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 39039 by maxmathsup by imad last updated on 01/Jul/18

let f(x) =(1/(1+∣sinx∣)) (2π periodic even) developp f at fourier serie .

$${let}\:{f}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+\mid{sinx}\mid}\:\:\:\left(\mathrm{2}\pi\:{periodic}\:{even}\right) \\ $$$${developp}\:{f}\:{at}\:{fourier}\:{serie}\:. \\ $$

Commented by abdo mathsup 649 cc 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 (1/(1 +∣sinx∣)) cos(nx)dx =(2/π) ∫_0 ^π ((cos(nx))/(1+sinx))dx ⇒(π/2)a_n = ∫_0 ^π ((cos(nx))/(1+sinx))dx but ∫_0 ^π ((cos(nx))/(1+sinx))dx =Re( ∫_0 ^π (e^(inx) /(1+sinx))dx) changement x=2t give ∫_0 ^π (e^(inx) /(1+sinx))dx = ∫_0 ^(2π) ((2e^(i2nt) )/(1+sin(2t)))dt also chang. e^(it) =z give ∫_0 ^(2π) ((2 e^(i2nt) )/(1+sin(2t)))dt = ∫_(∣z∣=1) ((2 z^(2n) )/(1+((z^2 −z^(−2) )/(2i)))) (dz/(iz)) = ∫_(∣z∣=1) ((4i z^(2n) )/(iz( 2i+z^2 −z^(−2) )))dz = ∫_(∣z∣=1) ((4z^(2n−1) )/(2i +z^2 −(1/z^2 ))) dz = ∫_(∣z∣=1) ((4 z^(2n+1) )/(2iz^2 +z^4 −1))dz let ϕ(z) = ((4z^(2n+1) )/(z^4 +2iz^2 −1)) ϕ(z) = ((4z^(2n+1) )/((z^2 +i)^2 )) = ((4 z^(2n+1) )/((z−(√(−i)))^2 (2+(√(−i)))^2 )) = ((4 z^(2n+1) )/((z −e^(−((iπ)/4)) )^2 (z +e^(−((iπ)/4)) )^2 )) so ϕ have a double poles +^(− ) e^(−((iπ)/4)) ∫_(∣z∣=1) ϕ(2)dz =2iπ{ Res(ϕ,e^(−((iπ)/4)) ) +Res(ϕ,−e^(−((iπ)/4)) )}

$${f}\left({x}\right)=\:\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} \:{cos}\left({nx}\right)\:\:{with} \\ $$$${a}_{{n}} =\:\frac{\mathrm{2}}{{T}}\:\int_{\left[{T}\right]} \:\:{f}\left({x}\right){cos}\left({nx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}\pi}\:\int_{−\pi} ^{\pi} \:\:\frac{\mathrm{1}}{\mathrm{1}\:+\mid{sinx}\mid}\:{cos}\left({nx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\pi}\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{cos}\left({nx}\right)}{\mathrm{1}+{sinx}}{dx}\:\Rightarrow\frac{\pi}{\mathrm{2}}{a}_{{n}} =\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{cos}\left({nx}\right)}{\mathrm{1}+{sinx}}{dx} \\ $$$${but}\:\int_{\mathrm{0}} ^{\pi} \:\frac{{cos}\left({nx}\right)}{\mathrm{1}+{sinx}}{dx}\:={Re}\left(\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{e}^{{inx}} }{\mathrm{1}+{sinx}}{dx}\right) \\ $$$${changement}\:{x}=\mathrm{2}{t}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{e}^{{inx}} }{\mathrm{1}+{sinx}}{dx}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{\mathrm{2}{e}^{{i}\mathrm{2}{nt}} }{\mathrm{1}+{sin}\left(\mathrm{2}{t}\right)}{dt}\:{also} \\ $$$${chang}.\:{e}^{{it}} ={z}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{\mathrm{2}\:{e}^{{i}\mathrm{2}{nt}} }{\mathrm{1}+{sin}\left(\mathrm{2}{t}\right)}{dt}\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{\mathrm{2}\:{z}^{\mathrm{2}{n}} }{\mathrm{1}+\frac{{z}^{\mathrm{2}} −{z}^{−\mathrm{2}} }{\mathrm{2}{i}}}\:\frac{{dz}}{{iz}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{\mathrm{4}{i}\:{z}^{\mathrm{2}{n}} }{{iz}\left(\:\mathrm{2}{i}+{z}^{\mathrm{2}} −{z}^{−\mathrm{2}} \right)}{dz} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{\mathrm{4}{z}^{\mathrm{2}{n}−\mathrm{1}} }{\mathrm{2}{i}\:+{z}^{\mathrm{2}} −\frac{\mathrm{1}}{{z}^{\mathrm{2}} }}\:{dz} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{\mathrm{4}\:{z}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{iz}^{\mathrm{2}} \:+{z}^{\mathrm{4}} \:−\mathrm{1}}{dz}\:\:{let} \\ $$$$\varphi\left({z}\right)\:=\:\frac{\mathrm{4}{z}^{\mathrm{2}{n}+\mathrm{1}} }{{z}^{\mathrm{4}} \:+\mathrm{2}{iz}^{\mathrm{2}} \:−\mathrm{1}} \\ $$$$\varphi\left({z}\right)\:=\:\frac{\mathrm{4}{z}^{\mathrm{2}{n}+\mathrm{1}} }{\left({z}^{\mathrm{2}} +{i}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{4}\:{z}^{\mathrm{2}{n}+\mathrm{1}} }{\left({z}−\sqrt{−{i}}\right)^{\mathrm{2}} \left(\mathrm{2}+\sqrt{−{i}}\right)^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{4}\:{z}^{\mathrm{2}{n}+\mathrm{1}} }{\left({z}\:−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}\:+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} }\:{so}\:\varphi\:{have}\:{a}\:{double} \\ $$$${poles}\:\overset{−\:} {+}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\varphi\left(\mathrm{2}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:+{Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$$ \\ $$

