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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 )) .
letAn=k=0n1sin(kπ2n)andBn=k=0n1cos(kπ2n)1)findAnandBnintermsofn2)calculatelimn+AnBn3)calculatelimn+Anlimn+Bn.
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 .
wehaveBn+iAn=k=0n1eikπ2n=k=0n1(eiπ2n)k=1eiπ21eiπ2n=1i1cos(π2n)isin(π2n)=1i2sin2(π4n)2isin(π4n)cos(π4n)=1i2isin(π4n)(cos(π4n)+isin(π4n))=i+12sin(π4n)eiπ4n=1+i2sin(π4n){cos(π4n)isin(π4n)}=cos(π4n)isin(π4n)+icos(π4n)+sin(π4n)2sin(π4n)Bn=cos(π4n)+sin(π4n)2sin(π4n)=12tan(π4n)+12An=cos(π4n)sin(π4n)2sin(π4n)=12tan(π4n)122)wehaveAnBn=cos(π4n)sin(π4n)cos(π4n)+sin(π4n)1π232n2π4n1π232n2+π4n1(n+)limn+AnBn=1.

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