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Question Number 74502 by mathmax by abdo last updated on 25/Nov/19

let f(x)=e^(−nx) ln(1+x^2 )  1)determine f^((n)) (x) and f^((n)) (0)  2)developp f at integr serie   (n integr natural)

letf(x)=enxln(1+x2)1)determinef(n)(x)andf(n)(0)2)developpfatintegrserie(nintegrnatural)

Commented by mathmax by abdo last updated on 25/Nov/19

1) f^((p)) (x)=Σ_(k=0) ^p  C_p ^k (ln(1+x^2 ))^((k))  (e^(−nx) )^((p−k))   =ln(1+x^2 )(−n)^p  e^(−nx)  +Σ_(k=1) ^p  C_p ^k (ln(x^2  +1))^((k)) (−n)^(p−k)  e^(−nx)   we have ln^′ (x^2 +1) =((2x)/(x^2 +1)) =(1/(x+i))+(1/(x−i)) ⇒  (ln(x^2 +1))^((k)) =((1/(x+i))+(1/(x−i)))^((k−1))  =(((−1)^(k−1) (k−1)!)/((x+i)^k )) +(((−1)^(k−1) (k−1)!)/((x−i)^k ))  =(−1)^(k−1) (k−1)!((1/((x+i)^k ))+(1/((x−i)^k )))=(−1)^(k−1) (k−1)!×(((x+i)^k +(x−i)^k )/((x^2  +1)^k ))  f^((p)) (x)=(−n)^p  e^(−nx) ln(x^2  +1)  +Σ_(k=1) ^p  C_p ^k (−1)^(k−1) (k−1)!×(((x+i)^k  +(x−i)^k )/((x^2  +1)^k ))(−n)^(p−k)  e^(−nx)

1)f(p)(x)=k=0pCpk(ln(1+x2))(k)(enx)(pk)=ln(1+x2)(n)penx+k=1pCpk(ln(x2+1))(k)(n)pkenxwehaveln(x2+1)=2xx2+1=1x+i+1xi(ln(x2+1))(k)=(1x+i+1xi)(k1)=(1)k1(k1)!(x+i)k+(1)k1(k1)!(xi)k=(1)k1(k1)!(1(x+i)k+1(xi)k)=(1)k1(k1)!×(x+i)k+(xi)k(x2+1)kf(p)(x)=(n)penxln(x2+1)+k=1pCpk(1)k1(k1)!×(x+i)k+(xi)k(x2+1)k(n)pkenx

Commented by mathmax by abdo last updated on 25/Nov/19

f^((n)) (x)=(−n)^n  e^(−nx) ln(x^2 +1)  +Σ_(k=1) ^n  C_n ^k (−1)^(k−1) (k−1)!×(((x+i)^k  +(x−i)^k )/((x^2  +1)^k ))(−n)^(n−k)  e^(−nx)   and   f^((n)) (0) =Σ_(k=1) ^n  C_n ^k (−1)^(k−1) (k−1)!×(i^k  +(−i)^k )(−n)^(n−k)  e^(−nx)   i^k  +(−i)^k  =2Re(i^k ) =2cos(((kπ)/2)) ⇒  f^((n)) (0) =2(−1)^(n−1) n^n  e^(−nx)  Σ_(k=1) ^n  (k−1)! C_n ^k  cos(((kπ)/2))

f(n)(x)=(n)nenxln(x2+1)+k=1nCnk(1)k1(k1)!×(x+i)k+(xi)k(x2+1)k(n)nkenxandf(n)(0)=k=1nCnk(1)k1(k1)!×(ik+(i)k)(n)nkenxik+(i)k=2Re(ik)=2cos(kπ2)f(n)(0)=2(1)n1nnenxk=1n(k1)!Cnkcos(kπ2)

Commented by mathmax by abdo last updated on 25/Nov/19

2)f(x)=Σ_(p=0) ^∞  ((f^((p)) (0))/(p!)) x^p   we have   f^((p)) (0) =Σ_(k=1) ^p  C_p ^k (−1)^(p−1) (k−1)!((2cos(((kπ)/2)))/1)(n)^(p−k)    =2Σ_(k=1) ^p  ((p!)/(k!(p−k)!))(−1)^(p−1) (k−1)! cos(((kπ)/2)) n^(p−k)   =2(−1)^(p−1)  n^p   Σ_(k=1) ^p    ((p!)/(k(p−k)!n^k )) cos(((kπ)/2)) ⇒  f(x)=2Σ_(p=0) ^∞ (−1)^(p−1) n^p { Σ_(k=1) ^p   ((cos(((kπ)/2)))/(k n^k (p−k)!))}x^p

