Question-118768

Question Number 118768 by mathdave last updated on 19/Oct/20

Answered by mindispower last updated on 23/Oct/20

\underset{m = 0}{\sum^{n}} e^{i m x} = \frac{1 - {(e^{i x})}^{n + 1}}{1 - e^{i x}} = \frac{e^{i \frac{n x}{2}} (e^{- i \frac{(n + 1) x}{2}} - e^{i \frac{(n + 1) x}{2}})}{e^{- i \frac{x}{2}} - e^{i \frac{x}{2}}}

= e^{i \frac{n x}{2}} . \frac{s i n (\frac{n + 1}{2} x)}{s i n (\frac{x}{2})}

\Rightarrow \frac{s i n (\frac{2 n + 1}{2} x)}{s i n (\frac{x}{2})} = e^{- i n x} \underset{m = 0}{\sum^{2 n}} e^{i m x}

\int_{- \infty}^{\infty} \frac{e^{i k x}}{1 + x^{2}} d x, k ⩾ 0

C_{R} = {{R e}^{i a}, a \in [0, π]}

\int_{C_{R}} \frac{e^{i k x}}{1 + x^{2}} d x = \int_{- \infty}^{\infty} \frac{e^{i k x}}{1 + x^{2}} d x + \int_{0}^{π} \frac{{i R e}^{- k s i n (x)} d x}{1 + R^{2} e^{2 i x}}

= 2 i π . \frac{e^{- k}}{2 i} = \frac{π}{e^{k}},

\Rightarrow \sum_{m ⩽ 2 n} \frac{π}{e^{∣ m - n ∣}} = \sum_{m ⩽ n} \frac{π}{e^{n - m}} + \sum_{m > n} \frac{π}{e^{m - n}}

= \frac{π}{e^{n}} (\frac{1 - e^{(n + 1)}}{1 - e}) + π e^{n} (e^{- (n + 1)} (\frac{1 - e^{- (n)}}{1 - e^{- 1}}))

l i m n \to 0

= - \frac{π e}{1 - e} + π e^{- 1} . \frac{1}{1 - e^{-}}

= \frac{π}{e - 1} (e + 1) = π \frac{\frac{e^{\frac{1}{2}}}{2} + \frac{e^{- \frac{1}{2}}}{2}}{\frac{1}{2} (e^{\frac{1}{2}} - e^{- \frac{1}{2}})} = π c o t h (\frac{1}{2})