n-0-sin-n-1-x-4-n-1-

Question Number 139826 by qaz last updated on 01/May/21

\underset{n = 0}{\sum^{\infty}} \frac{\sin [(n - 1) x]}{4^{n + 1}} = ?

Answered by mnjuly1970 last updated on 01/May/21

Ω := \underset{n = 0}{\sum^{\infty}} \frac{s i n ((n - 1) x)}{4^{n + 1}} = \frac{1}{4} I m \underset{n = 0}{\sum^{\infty}} \frac{e^{i (n - 1) x}}{4^{n}}

:= \frac{1}{4} I m {(e^{i x}) \underset{n = 0}{\sum^{\infty}} \frac{e^{i n x}}{4^{n}}}

:= \frac{1}{4} I m {(e^{i x}) \underset{n = 0}{\sum^{\infty}} {(\frac{e^{i x}}{4})}^{n}}

:= \frac{1}{4} I m {(e^{i x}) (\frac{1}{1 - \frac{e^{i x}}{4}})}

:= I m ((c o s (x) + i s i n (x)) . (\frac{1}{4 - c o s (x) - i s i n (x)}))

:= I m {(c o s (x) + i s i n (x)} . I m {\frac{4 - c o s (x) + i s i n (x)}{16 - 8 c o s (x) + 1}}

:= (s i n (x)) . (\frac{s i n (x)}{17 - 8 c o s (x)}) = \frac{{s i n}^{2} (x)}{17 - 8 c o s (x)} \dots

Answered by Dwaipayan Shikari last updated on 01/May/21

\underset{n = 1}{\sum^{\infty}} \frac{s i n (n x) c o s 2 x - c o s (n x) s i n (2 x)}{4^{n}}

= \frac{1}{2 i} c o s 2 x \underset{n = 1}{\sum^{\infty}} {(\frac{e^{i x}}{4})}^{n} - {(\frac{e^{- i x}}{4})}^{n} - \frac{1}{2} s i n (2 x) \underset{n = 1}{\sum^{\infty}} {(\frac{e^{i x}}{4})}^{n} + {(\frac{e^{- i x}}{4})}^{n}

= \frac{c o s (2 x)}{2 i} (\frac{1}{1 - \frac{e^{i x}}{4}} - \frac{1}{1 - \frac{e^{- i x}}{4}}) - \frac{s i n (2 x)}{2} (\frac{1}{1 - \frac{e^{i x}}{4}} + \frac{1}{1 - \frac{e^{- i x}}{4}})

= \frac{c o s (2 x)}{2 i} (\frac{4}{4 - e^{i x}} - \frac{4}{4 - e^{- i x}}) - \frac{s i n (2 x)}{2} (\frac{4}{4 - e^{i x}} + \frac{4}{4 - e^{- i x}})

= \frac{4 c o s (2 x)}{2 i} (\frac{2 i s i n (x)}{16 - 2 c o s x + 1}) - \frac{s i n (2 x)}{2} (\frac{8 - 2 c o s x}{16 - 2 c o s x + 1})

= \frac{4 c o s (2 x) s i n (x) - 4 s i n (2 x) + s i n (2 x) c o s (x)}{17 - 2 c o s x} \dots