2)f(x)=p=0f(p)(0)p!xpwehavef(p)(0)=k=1pCpk(1)p1(k1)!2cos(kπ2)1(n)pk=2k=1pp!k!(pk)!(1)p1(k1)!cos(kπ2)npk=2(1)p1npk=1pp!k(pk)!nkcos(kπ2)f(x)=2p=0(1)p1np{k=1pcos(kπ2)knk(pk)!}xp

Commented by mathmax by abdo last updated on 25/Nov/19

f^((n)) (0) =2(−1)^(n−1)  n^n  Σ_(k=1) ^n (k−1)!C_n ^k  cos(((kπ)/2)).

f(n)(0)=2(1)n1nnk=1n(k1)!Cnkcos(kπ2).

Answered by mind is power last updated on 25/Nov/19

ln^((k)) (1+x^2 )=ln(1+x^2 ),k=0  =((1/(x+i))+(1/(x−i)))^((k−1)) ,k≥1  =((((−1)^(k−1) (k−1)!)/((x+i)^((k)) ))+(((−1)^(k−1) (k−1)!)/((x−i)^((k)) )))   =(−1)^(k−1) (k−1)!.((((x−i)^((k−1)) +(x+i)^((k−1)) )/((x^2 +1)^k )))  =(((−1)^(k−1) (k−1)!.2Re((x+i)^((k−1)) ))/((x^2 +1)^k ))  x+i= { (((√(x^2 +1)).e^(iarctan((1/x))) ,x>0)),(((√(x^2 +1))e^(i((π/2)+arctan((1/x)))) ,x<0)) :}  x≥0  (d/dx^k ){ln(1+x^2 )}=(((−1)^(k−1) (k−1)!(x^2 +1)^((k−1)/2) 2Re{e^(i(k−1)arctan((1/x))) })/((x^2 +1)^k )),k≥1  f(x)=e^(−nx) ln(1+x^2 )  f^((n)) (x)=Σ_(k=0) ^n C_n ^k .(e^(−nx) )^((n−k)) .(ln^((k)) (1+x^2 ))  =Σ_(k=1) ^n C_n ^k (−n)^((n−k)) .e^(−nx) .(((−1)^(k−1) (k−1)!.cos((k−1)(arctan((1/x))))/((x^2 +1)^((k+1)/2) ))+(−n)^n e^(−nx) ln(1+x^2 )  f^((n)) (0)  =Σ_(k=1) ^n C_n ^k (−n)^(n−k) .(−1)^(k−1) .(k−1)!.cos((k−1)(π/2))  2)  f(x)=Σ_(j=0) ^(+∞)  ((f^j (0)x^j )/(j!))  f(x)=Σ_(j≥1) ((f^j (0)x^j )/(j!))=Σ_(j≥1) .(Σ_(k=1) ^j C_j ^k .(−n)^(j−k) (−1)^(k−1) cos(((k−1)/2)π))x^j .(1/(j!))

ln(k)(1+x2)=ln(1+x2),k=0=(1x+i+1xi)(k1),k1=((1)k1(k1)!(x+i)(k)+(1)k1(k1)!(xi)(k))=(1)k1(k1)!.((xi)(k1)+(x+i)(k1)(x2+1)k)=(1)k1(k1)!.2Re((x+i)(k1))(x2+1)kx+i={x2+1.eiarctan(1x),x>0x2+1ei(π2+arctan(1x)),x<0x0ddxk{ln(1+x2)}=(1)k1(k1)!(x2+1)k122Re{ei(k1)arctan(1x)}(x2+1)k,k1f(x)=enxln(1+x2)f(n)(x)=nk=0Cnk.(enx)(nk).(ln(k)(1+x2))=nk=1Cnk(n)(nk).enx.(1)k1(k1)!.cos((k1)(arctan(1x))(x2+1)k+12+(n)nenxln(1+x2)f(n)(0)=nk=1Cnk(n)nk.(1)k1.(k1)!.cos((k1)π2)2)f(x)=+j=0fj(0)xjj!f(x)=j1fj(0)xjj!=j1.(jk=1Cjk.(n)jk(1)k1cos(k12π))xj.1j!

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