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Question Number 68035 by mathmax by abdo last updated on 03/Sep/19
let f(x) =e^(−iαx) ,2π periodic .developp f at fourier serie.
$${let}\:{f}\left({x}\right)\:={e}^{−{i}\alpha{x}} \:\:\:\:,\mathrm{2}\pi\:\:{periodic}\:\:.{developp}\:{f}\:{at}\:{fourier}\:{serie}. \\ $$
Commented by mathmax by abdo last updated on 08/Sep/19
f(x) =Σ_(n=−∞) ^(+∞) a_n e^(inwx) with w =((2π)/T) =1 ⇒ f(x)=Σ_(n=−∞) ^(+∞) a_n e^(inx) and a_n =(1/T)∫_(−(T/2)) ^(T/2) f(x)e^(−inx) dx =(1/(2π)) ∫_(−π) ^π e^(−iαx) e^(−inx) dx =(1/(2π)) ∫_(−π) ^π e^(−i(α+n)x) dx ⇒ 2π a_n =∫_(−π) ^π e^(−i(α+n)x) dx =[(1/(−i(α+n)))e^(−i(α+n)x) ]_(−π) ^π =((−1)/(i(α+n))){ e^(−i(α+n)π) −e^(−i(α+n)(−π)) } =(i/(α+n)){ (−1)^n e^(−iπα) −(−1)^n e^(iπα) } =((−i(−1)^n )/(α+n)){ e^(iπα) −e^(−iπα) } =((−i(−1)^n )/(n+α)) (2isin(πα)) =((2(−1)^n sin(πα))/(n+α)) ⇒ e^(−iαx) =Σ_(n=−∞) ^(+∞) ((2(−1)^n sin(πα))/(n+α)) e^(inx) =2sin(πα) Σ_(n=−∞) ^(+∞) (((−1)^n )/(n+α)) e^(inx) .
$${f}\left({x}\right)\:=\sum_{{n}=−\infty} ^{+\infty} \:{a}_{{n}} {e}^{{inwx}} \:\:\:{with}\:{w}\:=\frac{\mathrm{2}\pi}{{T}}\:=\mathrm{1}\:\Rightarrow \\ $$$${f}\left({x}\right)=\sum_{{n}=−\infty} ^{+\infty} \:{a}_{{n}} {e}^{{inx}} \:\:\:\:\:{and}\:{a}_{{n}} =\frac{\mathrm{1}}{{T}}\int_{−\frac{{T}}{\mathrm{2}}} ^{\frac{{T}}{\mathrm{2}}} \:{f}\left({x}\right){e}^{−{inx}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\:\int_{−\pi} ^{\pi} \:{e}^{−{i}\alpha{x}} \:{e}^{−{inx}} {dx}\:=\frac{\mathrm{1}}{\mathrm{2}\pi}\:\int_{−\pi} ^{\pi} \:\:{e}^{−{i}\left(\alpha+{n}\right){x}} {dx}\:\Rightarrow \\ $$$$\mathrm{2}\pi\:{a}_{{n}} =\int_{−\pi} ^{\pi} \:{e}^{−{i}\left(\alpha+{n}\right){x}} {dx}\:=\left[\frac{\mathrm{1}}{−{i}\left(\alpha+{n}\right)}{e}^{−{i}\left(\alpha+{n}\right){x}} \right]_{−\pi} ^{\pi} \\ $$$$=\frac{−\mathrm{1}}{{i}\left(\alpha+{n}\right)}\left\{\:{e}^{−{i}\left(\alpha+{n}\right)\pi} −{e}^{−{i}\left(\alpha+{n}\right)\left(−\pi\right)} \right\} \\ $$$$=\frac{{i}}{\alpha+{n}}\left\{\:\:\left(−\mathrm{1}\right)^{{n}} \:{e}^{−{i}\pi\alpha} \:\:−\left(−\mathrm{1}\right)^{{n}} \:{e}^{{i}\pi\alpha} \right\} \\ $$$$=\frac{−{i}\left(−\mathrm{1}\right)^{{n}} }{\alpha+{n}}\left\{\:{e}^{{i}\pi\alpha} −{e}^{−{i}\pi\alpha} \right\}\:\:=\frac{−{i}\left(−\mathrm{1}\right)^{{n}} }{{n}+\alpha}\:\left(\mathrm{2}{isin}\left(\pi\alpha\right)\right) \\ $$$$=\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}} \:{sin}\left(\pi\alpha\right)}{{n}+\alpha}\:\Rightarrow \\ $$$${e}^{−{i}\alpha{x}} \:=\sum_{{n}=−\infty} ^{+\infty} \:\:\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}} {sin}\left(\pi\alpha\right)}{{n}+\alpha}\:{e}^{{inx}} \\ $$$$=\mathrm{2}{sin}\left(\pi\alpha\right)\:\sum_{{n}=−\infty} ^{+\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\alpha}\:{e}^{{inx}} \:\:. \\ $$
