let g(x) =((cosx +1)/(cos(2x)−3)) developp f at fourier serie


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 98944 by mathmax by abdo last updated on 17/Jun/20

$let g (x) = \frac{cosx + 1}{\cos (2 x) - 3} developp f at fourier serie$
	Answered by mathmax by abdo last updated on 18/Jun/20
	$g(x) =((cosx+1)/(cos(2x)−3)) =((((e^(ix) +e^(−ix) )/2)+1)/(((e^(2ix) +e^(−i2x) )/2)−3)) =((e^(ix) +e^(−ix) +2)/(e^(i2x) +e^(−i2x) −6)) changement e^(ix) =z give g(x) =((z+z^(−1) +2)/(z^2 +z^(−2) −6)) =((z^3 +z +2z^2 )/(z^4 +1−6z^2 )) =((z^3 +2z^2 +z)/(z^4 −6z^2 +1)) =f(z) let decompose f z^4 −6z^2 +1 =0→u^2 −6u +1 =0 →Δ^′ =9−1 =8 ⇒ u_1 =3+2(√2) and u_2 =3−2(√2) ⇒f(z) =((z^3 +2z^2 +z)/((z^2 −u_1 )(z^2 −u_2 ))) =(z^3 +2z^2 +z)((1/(z^2 −u_1 ))−(1/(z^2 −u_2 ))) =((z^3 +2z^2 +z)/(z^2 −u_1 ))−((z^3 +2z^2 +z)/(z^2 −u_2 )) =ϕ_1 (z)−ϕ_2 (z) ϕ_1 (z) =((z(z^2 −u_1 )+zu_1 +2z^2 +z)/(z^2 −u_1 )) =z +((2z^2 +(1+u_1 )z)/(z^2 −u_1 )) =z +((2(z^2 −u_1 )+2u_1 +(1+u_1 )z)/(z^2 −u_1 )) =z+2 +(((1+u_1 )z +2u_1 )/(z^2 −u_1 )) ϕ_2 (z) =z+2 +(((1+u_2 )z +2u_2 )/(z^2 −u_2 )) ⇒f(z) =(((1+u_1 )z +2u_1 )/(z^2 −u_1 ))−(((1+u_2 )z +2u_2 )/(z^2 −u_2 )) f(z) =(((1+u_2 )z +2u_2 )/(u_2 −z^2 ))−(((1+u_1 )z +2u_1 )/(u_1 −z^2 )) =(1/u_2 )×(((1+u_2 )z+2u_2 )/(1−((z/(√u_2 )))^2 )) −(1/u_1 )×(((1+u_1 )z+2u_1 )/(1−((z/(√u_1 )))^2 )) so if ∣z∣ <inf((√u_1 ),(√u_2 )) weget f(z) =(1/u_2 ){(1+u_2 )z +2u_2 }Σ_(n=0) ^∞ (z^(2n) /u_2 ^n ) −(1/u_1 ){ (1+u_1 )z +2u_1 }Σ_(n=0) ^∞ (z^(2n) /u_1 ^n ) =(1/(3−2(√2))){(4−2(√2))z +6−4(√2)}Σ_(n=0) ^∞ (e^(2inx) /((3−2(√2))^n )) −(1/(3+2(√2))){ (4+2(√2))+6+4(√2)}Σ_(n=0) ^∞ (e^(2inx) /((3+2(√2))^n ))$
	$g (x) = \frac{cosx + 1}{\cos (2 x) - 3} = \frac{\frac{e^{ix} + e^{- ix}}{2} + 1}{\frac{e^{2 ix} + e^{- i 2 x}}{2} - 3} = \frac{e^{ix} + e^{- ix} + 2}{e^{i 2 x} + e^{- i 2 x} - 6}$ $changement e^{ix} = z give g (x) = \frac{z + z^{- 1} + 2}{z^{2} + z^{- 2} - 6} = \frac{z^{3} + z + {2 z}^{2}}{z^{4} + 1 - {6 z}^{2}}$ $= \frac{z^{3} + {2 z}^{2} + z}{z^{4} - {6 z}^{2} + 1} = f (z) let decompose f$ $z^{4} - {6 z}^{2} + 1 = 0 \to u^{2} - 6 u + 1 = 0 \to Δ^{^{'}} = 9 - 1 = 8 \Rightarrow$ $u_{1} = 3 + 2 \sqrt{2} and u_{2} = 3 - 2 \sqrt{2} \Rightarrow f (z) = \frac{z^{3} + {2 z}^{2} + z}{(z^{2} - u_{1}) (z^{2} - u_{2})}$ $= (z^{3} + {2 z}^{2} + z) (\frac{1}{z^{2} - u_{1}} - \frac{1}{z^{2} - u_{2}}) = \frac{z^{3} + {2 z}^{2} + z}{z^{2} - u_{1}} - \frac{z^{3} + {2 z}^{2} + z}{z^{2} - u_{2}} = φ_{1} (z) - φ_{2} (z)$ $φ_{1} (z) = \frac{z (z^{2} - u_{1}) + {zu}_{1} + {2 z}^{2} + z}{z^{2} - u_{1}} = z + \frac{{2 z}^{2} + (1 + u_{1}) z}{z^{2} - u_{1}}$ $= z + \frac{2 (z^{2} - u_{1}) + {2 u}_{1} + (1 + u_{1}) z}{z^{2} - u_{1}} = z + 2 + \frac{(1 + u_{1}) z + {2 u}_{1}}{z^{2} - u_{1}}$ $φ_{2} (z) = z + 2 + \frac{(1 + u_{2}) z + {2 u}_{2}}{z^{2} - u_{2}} \Rightarrow f (z) = \frac{(1 + u_{1}) z + {2 u}_{1}}{z^{2} - u_{1}} - \frac{(1 + u_{2}) z + {2 u}_{2}}{z^{2} - u_{2}}$ $f (z) = \frac{(1 + u_{2}) z + {2 u}_{2}}{u_{2} - z^{2}} - \frac{(1 + u_{1}) z + {2 u}_{1}}{u_{1} - z^{2}} = \frac{1}{u_{2}} \times \frac{(1 + u_{2}) z + {2 u}_{2}}{1 - {(\frac{z}{\sqrt{u_{2}}})}^{2}}$ $- \frac{1}{u_{1}} \times \frac{(1 + u_{1}) z + {2 u}_{1}}{1 - {(\frac{z}{\sqrt{u_{1}}})}^{2}} so if ∣ z ∣ < \inf (\sqrt{u_{1}}, \sqrt{u_{2}}) weget$ $f (z) = \frac{1}{u_{2}} {(1 + u_{2}) z + {2 u}_{2}} \sum_{n = 0}^{\infty} \frac{z^{2 n}}{u_{2}^{n}} - \frac{1}{u_{1}} {(1 + u_{1}) z + {2 u}_{1}} \sum_{n = 0}^{\infty} \frac{z^{2 n}}{u_{1}^{n}}$ $= \frac{1}{3 - 2 \sqrt{2}} {(4 - 2 \sqrt{2}) z + 6 - 4 \sqrt{2}} \sum_{n = 0}^{\infty} \frac{e^{2 inx}}{{(3 - 2 \sqrt{2})}^{n}}$ $- \frac{1}{3 + 2 \sqrt{2}} {(4 + 2 \sqrt{2}) + 6 + 4 \sqrt{2}} \sum_{n = 0}^{\infty} \frac{e^{2 inx}}{{(3 + 2 \sqrt{2})}^{n}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum