let f(x)= (3/(2+cosx)) developp f ar fourier serie.


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Question Number 34913 by abdo imad last updated on 12/May/18

$l e t f (x) = \frac{3}{2 + c o s x} d e v e l o p p f a r f o u r i e r s e r i e .$
Commented by abdo imad last updated on 31/May/18
$we have cosx = ((e^(ix) +e^(−ix) )/2) let use the changement e^(ix) =z ⇒f(x)=ϕ(z)= (3/(2+((z+z^(−1) )/2))) = (6/(4 +z +z^(−1) )) = ((6z)/(4z +z^2 +1)) = ((6z)/(z^2 +4z +1)) polez of ϕ? roots of z^2 +4z +1 Δ^′ =2^2 −1=3 ⇒z_1 =−2+(√3) and z_2 =−2−(√3) ∣z_1 ∣<∣z_2 ∣ we consider the domain W ={z∈C/ ∣z_1 ∣<∣z∣<∣z_2 ∣} we have ϕ(z) = ((6z)/((z−z_1 )(z−z_2 ))) =((6z)/(z_1 −z_2 )){ (1/(z−z_1 )) −(1/(z−z_2 ))} = ((6z)/(2(√3))){ (1/(z−z_1 )) − (1/(z−z_2 ))}=z(√3){ (1/(z(1−(z_1 /z)))) −(1/(z_2 ((z/z_2 )−1)))} =z(√3){ (1/(z(1−(z_1 /z)))) + (1/(z_2 (1−(z/z_2 ))))} =(√3)Σ_(n=0) ^∞ ((z_1 /z))^n +(z/z_2 )(√3) Σ_(n=0) ^∞ ((z/z_2 ))^n (√3)Σ_(n=0) ^∞ (z_1 ^n /z^n ) + (z/z_1 ^− )(√3) Σ_(n=0) ^∞ (z^n /z_1 ^−^n ) ...be continued...$
$w e h a v e c o s x = \frac{e^{i x} + e^{- i x}}{2} l e t u s e t h e c h a n g e m e n t$ $e^{i x} = z \Rightarrow f (x) = φ (z) = \frac{3}{2 + \frac{z + z^{- 1}}{2}} = \frac{6}{4 + z + z^{- 1}}$ $= \frac{6 z}{4 z + z^{2} + 1} = \frac{6 z}{z^{2} + 4 z + 1} p o l e z o f φ ?$ $r o o t s o f z^{2} + 4 z + 1$ $Δ^{^{'}} = 2^{2} - 1 = 3 \Rightarrow z_{1} = - 2 + \sqrt{3} a n d z_{2} = - 2 - \sqrt{3}$ $∣ z_{1} ∣<∣ z_{2} ∣ w e c o n s i d e r t h e d o m a i n$ $W = {z \in C / ∣ z_{1} ∣<∣ z ∣<∣ z_{2} ∣} w e h a v e$ $φ (z) = \frac{6 z}{(z - z_{1}) (z - z_{2})} = \frac{6 z}{z_{1} - z_{2}} {\frac{1}{z - z_{1}} - \frac{1}{z - z_{2}}}$ $= \frac{6 z}{2 \sqrt{3}} {\frac{1}{z - z_{1}} - \frac{1}{z - z_{2}}} = z \sqrt{3} {\frac{1}{z (1 - \frac{z_{1}}{z})} - \frac{1}{z_{2} (\frac{z}{z_{2}} - 1)}}$ $= z \sqrt{3} {\frac{1}{z (1 - \frac{z_{1}}{z})} + \frac{1}{z_{2} (1 - \frac{z}{z_{2}})}}$ $= \sqrt{3} \sum_{n = 0}^{\infty} {(\frac{z_{1}}{z})}^{n} + \frac{z}{z_{2}} \sqrt{3} \sum_{n = 0}^{\infty} {(\frac{z}{z_{2}})}^{n}$ $\sqrt{3} \sum_{n = 0}^{\infty} \frac{z_{1}^{n}}{z^{n}} + \frac{z}{{\bar{z}}_{1}} \sqrt{3} \sum_{n = 0}^{\infty} \frac{z^{n}}{{\overset{-^{n}}{z}}_{1}} . . . b e c o n t i n u e d . . .$
Commented by abdo imad last updated on 31/May/18

$φ (z) = \sqrt{3} \sum_{n = 0}^{\infty} {(- 2 + \sqrt{3})}^{n} e^{- i n x} + \sqrt{3} \sum_{n = 0}^{\infty} {(\frac{1}{- 2 + \sqrt{3}})}^{n + 1} e^{i (n + 1) x}$ $= \sqrt{3} \sum_{n = 0}^{\infty} {(- 2 + \sqrt{3})}^{} e^{- i n x} + \sqrt{3} \sum_{n = 0}^{\infty} {(- 2 - \sqrt{3})}^{n + 1} e^{i (n + 1) x}$
Commented by abdo.msup.com last updated on 31/May/18

