Question Number 38204 by prof Abdo imad last updated on 22/Jun/18
$${if}\:\:\frac{\mathrm{1}}{{cosx}}\:=\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} {cos}\left({nx}\right) \\ $$$${calculate}\:{a}_{\mathrm{0}} \:{and}\:{a}_{{n}} \\ $$
Commented by math khazana by abdo last updated on 26/Jun/18
![let?developp at fourier?serie tbe?function f(x)=(1/(cosx)) (2π periodic even) f(x)=(a_0 /2) +Σ_(n=1) ^∞ a_n cos(nx) a_n =(2/T) ∫_([T]) f(x)cos(nx)dx=(2/(2π)) ∫_(−π) ^π ((cos(nx))/(cosx))dx ⇒ π a_n = ∫_(−π) ^π ((cos(nx))/(cosx))dx=Re( ∫_(−π) ^π (e^(inx) /(cosx))dx) changement e^(ix) =z give ∫_(−π) ^π (e^(inx) /(cosx))dx =∫_(∣z∣=1) 2(z^n /(z+z^(−1) )) (dz/(iz)) = ∫_(∣z∣=1) ((−2iz^n )/(z^2 +1))dz let ϕ(z)= ((−2iz^n )/(z^2 +1)) the poles of ϕ are i and −i ,residus theorem?give ∫_(∣z∣=1) ϕ(z)dz =2iπ Res(ϕ,i) =2iπ ((−2i.i^n )/(2i)) =−2iπi^n =−2π i^(n+1) ⇒ π a_(2n) =Re( −2π i^(2n+1) )=Re( −2πi(−1)^n )=0 π a_(2n+1) =Re(−2πi^(2n+2) )=Re( 2π(−1)^n ) =2π(−1)^n ⇒ a_(2n+1) = 2(−1)^n ⇒a_0 =0 and (1/(cosx)) = 2Σ_(n=0) ^∞ (−1)^n cos(2n+1)x](https://www.tinkutara.com/question/Q38504.png)
$${let}?{developp}\:{at}\:{fourier}?{serie}\:{tbe}?{function} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{{cosx}}\:\:\left(\mathrm{2}\pi\:{periodic}\:{even}\right) \\ $$$${f}\left({x}\right)=\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} {a}_{{n}} \:{cos}\left({nx}\right) \\ $$$${a}_{{n}} =\frac{\mathrm{2}}{{T}}\:\int_{\left[{T}\right]} {f}\left({x}\right){cos}\left({nx}\right){dx}=\frac{\mathrm{2}}{\mathrm{2}\pi}\:\int_{−\pi} ^{\pi} \:\frac{{cos}\left({nx}\right)}{{cosx}}{dx}\:\Rightarrow \\ $$$$\pi\:{a}_{{n}} =\:\int_{−\pi} ^{\pi} \:\:\frac{{cos}\left({nx}\right)}{{cosx}}{dx}={Re}\left(\:\int_{−\pi} ^{\pi} \:\frac{{e}^{{inx}} }{{cosx}}{dx}\right) \\ $$$${changement}\:{e}^{{ix}} ={z}\:{give} \\ $$$$\int_{−\pi} ^{\pi} \:\:\frac{{e}^{{inx}} }{{cosx}}{dx}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\mathrm{2}\frac{{z}^{{n}} }{{z}+{z}^{−\mathrm{1}} }\:\frac{{dz}}{{iz}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\frac{−\mathrm{2}{iz}^{{n}} }{{z}^{\mathrm{2}} \:+\mathrm{1}}{dz}\:{let}\:\varphi\left({z}\right)=\:\:\frac{−\mathrm{2}{iz}^{{n}} }{{z}^{\mathrm{2}} \:+\mathrm{1}}\:{the}\:{poles}\:{of} \\ $$$$\varphi\:{are}\:{i}\:{and}\:−{i}\:,{residus}\:{theorem}?{give} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$$=\mathrm{2}{i}\pi\:\:\frac{−\mathrm{2}{i}.{i}^{{n}} }{\mathrm{2}{i}}\:=−\mathrm{2}{i}\pi{i}^{{n}} \:=−\mathrm{2}\pi\:{i}^{{n}+\mathrm{1}} \:\Rightarrow \\ $$$$\pi\:{a}_{\mathrm{2}{n}} \:={Re}\left(\:−\mathrm{2}\pi\:{i}^{\mathrm{2}{n}+\mathrm{1}} \right)={Re}\left(\:−\mathrm{2}\pi{i}\left(−\mathrm{1}\right)^{{n}} \right)=\mathrm{0} \\ $$$$\pi\:{a}_{\mathrm{2}{n}+\mathrm{1}} ={Re}\left(−\mathrm{2}\pi{i}^{\mathrm{2}{n}+\mathrm{2}} \right)={Re}\left(\:\mathrm{2}\pi\left(−\mathrm{1}\right)^{{n}} \right) \\ $$$$=\mathrm{2}\pi\left(−\mathrm{1}\right)^{{n}} \:\Rightarrow\:{a}_{\mathrm{2}{n}+\mathrm{1}} =\:\mathrm{2}\left(−\mathrm{1}\right)^{{n}} \:\Rightarrow{a}_{\mathrm{0}} =\mathrm{0}\:{and} \\ $$$$\frac{\mathrm{1}}{{cosx}}\:=\:\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{cos}\left(\mathrm{2}{n}+\mathrm{1}\right){x} \\ $$$$ \\ $$$$ \\ $$