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Question Number 194638 by Erico last updated on 12/Jul/23

Prove that ∀n∈IN^∗           Σ_(k=1) ^(2^n −1)  (1/(sin^2 (((kπ)/2^(n+1) ))))= ((2^(2n+1) −2)/3)  Give in terms of n   Σ_(k=1) ^(2^n −1)  (1/(sin^4 (((kπ)/2^(n+1) ))))

ProvethatnIN2n1k=11sin2(kπ2n+1)=22n+123Giveintermsofn2n1k=11sin4(kπ2n+1)

Commented by witcher3 last updated on 12/Jul/23

(1/(sin^4 (x)))=(((sin^2 (x)+cos^2 (x))^2 )/(sin^4 (x)))=(1+(2/(tg^2 (x)))+(1/(tg^4 (x))))  Σ(1/(tg^2 (x)))..known By 1st Quation  Σtg^4 (x)...withe same methode Polynomial roots  p_4 =Σx_i ^4 ...use newtoon identity

1sin4(x)=(sin2(x)+cos2(x))2sin4(x)=(1+2tg2(x)+1tg4(x))Σ1tg2(x)..knownBy1stQuationΣtg4(x)...withesamemethodePolynomialrootsp4=Σxi4...usenewtoonidentity

Answered by witcher3 last updated on 12/Jul/23

Σ_(k=1) ^(2^n −1) (1/(sin^2 (((kπ)/2^(n+1) ))))=Σ_(k=1) ^(2^n −1) (1+(1/(tg^2 (((kπ)/2^(n+1) ))))),k→2^n −k  =Σ_(k=1) ^(2^n −1) (1+tg^2 (((kπ)/2^(n+1) ))),⇔Σ_(k=1) ^2^(n−1)  tg^2 (((kπ)/2^(n+1) ))=Σ_(k=1) ^(2^n −1) (1/(tg^2 (((kπ)/2^(n+1) ))))...I  Let   S=Σ_(k=1,k≠2^n ) ^(2^(n+1) −1) tg^2 (((kπ)/2^(n+1) ))=Σ_(k=1) ^(2^n −1) tg^2 (((kπ)/2^(n+1) ))+Σ_(2^n +1) ^(2^n −1) tg^2 (((kπ)/2^(n+1) ))  for the 2nd Sum k→2^n +k  ⇒S=Σ_(k=1) ^(2^n −1) tg^2 (((kπ)/2^(n+1) ))+tg^2 (((2^n +k)/2^(n+1) ))=Σ_(k=1) ^(2^n −1) tg^2 (((kπ)/2^(n+1) ))+(1/(tg^2 (((kπ)/2^(n+1) ))))  ⇒S=2Σ_(k=1) ^(2^n −1) tg^2 (((kπ)/2^(n+1) ))...A=[1,2^(n+1) −1]−{2^n }  let′s  find polynonial withe Roots tg(((kπ)/2^(n+1) )),k∈A  tg(2^(n+1) x)=0⇔2^(n+1) x=kπ⇔x=((kπ)/2^(n+1) ),k∈A∪{2^n ,0}  if we tack,((tg(2^(n+1) x))/(tg(x)))=0⇒x=((kπ)/2^(n+1) ),k∈A  we Know  tg(mx)=((Σ_(k=0) ^([((m−1)/2)]) (−1)^k  ((m),((2k+1)) )tg^(2k+1) (x))/(Σ_(k=0) ^([(m/2)]) (−1)^k  ((m),((2k)) ).tg^(2k) (x)_ ))  ⇒((tg(2^(n+1) x))/(tg(x)))=0⇔Σ_(k=0) ^(2^n −1) (−1)^k  ((2^(n+1) ),((2k+1)) )tg^(2k) (x)=p(tg(x))  Σ_(k=0) ^(2^n −1) (−1)^k  ((2^(n+1) ),((2k+1)) )t^(2k) =p(t),deg(p)=2^(n+1) −2  and root of P.are tan(((kπ)/2^(n+1) )),k∈A card (A)=2^(n+1) −2  so we have all root of P  if f(x)=Σ_(k=0) ^n a_k x^k   ;x_i root of P  Σx_i =−(a_(n−1) /a_n ),Σ_(1≤i<j≤n) x_i x_j =(a_(n−2) /a_n )  Σ_(k∈A) tg(((kπ)/2^(n+1) ))=0  Σ_(k_1 <k_2 ,(k_1 ,k_2 )∈A^2 ) tg(((k_1 π)/2^(n+1) ))tg(((k_2 π)/2^(n+1) ))=−( ((2^(n+1) ),((2^(n+1) −3)) )/ ((2^(n+1) ),((2^(n+1) −1)) ))  =−(1/6)(2^(n+1) −1)(2^(n+1) −2)  Σx_i ^2 =(Σx_i )^2 −2Σx_i x_j =(1/3)(2^(n+1) −1)(2^(n+1) −2)  S=2Σ_(k=1) ^(2^n −1) tg^2 (((kπ)/2^(n+1) ))  Σ_(k=1) ^(2^n −1) tg^2 (((kπ)/2^(n+1) ))=(1/6)(2^(2n+2) −3.2^(n+1) +2)=((2^(2n+1) −3.2^n +1)/3)  S_1 =Σ_(k=1) ^(2^n −1) (1/(sin^2 (((kπ)/2^(n+1) ))))=Σ_(k=1) ^(2^n −1) (1+(1/(tg(((kπ)/2^(n+1) )))))  =2^n −1+Σtg^2 (((kπ)/2^(m+1) ))=2^n −1+((2^(2n+1) −3.2^n +1)/3)  ⇔Σ_(k=1) ^(2^n −1) (1/(sin^2 (((kπ)/2^(n+1) ))))=((2^(2n+1) −2)/3)

