calculate-n-1-cos-2nx-n-

Question Number 67034 by mathmax by abdo last updated on 22/Aug/19

c a l c u l a t e \sum_{n = 1}^{\infty} \frac{c o s (2 n x)}{n}

Commented by mathmax by abdo last updated on 23/Aug/19

l e t S = \sum_{n = 1}^{\infty} \frac{c o s (2 n x)}{n} \Rightarrow S = R e (\sum_{n = 1}^{\infty} \frac{e^{i 2 n x}}{n})

l e t W (x) = \sum_{n = 1}^{\infty} \frac{e^{i 2 n x}}{n} \Rightarrow W^{^{'}} (x) = \sum_{n = 1}^{\infty} 2 n i \frac{e^{i 2 n x}}{n}

= 2 i \sum_{n = 1}^{\infty} {(e^{2 i x})}^{n} = 2 i e^{2 i x} \sum_{n = 1}^{\infty} {(e^{2 i x})}^{n - 1} = 2 i e^{2 i x} \sum_{n = 0}^{\infty} {(e^{2 i x})}^{n}

= 2 i e^{2 i x} \times \frac{1}{1 - e^{2 i x}} = \frac{2 i e^{2 i x}}{1 - e^{2 i x}} \Rightarrow

W (x) = - l n (1 - e^{2 i x}) w e h a v e 1 - e^{2 i x} = 1 - c o s (2 x) - i s i n (2 x)

= 2 {s i n}^{2} (x) - 2 i s i n (x) c o s (x) = - 2 i s i n (x) {c o s x + i s i n x}

= - 2 i s i n x e^{i x} \Rightarrow l n (1 - e^{2 i x}) = l n (- 2 i) + l n (s i n x) + i x

= l n (2) + l n (- i) + l n (s i n x) + i x = l n (2) + l n (e^{- \frac{i π}{2}}) + l n (s i n x) + i x

= l n (2) - \frac{i π}{2} + i x + l n (s i n x) = l n (2 s i n x) + i (x - \frac{π}{2}) \Rightarrow

\sum_{n = 1}^{\infty} \frac{c o s (2 n x)}{n} = l n (2 s i n x) .

Commented by mathmax by abdo last updated on 25/Aug/19

e r r o r a t f i n a l l i n e w e h a v e W (x) = - l n (1 - e^{2 i x}) \Rightarrow

W (x) = - l n (2 s i n x) - i (x - \frac{π}{2}) = - l n (2 s i n x) + i (\frac{π - 2 x}{2}) \Rightarrow

\sum_{n = 1}^{\infty} \frac{c o s (2 n x)}{n} = - l n (2 s i n x)