Menu Close

let-S-p-n-0-cos-npi-p-and-W-p-n-0-sin-npi-p-with-p-natural-integr-not0-1-find-a-simple-form-of-S-p-and-W-p-2-find-the-value-of-n-0-cos-npi-3-and-n-0-

Tinku Tara 0 Comments
Pin




Question Number 42285 by maxmathsup by imad last updated on 22/Aug/18
let S_p =Σ_(n=0) ^∞ cos(((nπ)/p)) and W_p =Σ_(n=0) ^∞ sin(((nπ)/p)) with p natural integr not0 1) find a simple form of S_p and W_p 2) find the value of Σ_(n=0) ^∞ cos(((nπ)/3)) and Σ_(n=0) ^∞ sin(((nπ)/3)) 3) find the value of Σ_(n=0) ^∞ cos(((nπ)/5)) and Σ_(n=0) ^∞ sin(((nπ)/5)) 4) calculate A =Σ_(n=0) ^∞ cos^2 (((nπ)/3)) and B =Σ_(n=0) ^∞ sin^2 (((nπ)/3)) .
$${let}\:\:{S}_{{p}} =\sum_{{n}=\mathrm{0}} ^{\infty} \:\:{cos}\left(\frac{{n}\pi}{{p}}\right)\:\:{and}\:\:{W}_{{p}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{sin}\left(\frac{{n}\pi}{{p}}\right)\:{with}\:{p}\:{natural}\:{integr}\:{not}\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{S}_{{p}} \:{and}\:{W}_{{p}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{cos}\left(\frac{{n}\pi}{\mathrm{3}}\right)\:{and}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{sin}\left(\frac{{n}\pi}{\mathrm{3}}\right) \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{cos}\left(\frac{{n}\pi}{\mathrm{5}}\right)\:{and}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{sin}\left(\frac{{n}\pi}{\mathrm{5}}\right) \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\:{A}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{cos}^{\mathrm{2}} \left(\frac{{n}\pi}{\mathrm{3}}\right)\:{and}\:{B}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{sin}^{\mathrm{2}} \left(\frac{{n}\pi}{\mathrm{3}}\right)\:. \\ $$
Commented by maxmathsup by imad last updated on 23/Aug/18
1) we have S_p +iW_p =Σ_(n=0) ^∞ e^(i((nπ)/p)) =Σ_(n=0) ^∞ (e^((iπ)/p) )^n =(1/(1−e^(i(π/p)) )) =(1/(1−cos((π/p))−isin((π/p)))) =(1/(2sin^2 ((π/(2p))) −2isin((π/(2p)))cos((π/(2p))))) = (1/(−2isin((π/(2p)))(cos((π/(2p)))+isin((π/(2p)))))) =(i/(2 sin((π/(2p))))) e^(−((iπ)/(2p))) =(i/(2sin((π/(2p))))){cos((π/(2p)))−i sin((π/(2p)))} =(i/2)cotan((π/(2p))) +(1/2) ⇒ S_p =(1/2) and W_p =(1/2)cotan((π/(2p))) 2) Σ_(n=0) ^∞ cos(((nπ)/3)) =S_3 =(1/2) and Σ_(n=0) ^∞ sin(((nπ)/3)) =W_3 =(1/2)cotan((π/6)) W_3 =(1/2) (√3)=((√3)/2) 3) Σ_(n=0) ^∞ cos(((nπ)/5)) =S_5 =(1/2) and Σ_(n=0) ^∞ sin(((nπ)/5)) =W_5 =(1/2)cotan((π/(10)))
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{S}_{{p}} \:+{iW}_{{p}} =\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{{i}\frac{{n}\pi}{{p}}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left({e}^{\frac{{i}\pi}{{p}}} \right)^{{n}} \:\:=\frac{\mathrm{1}}{\mathrm{1}−{e}^{{i}\frac{\pi}{{p}}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{cos}\left(\frac{\pi}{{p}}\right)−{isin}\left(\frac{\pi}{{p}}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}{p}}\right)\:−\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{2}{p}}\right){cos}\left(\frac{\pi}{\mathrm{2}{p}}\right)} \\ $$$$=\:\frac{\mathrm{1}}{−\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{2}{p}}\right)\left({cos}\left(\frac{\pi}{\mathrm{2}{p}}\right)+{isin}\left(\frac{\pi}{\mathrm{2}{p}}\right)\right)}\:\:=\frac{{i}}{\mathrm{2}\:{sin}\left(\frac{\pi}{\mathrm{2}{p}}\right)}\:{e}^{−\frac{{i}\pi}{\mathrm{2}{p}}} \\ $$$$=\frac{{i}}{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{2}{p}}\right)}\left\{{cos}\left(\frac{\pi}{\mathrm{2}{p}}\right)−{i}\:{sin}\left(\frac{\pi}{\mathrm{2}{p}}\right)\right\}\:=\frac{{i}}{\mathrm{2}}{cotan}\left(\frac{\pi}{\mathrm{2}{p}}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${S}_{{p}} =\frac{\mathrm{1}}{\mathrm{2}}\:\:{and}\:{W}_{{p}} =\frac{\mathrm{1}}{\mathrm{2}}{cotan}\left(\frac{\pi}{\mathrm{2}{p}}\right) \\ $$$$\left.\mathrm{2}\right)\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{cos}\left(\frac{{n}\pi}{\mathrm{3}}\right)\:={S}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\:\:{and}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{sin}\left(\frac{{n}\pi}{\mathrm{3}}\right)\:={W}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}{cotan}\left(\frac{\pi}{\mathrm{6}}\right) \\ $$$${W}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\:\sqrt{\mathrm{3}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\left.\mathrm{3}\right)\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{cos}\left(\frac{{n}\pi}{\mathrm{5}}\right)\:={S}_{\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{2}}\:\:{and}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{sin}\left(\frac{{n}\pi}{\mathrm{5}}\right)\:={W}_{\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{2}}{cotan}\left(\frac{\pi}{\mathrm{10}}\right) \\ $$
Commented by maxmathsup by imad last updated on 23/Aug/18
4) the Q is claculate A_n =Σ_(k=0) ^n cos^2 (((kπ)/3)) and B_n =Σ_(k=0) ^n sin^2 (((kπ)/3)).
$$\left.\mathrm{4}\right)\:{the}\:{Q}\:{is}\:{claculate}\:{A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{cos}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{3}}\right)\:{and}\:{B}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{3}}\right). \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *