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Question Number 33702 by math khazana by abdo last updated on 22/Apr/18

let p(x) =a_0  +a_1 x +a_2 x^2  +...a_n x^n   prove that  a_k = ((p^((k)) (0))/(k!))  ∀ k ∈[[0,n]] .

$${let}\:{p}\left({x}\right)\:={a}_{\mathrm{0}} \:+{a}_{\mathrm{1}} {x}\:+{a}_{\mathrm{2}} {x}^{\mathrm{2}} \:+...{a}_{{n}} {x}^{{n}} \\ $$$${prove}\:{that}\:\:{a}_{{k}} =\:\frac{{p}^{\left({k}\right)} \left(\mathrm{0}\right)}{{k}!}\:\:\forall\:{k}\:\in\left[\left[\mathrm{0},{n}\right]\right]\:. \\ $$

Commented by math khazana by abdo last updated on 29/Apr/18

we have p(x)=Σ_(k=0) ^n  a_k  x^k  ⇒  p^((p)) (x)= Σ_(k=0) ^n a_k  (x^k )^((p))   if p>k  p^((p)) (x=0  if p≤k    (x^k )^((p)) =k(k−1)...(k−p+1)x^(k−p)   p^((p)) (0)=a_p  (x^p )^((p)) =a_p p(p−1)....1=p! a_p  ⇒  a_p =((p^((p)) (0))/(p!)) .

$${we}\:{have}\:{p}\left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{a}_{{k}} \:{x}^{{k}} \:\Rightarrow \\ $$$${p}^{\left({p}\right)} \left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}} {a}_{{k}} \:\left({x}^{{k}} \right)^{\left({p}\right)} \:\:{if}\:{p}>{k}\:\:{p}^{\left({p}\right)} \left({x}=\mathrm{0}\right. \\ $$$${if}\:{p}\leqslant{k}\:\:\:\:\left({x}^{{k}} \right)^{\left({p}\right)} ={k}\left({k}−\mathrm{1}\right)...\left({k}−{p}+\mathrm{1}\right){x}^{{k}−{p}} \\ $$$${p}^{\left({p}\right)} \left(\mathrm{0}\right)={a}_{{p}} \:\left({x}^{{p}} \right)^{\left({p}\right)} ={a}_{{p}} {p}\left({p}−\mathrm{1}\right)....\mathrm{1}={p}!\:{a}_{{p}} \:\Rightarrow \\ $$$${a}_{{p}} =\frac{{p}^{\left({p}\right)} \left(\mathrm{0}\right)}{{p}!}\:. \\ $$

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