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Question Number 125760 by snipers237 last updated on 13/Dec/20

Let n≥1 and integer, P_n (X)=(1+X)^n −(1−X)^n    1) Factorize P_n   2)Deduce  S_n =Π_(k=1) ^n [4+cotan^2 (((kπ)/(2n+1))) ]

Letn1andinteger,Pn(X)=(1+X)n(1X)n1)FactorizePn2)DeduceSn=nk=1[4+cotan2(kπ2n+1)]

Answered by mathmax by abdo last updated on 13/Dec/20

1) P_n (x)=0 ⇔(((1−x)/(1+x)))^n  =1 =e^(i2kπ)  ⇒((1−x)/(1+x))=e^((i2kπ)/n)  ⇒  1−x=e^((i2kπ)/n) +e^((i2kπ)/n) x ⇒(1+e^((i2kπ)/n) )x=1−e^((i2kπ)/n)  ⇒so the roots are  x_k =((1−e^((i2kπ)/n) )/(1+e^((i2kπ)/n) )) =((1−cos(((2kπ)/n))−isin(((2kπ)/n)))/(1+cos(((2kπ)/n))+isin(((2kπ)/n))))  =((2sin^2 (((kπ)/n))−2isin(((kπ)/n))cos(((kπ)/n)))/(2cos^2 (((kπ)/n))+2isin(((kπ)/n))cos(((kπ)/n))))=((−isin(((kπ)/n))e^((ikπ)/n) )/(cos(((kπ)/n))e^((ikπ)/n) )) =−itan(((kπ)/n))  k∈[[o,n−1]] ⇒P_n (x)=λ Π_(k=0) ^(n−1) (x+itan(((kπ)/n))) let find λ  we have P_n (x)=(x+1)^n −(−1)^n (x−1)^n   =Σ_(k=0) ^n  C_n ^k  x^k  −(−1)^n  Σ_(k=0) ^n  C_n ^k  x^k (−1)^(n−k)   =Σ_(k=0) ^n  C_n ^k  x^k −Σ_(k0) ^n  C_n ^k (−1)^k  x^k   =Σ_(k=0) ^n C_n ^k (1−(−1)^k )x^k  =2Σ_(p=0) ^([((n−1)/2)])  C_n ^(2p+1)  x^(2p+1)   ⇒λ =2 C_n ^(2[((n−1)/2)]+1) ⇒P_n (x)=2 C_n ^(2[((n−1)/2)]+1)  Π_(k=0) ^(n−1) (x+itan(((kπ)/n)))  =2x C_n ^(2[((n−1)/2)]+1)  Π_(k=1) ^(n−1) (x+itan(((kπ)/n)))  P_n (2).P_n (−2) =−16λ_n ^2  Π_(k=1) ^n (2+itan(((kπ)/n)))Π_(k=1) ^n (−2+itan(((kπ)/n)))  =−16λ_n ^2 (−1)^n  Π_(k=1) ^n (2+itan(((kπ)/n)))(2−itan(((kπ)/n)))  =16λ_n ^2  (−1)^(n+1)  Π_(k=1) ^n (4+tan^2 (((kπ)/n)))...be continued...

1)Pn(x)=0(1x1+x)n=1=ei2kπ1x1+x=ei2kπn1x=ei2kπn+ei2kπnx(1+ei2kπn)x=1ei2kπnsotherootsarexk=1ei2kπn1+ei2kπn=1cos(2kπn)isin(2kπn)1+cos(2kπn)+isin(2kπn)=2sin2(kπn)2isin(kπn)cos(kπn)2cos2(kπn)+2isin(kπn)cos(kπn)=isin(kπn)eikπncos(kπn)eikπn=itan(kπn)k[[o,n1]]Pn(x)=λk=0n1(x+itan(kπn))letfindλwehavePn(x)=(x+1)n(1)n(x1)n=k=0nCnkxk(1)nk=0nCnkxk(1)nk=k=0nCnkxkk0nCnk(1)kxk=k=0nCnk(1(1)k)xk=2p=0[n12]Cn2p+1x2p+1λ=2Cn2[n12]+1Pn(x)=2Cn2[n12]+1k=0n1(x+itan(kπn))=2xCn2[n12]+1k=1n1(x+itan(kπn))Pn(2).Pn(2)=16λn2k=1n(2+itan(kπn))k=1n(2+itan(kπn))=16λn2(1)nk=1n(2+itan(kπn))(2itan(kπn))=16λn2(1)n+1k=1n(4+tan2(kπn))...becontinued...

