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Question Number 1120 by 123456 last updated on 17/Jun/15
given f:R→R  f(x)=a_n x^n +∙∙∙+a_0    with a_n ≠0, a_i ∈R,i∈{0,1,...,n},n∈N^∗   supose that x_1 ,...,x_n   are roots the n roots of f(x)  then proof or give a counter example that the root of  f^((n−1)) (x)  is  x=((x_1 +∙∙∙+x_n )/n)
$$\mathrm{given}\:{f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}\right)={a}_{{n}} {x}^{{n}} +\centerdot\centerdot\centerdot+{a}_{\mathrm{0}} \: \\ $$$$\mathrm{with}\:{a}_{{n}} \neq\mathrm{0},\:{a}_{{i}} \in\mathbb{R},{i}\in\left\{\mathrm{0},\mathrm{1},…,{n}\right\},{n}\in\mathbb{N}^{\ast} \\ $$$$\mathrm{supose}\:\mathrm{that}\:{x}_{\mathrm{1}} ,…,{x}_{{n}} \\ $$$$\mathrm{are}\:\mathrm{roots}\:\mathrm{the}\:{n}\:\mathrm{roots}\:\mathrm{of}\:{f}\left({x}\right) \\ $$$$\mathrm{then}\:\mathrm{proof}\:\mathrm{or}\:\mathrm{give}\:\mathrm{a}\:\mathrm{counter}\:\mathrm{example}\:\mathrm{that}\:\mathrm{the}\:\mathrm{root}\:\mathrm{of} \\ $$$${f}^{\left({n}−\mathrm{1}\right)} \left({x}\right) \\ $$$$\mathrm{is} \\ $$$${x}=\frac{{x}_{\mathrm{1}} +\centerdot\centerdot\centerdot+{x}_{{n}} }{{n}} \\ $$
Answered by prakash jain last updated on 17/Jun/15
f(x)=(x−x_1 )(x−x_2 )...(x−x_n )  f(x)=x^n −(x_1 +x_2 +..+x_n )x^(n−1) +...+x_1 x_2 ..x_n   f^(n−1) (x)=n(n−1)(n−2)...2x−(n−1)!(x_1 +x_2 +...+x_n )  Root of f^(n−1) (x)=((x_1 +x_2 +...+x_n )/n)
$${f}\left({x}\right)=\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{2}} \right)…\left({x}−{x}_{{n}} \right) \\ $$$${f}\left({x}\right)={x}^{{n}} −\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} +..+{x}_{{n}} \right){x}^{{n}−\mathrm{1}} +…+{x}_{\mathrm{1}} {x}_{\mathrm{2}} ..{x}_{{n}} \\ $$$${f}^{{n}−\mathrm{1}} \left({x}\right)={n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)…\mathrm{2}{x}−\left({n}−\mathrm{1}\right)!\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} +…+{x}_{{n}} \right) \\ $$$$\mathrm{Root}\:\mathrm{of}\:{f}^{{n}−\mathrm{1}} \left({x}\right)=\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +…+{x}_{{n}} }{{n}} \\ $$

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