let f(x)=e^(−x) cosx developp f at fourier serie 1) find the value of Σ_(n=−∞) ^(+∞) (((−1)^n )/(1+n^2 )) 2) calculate Σ


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
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
$f(x)=Σ_(n=−∞) ^(+∞) c_n e^(inx) with c_n =(1/T)∫_([T]) f(x)e^(−inx) dx = (1/(2π)) ∫_(−π) ^π e^((−1−in)x) cosx dx 2π c_n = Re( ∫_(−π) ^π e^((−1−in +i)x) dx) but ∫_(−π) ^π e^(−(1+(n−1)i)x) dx=[ (1/(−(1+(n−1)i)))e^(−(1+(n−1)i)x) ]_(−π) ^π =((−1)/(1+(n−1)i)) { e^(−π) (−1)^(n−1) −e^π (−1)^(n−1) } =(((−1)^(n−1) 2i sh(π)(1−(n−1)i))/(1+(n−1)^2 )) =((2ish(π)(−1)^(n−1) +2sh(π)(−1)^(n−1) )/((n−1)^2 +1)) ⇒ 2πc_n = ((2sh(π)(−1)^(n−1) )/((n−1)^2 +1)) ⇒ c_n = (((−1)^(n−1) sh(π))/(π{ (n−1)^2 +1})) ⇒ f(x)= Σ_(n=−∞) ^(+∞) (((−1)^(n−1) sh(π))/(π{ (n−1)^2 +1})) e^(inx) =((sh(π))/π) Σ_(n=−∞) ^(+∞) (((−1)^(n−1) )/((n−1)^2 +1)) e^(inx) =e^(−x) cosx$
$f (x) = \sum_{n = - \infty}^{+ \infty} c_{n} e^{i n x} w i t h$ $c_{n} = \frac{1}{T} \int_{[T]} f (x) e^{- i n x} d x$ $= \frac{1}{2 π} \int_{- π}^{π} e^{(- 1 - i n) x} c o s x d x$ $2 π c_{n} = R e (\int_{- π}^{π} e^{(- 1 - i n + i) x} d x) b u t$ $\int_{- π}^{π} e^{- (1 + (n - 1) i) x} d x = {[\frac{1}{- (1 + (n - 1) i)} e^{- (1 + (n - 1) i) x}]}_{- π}^{π}$ $= \frac{- 1}{1 + (n - 1) i} {e^{- π} {(- 1)}^{n - 1} - e^{π} {(- 1)}^{n - 1}}$ $= \frac{{(- 1)}^{n - 1} 2 i s h (π) (1 - (n - 1) i)}{1 + {(n - 1)}^{2}}$ $= \frac{2 i s h (π) {(- 1)}^{n - 1} + 2 s h (π) {(- 1)}^{n - 1}}{{(n - 1)}^{2} + 1} \Rightarrow$ $2 π c_{n} = \frac{2 s h (π) {(- 1)}^{n - 1}}{{(n - 1)}^{2} + 1} \Rightarrow$ $c_{n} = \frac{{(- 1)}^{n - 1} s h (π)}{π {{(n - 1)}^{2} + 1}} \Rightarrow$ $f (x) = \sum_{n = - \infty}^{+ \infty} \frac{{(- 1)}^{n - 1} s h (π)}{π {{(n - 1)}^{2} + 1}} e^{i n x}$ $= \frac{s h (π)}{π} \sum_{n = - \infty}^{+ \infty} \frac{{(- 1)}^{n - 1}}{{(n - 1)}^{2} + 1} e^{i n x} = e^{- x} c o s x$
Commented by math khazana by abdo last updated on 25/Jun/18

$2) f (0) = 1 = \frac{s h (π)}{π} \sum_{n = - \infty}^{+ \infty} \frac{{(- 1)}^{n - 1}}{{(n - 1)}^{2} + 1}$ $=_{n - 1 = p} \frac{s h (π)}{π} \sum_{p = - \infty}^{+ \infty} \frac{{(- 1)}^{p}}{p^{2} + 1} \Rightarrow$ $\sum_{p = - \infty}^{+ \infty} \frac{{(- 1)}^{p}}{p^{2} + 1} = \frac{π}{s h (π)} .$ $3) f (π) = - e^{- π} = - \frac{s h (π)}{π} \sum_{n = - \infty}^{+ \infty} \frac{1}{{(n - 1)}^{2} + 1}$ $\Rightarrow \sum_{n = - \infty}^{+ \infty} \frac{1}{n^{2} + 1} = \frac{π e^{- π}}{s h (π)} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum