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let-f-x-1-e-x-n-1-calculate-f-p-x-and-f-p-o-2-calculate-f-n-0-3-developp-f-at-integr-serie-




Question Number 37901 by math khazana by abdo last updated on 19/Jun/18
let f(x)= (1+e^(−x) )^n   1) calculate f^((p)) (x)  and f^((p)) (o)  2)calculate f^((n)) (0)  3)developp f at integr serie .
letf(x)=(1+ex)n1)calculatef(p)(x)andf(p)(o)2)calculatef(n)(0)3)developpfatintegrserie.
Commented by math khazana by abdo last updated on 21/Jun/18
we have f(x)=Σ_(k=0) ^n  C_n ^k   e^(−kx)  ⇒  f^((p)) (x)=Σ_(k=0) ^n  C_n ^k   (e^(−kx) )^((p))   but  (e^(−kx) )^((1)) =−k e^(−kx)  and (e^(−kx) )^((2)) =(−k^ )^2  e^(−kx)   (e^(−kx) )^((p)) =(−k)^p  e^(−kx)  ⇒  f^((p)) (x)= Σ_(k=0) ^n  C_n ^k   (−k)^p  e^(−kx)  ⇒  f^((p)) (0) = Σ_(k=0) ^n  (−k)^p  C_n ^k     2) f^((n)) (0)=Σ_(k=0) ^n (−k)^n  C_n ^k   3)f(x)=Σ_(p=0) ^∞  ((f^((p)) (0))/(p!)) x^p   =Σ_(p=0) ^∞    (1/(p!)){ Σ_(k=0) ^n (−k)^p  C_n ^k } x^(p )    .
wehavef(x)=k=0nCnkekxf(p)(x)=k=0nCnk(ekx)(p)but(ekx)(1)=kekxand(ekx)(2)=(k)2ekx(ekx)(p)=(k)pekxf(p)(x)=k=0nCnk(k)pekxf(p)(0)=k=0n(k)pCnk2)f(n)(0)=k=0n(k)nCnk3)f(x)=p=0f(p)(0)p!xp=p=01p!{k=0n(k)pCnk}xp.

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