let-f-x-e-x-cosx-developp-f-at-fourier-serie-1-find-the-value-of-n-1-n-1-n-2-2-calculate-n-0-1-n-2-1-

Question Number 38209 by prof Abdo imad last updated on 22/Jun/18

l e t f (x) = e^{- x} c o s x

d e v e l o p p f a t f o u r i e r s e r i e

1) f i n d t h e v a l u e o f \sum_{n = - \infty}^{+ \infty} \frac{{(- 1)}^{n}}{1 + n^{2}}

2) c a l c u l a t e \sum_{n = 0}^{\infty} \frac{1}{n^{2} + 1} .

Commented by math khazana by abdo last updated on 25/Jun/18

f (x) = \sum_{n = - \infty}^{+ \infty} c_{n} e^{i n x} w i t h

c_{n} = \frac{1}{T} \int_{[T]} f (x) e^{- i n x} d x

= \frac{1}{2 π} \int_{- π}^{π} e^{(- 1 - i n) x} c o s x d x

2 π c_{n} = R e (\int_{- π}^{π} e^{(- 1 - i n + i) x} d x) b u t

\int_{- π}^{π} e^{- (1 + (n - 1) i) x} d x = {[\frac{1}{- (1 + (n - 1) i)} e^{- (1 + (n - 1) i) x}]}_{- π}^{π}

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

= \frac{{(- 1)}^{n - 1} 2 i s h (π) (1 - (n - 1) i)}{1 + {(n - 1)}^{2}}

= \frac{2 i s h (π) {(- 1)}^{n - 1} + 2 s h (π) {(- 1)}^{n - 1}}{{(n - 1)}^{2} + 1} \Rightarrow

2 π c_{n} = \frac{2 s h (π) {(- 1)}^{n - 1}}{{(n - 1)}^{2} + 1} \Rightarrow

c_{n} = \frac{{(- 1)}^{n - 1} s h (π)}{π {{(n - 1)}^{2} + 1}} \Rightarrow

f (x) = \sum_{n = - \infty}^{+ \infty} \frac{{(- 1)}^{n - 1} s h (π)}{π {{(n - 1)}^{2} + 1}} e^{i n x}

= \frac{s h (π)}{π} \sum_{n = - \infty}^{+ \infty} \frac{{(- 1)}^{n - 1}}{{(n - 1)}^{2} + 1} e^{i n x} = e^{- x} c o s x

Commented by math khazana by abdo last updated on 25/Jun/18

2) f (0) = 1 = \frac{s h (π)}{π} \sum_{n = - \infty}^{+ \infty} \frac{{(- 1)}^{n - 1}}{{(n - 1)}^{2} + 1}

=_{n - 1 = p} \frac{s h (π)}{π} \sum_{p = - \infty}^{+ \infty} \frac{{(- 1)}^{p}}{p^{2} + 1} \Rightarrow

\sum_{p = - \infty}^{+ \infty} \frac{{(- 1)}^{p}}{p^{2} + 1} = \frac{π}{s h (π)} .

3) f (π) = - e^{- π} = - \frac{s h (π)}{π} \sum_{n = - \infty}^{+ \infty} \frac{1}{{(n - 1)}^{2} + 1}

\Rightarrow \sum_{n = - \infty}^{+ \infty} \frac{1}{n^{2} + 1} = \frac{π e^{- π}}{s h (π)} .