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Question Number 68034 by mathmax by abdo last updated on 03/Sep/19
$${let}\:{f}\left({x}\right)\:={e}^{−{x}} \:\:,\:\:\mathrm{2}\pi\:\:{periodic}\:\:{developp}\:{f}\:{at}\:{fourier}\:{serie}. \\ $$
Commented by mathmax by abdo last updated on 07/Sep/19
![f(x) =Σ_(n=−∞) ^∞ a_n e^(inwx) with w=((2π)/T) =1 ⇒ f(x) =Σ_(n=−∞) ^(+∞) a_n e^(inx) with a_n =(1/T) ∫_(−(T/2)) ^(T/2) f(x)e^(−inx) dx a_n =(1/(2π))∫_(−π) ^π e^(−x) e^(−inx) dx ⇒2π a_n =∫_(−π) ^π e^(−(1+in)x) dx =[((−1)/(1+in)) e^(−(1+in)x) ]_(−π) ^π =−(1/(1+in)){ e^(−(1+in)π) −e^(−(1+in)(−π)) } =−(1/(1+in)){ (−1)^n e^(−π) −e^π (−1)^n } =(((−1)^n )/(1+in)){ e^π −e^(−π) } =((2(−1)^n )/(1+in)) sh(π) ⇒2πa_n =((2(−1)^n )/(1+in))sh(π) ⇒a_n =((sh(π))/π)(((−1)^n )/(1+in)) ⇒ f(x) =Σ_(n=−∞) ^(+∞) ((sh(π))/π)(((−1)^n )/(1+in)) e^(inx) =e^(−x) but Σ_(n=−∞) ^(+∞ ) a_n e^(inx) =Σ_(n=−∞) ^0 a_n e^(inx) +Σ_(n=1) ^∞ a_n e^(inx) =a_0 +Σ_(n=1) ^∞ a_(−n) e^(−nx) +Σ_(n=1) ^∞ a_n e^(inx) =a_0 +Σ_(n=1) ^∞ a_(−n) {cos(nx)−isin(nx)} +Σ_(n=1) ^∞ a_n {cos(nx)+isin)nx)} =a_0 +Σ_(n=1) ^∞ (a_n +a_(−n) )cos(nx) +i Σ_(n=1) ^∞ (a_n +a_(−n) )sin(nx) a_n +a_(−n) =(−1)^n ((sh(π))/π){ (1/(1+in)) +(1/(1−in))} =(−1)^n ((sh(π))/π)×(2/(1+n^2 )) a_0 =(1/(2π))∫_(−π) ^π e^(−x) dx =−(1/(2π))[e^(−x) ]_(−π) ^π =−(1/(2π)){e^(−π) −e^π } =(1/(2π)){ e^π −e^(−π) } =((2sh(π))/(2π)) =((sh(π))/π) ⇒ e^(−x) =((sh(π))/π) +((2sh(π))/π)Σ_(n=1) ^∞ (((−1)^n )/(n^2 +1))cos(nx)+((2i sh(π))/π)Σ_(n=1) ^∞ (((−1)^n )/(n^2 +1))sin(nx)](Q68256.png)
$${f}\left({x}\right)\:=\sum_{{n}=−\infty} ^{\infty} \:{a}_{{n}} {e}^{{inwx}} \:\:\:\:{with}\:{w}=\frac{\mathrm{2}\pi}{{T}}\:=\mathrm{1}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\sum_{{n}=−\infty} ^{+\infty} \:{a}_{{n}} {e}^{{inx}} \:\:{with}\:{a}_{{n}} \:=\frac{\mathrm{1}}{{T}}\:\int_{−\frac{{T}}{\mathrm{2}}} ^{\frac{{T}}{\mathrm{2}}} \:{f}\left({x}\right){e}^{−{inx}} {dx} \\ $$$${a}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{−\pi} ^{\pi} {e}^{−{x}} \:{e}^{−{inx}} {dx}\:\Rightarrow\mathrm{2}\pi\:{a}_{{n}} =\int_{−\pi} ^{\pi} \:{e}^{−\left(\mathrm{1}+{in}\right){x}} {dx} \\ $$$$=\left[\frac{−\mathrm{1}}{\mathrm{1}+{in}}\:{e}^{−\left(\mathrm{1}+{in}\right){x}} \right]_{−\pi} ^{\pi} \:=−\frac{\mathrm{1}}{\mathrm{1}+{in}}\left\{\:{e}^{−\left(\mathrm{1}+{in}\right)\pi} −{e}^{−\left(\mathrm{1}+{in}\right)\left(−\pi\right)} \right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{1}+{in}}\left\{\:\left(−\mathrm{1}\right)^{{n}} {e}^{−\pi} −{e}^{\pi} \left(−\mathrm{1}\right)^{{n}} \right\}\:=\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{1}+{in}}\left\{\:{e}^{\pi} −{e}^{−\pi} \right\} \\ $$$$=\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}} }{\mathrm{1}+{in}}\:{sh}\left(\pi\right)\:\Rightarrow\mathrm{2}\pi{a}_{{n}} =\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}} }{\mathrm{1}+{in}}{sh}\left(\pi\right)\:\Rightarrow{a}_{{n}} =\frac{{sh}\left(\pi\right)}{\pi}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{1}+{in}}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\sum_{{n}=−\infty} ^{+\infty} \:\frac{{sh}\left(\pi\right)}{\pi}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{1}+{in}}\:{e}^{{inx}} \:\:\:={e}^{−{x}} \:\:{but} \\ $$$$\sum_{{n}=−\infty} ^{+\infty\:} \:{a}_{{n}} {e}^{{inx}} \:=\sum_{{n}=−\infty} ^{\mathrm{0}} \:\:{a}_{{n}} {e}^{{inx}} \:+\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} {e}^{{inx}} \\ $$$$={a}_{\mathrm{0}} \:+\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{−{n}} {e}^{−{nx}} \:+\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} {e}^{{inx}} \\ $$$$\left.=\left.{a}_{\mathrm{0}} \:+\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{−{n}} \left\{{cos}\left({nx}\right)−{isin}\left({nx}\right)\right\}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} \left\{{cos}\left({nx}\right)+{isin}\right){nx}\right)\right\} \\ $$$$={a}_{\mathrm{0}} \:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\left({a}_{{n}} +{a}_{−{n}} \right){cos}\left({nx}\right)\:+{i}\:\sum_{{n}=\mathrm{1}} ^{\infty} \left({a}_{{n}} +{a}_{−{n}} \right){sin}\left({nx}\right) \\ $$$${a}_{{n}} +{a}_{−{n}} =\left(−\mathrm{1}\right)^{{n}} \:\frac{{sh}\left(\pi\right)}{\pi}\left\{\:\frac{\mathrm{1}}{\mathrm{1}+{in}}\:+\frac{\mathrm{1}}{\mathrm{1}−{in}}\right\} \\ $$$$=\left(−\mathrm{1}\right)^{{n}} \:\frac{{sh}\left(\pi\right)}{\pi}×\frac{\mathrm{2}}{\mathrm{1}+{n}^{\mathrm{2}} } \\ $$$${a}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{−\pi} ^{\pi} \:{e}^{−{x}} \:{dx}\:=−\frac{\mathrm{1}}{\mathrm{2}\pi}\left[{e}^{−{x}} \right]_{−\pi} ^{\pi} \:=−\frac{\mathrm{1}}{\mathrm{2}\pi}\left\{{e}^{−\pi} −{e}^{\pi} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\left\{\:{e}^{\pi} −{e}^{−\pi} \right\}\:=\frac{\mathrm{2}{sh}\left(\pi\right)}{\mathrm{2}\pi}\:=\frac{{sh}\left(\pi\right)}{\pi}\:\Rightarrow \\ $$$${e}^{−{x}} \:=\frac{{sh}\left(\pi\right)}{\pi}\:+\frac{\mathrm{2}{sh}\left(\pi\right)}{\pi}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \:+\mathrm{1}}{cos}\left({nx}\right)+\frac{\mathrm{2}{i}\:{sh}\left(\pi\right)}{\pi}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \:+\mathrm{1}}{sin}\left({nx}\right) \\ $$