Answered by mindispower last updated on 13/Dec/20

P_n (X)=0  ⇒(((1+x)/(1−x)))^n =1  ((1+x)/(1−x))=e^(2ikπ/n)   k∈[0,n[  x=((1−e^((2ikπ)/n) )/(1+e^((2ikπ)/n) ))=−((e^((ikπ)/n) −e^(−((ikπ)/n)) )/(e^((ikπ)/n) +e^(−i((kπ)/n)) ))=−itg(((kπ)/n))  k∈[0,n−1]  if n=2m,k≠m  P_n (x)=aΠ_(k∈[0,n−1]) (X+itg(((kπ)/n)))if n=2m+1  =Π_(k∈[0,n−1]≠(n/2)) (X+itg(((kπ)/n))),n=2m  one easy way too see this  if n=2m,degp_(2m) ≤2m−1  number of roots≤2m−1  2)  P_(2n+1) =2Π_(0≤k≤2n) (X+itg(((kπ)/(2n+1))))  P_(2n+1) =2x.Π_(k=1) ^n (X+itg(((kπ)/(2n+1))).Π_(k=n+1) ^(2n) (X+itg(((kπ)/(2n+1))))  =2xΠ_(k=1) ^n (X+itg(((kπ)/(2n+1))))Π_(k=1) ^n (X+itg((((2n+1−k)π)/(2n+1))))  =2xΠ_(k=1) ^n (X+itg(((kπ)/(2n+1))))Π_(k=1) ^n (X+itg(π−((kπ)/(2n+1))))  =2xΠ_(k=1) ^n (X+itg(((kπ)/(2n+1))))Π_(k=1) ^n (X−itg(((kπ)/(2n+1))))  =2xΠ_(k=1) ^n (X^2 +tg^2 (((kπ)/(2n+1))))=p_(2n+1) (x)  x=(1/2)  ⇒p_(2n+1) ((1/2))=Π_(k=1) ^n ((1/4)+(1/(cotan^2 (((kπ)/(2n+1))))))..E  =Π_(k=1) ^n (1/(4cotan^2 (((kπ)/(2n+1)))=T)) .Π_(k=1) ^n (4+cotan^2 (((kπ)/(2n+1)))_(=S_n )   T_n =(1/4^n )Π_(k=1) ^n tg^2 (((kπ)/(2n+1)))    2xΠ_(k=1) ^n (X^2 +tg^2 (((kπ)/(2n+1))))=p_(2n+1) (x)  ⇒Π_(k=1) ^n (X^2 +tg^2 (((kπ)/(2n+1))))=((P_(2n+1) (x))/(2x))  ⇒lim_(x→0) Π(X^2 +tg^2 (((kπ)/(2n+1)))=(1/2).lim_(x→0) ((P_(2n+1) (x))/x)  =(1/2)P_(2n+1) ^′ (0)=(1/2)((2n+1)+(2n+1))=2n+1  ⇒Π_(k=1) ^n tg^2 (((kπ)/(2n+1)))=2n+1  T_n =(1/(4^n (2n+1)))    E⇔P_(2n+1) ((1/2))=(1/(4^n (2n+1))).S_n =((3/2))^n −(1/2^n )  S_n =(2n+1)(6^n −2^n )

Pn(X)=0(1+x1x)n=11+x1x=e2ikπ/nk[0,n[x=1e2ikπn1+e2ikπn=eikπneikπneikπn+eikπn=itg(kπn)k[0,n1]ifn=2m,kmPn(x)=ak[0,n1](X+itg(kπn))ifn=2m+1=k[0,n1]n2(X+itg(kπn)),n=2moneeasywaytooseethisifn=2m,degp2m2m1numberofroots2m12)P2n+1=20k2n(X+itg(kπ2n+1))P2n+1=2x.nk=1(X+itg(kπ2n+1).2nk=n+1(X+itg(kπ2n+1))=2xnk=1(X+itg(kπ2n+1))nk=1(X+itg((2n+1k)π2n+1))=2xnk=1(X+itg(kπ2n+1))nk=1(X+itg(πkπ2n+1))=2xnk=1(X+itg(kπ2n+1))nk=1(Xitg(kπ2n+1))=2xnk=1(X2+tg2(kπ2n+1))=p2n+1(x)x=12p2n+1(12)=nk=1(14+1cotan2(kπ2n+1))..EMissing \left or extra \rightTn=14nnk=1tg2(kπ2n+1)2xnk=1(X2+tg2(kπ2n+1))=p2n+1(x)nk=1(X2+tg2(kπ2n+1))=P2n+1(x)2xlimx0Π(X2+tg2(kπ2n+1)=12.limx0P2n+1(x)x=12P2n+1(0)=12((2n+1)+(2n+1))=2n+1nk=1tg2(kπ2n+1)=2n+1Tn=14n(2n+1)EP2n+1(12)=14n(2n+1).Sn=(32)n12nSn=(2n+1)(6n2n)

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