let f(x) =Σ_(n=1) ^∞ ((sin(nx))/n) x^n with −1<x<1 1) find a explicite form of f(x) 2) find the value of Σ


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Question Number 42089 by maxmathsup by imad last updated on 17/Aug/18

$l e t f (x) = \sum_{n = 1}^{\infty} \frac{s i n (n x)}{n} x^{n} w i t h - 1 < x < 1$ $1) f i n d a e x p l i c i t e f o r m o f f (x)$ $2) f i n d t h e v a l u e o f \sum_{n = 1}^{\infty} \frac{1}{n 2^{n}} s i n (\frac{n}{2})$
Commented bymaxmathsup by imad last updated on 06/Nov/18
$1) we have f(x)=Im(Σ_(n=1) ^∞ ((e^(inx) x^n )/n)) =Im(Σ_(n=1) ^∞ (((xe^(ix) )^n )/n)) let z =x e^(ix) and W(z) =Σ_(n=1) ^∞ (z^n /n) ⇒(dW/dz)(z)=Σ_(n=1) ^∞ z^(n−1) =Σ_(n=0) ^∞ z^n =(1/(1−z)) ⇒ W(z)=−ln(1−z) +c but c=W(0)=0 ⇒W(z)=−ln(1−z) =−ln(1−xe^(ix) )=−ln(1−xcosx −ix sinx)=a+ib ⇒ e^(a+ib) = (1/(1−xcosx−ixsinx)) =((1−xcosx+ixsinx)/((1−xcosx)^2 +x^2 sin^2 x)) =((1−xcosx+ixsnx)/(1−2xcosx +x^2 cos^2 x +x^2 sin^2 x)) =((1−xcosx+ixsinx)/(1−2xcosx +x^2 )) ⇒ e^a (cosb +isinb) =((1−xcosx)/(1−2xcosx +x^2 )) +i((xsinx)/(1−2xcosx +x^2 )) ⇒ e^a cosb =((1−xcosx)/(1−2xcosx +x^2 )) and e^a sinb =((x sinx)/(1−2xcosx +x^2 )) ⇒ tanb =((xsinx)/(1−xcosx)) ⇒b =arctan(((xsinx)/(1−x cosx))) ⇒f(x)=Im(W(z)) =arctan(((xsinx)/(1−xcosx))) 2) Σ_(n=1) ^∞ (1/(n2^n ))sin((n/2)) =f((1/2)) =arctan(((sin((1/2)))/(2(1−(1/2)cos((1/2))))) =arctan{ ((sin(2^(−1) ))/(2−cos(2^(−1) )))} .$
$1) w e h a v e f (x) = I m (\sum_{n = 1}^{\infty} \frac{e^{i n x} x^{n}}{n}) = I m (\sum_{n = 1}^{\infty} \frac{{({x e}^{i x})}^{n}}{n}) l e t$ $z = x e^{i x} a n d W (z) = \sum_{n = 1}^{\infty} \frac{z^{n}}{n} \Rightarrow \frac{d W}{d z} (z) = \sum_{n = 1}^{\infty} z^{n - 1} = \sum_{n = 0}^{\infty} z^{n} = \frac{1}{1 - z} \Rightarrow$ $W (z) = - l n (1 - z) + c b u t c = W (0) = 0 \Rightarrow W (z) = - l n (1 - z)$ $= - l n (1 - {x e}^{i x}) = - l n (1 - x c o s x - i x s i n x) = a + i b \Rightarrow$ $e^{a + i b} = \frac{1}{1 - x c o s x - i x s i n x} = \frac{1 - x c o s x + i x s i n x}{{(1 - x c o s x)}^{2} + x^{2} {s i n}^{2} x}$ $= \frac{1 - x c o s x + i x s n x}{1 - 2 x c o s x + x^{2} {c o s}^{2} x + x^{2} {s i n}^{2} x} = \frac{1 - x c o s x + i x s i n x}{1 - 2 x c o s x + x^{2}} \Rightarrow$ $e^{a} (c o s b + i s i n b) = \frac{1 - x c o s x}{1 - 2 x c o s x + x^{2}} + i \frac{x s i n x}{1 - 2 x c o s x + x^{2}} \Rightarrow$ $e^{a} c o s b = \frac{1 - x c o s x}{1 - 2 x c o s x + x^{2}} a n d e^{a} s i n b = \frac{x s i n x}{1 - 2 x c o s x + x^{2}} \Rightarrow$ $t a n b = \frac{x s i n x}{1 - x c o s x} \Rightarrow b = a r c t a n (\frac{x s i n x}{1 - x c o s x}) \Rightarrow f (x) = I m (W (z))$ $= a r c t a n (\frac{x s i n x}{1 - x c o s x})$ $2) \sum_{n = 1}^{\infty} \frac{1}{n 2^{n}} s i n (\frac{n}{2}) = f (\frac{1}{2}) = a r c t a n (\frac{s i n (\frac{1}{2})}{2 (1 - \frac{1}{2} c o s (\frac{1}{2})})$ $= a r c t a n {\frac{s i n (2^{- 1})}{2 - c o s (2^{- 1})}} .$
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