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Question Number 31460 by abdo imad last updated on 08/Mar/18

find in terms of  n the value of  A_n = ∫_0 ^1  Π_(k=1) ^(n−1) (x^2  −2xcos(((kπ)/n)) +1)dx   with n from N^★ .

findintermsofnthevalueofAn=01k=1n1(x22xcos(kπn)+1)dxwithnfromN.

Commented by abdo imad last updated on 10/Mar/18

let decompose inside C[x] p(x)=x^(2n)  −1 the roots of  p(x)are z_k =e^(i((kπ)/n))   with k∈[0,2n−1 ]] we have  z_0 =1 , z_1 =e^(i(π/n))   ,z_2 =e^(i((2π)/n))  ,...z_(n−1) =e^(i(((n−1)π)/n))  ,z_n =−1 ,  z_(n+1) = e^(i(((n+1)π)/n)) =  z_(n−1) ^−    , z_(n+2) =  z_(n−2) ^−  ,  z_(2n−1) =z_1 ^−  ⇒  p(x)=Π_(k=0) ^(2n−1)  (x−z_k )=(x^2 −1)Π_(k=1) ^(n−1) (x−z_k )(x−z_k ^− )  =(x^2 −1) Π_(k=1) ^(n−1)  (x^2  −2Re(z_k )x  +∣z_k ∣^2 )  =(x^2  −1) Π_(k=1) ^(n−1)  (x^2  −2cos(((kπ)/n))x +1) ⇒  Π_(k=1) ^(n−1)  (x^2  −2cos(((kπ)/n))x +1)= ((x^(2n)  −1)/(x^2  −1)) ⇒  A_n = ∫_0 ^1   ((x^(2n)  −1)/(x^2 −1))dx= ∫_0 ^1  (1+x^2  +x^4  +...+x^(2n−2) )dx  A_n  = [x +(1/3) x^3  +(1/5)x^5      +....+(1/(2n−1))x^(2n−1) ]_0 ^1   =1 +(1/3) +(1/5) +.....+(1/(2n−1)) let find A_n  interms of H_n   A_n =1+(1/2) +(1/3) +(1/5) +...+(1/(2n−1)) +(1/(2n)) −(1/2) −(1/4) −...−(1/(2n))  A_n =H_(2n)  −(1/2) H_(n )     with H_n =Σ_(k=1) ^n  (1/k)  .

letdecomposeinsideC[x]p(x)=x2n1therootsofp(x)arezk=eikπnwithk[0,2n1]]wehavez0=1,z1=eiπn,z2=ei2πn,...zn1=ei(n1)πn,zn=1,zn+1=ei(n+1)πn=zn1,zn+2=zn2,z2n1=z1p(x)=k=02n1(xzk)=(x21)k=1n1(xzk)(xzk)=(x21)k=1n1(x22Re(zk)x+zk2)=(x21)k=1n1(x22cos(kπn)x+1)k=1n1(x22cos(kπn)x+1)=x2n1x21An=01x2n1x21dx=01(1+x2+x4+...+x2n2)dxAn=[x+13x3+15x5+....+12n1x2n1]01=1+13+15+.....+12n1letfindAnintermsofHnAn=1+12+13+15+...+12n1+12n1214...12nAn=H2n12HnwithHn=k=1n1k.

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