let-f-x-n-1-sin-nx-n-x-n-with-1-lt-x-lt-1-1-find-a-explicite-form-of-f-x-2-find-the-value-of-n-1-1-n2-n-sin-n-2-

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 by maxmathsup by imad last updated on 06/Nov/18

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})}} .