developp at fourier serie f(x)=(3/(1+2cosx)) by use of two methods


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Question Number 143380 by Mathspace last updated on 13/Jun/21

$d e v e l o p p a t f o u r i e r s e r i e$ $f (x) = \frac{3}{1 + 2 c o s x}$ $b y u s e o f t w o m e t h o d s$
	Answered by mathmax by abdo last updated on 14/Jun/21
	$method 1 f(x)=(3/(1+2cosx))=(3/(1+2((e^(ix) +e^(−ix) )/2)))=(3/(1+e^(ix) +e^(−ix) )) =_(e^(ix) =z) (3/(1+z +z^(−1) ))=((3z)/(z+z^2 +1))=((3z)/(z^2 +z+1))=Ψ(z) z^2 +z+1=0→Δ=−3 ⇒z_1 =((−1+i(√3))/2)=e^((2iπ)/3) and z_2 =((−1−i(√3))/2)=e^(−((2iπ)/3)) ⇒Ψ(z)=((3z)/((z−z_1 )(z−z_2 ))) =((3z)/(z_1 −z_2 ))((1/(z−z_1 ))−(1/(z−z_2 ))) =(3/(i(√3)))((z/(z−z_1 ))−(z/(z−z_2 )))=−i(√3)(((z−z_1 +z_1 )/(z−z_1 ))−((z−z_2 +z_2 )/(z−z_2 ))) =−i(√3)((z_1 /(z−z_1 ))−(z_2 /(z−z_2 )))=i(√3)((z_1 /(z_1 −z))−(z_2 /(z_2 −z))) =i(√3){(1/(1−(z/z_1 )))−(1/(1−(z/z_2 )))} we have ∣(z/z_1 )∣=∣(z/z_2 )∣=1 ⇒ Ψ(z)=i(√3)(Σ_(n=0) ^∞ (z^n /z_1 ^n )−Σ_(n=0) ^∞ (z^n /z_2 ^n )) =i(√3)Σ_(n=0) ^∞ (e^(−((2inπ)/3)) +e^((2inπ)/3) )z^n =i(√3)Σ_(n=0) ^∞ 2cos(((2nπ)/3))z^n =2i(√3)Σ_(n=0) ^∞ cos(((2nπ)/3))e^(inx) =2i(√3)Σ_(n=0) ^∞ cos(((2nπ)/3))(cos(nx)+isin(nx)) =2i(√3)Σ_(n=0) ^∞ cos(((2nπ)/3))cos(nx)−2(√3)Σ_(n=0) ^∞ cos(((2nπ)/3))sin(nx) butΨ(z)=f(x) real ⇒f(x)=−2(√3)Σ_(n=0) ^∞ cos(((2nπ)/3))sin(nx)$
	$method 1 f (x) = \frac{3}{1 + 2 cosx} = \frac{3}{1 + 2 \frac{e^{ix} + e^{- ix}}{2}} = \frac{3}{1 + e^{ix} + e^{- ix}}$ $=_{e^{ix} = z} \frac{3}{1 + z + z^{- 1}} = \frac{3 z}{z + z^{2} + 1} = \frac{3 z}{z^{2} + z + 1} = Ψ (z)$ $z^{2} + z + 1 = 0 \to Δ = - 3 \Rightarrow z_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{2 i π}{3}} and z_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{2 i π}{3}}$ $\Rightarrow Ψ (z) = \frac{3 z}{(z - z_{1}) (z - z_{2})} = \frac{3 z}{z_{1} - z_{2}} (\frac{1}{z - z_{1}} - \frac{1}{z - z_{2}})$ $= \frac{3}{i \sqrt{3}} (\frac{z}{z - z_{1}} - \frac{z}{z - z_{2}}) = - i \sqrt{3} (\frac{z - z_{1} + z_{1}}{z - z_{1}} - \frac{z - z_{2} + z_{2}}{z - z_{2}})$ $= - i \sqrt{3} (\frac{z_{1}}{z - z_{1}} - \frac{z_{2}}{z - z_{2}}) = i \sqrt{3} (\frac{z_{1}}{z_{1} - z} - \frac{z_{2}}{z_{2} - z})$ $= i \sqrt{3} {\frac{1}{1 - \frac{z}{z_{1}}} - \frac{1}{1 - \frac{z}{z_{2}}}} we have ∣ \frac{z}{z_{1}} ∣=∣ \frac{z}{z_{2}} ∣= 1 \Rightarrow$ $Ψ (z) = i \sqrt{3} (\sum_{n = 0}^{\infty} \frac{z^{n}}{z_{1}^{n}} - \sum_{n = 0}^{\infty} \frac{z^{n}}{z_{2}^{n}})$ $= i \sqrt{3} \sum_{n = 0}^{\infty} (e^{- \frac{2 in π}{3}} + e^{\frac{2 in π}{3}}) z^{n}$ $= i \sqrt{3} \sum_{n = 0}^{\infty} 2 \cos (\frac{2 n π}{3}) z^{n} = 2 i \sqrt{3} \sum_{n = 0}^{\infty} \cos (\frac{2 n π}{3}) e^{inx}$ $= 2 i \sqrt{3} \sum_{n = 0}^{\infty} \cos (\frac{2 n π}{3}) (\cos (nx) + isin (nx))$ $= 2 i \sqrt{3} \sum_{n = 0}^{\infty} \cos (\frac{2 n π}{3}) \cos (nx) - 2 \sqrt{3} \sum_{n = 0}^{\infty} \cos (\frac{2 n π}{3}) \sin (nx)$ $but Ψ (z) = f (x) real \Rightarrow f (x) = - 2 \sqrt{3} \sum_{n = 0}^{\infty} \cos (\frac{2 n π}{3}) \sin (nx)$
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