let f(x)= ((cos(αx))/(cosx)) (2π periodic even) developp f at fourier serie.


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Question Number 39025 by maxmathsup by imad last updated on 01/Jul/18

$l e t f (x) = \frac{c o s (α x)}{c o s x} (2 π p e r i o d i c e v e n)$ $d e v e l o p p f a t f o u r i e r s e r i e .$
Commented by math khazana by abdo last updated on 05/Jul/18
$f(x) =(a_0 /2) +Σ_(n=1) ^∞ a_n cos(nx) with a_n = (2/T) ∫_([T]) f(x)cos(nx) dx =(2/(2π)) ∫_(−π) ^π ((cos(αx)cos(nx))/(cosx))dx ⇒ π a_n = (1/2) ∫_(−π) ^π ((cos(n+α)x +cos(n−α)x)/(cosx))dx 2π a_n = ∫_(−π) ^π ((cos(n+α)x)/(cosx))dx +∫_(−π) ^π ((cos(n−α)x)/(cosx))dx let find first I_λ =∫_(−π) ^π ((cos(λx))/(cosx))dx I_λ = Re( ∫_(−π) ^π (e^(iλx) /(cosx))dx) changement e^(ix) =z give ∫_(−π) ^π (e^(iλx) /(cosx)) dx = ∫_(∣z∣=1) ((2z^λ )/(z+z^(−1) )) (dz/(iz)) = ∫_(∣z∣=1) ((−2i z^λ )/(z^2 +1))dz let ϕ(z) = ((−2iz^λ )/(z^(2 ) +1)) ∫_(∣z∣=1) ϕ(z)dz =2iπ{ Res(ϕ,i) +Res(ϕ,−i)} Res(ϕ,i)= ((−2i i^λ )/(2i)) = −i^λ Res(ϕ^� ,−i) =((−2i (−i)^λ )/(−2i)) =(−i)^λ ⇒ ∫_(∣z∣=1) ϕ(z)dz =2iπ{ −i^λ +(−i)^λ } =−2iπ{ i^λ −(−i)^λ } =−2iπ 2i Im(i^λ ) =4π Im( cos(((λπ)/2)) +isin(((λπ)/2))} =4π sin(((λπ)/2)) ⇒ I_λ =4π sin(((λπ)/2)) ⇒ 2π a_n = 4πsin((((n+α)π)/2)) +4π sin((((n−α)π)/2))⇒ a_n = (1/2){ sin((((n+α)π)/2)) +sin((((n−α)π)/2))} a_0 =(1/π) ∫_(−π) ^π ((cos(αx))/(cosx))dx =(1/π)4π sin(((απ)/2)) =4 sin((α/2)) ⇒(a_0 /2) = 2 sin((α/2)) ⇒ f(x) = 2sin((α/2)) +(1/2)Σ_(n=1) ^∞ {sin((((n+α)π)/2))+sin((((n−α)π)/2))}$
$f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} c o s (n x) w i t h$ $a_{n} = \frac{2}{T} \int_{[T]} f (x) c o s (n x) d x$ $= \frac{2}{2 π} \int_{- π}^{π} \frac{c o s (α x) c o s (n x)}{c o s x} d x \Rightarrow$ $π a_{n} = \frac{1}{2} \int_{- π}^{π} \frac{c o s (n + α) x + c o s (n - α) x}{c o s x} d x$ $2 π a_{n} = \int_{- π}^{π} \frac{c o s (n + α) x}{c o s x} d x + \int_{- π}^{π} \frac{c o s (n - α) x}{c o s x} d x$ $l e t f i n d f i r s t I_{λ} = \int_{- π}^{π} \frac{c o s (λ x)}{c o s x} d x$ $I_{λ} = R e (\int_{- π}^{π} \frac{e^{i λ x}}{c o s x} d x) c h a n g e m e n t e^{i x} = z g i v e$ $\int_{- π}^{π} \frac{e^{i λ x}}{c o s x} d x = \int_{∣ z ∣= 1} \frac{2 z^{λ}}{z + z^{- 1}} \frac{d z}{i z}$ $= \int_{∣ z ∣= 1} \frac{- 2 i z^{λ}}{z^{2} + 1} d z l e t φ (z) = \frac{- 2 {i z}^{λ}}{z^{2} + 1}$ $\int_{∣ z ∣= 1} φ (z) d z = 2 i π {R e s (φ, i) + R e s (φ, - i)}$ $R e s (φ, i) = \frac{- 2 i i^{λ}}{2 i} = - i^{λ}$ $R e s (\bar{φ}, - i) = \frac{- 2 i {(- i)}^{λ}}{- 2 i} = {(- i)}^{λ} \Rightarrow$ $\int_{∣ z ∣= 1} φ (z) d z = 2 i π {- i^{λ} + {(- i)}^{λ}}$ $= - 2 i π {i^{λ} - {(- i)}^{λ}} = - 2 i π 2 i I m (i^{λ})$ $= 4 π I m (c o s (\frac{λ π}{2}) + i s i n (\frac{λ π}{2})} = 4 π s i n (\frac{λ π}{2}) \Rightarrow$ $I_{λ} = 4 π s i n (\frac{λ π}{2}) \Rightarrow$ $2 π a_{n} = 4 π s i n (\frac{(n + α) π}{2}) + 4 π s i n (\frac{(n - α) π}{2}) \Rightarrow$ $a_{n} = \frac{1}{2} {s i n (\frac{(n + α) π}{2}) + s i n (\frac{(n - α) π}{2})}$ $a_{0} = \frac{1}{π} \int_{- π}^{π} \frac{c o s (α x)}{c o s x} d x = \frac{1}{π} 4 π s i n (\frac{α π}{2})$ $= 4 s i n (\frac{α}{2}) \Rightarrow \frac{a_{0}}{2} = 2 s i n (\frac{α}{2}) \Rightarrow$ $f (x) = 2 s i n (\frac{α}{2}) + \frac{1}{2} \sum_{n = 1}^{\infty} {s i n (\frac{(n + α) π}{2}) + s i n (\frac{(n - α) π}{2})}$
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