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