Question Number 42285 by maxmathsup by imad last updated on 22/Aug/18 | ||
$${let}\:\:{S}_{{p}} =\sum_{{n}=\mathrm{0}} ^{\infty} \:\:{cos}\left(\frac{{n}\pi}{{p}}\right)\:\:{and}\:\:{W}_{{p}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{sin}\left(\frac{{n}\pi}{{p}}\right)\:{with}\:{p}\:{natural}\:{integr}\:{not}\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{S}_{{p}} \:{and}\:{W}_{{p}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{cos}\left(\frac{{n}\pi}{\mathrm{3}}\right)\:{and}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{sin}\left(\frac{{n}\pi}{\mathrm{3}}\right) \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{cos}\left(\frac{{n}\pi}{\mathrm{5}}\right)\:{and}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{sin}\left(\frac{{n}\pi}{\mathrm{5}}\right) \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\:{A}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{cos}^{\mathrm{2}} \left(\frac{{n}\pi}{\mathrm{3}}\right)\:{and}\:{B}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{sin}^{\mathrm{2}} \left(\frac{{n}\pi}{\mathrm{3}}\right)\:. \\ $$ | ||
Commented by maxmathsup by imad last updated on 23/Aug/18 | ||
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{S}_{{p}} \:+{iW}_{{p}} =\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{{i}\frac{{n}\pi}{{p}}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left({e}^{\frac{{i}\pi}{{p}}} \right)^{{n}} \:\:=\frac{\mathrm{1}}{\mathrm{1}−{e}^{{i}\frac{\pi}{{p}}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{cos}\left(\frac{\pi}{{p}}\right)−{isin}\left(\frac{\pi}{{p}}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}{p}}\right)\:−\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{2}{p}}\right){cos}\left(\frac{\pi}{\mathrm{2}{p}}\right)} \\ $$$$=\:\frac{\mathrm{1}}{−\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{2}{p}}\right)\left({cos}\left(\frac{\pi}{\mathrm{2}{p}}\right)+{isin}\left(\frac{\pi}{\mathrm{2}{p}}\right)\right)}\:\:=\frac{{i}}{\mathrm{2}\:{sin}\left(\frac{\pi}{\mathrm{2}{p}}\right)}\:{e}^{−\frac{{i}\pi}{\mathrm{2}{p}}} \\ $$$$=\frac{{i}}{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{2}{p}}\right)}\left\{{cos}\left(\frac{\pi}{\mathrm{2}{p}}\right)−{i}\:{sin}\left(\frac{\pi}{\mathrm{2}{p}}\right)\right\}\:=\frac{{i}}{\mathrm{2}}{cotan}\left(\frac{\pi}{\mathrm{2}{p}}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${S}_{{p}} =\frac{\mathrm{1}}{\mathrm{2}}\:\:{and}\:{W}_{{p}} =\frac{\mathrm{1}}{\mathrm{2}}{cotan}\left(\frac{\pi}{\mathrm{2}{p}}\right) \\ $$$$\left.\mathrm{2}\right)\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{cos}\left(\frac{{n}\pi}{\mathrm{3}}\right)\:={S}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\:\:{and}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{sin}\left(\frac{{n}\pi}{\mathrm{3}}\right)\:={W}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}{cotan}\left(\frac{\pi}{\mathrm{6}}\right) \\ $$$${W}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\:\sqrt{\mathrm{3}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\left.\mathrm{3}\right)\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{cos}\left(\frac{{n}\pi}{\mathrm{5}}\right)\:={S}_{\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{2}}\:\:{and}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{sin}\left(\frac{{n}\pi}{\mathrm{5}}\right)\:={W}_{\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{2}}{cotan}\left(\frac{\pi}{\mathrm{10}}\right) \\ $$ | ||
Commented by maxmathsup by imad last updated on 23/Aug/18 | ||
$$\left.\mathrm{4}\right)\:{the}\:{Q}\:{is}\:{claculate}\:{A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{cos}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{3}}\right)\:{and}\:{B}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{3}}\right). \\ $$ | ||