Menu Close

Question-141289

Tinku Tara 0 Comments
Pin




Question Number 141289 by ajfour last updated on 17/May/21
Commented by ajfour last updated on 17/May/21
x^3 −x−c=0 . To find excess!
$$\:\:{x}^{\mathrm{3}} −{x}−{c}=\mathrm{0}\:.\:{To}\:{find}\:{excess}! \\ $$
Answered by ajfour last updated on 18/May/21
y=x^3 −x R^2 =x^2 +y^2 (R^2 −c^2 )(R^2 −c^2 −1)^2 =c^2 (R^2 )^3 −(3c^2 +2)(R^2 )^2 +{(c^2 +1)^2 +2c^2 (c^2 +1)}R^2 −c^2 (c^2 +1)^2 −c^2 =0 let roots of x^3 −x=c be α, −β, −γ and corresponding radii p, q, r. p^2 =α^2 +c^2 q^2 =β^2 +c^2 r^2 =γ^2 +c^2 2prcos (θ+δ)+(α−γ)^2 +4c^2 = p^2 +r^2 pcos θ=α, rcos δ=γ psin θ=c, rsin δ=c ⇒ 2pr(((αγ−c^2 )/(pr)))+(α−γ)^2 =p^2 +r^2 −4c^2 ⇒ 2(αγ−c^2 )+(α−γ)^2 = p^2 +r^2 −4c^2 p^2 +q^2 +r^2 =3c^2 +2 p^2 r^2 +q^2 (p^2 +r^2 )=(c^2 +1)(3c^2 +1) p^2 q^2 r^2 = c^2 (c^2 +1)^2 +c^2 And αγ+1=(α+γ)^2 (α−γ)^2 =1−3γα αγ(α+γ)=c ⇒ αγ+(αγ)^2 =c(α+γ) ⇒ p^2 +r^2 =2c^2 +1−αγ α^2 +γ^2 +αγ=1 ...
$$\:\:{y}={x}^{\mathrm{3}} −{x} \\ $$$${R}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$$\:\left({R}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)\left({R}^{\mathrm{2}} −{c}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$$\left({R}^{\mathrm{2}} \right)^{\mathrm{3}} −\left(\mathrm{3}{c}^{\mathrm{2}} +\mathrm{2}\right)\left({R}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\:\:+\left\{\left({c}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} \left({c}^{\mathrm{2}} +\mathrm{1}\right)\right\}{R}^{\mathrm{2}} \\ $$$$\:\:−{c}^{\mathrm{2}} \left({c}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −{c}^{\mathrm{2}} =\mathrm{0} \\ $$$${let}\:{roots}\:{of}\:{x}^{\mathrm{3}} −{x}={c} \\ $$$${be}\:\alpha,\:−\beta,\:−\gamma\:\:{and}\:{corresponding} \\ $$$${radii}\:\:{p},\:{q},\:{r}. \\ $$$$\:\:{p}^{\mathrm{2}} =\alpha^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\:\:{q}^{\mathrm{2}} =\beta^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\:\:{r}^{\mathrm{2}} =\gamma^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\mathrm{2}{pr}\mathrm{cos}\:\left(\theta+\delta\right)+\left(\alpha−\gamma\right)^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} \\ $$$$\:\:=\:{p}^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$${p}\mathrm{cos}\:\theta=\alpha,\:\:{r}\mathrm{cos}\:\delta=\gamma \\ $$$${p}\mathrm{sin}\:\theta={c},\:{r}\mathrm{sin}\:\delta={c} \\ $$$$\Rightarrow\:\mathrm{2}{pr}\left(\frac{\alpha\gamma−{c}^{\mathrm{2}} }{{pr}}\right)+\left(\alpha−\gamma\right)^{\mathrm{2}} \\ $$$$\:\:\:={p}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{4}{c}^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{2}\left(\alpha\gamma−{c}^{\mathrm{2}} \right)+\left(\alpha−\gamma\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\:{p}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{4}{c}^{\mathrm{2}} \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} =\mathrm{3}{c}^{\mathrm{2}} +\mathrm{2} \\ $$$${p}^{\mathrm{2}} {r}^{\mathrm{2}} +{q}^{\mathrm{2}} \left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)=\left({c}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}{c}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${p}^{\mathrm{2}} {q}^{\mathrm{2}} {r}^{\mathrm{2}} =\:{c}^{\mathrm{2}} \left({c}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$${And}\:\:\alpha\gamma+\mathrm{1}=\left(\alpha+\gamma\right)^{\mathrm{2}} \\ $$$$\left(\alpha−\gamma\right)^{\mathrm{2}} =\mathrm{1}−\mathrm{3}\gamma\alpha \\ $$$$\:\:\alpha\gamma\left(\alpha+\gamma\right)={c} \\ $$$$\:\:\Rightarrow\:\:\alpha\gamma+\left(\alpha\gamma\right)^{\mathrm{2}} ={c}\left(\alpha+\gamma\right) \\ $$$$\Rightarrow\:\:{p}^{\mathrm{2}} +{r}^{\mathrm{2}} =\mathrm{2}{c}^{\mathrm{2}} +\mathrm{1}−\alpha\gamma \\ $$$$\:\:\:\alpha^{\mathrm{2}} +\gamma^{\mathrm{2}} +\alpha\gamma=\mathrm{1} \\ $$$$… \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *