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Question Number 1120 by 123456 last updated on 17/Jun/15
$$\mathrm{given}\:{f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}\right)={a}_{{n}} {x}^{{n}} +\centerdot\centerdot\centerdot+{a}_{\mathrm{0}} \: \\ $$$$\mathrm{with}\:{a}_{{n}} \neq\mathrm{0},\:{a}_{{i}} \in\mathbb{R},{i}\in\left\{\mathrm{0},\mathrm{1},...,{n}\right\},{n}\in\mathbb{N}^{\ast} \\ $$$$\mathrm{supose}\:\mathrm{that}\:{x}_{\mathrm{1}} ,...,{x}_{{n}} \\ $$$$\mathrm{are}\:\mathrm{roots}\:\mathrm{the}\:{n}\:\mathrm{roots}\:\mathrm{of}\:{f}\left({x}\right) \\ $$$$\mathrm{then}\:\mathrm{proof}\:\mathrm{or}\:\mathrm{give}\:\mathrm{a}\:\mathrm{counter}\:\mathrm{example}\:\mathrm{that}\:\mathrm{the}\:\mathrm{root}\:\mathrm{of} \\ $$$${f}^{\left({n}−\mathrm{1}\right)} \left({x}\right) \\ $$$$\mathrm{is} \\ $$$${x}=\frac{{x}_{\mathrm{1}} +\centerdot\centerdot\centerdot+{x}_{{n}} }{{n}} \\ $$
Answered by prakash jain last updated on 17/Jun/15
$${f}\left({x}\right)=\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{2}} \right)...\left({x}−{x}_{{n}} \right) \\ $$$${f}\left({x}\right)={x}^{{n}} −\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} +..+{x}_{{n}} \right){x}^{{n}−\mathrm{1}} +...+{x}_{\mathrm{1}} {x}_{\mathrm{2}} ..{x}_{{n}} \\ $$$${f}^{{n}−\mathrm{1}} \left({x}\right)={n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)...\mathrm{2}{x}−\left({n}−\mathrm{1}\right)!\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} +...+{x}_{{n}} \right) \\ $$$$\mathrm{Root}\:\mathrm{of}\:{f}^{{n}−\mathrm{1}} \left({x}\right)=\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +...+{x}_{{n}} }{{n}} \\ $$