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Finding $cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+…+cos(theta+nalpha)$ with complex variable analysis

2

$begingroup$

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

share|cite|improve this question

edited 9 mins ago

YuiTo Cheng

2,58641037

asked 1 hour ago

UnbelievableUnbelievable

1163

$endgroup$

add a comment | 

2

$begingroup$

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

share|cite|improve this question

edited 9 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?

$U=frac{sin(frac{n+1}{2}alpha)}{sin(frac{1}{2}alpha)}cos(theta+frac{1}{2}nalpha)$

sequences-and-series complex-analysis complex-numbers

share|cite|improve this question

edited 9 mins ago

YuiTo Cheng

2,58641037

asked 1 hour ago

UnbelievableUnbelievable

1163

$endgroup$

share|cite|improve this question

edited 9 mins ago

YuiTo Cheng

2,58641037

asked 1 hour ago

UnbelievableUnbelievable

1163

share|cite|improve this question

edited 9 mins ago

YuiTo Cheng

2,58641037

asked 1 hour ago

UnbelievableUnbelievable

1163

edited 9 mins ago

YuiTo Cheng

2,58641037

edited 9 mins ago

YuiTo Cheng

2,58641037

asked 1 hour ago

UnbelievableUnbelievable

1163

asked 1 hour ago

UnbelievableUnbelievable

1163

1 Answer
1

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5

$begingroup$

Use the fact that this is almost a geometric series. $$begin{align}U&=mathfrak R[e^{itheta} +e^{itheta+ialpha}+cdots+e^{itheta+inalpha}]\&=mathfrak Rleft[e^{itheta}sum_{j=0}^n e^{ijalpha}right]\&=mathfrak Rleft[e^{itheta}frac{1-e^{i(n+1)alpha}}{1-e^{ialpha}}right]\&=mathfrak Rleft[e^{itheta}frac{e^{-i(n+1)alpha/2}-e^{i(n+1)alpha/2}}{e^{-ialpha/2}-e^{ialpha/2}}e^{inalpha/2}right]\&=mathfrak Rleft[e^{i(nalpha/2+theta)}frac{sin[(n+1)alpha/2]}{sin[alpha/2]}right]\&=cos(theta+tfrac{nalpha}{2})frac{sinleft(frac12(n+1)alpharight)}{sinleft(frac12alpharight)}end{align}$$

share|cite|improve this answer

answered 53 mins ago

dialogdialog

1197

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5

$begingroup$

Use the fact that this is almost a geometric series. $$begin{align}U&=mathfrak R[e^{itheta} +e^{itheta+ialpha}+cdots+e^{itheta+inalpha}]\&=mathfrak Rleft[e^{itheta}sum_{j=0}^n e^{ijalpha}right]\&=mathfrak Rleft[e^{itheta}frac{1-e^{i(n+1)alpha}}{1-e^{ialpha}}right]\&=mathfrak Rleft[e^{itheta}frac{e^{-i(n+1)alpha/2}-e^{i(n+1)alpha/2}}{e^{-ialpha/2}-e^{ialpha/2}}e^{inalpha/2}right]\&=mathfrak Rleft[e^{i(nalpha/2+theta)}frac{sin[(n+1)alpha/2]}{sin[alpha/2]}right]\&=cos(theta+tfrac{nalpha}{2})frac{sinleft(frac12(n+1)alpharight)}{sinleft(frac12alpharight)}end{align}$$

share|cite|improve this answer

answered 53 mins ago

dialogdialog

1197

$endgroup$

add a comment | 

5

$begingroup$

Use the fact that this is almost a geometric series. $$begin{align}U&=mathfrak R[e^{itheta} +e^{itheta+ialpha}+cdots+e^{itheta+inalpha}]\&=mathfrak Rleft[e^{itheta}sum_{j=0}^n e^{ijalpha}right]\&=mathfrak Rleft[e^{itheta}frac{1-e^{i(n+1)alpha}}{1-e^{ialpha}}right]\&=mathfrak Rleft[e^{itheta}frac{e^{-i(n+1)alpha/2}-e^{i(n+1)alpha/2}}{e^{-ialpha/2}-e^{ialpha/2}}e^{inalpha/2}right]\&=mathfrak Rleft[e^{i(nalpha/2+theta)}frac{sin[(n+1)alpha/2]}{sin[alpha/2]}right]\&=cos(theta+tfrac{nalpha}{2})frac{sinleft(frac12(n+1)alpharight)}{sinleft(frac12alpharight)}end{align}$$

share|cite|improve this answer

answered 53 mins ago

dialogdialog

1197

$endgroup$

add a comment | 

$begingroup$

Use the fact that this is almost a geometric series. $$begin{align}U&=mathfrak R[e^{itheta} +e^{itheta+ialpha}+cdots+e^{itheta+inalpha}]\&=mathfrak Rleft[e^{itheta}sum_{j=0}^n e^{ijalpha}right]\&=mathfrak Rleft[e^{itheta}frac{1-e^{i(n+1)alpha}}{1-e^{ialpha}}right]\&=mathfrak Rleft[e^{itheta}frac{e^{-i(n+1)alpha/2}-e^{i(n+1)alpha/2}}{e^{-ialpha/2}-e^{ialpha/2}}e^{inalpha/2}right]\&=mathfrak Rleft[e^{i(nalpha/2+theta)}frac{sin[(n+1)alpha/2]}{sin[alpha/2]}right]\&=cos(theta+tfrac{nalpha}{2})frac{sinleft(frac12(n+1)alpharight)}{sinleft(frac12alpharight)}end{align}$$

share|cite|improve this answer

answered 53 mins ago

dialogdialog

1197

$endgroup$

share|cite|improve this answer

answered 53 mins ago

dialogdialog

1197

answered 53 mins ago

dialogdialog

1197

answered 53 mins ago

dialogdialog

1197