Σ_1 ^∞ (((sin (x))/x))=?


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Question Number 122883 by kolos last updated on 20/Nov/20

$\sum_{1}^{\infty} (\frac{\sin (x)}{x}) = ?$
Commented by Dwaipayan Shikari last updated on 20/Nov/20

$\underset{n = 1}{\sum^{\infty}} \frac{s i n x}{x} = \frac{1}{2 i} \underset{n = 1}{\sum^{\infty}} \frac{e^{i x}}{x} - \frac{1}{2 i} \sum^{\infty} \frac{e^{- i x}}{x}$ $= \frac{1}{2 i} (- l o g (1 - e^{i}) + l o g (1 - e^{- i})) = \frac{1}{2 i} l o g (\frac{1 - e^{- i}}{1 - e^{i}})$ $= \frac{1}{2 i} l o g (- e^{- i}) = \frac{1}{2 i} l o g (e^{i π}) + \frac{1}{2 i} ({l o g e}^{- i}) = \frac{π}{2} - \frac{1}{2}$
Commented by kolos last updated on 21/Nov/20

$t h a n k s o m u c h, e v e r y s i n g l e o n e$
	Answered by TANMAY PANACEA last updated on 20/Nov/20
	$i think q=Σ_(x=1) ^∞ ((sinx)/x) p=Σ_(x=1) ^∞ ((cosx)/x) p+iq=Σ_(x=1) ^∞ (e^(ix) /x)=(e^i /1)+(e^(2i) /2)+(e^(3i) /3)+...∞ p+iq=−ln(1−e^i )=−ln(1−cos1−isin1) p+iq=−ln(2sin^2 (1/2)−i×2sin(1/2)xcos(1/2)) p+iq=−ln{−2sin(1/2)(sin(1/2)+icos(1/2))} p+iq=−ln{2sin(π+(1/2))(cos((π/2)−(1/2))+isin((π/2)−(1/2))} p+iaq=−ln(2sin(π+(1/2)))−ln(e^(i((π/2)−(1/2))) p+iq=−ln(2sin(π+(1/2))−i((π/2)−(1/2)) q=−(((π−1)/2)) ln(1−t)=−t−(t^2 /2)−(t^3 /3)−(t^4 /4)...∞ −ln(1−t)=t+(t^2 /2)+(t^3 /3)+..∞$
	$i t h i n k$ $q = \underset{x = 1}{\sum^{\infty}} \frac{s i n x}{x}$ $p = \underset{x = 1}{\sum^{\infty}} \frac{c o s x}{x}$ $p + i q = \underset{x = 1}{\sum^{\infty}} \frac{e^{i x}}{x} = \frac{e^{i}}{1} + \frac{e^{2 i}}{2} + \frac{e^{3 i}}{3} + . . . \infty$ $p + i q = - l n (1 - e^{i}) = - l n (1 - c o s 1 - i s i n 1)$ $p + i q = - l n (2 {s i n}^{2} \frac{1}{2} - i \times 2 s i n \frac{1}{2} x c o s \frac{1}{2})$ $p + i q = - l n {- 2 s i n \frac{1}{2} (s i n \frac{1}{2} + i c o s \frac{1}{2})}$ $p + i q = - l n {2 s i n (π + \frac{1}{2}) (c o s (\frac{π}{2} - \frac{1}{2}) + i s i n (\frac{π}{2} - \frac{1}{2})}$ $p + i a q = - l n (2 s i n (π + \frac{1}{2})) - l n (e^{i (\frac{π}{2} - \frac{1}{2})}$ $p + i q = - l n (2 s i n (π + \frac{1}{2}) - i (\frac{π}{2} - \frac{1}{2})$ $q = - (\frac{π - 1}{2})$ $l n (1 - t) = - t - \frac{t^{2}}{2} - \frac{t^{3}}{3} - \frac{t^{4}}{4} . . . \infty$ $- l n (1 - t) = t + \frac{t^{2}}{2} + \frac{t^{3}}{3} + . . \infty$
	Answered by Dwaipayan Shikari last updated on 20/Nov/20

	$I n t h e s a m e w a y$ $\underset{n = 1}{\sum^{\infty}} \frac{c o s x}{x} = \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{e^{i x}}{x} + \frac{e^{- i x}}{x} = \frac{1}{2} (l o g (\frac{1}{(1 - e^{i}) (1 - e^{- i})})$ $= \frac{1}{2} l o g (\frac{1}{1 - (e^{i} + e^{- i}) + 1}) = \frac{1}{2} l o g (\frac{1}{2 - 2 c o s (1)})$
	Answered by mathmax by abdo last updated on 21/Nov/20

	$i think the Q is \sum_{n = 1}^{\infty} \frac{\sin (n)}{n}$ $let S (x) = \sum_{n = 1}^{\infty} \frac{\sin (nx)}{n} \Rightarrow S (x) = Im (\sum_{n = 1}^{\infty} \frac{e^{inx}}{n})$ $w (x) = \sum_{n = 1}^{\infty} \frac{e^{inx}}{n} \Rightarrow w^{^{'}} (x) = \sum_{n = 1}^{\infty} \frac{in e^{inx}}{n} = i \sum_{n = 1}^{\infty} {(e^{ix})}^{n}$ $= i \times \frac{1}{1 - e^{ix}} \Rightarrow w (x) = - \ln (1 - e^{ix}) + c with w) 0) = 0 = c \Rightarrow$ $w (x) = - \ln (1 - e^{ix}) \Rightarrow S (x) = Im (- \ln (1 - e^{ix})) we have$ $\ln (1 - e^{ix}) = \ln (1 - cosx - isinx) = \ln ({2 \sin}^{2} (\frac{x}{2}) - 2 isin (\frac{x}{2}) \cos (\frac{x}{2}))$ $= \ln (2) + \ln (- isin (\frac{x}{2}) (\cos (\frac{x}{2}) + isin (\frac{x}{2})) = \ln (2) + \ln (- i) + \ln (\sin (\frac{x}{2})) + \ln (e^{\frac{ix}{2}})$ $= \ln (2) - \frac{i π}{2} + \ln (\sin (\frac{x}{2})) + \frac{ix}{2} = \ln (2 \sin (\frac{x}{2})) + \frac{x - π}{2} i \Rightarrow$ $S (x) = \frac{π - x}{2} \Rightarrow \sum_{n = 1}^{\infty} \frac{\sin (n)}{n} = S (1) = \frac{π - 1}{2}$
	Answered by mnjuly1970 last updated on 21/Nov/20

	$Φ = \underset{n = 1}{\sum^{\infty}} (\frac{s i n (n)}{n}) = \frac{1}{2 i} \underset{n = 1}{\sum^{\infty}} (\frac{e^{i n} - e^{- i n}}{n})$ $= \frac{1}{2 i} [- l n (1 - e^{i}) + l n (1 - e^{- i})]$ $= \frac{1}{2 i} l n (\frac{1 - e^{- i}}{1 - e^{i}}) = \frac{1}{2 i} l n (- e^{- i})$ $= \frac{1}{2 i} [i π - i] = \frac{π - 1}{2}$
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