Question Number 45968 by maxmathsup by imad last updated on 19/Oct/18
$$\left.\mathrm{1}\right){find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{cos}\left({nx}\right)}{{n}}\:{and}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left({nx}\right)}{{n}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}}{cos}\left(\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\right)\:{and}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}}{sin}\left(\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\right) \\ $$
Commented by maxmathsup by imad last updated on 23/Oct/18
$$\left.\mathrm{1}\right)\:{let}\:{f}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left({nx}\right)}{{n}}\:{and}\:{g}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left({nx}\right)}{{n}}\:{we}\:{have} \\ $$$${f}\left({x}\right)+{ig}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{inx}} }{{n}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left({e}^{{ix}} \right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{1}−{e}^{{ix}} \right) \\ $$$$=−{ln}\left(\mathrm{1}−{cosx}\:−{isinx}\right)=−{ln}\left(\mathrm{2}\:{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{2}{isin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)\right) \\ $$$$=−{ln}\left(−{i}\:{sin}\left(\frac{{x}}{\mathrm{2}}\right)\left(\:{e}^{{i}\frac{{x}}{\mathrm{2}}} \right)\right)=−{ln}\left(−{i}\right)\:+{ln}\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)\right)−{i}\frac{{x}}{\mathrm{2}} \\ $$$$=−{ln}\left({e}^{−{i}\frac{\pi}{\mathrm{2}}} \right)−{i}\frac{{x}}{\mathrm{2}}\:+{ln}\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)\right)=\frac{{i}\pi}{\mathrm{2}}−{i}\frac{{x}}{\mathrm{2}}\:+{ln}\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)\right) \\ $$$$={i}\frac{\pi−{x}}{\mathrm{2}}\:+{ln}\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)\right)\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{cos}\left({nx}\right)}{{n}}\:={ln}\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)\right)\:{and} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left({nx}\right)}{{n}}\:=\frac{\pi−{x}}{\mathrm{2}}\:. \\ $$
Commented by maxmathsup by imad last updated on 23/Oct/18
$$\left.\mathrm{2}\right){we}\:{have}\:{proved}\:{that}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left({nx}\right)}{{n}}={ln}\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)\:{for}\:{x}=\frac{\mathrm{2}\pi}{\mathrm{3}}\:\Rightarrow\right. \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left(\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\right)}{{n}}\:={ln}\left({sin}\left(\frac{\pi}{\mathrm{3}}\right)\right)={ln}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{3}\right)−{ln}\left(\mathrm{2}\right)\:{also}\:{we}\:{have} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{sin}\left({nx}\right)}{{n}}\:=\frac{\pi−{x}}{\mathrm{2}}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left(\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\right)}{{n}}\:=\frac{\pi−\frac{\mathrm{2}\pi}{\mathrm{3}}}{\mathrm{2}}\:=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{3}}\:=\frac{\pi}{\mathrm{6}}\:. \\ $$
Answered by Smail last updated on 22/Oct/18
$${p}\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{{inx}} }{{n}} \\ $$$${ln}\left(\mathrm{1}−{x}\right)=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}}\:\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$$${So}\:\:{p}\left({x}\right)=−\left(−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({e}^{{ix}} \right)^{{n}} }{{n}}\right)=−{ln}\left(\mathrm{1}−{e}^{{ix}} \right) \\ $$$${p}\left({x}\right)={a}+{ib}=−{ln}\left(\mathrm{1}−{e}^{{ix}} \right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{{inx}} }{{n}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left({nx}\right)}{{n}}+{i}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({nx}\right)}{{n}} \\ $$$${e}^{{a}+{ib}} =\frac{\mathrm{1}}{\mathrm{1}−{cos}\left({x}\right)−{isin}\left({x}\right)} \\ $$$${e}^{{a}} {cos}\left({b}\right)+{ie}^{{a}} {sin}\left({b}\right)=\frac{\mathrm{1}−{cos}\left({x}\right)+{isin}\left({x}\right)}{\mathrm{2}\left(\mathrm{1}−{cos}\left({x}\right)\right)} \\ $$$${e}^{{a}} {cos}\left({b}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:{e}^{{a}} {sin}\left({b}\right)=\frac{{sin}\left({x}\right)}{\mathrm{2}\left(\mathrm{1}−{cos}\left({x}\right)\right)} \\ $$$${tan}\left({b}\right)=\frac{{sin}\left({x}\right)}{\mathrm{1}−{cos}\left({x}\right)} \\ $$$${b}={tan}^{−\mathrm{1}} \left(\frac{{sin}\left({x}\right)}{\mathrm{1}−{cos}\left({x}\right)}\right) \\ $$$${which}\:{is}\:{b}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({nx}\right)}{{n}}={tan}^{−\mathrm{1}} \left(\frac{{sinx}}{\mathrm{1}−{cosx}}\right) \\ $$$${and}\:\:{e}^{{a}} {cos}\left({b}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${e}^{{a}} =\frac{\mathrm{1}}{\mathrm{2}{cos}\left({b}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \left({b}\right)} \\ $$$${e}^{{a}} =\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}+\frac{{sin}^{\mathrm{2}} {x}}{\left(\mathrm{1}−{cosx}\right)^{\mathrm{2}} }}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{2}\left(\mathrm{1}−{cos}\left({x}\right)\right)}{\left(\mathrm{1}−{cosx}\right)^{\mathrm{2}} }} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{1}−{cosx}}}\:{So}\:\:{a}={ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)+{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{cosx}}}\right) \\ $$$${a}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left({nx}\right)}{{n}}=−\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{1}−{cosx}\right)\right) \\ $$
Commented by Smail last updated on 22/Oct/18
$${a}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{1}−{cosx}\right)\right) \\ $$$${a}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left(\mathrm{2}{n}\pi/\mathrm{3}\right)}{{n}}=−\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{1}−{cos}\left(\mathrm{2}\pi/\mathrm{3}\right)\right)\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right)=−\frac{{ln}\mathrm{3}}{\mathrm{2}} \\ $$$${b}\left({x}\right)={tan}^{−\mathrm{1}} \left(\frac{{sinx}}{\mathrm{1}−{cosx}}\right) \\ $$$${b}\left(\mathrm{2}\pi/\mathrm{3}\right)={tan}^{−\mathrm{1}} \left(\frac{{sin}\left(\mathrm{2}\pi/\mathrm{3}\right)}{\mathrm{1}−{cos}\left(\mathrm{2}\pi/\mathrm{3}\right)}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\right)=\frac{\pi}{\mathrm{6}}+{k}\pi \\ $$$${so}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left(\mathrm{2}{n}\pi/\mathrm{3}\right)}{{n}}=\frac{\pi}{\mathrm{6}} \\ $$