2n1k=11sin2(kπ2n+1)=2n1k=1(1+1tg2(kπ2n+1)),k2nk=2n1k=1(1+tg2(kπ2n+1)),2n1k=1tg2(kπ2n+1)=2n1k=11tg2(kπ2n+1)...ILetS=2n+11k=1,k2ntg2(kπ2n+1)=2n1k=1tg2(kπ2n+1)+2n12n+1tg2(kπ2n+1)forthe2ndSumk2n+kS=2n1k=1tg2(kπ2n+1)+tg2(2n+k2n+1)=2n1k=1tg2(kπ2n+1)+1tg2(kπ2n+1)S=22n1k=1tg2(kπ2n+1)...A=[1,2n+11]{2n}letsfindpolynonialwitheRootstg(kπ2n+1),kAtg(2n+1x)=02n+1x=kπx=kπ2n+1,kA{2n,0}ifwetack,tg(2n+1x)tg(x)=0x=kπ2n+1,kAweKnowtg(mx)=[m12]k=0(1)k(m2k+1)tg2k+1(x)[m2]k=0(1)k(m2k).tg2k(x)tg(2n+1x)tg(x)=02n1k=0(1)k(2n+12k+1)tg2k(x)=p(tg(x))2n1k=0(1)k(2n+12k+1)t2k=p(t),deg(p)=2n+12androotofP.aretan(kπ2n+1),kAcard(A)=2n+12sowehaveallrootofPiff(x)=nk=0akxk;xirootofPΣxi=an1an,1i<jnxixj=an2ankAtg(kπ2n+1)=0k1<k2,(k1,k2)A2tg(k1π2n+1)tg(k2π2n+1)=(2n+12n+13)(2n+12n+11)=16(2n+11)(2n+12)Σxi2=(Σxi)22Σxixj=13(2n+11)(2n+12)S=22n1k=1tg2(kπ2n+1)2n1k=1tg2(kπ2n+1)=16(22n+23.2n+1+2)=22n+13.2n+13S1=2n1k=11sin2(kπ2n+1)=2n1k=1(1+1tg(kπ2n+1))=2n1+Σtg2(kπ2m+1)=2n1+22n+13.2n+132n1k=11sin2(kπ2n+1)=22n+123

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