$e r r o r i n t h e f i n a l l i n e w e h a v e$ $z_{1} . z_{2} = (2 - \sqrt{3}) (2 + \sqrt{3}) = 1 \Rightarrow z_{2} = \frac{1}{z_{1}} s o$ $φ (z) = \sqrt{3} \sum_{n = 0}^{\infty} \frac{z_{1}^{n}}{z^{n}} + z z_{1} \sum_{n = 0}^{\infty} z_{1}^{n} z^{n}$ $= \sqrt{3} \sum_{n = 0}^{\infty} {(\sqrt{3} - 2)}^{n} e^{- i n x} + \sum_{n = 0}^{\infty} {(\sqrt{3} - 2)}^{n + 1} e^{i (n + 1) x}$ $=$
Commented by prof Abdo imad last updated on 01/Jun/18

$m e t h o d o f R e s i d u s f i s 2 π p e r i o d i c e v e n s o$ $f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} c o s (n x) w i t h$ $a_{n} = \frac{2}{T} \int_{[T]} f (x) c o s (n x) d x$ $= \frac{2}{2 π} \int_{0}^{2 π} \frac{3 c o s (n x)}{2 + c o s x} d x = \frac{3}{π} \int_{0}^{2 π} \frac{c o s (n x)}{2 + c o s x} d x \Rightarrow$ $\frac{π}{3} a_{n} = \int_{0}^{2 π} \frac{c o s (n x)}{2 + c o s x} d x c h a n g e m e n t e^{i x} = z g i v e$ $\frac{π}{3} a_{n} = R e (\int_{0}^{2 π} \frac{e^{i n x}}{2 + c o s x} d x) = R e (A_{n}) b u t$ $A_{n} = \int_{∣ z ∣= 1} \frac{z^{n}}{2 + \frac{z + z^{- 1}}{2}} \frac{d z}{i z}$ $= \int_{∣ z ∣= 1} \frac{2 z^{n}}{i z (4 + z + z^{- 1})} d z$ $= \int_{∣ z ∣= 1} \frac{- 2 {i z}^{n}}{4 z + z^{2} + 1} d z$ $= \int_{∣ z ∣= 1} \frac{- 2 i z^{n}}{z^{2} + 4 z + 1} d z l e t i n t r o d u c e t h e c o m p l e x$ $f u n c t i o n φ (z) = \frac{- 2 i z^{n}}{z^{2} + 4 z ? + 1} p o l e s o f φ ?$ $r o o t s o f z^{2} + 4 z + 1$ $Δ^{^{'}} = 2^{2} - 1 = 3 \Rightarrow z_{1} = - 2 + \sqrt{3} a n d z_{2} = - 2 - \sqrt{3}$ $∣ z_{1} ∣ - 1 = 2 - \sqrt{3} - 1 = 1 - \sqrt{3} < 0 \Rightarrow ∣ z_{1} ∣< 1$ $∣ z_{2} ∣ - 1 = 2 + \sqrt{3} - 1 = 1 + \sqrt{3} > 0 \Rightarrow ∣ z_{2} ∣> 1 t o$ $e l i m i n a t e f r o m r e s i d u s$
Commented by prof Abdo imad last updated on 01/Jun/18

$\int_{∣ z ∣= 1} φ (z) d z = 2 i π R e s (φ, z_{1}) b u t$ $φ (z) = \frac{- 2 i z^{n}}{(z - z_{1}) (z - z_{2})}$ $R e s (φ, z_{1}) = \frac{- 2 i z_{1}^{n}}{z_{1} - z_{2}} = \frac{- 2 i {(- 2 + \sqrt{3})}^{n}}{2 \sqrt{3}} \Rightarrow$ $\int_{∣ z ∣= 1} φ (z) d z = 2 i π \frac{- 2 i {(- 2 + \sqrt{3})}^{n}}{2 \sqrt{3}}$ $= \frac{2 π}{\sqrt{3}} {(- 2 + \sqrt{3})}^{n}$
Commented by abdo.msup.com last updated on 01/Jun/18

$\frac{π}{3} a_{n} = \frac{2 π}{\sqrt{3}} {(- 2 + \sqrt{3})}^{n} \Rightarrow$ $a_{n} = \frac{6}{\sqrt{3}} {(- 2 + \sqrt{3})}^{n} \Rightarrow a_{0} = \frac{6}{\sqrt{3}} a n d$ $\frac{a_{0}}{2} = \sqrt{3} \Rightarrow f (x) = \sqrt{3} + \frac{6}{\sqrt{3}} \sum_{n = 1}^{\infty} {(- 2 + \sqrt{3})}^{n} c o s (n x) .$
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