developpf(x)=(2/(3+cosx)) at fourier serie


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Question Number 142228 by mathmax by abdo last updated on 28/May/21

$developpf (x) = \frac{2}{3 + cosx} at fourier serie$
	Answered by mathmax by abdo last updated on 28/May/21
	$another way f is even ⇒f(x)=(a_o /2) +Σ_(n=1) ^∞ a_n cos(nx) a_n =(2/T)∫_([T]) f(x)cos(nx) =(2/(2π))∫_0 ^(2π) (2/(3+cosx))cos(nx)dx =(2/π)∫_0 ^(2π) ((cos(nx))/(3+cosx))dx ⇒(π/2)a_n =∫_0 ^(2π) ((cos(nx))/(3+cosx))dx =_(e^(ix) =z) ∫_(∣z∣=1) (((z^n +z^(−n) )/2)/(3+((z+z^(−1) )/2)))(dz/(iz))=−2i∫_(∣z∣=1) ((z^n +z^(−n) )/((6+z+z^(−1) )z))dz =−2i∫_(∣z∣=1) ((z^n +z^(−n) )/(z^2 +6z +1))dz let ϕ(z)=((z^n +z^(−n) )/(z^2 +6z +1)) poles of ϕ! Δ^′ =8 ⇒z_1 =−3+2(√2) and z_2 =−3−2(√2) ⇒ϕ(z)=((z^n +z^(−n) )/((z−z_1 )(z−z_2 ))) ∣z_1 ∣−1 =3−2(√2)−1 =2−2(√2)<0 ⇒∣z_1 ∣<1 ∣z_2 ∣−1=3+2(√2)−1 =2+2(√2)>0 (out of circle) ⇒ ∫_(∣z∣=1) ϕ(z)dz =2iπ Res(ϕ,z_1 ) =2iπ((z_1 ^n +z_1 ^(−n) )/(z_1 −z_2 )) =((2iπ)/(4(√2))){ (−3+2(√2))^n +(−3+2(√2))^(−n) } =((iπ)/(2(√2))){ (−3+2(√2))^n +(−3+2(√2))^(−n) } ⇒ (π/2)a_n =−2i×((iπ)/(2(√2))){(−3+2(√2))^n +(−3+2(√2))^(−n) } =(π/( (√2))){(−3+2(√2))^n [+(−3+2(√2))^n } ⇒a_n =(√2)(−1)^n { (3−2(√2))^n +(3−2(√2))^(−n) } a_o =2(√2)=(2/π)∫_0 ^(2π) (dx/(3+cosx)) ⇒ (2/(3+cosx))=(√2)+(√2)Σ_(n=1) ^∞ (−1)^n {(3−2(√2))^n +(3−2(√2))^(−n) }cos(nx)$
	$another way f is even \Rightarrow f (x) = \frac{a_{o}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos (nx)$ $a_{n} = \frac{2}{T} \int_{[T]} f (x) \cos (nx) = \frac{2}{2 π} \int_{0}^{2 π} \frac{2}{3 + cosx} \cos (nx) dx$ $= \frac{2}{π} \int_{0}^{2 π} \frac{\cos (nx)}{3 + cosx} dx \Rightarrow \frac{π}{2} a_{n} = \int_{0}^{2 π} \frac{\cos (nx)}{3 + cosx} dx$ $=_{e^{ix} = z} \int_{∣ z ∣= 1} \frac{\frac{z^{n} + z^{- n}}{2}}{3 + \frac{z + z^{- 1}}{2}} \frac{dz}{iz} = - 2 i \int_{∣ z ∣= 1} \frac{z^{n} + z^{- n}}{(6 + z + z^{- 1}) z} dz$ $= - 2 i \int_{∣ z ∣= 1} \frac{z^{n} + z^{- n}}{z^{2} + 6 z + 1} dz let φ (z) = \frac{z^{n} + z^{- n}}{z^{2} + 6 z + 1} poles of φ!$ $Δ^{^{'}} = 8 \Rightarrow z_{1} = - 3 + 2 \sqrt{2} and z_{2} = - 3 - 2 \sqrt{2} \Rightarrow φ (z) = \frac{z^{n} + z^{- n}}{(z - z_{1}) (z - z_{2})}$ $∣ z_{1} ∣ - 1 = 3 - 2 \sqrt{2} - 1 = 2 - 2 \sqrt{2} < 0 \Rightarrow∣ z_{1} ∣< 1$ $∣ z_{2} ∣ - 1 = 3 + 2 \sqrt{2} - 1 = 2 + 2 \sqrt{2} > 0 (out of circle) \Rightarrow$ $\int_{∣ z ∣= 1} φ (z) dz = 2 i π Res (φ, z_{1}) = 2 i π \frac{z_{1}^{n} + z_{1}^{- n}}{z_{1} - z_{2}}$ $= \frac{2 i π}{4 \sqrt{2}} {{(- 3 + 2 \sqrt{2})}^{n} + {(- 3 + 2 \sqrt{2})}^{- n}}$ $= \frac{i π}{2 \sqrt{2}} {{(- 3 + 2 \sqrt{2})}^{n} + {(- 3 + 2 \sqrt{2})}^{- n}} \Rightarrow$ $\frac{π}{2} a_{n} = - 2 i \times \frac{i π}{2 \sqrt{2}} {{(- 3 + 2 \sqrt{2})}^{n} + {(- 3 + 2 \sqrt{2})}^{- n}}$ $= \frac{π}{\sqrt{2}} {{(- 3 + 2 \sqrt{2})}^{n} [+ {(- 3 + 2 \sqrt{2})}^{n}}$ $\Rightarrow a_{n} = \sqrt{2} {(- 1)}^{n} {{(3 - 2 \sqrt{2})}^{n} + {(3 - 2 \sqrt{2})}^{- n}}$ $a_{o} = 2 \sqrt{2} = \frac{2}{π} \int_{0}^{2 π} \frac{dx}{3 + cosx} \Rightarrow$ $\frac{2}{3 + cosx} = \sqrt{2} + \sqrt{2} \sum_{n = 1}^{\infty} {(- 1)}^{n} {{(3 - 2 \sqrt{2})}^{n} + {(3 - 2 \sqrt{2})}^{- n}} \cos (nx)$
	Commented by Mathspace last updated on 28/May/21
	$(2/(3+cosx))=(√2)+(√2)Σ_(n=1) ^∞ (−1)^n {(3−2(√2))^n +(3+2(√2))^(−n) }cos(nx) a_n =(√2)(−1)^n {(3−2(√2))^n +(3+2(√2))^(−n) }$
	$\frac{2}{3 + c o s x} = \sqrt{2} + \sqrt{2} \sum_{n = 1}^{\infty} {(- 1)}^{n} {{(3 - 2 \sqrt{2})}^{n} + {(3 + 2 \sqrt{2})}^{- n}} c o s (n x)$ $a_{n} = \sqrt{2} {(- 1)}^{n} {{(3 - 2 \sqrt{2})}^{n} + {(3 + 2 \sqrt{2})}^{- n}}$
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