Question Number 42222 by maxmathsup by imad last updated on 20/Aug/18
$${let}\:{f}\left({x}\right)\:={e}^{−\mid{x}\mid} \:,\:\:\mathrm{2}\pi\:{periodic}\:{even}\:\:{developp}\:{f}\:{at}\:{fourier}\:{serie}\:. \\ $$
Answered by maxmathsup by imad last updated on 21/Aug/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 e^(−∣x∣) cos(nx) dx =(2/π) ∫_0 ^π e^(−x) cos(nx)dx ⇒ (π/2) a_n = ∫_0 ^π e^(−x) cos(nx)dx =Re( ∫_0 ^π e^(−x) e^(inx) dx) but ∫_0 ^π e^((−1+in)x) dx =[(1/(−1+in)) e^((−1+in)x) ]_0 ^π =((−1)/(1−in)){ e^((−1+in)π) −1} =((−1)/(1−in)){ e^(−π) (−1)^n −1} = (1/(1−in)){1−(−1)^n e^(−π) } =(((1−(−1)^n e^(−π) )(1+in))/(1+n^2 )) ⇒(π/2) a_n =((1−(−1)^n e^(−π) )/(n^2 +1)) ⇒ a_n =(2/π) ((1−(−1)^n e^(−π) )/(n^2 +1)) we have (π/2)a_0 =∫_0 ^π e^(−x) dx =[−e^(−x) ]_0 ^π ⇒ =1−e^(−π) ⇒ a_0 =(2/π)(1−e^(−π) ) ⇒(a_0 /2) = ((1−e^(−π) )/π) ⇒ e^(−∣x∣) =((1−e^(−π) )/π) +(2/π) Σ_(n=1) ^∞ ((1−(−1)^n e^(−π) )/(n^2 +1)) cos(nx) .](https://www.tinkutara.com/question/Q42265.png)
$${f}\left({x}\right)=\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} {cos}\left({nx}\right)\:\:{with}\:{a}_{{n}} =\frac{\mathrm{2}}{{T}}\:\int_{\left[{T}\right]} {f}\left({x}\right){cos}\left({nx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}\pi}\:\int_{−\pi} ^{\pi} \:{e}^{−\mid{x}\mid} \:{cos}\left({nx}\right)\:{dx}\:=\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:\:{e}^{−{x}} \:{cos}\left({nx}\right){dx}\:\Rightarrow \\ $$$$\frac{\pi}{\mathrm{2}}\:{a}_{{n}} \:=\:\int_{\mathrm{0}} ^{\pi} \:\:{e}^{−{x}} \:{cos}\left({nx}\right){dx}\:={Re}\left(\:\int_{\mathrm{0}} ^{\pi} \:{e}^{−{x}} {e}^{{inx}} {dx}\right)\:{but} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:{e}^{\left(−\mathrm{1}+{in}\right){x}} {dx}\:=\left[\frac{\mathrm{1}}{−\mathrm{1}+{in}}\:{e}^{\left(−\mathrm{1}+{in}\right){x}} \right]_{\mathrm{0}} ^{\pi} \:=\frac{−\mathrm{1}}{\mathrm{1}−{in}}\left\{\:{e}^{\left(−\mathrm{1}+{in}\right)\pi} \:−\mathrm{1}\right\} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{1}−{in}}\left\{\:{e}^{−\pi} \left(−\mathrm{1}\right)^{{n}} −\mathrm{1}\right\}\:=\:\frac{\mathrm{1}}{\mathrm{1}−{in}}\left\{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\pi} \right\} \\ $$$$=\frac{\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} {e}^{−\pi} \right)\left(\mathrm{1}+{in}\right)}{\mathrm{1}+{n}^{\mathrm{2}} }\:\Rightarrow\frac{\pi}{\mathrm{2}}\:{a}_{{n}} =\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\pi} }{{n}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${a}_{{n}} =\frac{\mathrm{2}}{\pi}\:\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\pi} }{{n}^{\mathrm{2}} \:+\mathrm{1}}\:\:\:\:\:{we}\:{have}\:\:\frac{\pi}{\mathrm{2}}{a}_{\mathrm{0}} =\int_{\mathrm{0}} ^{\pi} \:{e}^{−{x}} \:{dx}\:=\left[−{e}^{−{x}} \right]_{\mathrm{0}} ^{\pi} \:\Rightarrow \\ $$$$=\mathrm{1}−{e}^{−\pi} \:\:\Rightarrow\:{a}_{\mathrm{0}} =\frac{\mathrm{2}}{\pi}\left(\mathrm{1}−{e}^{−\pi} \right)\:\Rightarrow\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:=\:\frac{\mathrm{1}−{e}^{−\pi} }{\pi}\:\Rightarrow \\ $$$${e}^{−\mid{x}\mid} \:\:\:=\frac{\mathrm{1}−{e}^{−\pi} }{\pi}\:\:+\frac{\mathrm{2}}{\pi}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\pi} }{{n}^{\mathrm{2}} \:+\mathrm{1}}\:{cos}\left({nx}\right)\:. \\ $$