Σ_(n=0) ^∞ ((sin [(n−1)x])/4^(n+1) )=?


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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
	$Ω:=Σ_(n=0) ^∞ ((sin((n−1)x))/4^(n+1) )=(1/4)ImΣ_(n=0) ^∞ (e^(i(n−1)x) /4^n ) :=(1/4)Im{(e^(ix) )Σ_(n=0) ^∞ (e^(inx) /4^n )} := (1/4)Im{(e^(ix) )Σ_(n=0) ^∞ ((e^(ix) /4))^n } :=(1/4) Im{(e^(ix) )((1/(1−(e^(ix) /4))))} :=Im((cos(x)+isin(x)).((1/(4−cos(x)−isin(x))))) :=Im{(cos(x)+isin(x)}.Im{((4−cos(x)+isin(x))/(16−8cos(x)+1))} := (sin(x)).(((sin(x))/(17−8cos(x))))=((sin^2 (x))/(17−8cos(x))) ...$
	$Ω := \underset{n = 0}{\sum^{\infty}} \frac{s i n ((n - 1) x)}{4^{n + 1}} = \frac{1}{4} I m \underset{n = 0}{\sum^{\infty}} \frac{e^{i (n - 1) x}}{4^{n}}$ $:= \frac{1}{4} I m {(e^{i x}) \underset{n = 0}{\sum^{\infty}} \frac{e^{i n x}}{4^{n}}}$ $:= \frac{1}{4} I m {(e^{i x}) \underset{n = 0}{\sum^{\infty}} {(\frac{e^{i x}}{4})}^{n}}$ $:= \frac{1}{4} I m {(e^{i x}) (\frac{1}{1 - \frac{e^{i x}}{4}})}$ $:= I m ((c o s (x) + i s i n (x)) . (\frac{1}{4 - c o s (x) - i s i n (x)}))$ $:= I m {(c o s (x) + i s i n (x)} . I m {\frac{4 - c o s (x) + i s i n (x)}{16 - 8 c o s (x) + 1}}$ $:= (s i n (x)) . (\frac{s i n (x)}{17 - 8 c o s (x)}) = \frac{{s i n}^{2} (x)}{17 - 8 c o s (x)} . . .$
	Answered by Dwaipayan Shikari last updated on 01/May/21

	$\underset{n = 1}{\sum^{\infty}} \frac{s i n (n x) c o s 2 x - c o s (n x) s i n (2 x)}{4^{n}}$ $= \frac{1}{2 i} c o s 2 x \underset{n = 1}{\sum^{\infty}} {(\frac{e^{i x}}{4})}^{n} - {(\frac{e^{- i x}}{4})}^{n} - \frac{1}{2} s i n (2 x) \underset{n = 1}{\sum^{\infty}} {(\frac{e^{i x}}{4})}^{n} + {(\frac{e^{- i x}}{4})}^{n}$ $= \frac{c o s (2 x)}{2 i} (\frac{1}{1 - \frac{e^{i x}}{4}} - \frac{1}{1 - \frac{e^{- i x}}{4}}) - \frac{s i n (2 x)}{2} (\frac{1}{1 - \frac{e^{i x}}{4}} + \frac{1}{1 - \frac{e^{- i x}}{4}})$ $= \frac{c o s (2 x)}{2 i} (\frac{4}{4 - e^{i x}} - \frac{4}{4 - e^{- i x}}) - \frac{s i n (2 x)}{2} (\frac{4}{4 - e^{i x}} + \frac{4}{4 - e^{- i x}})$ $= \frac{4 c o s (2 x)}{2 i} (\frac{2 i s i n (x)}{16 - 2 c o s x + 1}) - \frac{s i n (2 x)}{2} (\frac{8 - 2 c o s x}{16 - 2 c o s x + 1})$ $= \frac{4 c o s (2 x) s i n (x) - 4 s i n (2 x) + s i n (2 x) c o s (x)}{17 - 2 c o s x} . . .$
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