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Question Number 125760 by snipers237 last updated on 13/Dec/20
Letn⩾1andinteger,Pn(X)=(1+X)n−(1−X)n1)FactorizePn2)DeduceSn=∏nk=1[4+cotan2(kπ2n+1)]
Answered by mathmax by abdo last updated on 13/Dec/20
1)Pn(x)=0⇔(1−x1+x)n=1=ei2kπ⇒1−x1+x=ei2kπn⇒1−x=ei2kπn+ei2kπnx⇒(1+ei2kπn)x=1−ei2kπn⇒sotherootsarexk=1−ei2kπn1+ei2kπn=1−cos(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,n−1]]⇒Pn(x)=λ∏k=0n−1(x+itan(kπn))letfindλwehavePn(x)=(x+1)n−(−1)n(x−1)n=∑k=0nCnkxk−(−1)n∑k=0nCnkxk(−1)n−k=∑k=0nCnkxk−∑k0nCnk(−1)kxk=∑k=0nCnk(1−(−1)k)xk=2∑p=0[n−12]Cn2p+1x2p+1⇒λ=2Cn2[n−12]+1⇒Pn(x)=2Cn2[n−12]+1∏k=0n−1(x+itan(kπn))=2xCn2[n−12]+1∏k=1n−1(x+itan(kπn))Pn(2).Pn(−2)=−16λn2∏k=1n(2+itan(kπn))∏k=1n(−2+itan(kπn))=−16λn2(−1)n∏k=1n(2+itan(kπn))(2−itan(kπn))=16λn2(−1)n+1∏k=1n(4+tan2(kπn))...becontinued...
Answered by mindispower last updated on 13/Dec/20
Pn(X)=0⇒(1+x1−x)n=11+x1−x=e2ikπ/nk∈[0,n[x=1−e2ikπn1+e2ikπn=−eikπn−e−ikπneikπn+e−ikπn=−itg(kπn)k∈[0,n−1]ifn=2m,k≠mPn(x)=a∏k∈[0,n−1](X+itg(kπn))ifn=2m+1=∏k∈[0,n−1]≠n2(X+itg(kπn)),n=2moneeasywaytooseethisifn=2m,degp2m⩽2m−1numberofroots⩽2m−12)P2n+1=2∏0⩽k⩽2n(X+itg(kπ2n+1))P2n+1=2x.∏nk=1(X+itg(kπ2n+1).∏2nk=n+1(X+itg(kπ2n+1))=2x∏nk=1(X+itg(kπ2n+1))∏nk=1(X+itg((2n+1−k)π2n+1))=2x∏nk=1(X+itg(kπ2n+1))∏nk=1(X+itg(π−kπ2n+1))=2x∏nk=1(X+itg(kπ2n+1))∏nk=1(X−itg(kπ2n+1))=2x∏nk=1(X2+tg2(kπ2n+1))=p2n+1(x)x=12⇒p2n+1(12)=∏nk=1(14+1cotan2(kπ2n+1))..EMissing \left or extra \rightMissing \left or extra \rightTn=14n∏nk=1tg2(kπ2n+1)2x∏nk=1(X2+tg2(kπ2n+1))=p2n+1(x)⇒∏nk=1(X2+tg2(kπ2n+1))=P2n+1(x)2x⇒limx→0Π(X2+tg2(kπ2n+1)=12.limx→0P2n+1(x)x=12P2n+1′(0)=12((2n+1)+(2n+1))=2n+1⇒∏nk=1tg2(kπ2n+1)=2n+1Tn=14n(2n+1)E⇔P2n+1(12)=14n(2n+1).Sn=(32)n−12nSn=(2n+1)(6n−2n)
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