The answer of a series with complex variable 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?
$U=frac{sin(frac{n+1}{2}alpha)}{sin(frac{1}{2}alpha)}cos(theta+frac{1}{2}nalpha)$
sequences-and-series complex-analysis complex-numbers
edited 37 mins ago
John Doe
12.1k11340
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?
$U=frac{sin(frac{n+1}{2}alpha)}{sin(frac{1}{2}alpha)}cos(theta+frac{1}{2}nalpha)$
sequences-and-series complex-analysis complex-numbers
edited 37 mins ago
John Doe
12.1k11340
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?
$U=frac{sin(frac{n+1}{2}alpha)}{sin(frac{1}{2}alpha)}cos(theta+frac{1}{2}nalpha)$
sequences-and-series complex-analysis complex-numbers
edited 37 mins ago
John Doe
12.1k11340
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?
$U=frac{sin(frac{n+1}{2}alpha)}{sin(frac{1}{2}alpha)}cos(theta+frac{1}{2}nalpha)$
sequences-and-series complex-analysis complex-numbers
sequences-and-series complex-analysis complex-numbers
edited 37 mins ago
John Doe
12.1k11340
asked 1 hour ago
UnbelievableUnbelievable
1163
edited 37 mins ago
John Doe
12.1k11340
asked 1 hour ago
UnbelievableUnbelievable
1163
edited 37 mins ago
John Doe
12.1k11340
edited 37 mins ago
John Doe
12.1k11340
edited 37 mins ago
John Doe
12.1k11340
12.1k11340
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. $$begin{align}U&=mathfrak R[e^{itheta} +e^{itheta+ialpha}+cdots+e^{itheta+inalpha}]\&=mathfrak Rleft[e^{itheta}sum_{j=0}^n e^{ijalpha}right]\&=mathfrak Rleft[e^{itheta}frac{1-e^{i(n+1)alpha}}{1-e^{ialpha}}right]\&=mathfrak Rleft[e^{itheta}frac{e^{-i(n+1)alpha/2}-e^{i(n+1)alpha/2}}{e^{-ialpha/2}-e^{ialpha/2}}e^{inalpha/2}right]\&=mathfrak Rleft[e^{i(nalpha/2+theta)}frac{sin[(n+1)alpha/2]}{sin[alpha/2]}right]\&=cos(theta+tfrac{nalpha}{2})frac{sinleft(frac12(n+1)alpharight)}{sinleft(frac12alpharight)}end{align}$$
answered 39 mins ago
dialogdialog
997
$endgroup$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use the fact that this is almost a geometric series. $$begin{align}U&=mathfrak R[e^{itheta} +e^{itheta+ialpha}+cdots+e^{itheta+inalpha}]\&=mathfrak Rleft[e^{itheta}sum_{j=0}^n e^{ijalpha}right]\&=mathfrak Rleft[e^{itheta}frac{1-e^{i(n+1)alpha}}{1-e^{ialpha}}right]\&=mathfrak Rleft[e^{itheta}frac{e^{-i(n+1)alpha/2}-e^{i(n+1)alpha/2}}{e^{-ialpha/2}-e^{ialpha/2}}e^{inalpha/2}right]\&=mathfrak Rleft[e^{i(nalpha/2+theta)}frac{sin[(n+1)alpha/2]}{sin[alpha/2]}right]\&=cos(theta+tfrac{nalpha}{2})frac{sinleft(frac12(n+1)alpharight)}{sinleft(frac12alpharight)}end{align}$$
answered 39 mins ago
dialogdialog
997
$endgroup$
add a comment |
$begingroup$
Use the fact that this is almost a geometric series. $$begin{align}U&=mathfrak R[e^{itheta} +e^{itheta+ialpha}+cdots+e^{itheta+inalpha}]\&=mathfrak Rleft[e^{itheta}sum_{j=0}^n e^{ijalpha}right]\&=mathfrak Rleft[e^{itheta}frac{1-e^{i(n+1)alpha}}{1-e^{ialpha}}right]\&=mathfrak Rleft[e^{itheta}frac{e^{-i(n+1)alpha/2}-e^{i(n+1)alpha/2}}{e^{-ialpha/2}-e^{ialpha/2}}e^{inalpha/2}right]\&=mathfrak Rleft[e^{i(nalpha/2+theta)}frac{sin[(n+1)alpha/2]}{sin[alpha/2]}right]\&=cos(theta+tfrac{nalpha}{2})frac{sinleft(frac12(n+1)alpharight)}{sinleft(frac12alpharight)}end{align}$$
answered 39 mins ago
dialogdialog
997
$endgroup$
add a comment |
$begingroup$
Use the fact that this is almost a geometric series. $$begin{align}U&=mathfrak R[e^{itheta} +e^{itheta+ialpha}+cdots+e^{itheta+inalpha}]\&=mathfrak Rleft[e^{itheta}sum_{j=0}^n e^{ijalpha}right]\&=mathfrak Rleft[e^{itheta}frac{1-e^{i(n+1)alpha}}{1-e^{ialpha}}right]\&=mathfrak Rleft[e^{itheta}frac{e^{-i(n+1)alpha/2}-e^{i(n+1)alpha/2}}{e^{-ialpha/2}-e^{ialpha/2}}e^{inalpha/2}right]\&=mathfrak Rleft[e^{i(nalpha/2+theta)}frac{sin[(n+1)alpha/2]}{sin[alpha/2]}right]\&=cos(theta+tfrac{nalpha}{2})frac{sinleft(frac12(n+1)alpharight)}{sinleft(frac12alpharight)}end{align}$$
answered 39 mins ago
dialogdialog
997
$endgroup$
Use the fact that this is almost a geometric series. $$begin{align}U&=mathfrak R[e^{itheta} +e^{itheta+ialpha}+cdots+e^{itheta+inalpha}]\&=mathfrak Rleft[e^{itheta}sum_{j=0}^n e^{ijalpha}right]\&=mathfrak Rleft[e^{itheta}frac{1-e^{i(n+1)alpha}}{1-e^{ialpha}}right]\&=mathfrak Rleft[e^{itheta}frac{e^{-i(n+1)alpha/2}-e^{i(n+1)alpha/2}}{e^{-ialpha/2}-e^{ialpha/2}}e^{inalpha/2}right]\&=mathfrak Rleft[e^{i(nalpha/2+theta)}frac{sin[(n+1)alpha/2]}{sin[alpha/2]}right]\&=cos(theta+tfrac{nalpha}{2})frac{sinleft(frac12(n+1)alpharight)}{sinleft(frac12alpharight)}end{align}$$
answered 39 mins ago
dialogdialog
997
answered 39 mins ago
dialogdialog
997
answered 39 mins ago
dialogdialog
997
answered 39 mins ago
dialogdialog
997
997
add a comment |
add a comment |
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