Menu Close

let-A-n-k-0-n-1-sin-kpi-2n-and-B-n-k-0-n-1-cos-kpi-2n-1-find-A-n-and-B-n-interms-of-n-2-calculate-lim-n-A-n-B-n-3-calculate-lim-n-A-n-lim-n-




Question Number 47857 by maxmathsup by imad last updated on 15/Nov/18
let A_n =Σ_(k=0) ^(n−1)  sin(((kπ)/(2n))) and B_n =Σ_(k=0) ^(n−1)  cos(((kπ)/(2n)))  1) find A_n  and B_n  interms of n  2)calculate lim_(n→+∞)   (A_n /B_n )  3)calculate ((lim_(n→+∞) A_n )/(lim_(n→+∞  ) B_n )) .
$${let}\:{A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)\:{and}\:{B}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{cos}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right) \\ $$$$\left.\mathrm{1}\right)\:{find}\:{A}_{{n}} \:{and}\:{B}_{{n}} \:{interms}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right){calculate}\:{lim}_{{n}\rightarrow+\infty} \:\:\frac{{A}_{{n}} }{{B}_{{n}} } \\ $$$$\left.\mathrm{3}\right){calculate}\:\frac{{lim}_{{n}\rightarrow+\infty} {A}_{{n}} }{{lim}_{{n}\rightarrow+\infty\:\:} {B}_{{n}} }\:. \\ $$
Commented by maxmathsup by imad last updated on 17/Nov/18
we have B_n +iA_n =Σ_(k=0) ^(n−1)  e^(i((kπ)/(2n)))  =Σ_(k=0) ^(n−1)  (e^(i(π/(2n))) )^k  = ((1−e^((iπ)/2) )/(1−e^((iπ)/(2n)) ))  =((1−i)/(1−cos((π/(2n)))−isin((π/(2n))))) =((1−i)/(2sin^2 ((π/(4n)))−2isin((π/(4n)))cos((π/(4n)))))  =((1−i)/(−2i sin((π/(4n)))(cos((π/(4n)))+i sin((π/(4n))))))  =((i+1)/(2sin((π/(4n))))) e^(−((iπ)/(4n)))  =((1+i)/(2sin((π/(4n))))){cos((π/(4n)))−i sin((π/(4n)))}  =((cos((π/(4n)))−i sin((π/(4n))) +i cos((π/(4n))) +sin((π/(4n))))/(2sin((π/(4n))))) ⇒  B_n =((cos((π/(4n)))+sin((π/(4n))))/(2sin((π/(4n))))) =(1/(2tan((π/(4n))))) +(1/2)  A_n =((cos((π/(4n)))−sin((π/(4n))))/(2sin((π/(4n))))) = (1/(2 tan((π/(4n))))) −(1/2)  2) we have (A_n /B_n ) =((cos((π/(4n)))−sin((π/(4n))))/(cos((π/(4n)))+sin((π/(4n))))) ∼((1−(π^2 /(32n^2 ))−(π/(4n)))/(1−(π^2 /(32n^2 ))+(π/(4n)))) →1 (n→+∞) ⇒  lim_(n→+∞)    (A_n /B_n ) =1 .
$${we}\:{have}\:{B}_{{n}} +{iA}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{{i}\frac{{k}\pi}{\mathrm{2}{n}}} \:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({e}^{{i}\frac{\pi}{\mathrm{2}{n}}} \right)^{{k}} \:=\:\frac{\mathrm{1}−{e}^{\frac{{i}\pi}{\mathrm{2}}} }{\mathrm{1}−{e}^{\frac{{i}\pi}{\mathrm{2}{n}}} } \\ $$$$=\frac{\mathrm{1}−{i}}{\mathrm{1}−{cos}\left(\frac{\pi}{\mathrm{2}{n}}\right)−{isin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\:=\frac{\mathrm{1}−{i}}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}{n}}\right)−\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{4}{n}}\right){cos}\left(\frac{\pi}{\mathrm{4}{n}}\right)} \\ $$$$=\frac{\mathrm{1}−{i}}{−\mathrm{2}{i}\:{sin}\left(\frac{\pi}{\mathrm{4}{n}}\right)\left({cos}\left(\frac{\pi}{\mathrm{4}{n}}\right)+{i}\:{sin}\left(\frac{\pi}{\mathrm{4}{n}}\right)\right)} \\ $$$$=\frac{{i}+\mathrm{1}}{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{4}{n}}\right)}\:{e}^{−\frac{{i}\pi}{\mathrm{4}{n}}} \:=\frac{\mathrm{1}+{i}}{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{4}{n}}\right)}\left\{{cos}\left(\frac{\pi}{\mathrm{4}{n}}\right)−{i}\:{sin}\left(\frac{\pi}{\mathrm{4}{n}}\right)\right\} \\ $$$$=\frac{{cos}\left(\frac{\pi}{\mathrm{4}{n}}\right)−{i}\:{sin}\left(\frac{\pi}{\mathrm{4}{n}}\right)\:+{i}\:{cos}\left(\frac{\pi}{\mathrm{4}{n}}\right)\:+{sin}\left(\frac{\pi}{\mathrm{4}{n}}\right)}{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{4}{n}}\right)}\:\Rightarrow \\ $$$${B}_{{n}} =\frac{{cos}\left(\frac{\pi}{\mathrm{4}{n}}\right)+{sin}\left(\frac{\pi}{\mathrm{4}{n}}\right)}{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{4}{n}}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}{tan}\left(\frac{\pi}{\mathrm{4}{n}}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${A}_{{n}} =\frac{{cos}\left(\frac{\pi}{\mathrm{4}{n}}\right)−{sin}\left(\frac{\pi}{\mathrm{4}{n}}\right)}{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{4}{n}}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}\:{tan}\left(\frac{\pi}{\mathrm{4}{n}}\right)}\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:\frac{{A}_{{n}} }{{B}_{{n}} }\:=\frac{{cos}\left(\frac{\pi}{\mathrm{4}{n}}\right)−{sin}\left(\frac{\pi}{\mathrm{4}{n}}\right)}{{cos}\left(\frac{\pi}{\mathrm{4}{n}}\right)+{sin}\left(\frac{\pi}{\mathrm{4}{n}}\right)}\:\sim\frac{\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{32}{n}^{\mathrm{2}} }−\frac{\pi}{\mathrm{4}{n}}}{\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{32}{n}^{\mathrm{2}} }+\frac{\pi}{\mathrm{4}{n}}}\:\rightarrow\mathrm{1}\:\left({n}\rightarrow+\infty\right)\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:\:\:\frac{{A}_{{n}} }{{B}_{{n}} }\:=\mathrm{1}\:. \\ $$

Leave a Reply

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