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Question Number 54939 by gunawan last updated on 15/Feb/19

coordinate x^2 to basis {x^2 +x, x+1, x^2 +1} at P_2 is...

$$\mathrm{coordinate}\:{x}^{\mathrm{2}} \:\mathrm{to}\:\mathrm{basis}\:\left\{{x}^{\mathrm{2}} +{x},\:{x}+\mathrm{1},\:{x}^{\mathrm{2}} +\mathrm{1}\right\} \\ $$$$\mathrm{at}\:\mathrm{P}_{\mathrm{2}} \:\mathrm{is}... \\ $$

Commented by Abdo msup. last updated on 15/Feb/19

let find α ,β and γ with verify x^2 =α(x^2 +x)+β(x+1)+γ(x^2 +1) ∀x ⇒ x^2 =(α+γ)x^(2 ) +(α+β)x +β+γ ⇒ { ((α+γ =1)),((α+β=0 and β+γ =0 ⇒α−γ =0 ⇒γ=α ⇒α=(1/2))) :} =γ and β=−γ =−(1/2) so x^2 =(1/2)(x^2 +x)−(1/2)(x+1)+(1/2)(x^2 +1) so x^2 ((1/2),−(1/2),(1/2)) in that basis .

$${let}\:{find}\:\:\alpha\:,\beta\:{and}\:\:\gamma\:{with}\:{verify}\: \\ $$$${x}^{\mathrm{2}} =\alpha\left({x}^{\mathrm{2}} \:+{x}\right)+\beta\left({x}+\mathrm{1}\right)+\gamma\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\:\:\forall{x}\:\Rightarrow \\ $$$${x}^{\mathrm{2}} =\left(\alpha+\gamma\right){x}^{\mathrm{2}\:} \:+\left(\alpha+\beta\right){x}\:+\beta+\gamma\:\Rightarrow \\ $$$$\begin{cases}{\alpha+\gamma\:=\mathrm{1}}\\{\alpha+\beta=\mathrm{0}\:\:\:{and}\:\beta+\gamma\:=\mathrm{0}\:\Rightarrow\alpha−\gamma\:=\mathrm{0}\:\Rightarrow\gamma=\alpha\:\Rightarrow\alpha=\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$$$=\gamma\:\:{and}\:\beta=−\gamma\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\:{so}\: \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} \:+{x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right) \\ $$$${so}\:{x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}},−\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)\:{in}\:{that}\:{basis}\:. \\ $$

Commented by gunawan last updated on 15/Feb/19

thank you very much Sir

$$\mathrm{thank}\:{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir} \\ $$

Commented by Abdo msup. last updated on 15/Feb/19

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

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