Commented by abdo mathsup 649 cc last updated on 05/Jul/18

Res(ϕ, e^(−((iπ)/4)) )=lim_(z→ e^(−((iπ)/4)) ) { (z −e^(−((iπ)/4)) )^2 ϕ(z)}^((1)) =lim_(z→ e^(−((iπ)/4)) ) { ((4 z^(2n+1) )/((z+e^(−((iπ)/4)) )^2 ))}^((1)) =lim_(z→e^(−((iπ)/4)) ) 4 (((2n+1)z^(2n) (z +e^(−((iπ)/4)) )^2 −2(z+e^(−((iπ)/4)) )z^(2n+1) )/((z +e^(−((iπ)/4)) )^4 )) =lim_(z→e^(−((iπ)/4)) ) 4 (((2n+1)z^(2n) (z +e^(−((iπ)/4)) )−2 z^(2n+1) )/((z +e^(−((iπ)/4)) )^3 )) =4(((2n+1) e^(−((inπ)/2)) (2 e^(−((iπ)/4)) ) −2 e^(−i(((2n+1)π)/4)) )/((2e^(−((iπ)/4)) )^3 )) =8(((2n+1) e^(−i((n/2) +(1/4))π) −e^(−i(((2n+1)π)/4)) )/(8 e^(−i((3π)/4)) )) = ((2n e^(−i(((2n+1)π)/4)) )/e^(−i((3π)/4)) ) =2n e^(−i{((2n+1)/4)+(3/4)}π) =2n e^(−i (((n+2)π)/2)) =−2n e^(−((inπ)/2)) =−2n{cos(((nπ)/2))−i sin(((nπ)/2))}

$${Res}\left(\varphi,\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)={lim}_{{z}\rightarrow\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} } \:\left\{\:\left({z}\:−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\left\{\:\:\frac{\mathrm{4}\:{z}^{\mathrm{2}{n}+\mathrm{1}} }{\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{e}^{−\frac{{i}\pi}{\mathrm{4}}} } \:\mathrm{4}\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right){z}^{\mathrm{2}{n}} \left({z}\:+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \:−\mathrm{2}\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right){z}^{\mathrm{2}{n}+\mathrm{1}} }{\left({z}\:+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{e}^{−\frac{{i}\pi}{\mathrm{4}}} } \:\:\mathrm{4}\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right){z}^{\mathrm{2}{n}} \left({z}\:+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)−\mathrm{2}\:{z}^{\mathrm{2}{n}+\mathrm{1}} }{\left({z}\:+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{3}} } \\ $$$$=\mathrm{4}\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\:{e}^{−\frac{{in}\pi}{\mathrm{2}}} \:\left(\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:−\mathrm{2}\:{e}^{−{i}\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{4}}} }{\left(\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{3}} } \\ $$$$=\mathrm{8}\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\:{e}^{−{i}\left(\frac{{n}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}\right)\pi} \:\:−{e}^{−{i}\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{4}}} }{\mathrm{8}\:{e}^{−{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} } \\ $$$$=\:\frac{\mathrm{2}{n}\:{e}^{−{i}\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{4}}} }{{e}^{−{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} }\:=\mathrm{2}{n}\:\:{e}^{−{i}\left\{\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}\right\}\pi} \\ $$$$=\mathrm{2}{n}\:\:{e}^{−{i}\:\frac{\left({n}+\mathrm{2}\right)\pi}{\mathrm{2}}} \:\:=−\mathrm{2}{n}\:{e}^{−\frac{{in}\pi}{\mathrm{2}}} \\ $$$$=−\mathrm{2}{n}\left\{{cos}\left(\frac{{n}\pi}{\mathrm{2}}\right)−{i}\:{sin}\left(\frac{{n}\pi}{\mathrm{2}}\right)\right\} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com