Commented by mathmax by abdo last updated on 08/Sep/19
2π a_n =((2(−1)^n sin(πα))/(n+α)) ⇒ a_n =((sin(πα))/π) (((−1)^n )/(n+α)) ⇒ e^(−iαx) =((sin(πα))/π)Σ_(n=−∞) ^(+∞) (((−1)^n )/(n+α)) e^(inx)
$$\mathrm{2}\pi\:{a}_{{n}} =\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}} {sin}\left(\pi\alpha\right)}{{n}+\alpha}\:\Rightarrow\:{a}_{{n}} =\frac{{sin}\left(\pi\alpha\right)}{\pi}\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\alpha}\:\Rightarrow \\ $$$${e}^{−{i}\alpha{x}} \:\:=\frac{{sin}\left(\pi\alpha\right)}{\pi}\sum_{{n}=−\infty} ^{+\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\alpha}\:{e}^{{inx}} \:\: \\ $$
Commented by mathmax by abdo last updated on 08/Sep/19
remark we have Σ_(n=−∞) ^(+∞) (((−1)^n )/(n+α))e^(inx) =Σ_(n=−∞) ^o (((−1)^n )/(n+α))e^(inx) +Σ_(n=1) ^∞ (((−1)^n )/(n+α)) e^(inx) =(1/α) +Σ_(n=1) ^∞ (((−1)^n )/(α−n))e^(−inx) +Σ_(n=1) ^∞ (((−1)^n )/(n+α)) e^(inx) =(1/α) +Σ_(n=1) ^∞ (((−1)^n )/(α−n))(cos(nx)−isin(nx))+Σ_(n=1) ^∞ (((−1)^n )/(n+α))(cos(nx)+isin(nx)) =(1/α)+Σ_(n=1) ^∞ (−1)^n cos(nx){(1/(α−n))+(1/(α+n))} +iΣ_(n=1) ^∞ (−1)^n sin(nx){(1/(α+n))−(1/(α−n))} =(1/α) +Σ_(n=1) ^∞ (((−1)^n 2α)/(α^2 −n^2 )) cos(nx) −i Σ_(n=1) ^∞ (((−1)^n (2n))/(α^2 −n^2 )) =cos(αx)−isin(αx) ⇒ cos(αx) =(1/α) +2α Σ_(n=1) ^∞ (((−1)^n cos(nx))/(α^2 −n^2 )) and sin(αx) =2 Σ_(n=1) ^∞ ((n(−1)^n )/(α^2 −n^2 ))
$${remark}\:{we}\:{have}\:\sum_{{n}=−\infty} ^{+\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\alpha}{e}^{{inx}} \\ $$$$=\sum_{{n}=−\infty} ^{{o}} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\alpha}{e}^{{inx}} \:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\alpha}\:{e}^{{inx}} \\ $$$$=\frac{\mathrm{1}}{\alpha}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\alpha−{n}}{e}^{−{inx}} \:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\alpha}\:{e}^{{inx}} \\ $$$$=\frac{\mathrm{1}}{\alpha}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\alpha−{n}}\left({cos}\left({nx}\right)−{isin}\left({nx}\right)\right)+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\alpha}\left({cos}\left({nx}\right)+{isin}\left({nx}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\alpha}+\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} {cos}\left({nx}\right)\left\{\frac{\mathrm{1}}{\alpha−{n}}+\frac{\mathrm{1}}{\alpha+{n}}\right\} \\ $$$$+{i}\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} {sin}\left({nx}\right)\left\{\frac{\mathrm{1}}{\alpha+{n}}−\frac{\mathrm{1}}{\alpha−{n}}\right\} \\ $$$$=\frac{\mathrm{1}}{\alpha}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \mathrm{2}\alpha}{\alpha^{\mathrm{2}} −{n}^{\mathrm{2}} }\:{cos}\left({nx}\right)\:−{i}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}\right)}{\alpha^{\mathrm{2}} −{n}^{\mathrm{2}} } \\ $$$$={cos}\left(\alpha{x}\right)−{isin}\left(\alpha{x}\right)\:\Rightarrow \\ $$$${cos}\left(\alpha{x}\right)\:=\frac{\mathrm{1}}{\alpha}\:+\mathrm{2}\alpha\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{cos}\left({nx}\right)}{\alpha^{\mathrm{2}} −{n}^{\mathrm{2}} }\:\:{and} \\ $$$${sin}\left(\alpha{x}\right)\:=\mathrm{2}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{n}\left(−\mathrm{1}\right)^{{n}} }{\alpha^{\mathrm{2}} −{n}^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 08/Sep/19
error at final line e^(−iαx) =((sin(απ))/π){(1/α)+2α Σ_(n=1) ^∞ (((−1)^n cos(nx))/(𝛂^2 −n^2 ))} −i((sin(απ))/π) Σ_(n=1) ^∞ (((−1)^n (2n))/(α^2 −n^2 )) =cos(αx)−isin(αx) ⇒ cos(αx) =((sin(απ))/(απ)) +((2α)/π) sin(απ)Σ_(n=1) ^∞ (((−1)^n cos(nx))/(α^2 −n^2 )) sin(αx)=((2sin(απ))/π)Σ_(n=1) ^∞ (((−1)^n )/(α^2 −n^2 ))
$${error}\:{at}\:{final}\:{line}\: \\ $$$${e}^{−{i}\alpha{x}} \:=\frac{{sin}\left(\alpha\pi\right)}{\pi}\left\{\frac{\mathrm{1}}{\alpha}+\mathrm{2}\alpha\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} \:\boldsymbol{{cos}}\left(\boldsymbol{{nx}}\right)}{\boldsymbol{\alpha}^{\mathrm{2}} −{n}^{\mathrm{2}} }\right\} \\ $$$$−{i}\frac{{sin}\left(\alpha\pi\right)}{\pi}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}\right)}{\alpha^{\mathrm{2}} −{n}^{\mathrm{2}} }\:={cos}\left(\alpha{x}\right)−{isin}\left(\alpha{x}\right)\:\Rightarrow \\ $$$${cos}\left(\alpha{x}\right)\:=\frac{{sin}\left(\alpha\pi\right)}{\alpha\pi}\:+\frac{\mathrm{2}\alpha}{\pi}\:{sin}\left(\alpha\pi\right)\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{cos}\left({nx}\right)}{\alpha^{\mathrm{2}} −{n}^{\mathrm{2}} } \\ $$$${sin}\left(\alpha{x}\right)=\frac{\mathrm{2}{sin}\left(\alpha\pi\right)}{\pi}\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\alpha^{\mathrm{2}} −{n}^{\mathrm{2}} } \\ $$

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