let a_k =∫_(−(π/(2 )) +kπ) ^(−(π/2) +(k+1)π) e^(−t) cost dt 1) calculate a_k 2) find lim_(n→+∞) Σ_(k=0) ^n ∣a


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 40138 by maxmathsup by imad last updated on 16/Jul/18

$l e t a_{k} = \int_{- \frac{π}{2} + k π}^{- \frac{π}{2} + (k + 1) π} e^{- t} c o s t d t$ $1) c a l c u l a t e a_{k}$ $2) f i n d {l i m}_{n \to + \infty} \sum_{k = 0}^{n} ∣ a_{k} ∣ .$
Commented by maxmathsup by imad last updated on 20/Jul/18
$1) we have a_k =_(t=kπ +x) ∫_(−(π/2)) ^(π/2) e^(−(kπ+x)) cos(x+kπ)dx = ∫_(−(π/2)) ^(π/2) e^(−kπ) e^(−x) (−1)^k cosx dx =(−1)^k e^(−kπ) ∫_(−(π/2)) ^(π/2) e^(−x) cosxdx but ∫_(−(π/2)) ^(π/2) e^(−x) cosx dx=Re( ∫_(−(π/2)) ^(π/2) e^(−x) e^(ix) ) =Re( ∫_(−(π/2)) ^(π/2) e^((−1+i)x) dx) =Re( [(1/(−1+i)) e^((−1+i)x) ]_(−(π/2)) ^(π/2) =Re( (1/(−1+i))( e^((−1+i)(π/2)) −e^(−(−1+i)(π/2)) )) =Re{ ((−1)/(1−i))( e^(−(π/2)) i + e^(π/2) i}=Re { ((−i)/(1−i))( e^(π/2) +e^(−(π/2)) )} =Re( ((−i(1+i))/2)( e^(π/2) +e^(−(π/2)) ))= Re( ((−i+1)/2)(e^(π/2) +e^(−(π/2)) )) =((e^(π/2) +e^(−(π/2)) )/2) =ch((π/2)) ⇒ a_k =(−1)^k e^(−kπ) ch((π/2)) .$
$1) w e h a v e a_{k} =_{t = k π + x} \int_{- \frac{π}{2}}^{\frac{π}{2}} e^{- (k π + x)} c o s (x + k π) d x$ $= \int_{- \frac{π}{2}}^{\frac{π}{2}} e^{- k π} e^{- x} {(- 1)}^{k} c o s x d x$ $= {(- 1)}^{k} e^{- k π} \int_{- \frac{π}{2}}^{\frac{π}{2}} e^{- x} c o s x d x b u t \int_{- \frac{π}{2}}^{\frac{π}{2}} e^{- x} c o s x d x = R e (\int_{- \frac{π}{2}}^{\frac{π}{2}} e^{- x} e^{i x})$ $= R e (\int_{- \frac{π}{2}}^{\frac{π}{2}} e^{(- 1 + i) x} d x) = R e ({[\frac{1}{- 1 + i} e^{(- 1 + i) x}]}_{- \frac{π}{2}}^{\frac{π}{2}}$ $= R e (\frac{1}{- 1 + i} (e^{(- 1 + i) \frac{π}{2}} - e^{- (- 1 + i) \frac{π}{2}}))$ $= R e {\frac{- 1}{1 - i} (e^{- \frac{π}{2}} i + e^{\frac{π}{2}} i} = R e {\frac{- i}{1 - i} (e^{\frac{π}{2}} + e^{- \frac{π}{2}})}$ $= R e (\frac{- i (1 + i)}{2} (e^{\frac{π}{2}} + e^{- \frac{π}{2}})) = R e (\frac{- i + 1}{2} (e^{\frac{π}{2}} + e^{- \frac{π}{2}}))$ $= \frac{e^{\frac{π}{2}} + e^{- \frac{π}{2}}}{2} = c h (\frac{π}{2}) \Rightarrow$ $a_{k} = {(- 1)}^{k} e^{- k π} c h (\frac{π}{2}) .$
Commented by maxmathsup by imad last updated on 20/Jul/18

$2) l e t S_{n} = \sum_{k = 0}^{n} ∣ a_{k} ∣$ $S_{n} = \sum_{k = 0}^{n} c h (\frac{π}{2}) {(e^{- π})}^{k} = c h (\frac{π}{2}) \sum_{k = 0}^{n} {(e^{- π})}^{k}$ $= c h (\frac{π}{2}) \frac{1 - e^{- (n + 1) π}}{1 - e^{- π}} \Rightarrow {l i m}_{n \to + \infty} S_{n} = \frac{c h (\frac{π}{2})}{1 - e^{- π}} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum