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

$$letp\left(x\right)=a_0+a_{1}x+a_2{x}^2+...a_n{x}^n$$$$provethat{a}_k=\frac{p^{\left(k\right)}\left(0\right)}{k!}\mathrm{\forall }k\in \left[[0,n]\right].$$

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

$$wehavep\left(x\right)=\sum _{k=0}^n{a}_k{x}^k\Rightarrow$$$$p^{\left(p\right)}\left(x\right)=\sum _{k=0}^n{a}_k{\left(x^k\right)}^{\left(p\right)}ifp>k{p}^{\left(p\right)}(x=0$$$$ifp⩽k{\left(x^k\right)}^{\left(p\right)}=k(k-1)...(k-p+1)x^{k-p}$$$$p^{\left(p\right)}\left(0\right)=a_p{\left(x^p\right)}^{\left(p\right)}=a_{p}p(p-1)....1=p!a_p\Rightarrow$$$$a_p=\frac{p^{\left(p\right)}\left(0\right)}{p!}.$$

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