Menu Close

let-p-x-x-i-3-n-x-i-3-n-with-x-real-1-simlify-p-x-2-find-the-roots-of-P-x-3-decompose-inside-C-x-p-x-4-calculate-0-1-p-x-dx-




Question Number 61651 by maxmathsup by imad last updated on 05/Jun/19
let p(x) =(x+i(√3))^n +(x−i(√3))^n     with x real  1) simlify p(x)  2) find the roots of P(x)  3)decompose inside C[x]  p(x)  4) calculate ∫_0 ^1 p(x)dx
letp(x)=(x+i3)n+(xi3)nwithxreal1)simlifyp(x)2)findtherootsofP(x)3)decomposeinsideC[x]p(x)4)calculate01p(x)dx
Commented by maxmathsup by imad last updated on 06/Jun/19
1)  we have P(x) =Σ_(k=0) ^n  C_n ^k (i(√3))^k  x^(n−k)  +Σ_(k=0) ^n  C_n ^k (−i(√3))^k  x^(n−k)   =Σ_(k=0) ^n  C_n ^k { i^k  +(−i)^k )3^(k/2)  x^(n−k)  =Σ_(k=2p) (...) +Σ_(k=2p+1) (...)  =2Σ_(p=0) ^([(n/2)])   C_n ^(2p)   (−1)^p  3^p  x^(n−2p)  = 2 Σ_(p=0) ^([(n/2)]) (−3)^p  C_n ^(2p)  x^(n−2p)   another way with arctan  we have P(x) = 2 Re((x+i(√3))^n )   but    ∣x+i(√3)∣ =(√(x^2  +3)) ⇒  x+i(√3)=(√(x^2  +3))((x/( (√(x^2  +3)))) +i ((√3)/( (√(x^(2 ) +3))))) =r e^(iθ)  ⇒r =(√(x^2  +3))  and  tanθ =((√3)/x)  (   we suppose x≠0) ⇒ θ arctan(((√3)/x)) ⇒x+i(√3)=(√(x^2  +3)) e^(iarctan(((√3)/x)))  ⇒  (x+i(√3))^n  =(x^2  +3)^(n/2)   e^(in arctan(((√3)/x)))  ⇒P(x) =2(x^2  +3)^(n/2)  cos(narctan(((√3)/x)))
1)wehaveP(x)=k=0nCnk(i3)kxnk+k=0nCnk(i3)kxnk=k=0nCnk{ik+(i)k)3k2xnk=k=2p()+k=2p+1()=2p=0[n2]Cn2p(1)p3pxn2p=2p=0[n2](3)pCn2pxn2panotherwaywitharctanwehaveP(x)=2Re((x+i3)n)butx+i3=x2+3x+i3=x2+3(xx2+3+i3x2+3)=reiθr=x2+3andtanθ=3x(wesupposex0)θarctan(3x)x+i3=x2+3eiarctan(3x)(x+i3)n=(x2+3)n2einarctan(3x)P(x)=2(x2+3)n2cos(narctan(3x))
Commented by maxmathsup by imad last updated on 06/Jun/19
2) P(x)=0 ⇔ (x−i(√3))^n  =−(x+i(√3))^n  ⇔(((x−i(√3))/(x+i(√3))))^n  =−1 ⇔ Z^n  =−1 with  Z =((x−i(√3))/(x+i(√3))) ⇒Zx+i(√3)Z =x−i(√3) ⇒(Z−1)x =−i(√3)Z −i(√3) ⇒  (Z−1)x =−i(√3)(Z+1) ⇒x =i(√3)((1+Z)/(1−Z))  Z^n  =−1 ⇒ Z^n  =e^(i(2k+1)π)  ⇒Z_k =e^(i(((2k+1)π)/n))    and  k∈[[0,n−1]] ⇒the roots of P(x)  are x_k =i(√3)  ((1+e^((i(2k+1)π)/n) )/(1−e^((i(2k+1)π)/n) ))  let skmplify  x_k   x_k =i(√3)((1+cos((2k+1)(π/n))+2i sin((2k+1)(π/(2n)))cos((2k+1)(π/(2n))))/(1−cos((2k+1)(π/n))−2i sin(2k+1)(π/(2n)))cos((2k+1)(π/(2n)))))  =i(√3)((2cos^2 ((2k+1)(π/(2n))) +2i sin((2k+1)(π/(2n)))cos((2k+1)(π/(2n))))/(2sin^2 (l2k+1)(π/(2n))−2i sin(2k+1)(π/(2n))) cos(2k+1)(π/(2n))))  =i(√3)((cos(2k+1)(π/(2n))(e^(i(2k+1)(π/(2n))) ))/(−isin(2k+1)(π/(2n))( e^(i(2k+1)(π/(2n))) ))) =−(√3)cotan((2k+1)(π/(2n)))   k∈[[0,n−1]]  ×
2)P(x)=0(xi3)n=(x+i3)n(xi3x+i3)n=1Zn=1withZ=xi3x+i3Zx+i3Z=xi3(Z1)x=i3Zi3(Z1)x=i3(Z+1)x=i31+Z1ZZn=1Zn=ei(2k+1)πZk=ei(2k+1)πnandk[[0,n1]]therootsofP(x)arexk=i31+ei(2k+1)πn1ei(2k+1)πnletskmplifyxkxk=i31+cos((2k+1)πn)+2isin((2k+1)π2n)cos((2k+1)π2n)1cos((2k+1)πn)2isin(2k+1)π2n)cos((2k+1)π2n)=i32cos2((2k+1)π2n)+2isin((2k+1)π2n)cos((2k+1)π2n)2sin2(l2k+1)π2n2isin(2k+1)π2n)cos(2k+1)π2n=i3cos(2k+1)π2n(ei(2k+1)π2n)isin(2k+1)π2n(ei(2k+1)π2n)=3cotan((2k+1)π2n)k[[0,n1]]×
Commented by maxmathsup by imad last updated on 06/Jun/19
3) P(x) =λΠ_(k=0) ^(n−1) (x−Z_k ) =λ Π_(k=0) ^(n−1) (x+(√3)cotan((2k+1)(π/(2n))))  λ is thedominent coefficient  of P(x).
3)P(x)=λk=0n1(xZk)=λk=0n1(x+3cotan((2k+1)π2n))λisthedominentcoefficientofP(x).
Commented by maxmathsup by imad last updated on 06/Jun/19
4) we have P(x) =2 Σ_(p=0) ^([(n/2)]) (−3)^p  C_n ^(2p)  x^(n−2p)  ⇒  ∫_0 ^1   P(x)dx =2 Σ_(p=0) ^([(n/2)]) (−3)^p  C_n ^(2p)    [(1/(n−2p +1))x^(n−2p+1) ]_0 ^1   = 2 Σ_(p=0) ^([(n/2)])    (((−3)^p  C_n ^(2p) )/(n−2p+1)) .
4)wehaveP(x)=2p=0[n2](3)pCn2pxn2p01P(x)dx=2p=0[n2](3)pCn2p[1n2p+1xn2p+1]01=2p=0[n2](3)pCn2pn2p+1.

Leave a Reply

Your email address will not be published. Required fields are marked *