Question-118791

Question Number 118791 by Algoritm last updated on 19/Oct/20

Answered by Olaf last updated on 19/Oct/20

f_{p} (a) = \underset{n = 1}{\sum^{p}} \frac{\sin (2 n * a)}{2^{n}}

f_{p} (a) = \underset{n = 1}{\sum^{p}} \frac{e^{i 2 n a} - e^{- i 2 n a}}{2^{n + 1} i}

f_{p} (a) = \frac{1}{2 i} \underset{n = 1}{\sum^{p}} {(\frac{e^{2 i a}}{2})}^{n} - \frac{1}{2 i} \underset{n = 1}{\sum^{p}} {(\frac{e^{- 2 i a}}{2})}^{n}

f_{p} (a) = \frac{1}{2 i} . \frac{e^{2 i a}}{2} . \frac{1 - \frac{e^{2 i a p}}{2^{p}}}{1 - \frac{e^{2 i a}}{2}} - \frac{1}{2 i} . \frac{e^{- 2 i a}}{2} . \frac{1 - \frac{e^{- 2 i a p}}{2^{p}}}{1 - \frac{e^{- 2 i a}}{2}}

f_{\infty} (a) = \frac{1}{2 i} . \frac{e^{2 i a}}{2} . \frac{1}{1 - \frac{e^{2 i a}}{2}} - \frac{1}{2 i} . \frac{e^{- 2 i a}}{2} . \frac{1}{1 - \frac{e^{- 2 i a}}{2}}

f_{\infty} (a) = \frac{1}{2 i} . \frac{e^{2 i a}}{2 - e^{2 i a}} - \frac{1}{2 i} . \frac{e^{- 2 i a}}{2 - e^{- 2 i a}}

f_{\infty} (a) = \frac{1}{2 i} [\frac{e^{2 i a}}{2 - e^{2 i a}} - \frac{e^{- 2 i a}}{2 - e^{- 2 i a}}]

f_{\infty} (a) = \frac{1}{2 i} [\frac{e^{2 i a} (2 - e^{- 2 i a}) - e^{- 2 i a} (2 - e^{2 i a})}{(2 - e^{2 i a}) (2 - e^{- 2 i a})}]

f_{\infty} (a) = \frac{1}{2 i} [\frac{2 (e^{2 i a} - e^{- 2 i a})}{4 - 2 (e^{2 i a} + e^{- 2 i a}) + 1}]

f_{\infty} (a) = \frac{2 \sin (2 a)}{5 - 4 \cos (2 a)}

Answered by mathmax by abdo last updated on 19/Oct/20

\sum_{n = 1}^{\infty} \frac{\sin (2 na)}{2^{n}} = Im (\sum_{n = 0}^{\infty} \frac{e^{2 ina}}{2^{n}}) we have

\sum_{n = 0}^{\infty} \frac{e^{2 ina}}{2^{n}} = \sum_{n = 0}^{\infty} {(\frac{e^{2 ia}}{2})}^{n} = \frac{1}{1 - \frac{e^{2 ia}}{2}} (∣ \frac{e^{2 ia}}{2} ∣= \frac{1}{2} < 1)

= \frac{2}{2 - e^{2 ia}} = \frac{2}{2 - \cos (2 a) - isin (2 a)} = \frac{2 (2 - \cos (2 a) + isin (2 a))}{{(2 - \cos (2 a))}^{2} + \sin^{2} (2 a)}

= \frac{4 - 2 \cos (2 a) + 2 isin (2 a)}{4 - 4 \cos (2 a) + \cos^{2} (2 a) + \sin^{2} (2 a)} = \frac{4 - 2 \cos (2 a) + 2 isin (2 a)}{5 - 4 \cos (2 a)}

\Rightarrow \sum_{n = 1}^{\infty} \frac{\sin (2 na)}{2^{n}} = \frac{2 \sin (2 a)}{5 - 4 \cos (2 a)}