numbers
edited 5 mins ago
YuiTo Cheng
2,58641037
asked 1 hour ago
UnbelievableUnbelievable
1163
$endgroup$
add a comment |
$begingroup$
We have a series as
$cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+...+cos(theta+nalpha)=U$
How can we make use of complex variable analysis to arrive at the term below which is equivalent to the above series?
$U=fracsin(fracn+12alpha)sin(frac12alpha)cos(theta+frac12nalpha)$
sequences-and-series complex-analysis complex-numbers
edited 5 mins ago
YuiTo Cheng
2,58641037
asked 1 hour ago
UnbelievableUnbelievable
1163
$endgroup$
We have a series as
$cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+...+cos(theta+nalpha)=U$
How can we make use of complex variable analysis to arrive at the term below which is equivalent to the above series?
$U=fracsin(fracn+12alpha)sin(frac12alpha)cos(theta+frac12nalpha)$
sequences-and-series complex-analysis complex-numbers
sequences-and-series complex-analysis complex-numbers
edited 5 mins ago
YuiTo Cheng
2,58641037
asked 1 hour ago
UnbelievableUnbelievable
1163
edited 5 mins ago
YuiTo Cheng
2,58641037
asked 1 hour ago
UnbelievableUnbelievable
1163
edited 5 mins ago
YuiTo Cheng
2,58641037
edited 5 mins ago
YuiTo Cheng
2,58641037
edited 5 mins ago
YuiTo Cheng
2,58641037
2,58641037
asked 1 hour ago
UnbelievableUnbelievable
1163
asked 1 hour ago
UnbelievableUnbelievable
1163
asked 1 hour ago
UnbelievableUnbelievable
1163
1163
add a comment |
add a comment |
1 Answer
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$begingroup$
Use the fact that this is almost a geometric series. $$beginalignU&=mathfrak R[e^itheta +e^itheta+ialpha+cdots+e^itheta+inalpha]\&=mathfrak Rleft[e^ithetasum_j=0^n e^ijalpharight]\&=mathfrak Rleft[e^ithetafrac1-e^i(n+1)alpha1-e^ialpharight]\&=mathfrak Rleft[e^ithetafrace^-i(n+1)alpha/2-e^i(n+1)alpha/2e^-ialpha/2-e^ialpha/2e^inalpha/2right]\&=mathfrak Rleft[e^i(nalpha/2+theta)fracsin[(n+1)alpha/2]sin[alpha/2]right]\&=cos(theta+tfracnalpha2)fracsinleft(frac12(n+1)alpharight)sinleft(frac12alpharight)endalign$$
answered 48 mins ago
dialogdialog
1197
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
Use the fact that this is almost a geometric series. $$beginalignU&=mathfrak R[e^itheta +e^itheta+ialpha+cdots+e^itheta+inalpha]\&=mathfrak Rleft[e^ithetasum_j=0^n e^ijalpharight]\&=mathfrak Rleft[e^ithetafrac1-e^i(n+1)alpha1-e^ialpharight]\&=mathfrak Rleft[e^ithetafrace^-i(n+1)alpha/2-e^i(n+1)alpha/2e^-ialpha/2-e^ialpha/2e^inalpha/2right]\&=mathfrak Rleft[e^i(nalpha/2+theta)fracsin[(n+1)alpha/2]sin[alpha/2]right]\&=cos(theta+tfracnalpha2)fracsinleft(frac12(n+1)alpharight)sinleft(frac12alpharight)endalign$$
answered 48 mins ago
dialogdialog
1197
$endgroup$
add a comment |
$begingroup$
Use the fact that this is almost a geometric series. $$beginalignU&=mathfrak R[e^itheta +e^itheta+ialpha+cdots+e^itheta+inalpha]\&=mathfrak Rleft[e^ithetasum_j=0^n e^ijalpharight]\&=mathfrak Rleft[e^ithetafrac1-e^i(n+1)alpha1-e^ialpharight]\&=mathfrak Rleft[e^ithetafrace^-i(n+1)alpha/2-e^i(n+1)alpha/2e^-ialpha/2-e^ialpha/2e^inalpha/2right]\&=mathfrak Rleft[e^i(nalpha/2+theta)fracsin[(n+1)alpha/2]sin[alpha/2]right]\&=cos(theta+tfracnalpha2)fracsinleft(frac12(n+1)alpharight)sinleft(frac12alpharight)endalign$$
answered 48 mins ago
dialogdialog
1197
$endgroup$
add a comment |
$begingroup$
Use the fact that this is almost a geometric series. $$beginalignU&=mathfrak R[e^itheta +e^itheta+ialpha+cdots+e^itheta+inalpha]\&=mathfrak Rleft[e^ithetasum_j=0^n e^ijalpharight]\&=mathfrak Rleft[e^ithetafrac1-e^i(n+1)alpha1-e^ialpharight]\&=mathfrak Rleft[e^ithetafrace^-i(n+1)alpha/2-e^i(n+1)alpha/2e^-ialpha/2-e^ialpha/2e^inalpha/2right]\&=mathfrak Rleft[e^i(nalpha/2+theta)fracsin[(n+1)alpha/2]sin[alpha/2]right]\&=cos(theta+tfracnalpha2)fracsinleft(frac12(n+1)alpharight)sinleft(frac12alpharight)endalign$$
answered 48 mins ago
dialogdialog
1197
$endgroup$
Use the fact that this is almost a geometric series. $$beginalignU&=mathfrak R[e^itheta +e^itheta+ialpha+cdots+e^itheta+inalpha]\&=mathfrak Rleft[e^ithetasum_j=0^n e^ijalpharight]\&=mathfrak Rleft[e^ithetafrac1-e^i(n+1)alpha1-e^ialpharight]\&=mathfrak Rleft[e^ithetafrace^-i(n+1)alpha/2-e^i(n+1)alpha/2e^-ialpha/2-e^ialpha/2e^inalpha/2right]\&=mathfrak Rleft[e^i(nalpha/2+theta)fracsin[(n+1)alpha/2]sin[alpha/2]right]\&=cos(theta+tfracnalpha2)fracsinleft(frac12(n+1)alpharight)sinleft(frac12alpharight)endalign$$
answered 48 mins ago
dialogdialog
1197
answered 48 mins ago
dialogdialog
1197
answered 48 mins ago
dialogdialog
1197
answered 48 mins ago
dialogdialog
1197
1197
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