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Question Number 114263 by Khalmohmmad last updated on 18/Sep/20

$${3\mathrm{x}}^3-3\mathrm{x}-\frac{1}{2}=0$$$${\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3?$$

Answered by MJS_new last updated on 18/Sep/20

$$x^3-x-\frac{1}{6}=0$$$$p=-1\wedge q=-\frac{1}{6}$$$$D=\frac{p^3}{27}+\frac{q^2}{4}=-\frac{13}{432}<0\Rightarrow \mathrm{trigonometric}\mathrm{solution}$$$$x_{j+1}=\frac{2\sqrt{-3p}}{3}\sin (\frac{2\pi j}{3}+\frac{1}{3}\arcsin \frac{3\sqrt{3}q}{2\sqrt{-p^3}}))\mathrm{with}j=0,1,2$$$$\mathrm{in}\mathrm{this}\mathrm{case}$$$$x_1=-\frac{2\sqrt{3}}{3}\sin \left(\frac{1}{3}\arcsin \frac{\sqrt{3}}{4}\right)\approx -.171731$$$$x_2=\frac{2\sqrt{3}}{3}\sin (\frac{\pi }{3}+\frac{1}{3}\arcsin \frac{\sqrt{3}}{4})\approx 1.07474$$$$x_3=-\frac{2\sqrt{3}}{3}\cos (\frac{\pi }{6}+\frac{1}{3}\arcsin \frac{\sqrt{3}}{4})\approx -.903013$$

Commented by Khalmohmmad last updated on 19/Sep/20

$$\mathrm{you}\mathrm{can}\mathrm{explain}\mathrm{the}\mathrm{formula}$$$$\mathrm{please}$$

Commented by MJS_new last updated on 19/Sep/20

$$\left(1\right)x^3+px+q=0$$$$D=\frac{p^3}{27}+\frac{q^2}{4}$$$$\mathrm{if}D⩾0\Rightarrow {\mathrm{Cardano}}^{\prime }\mathrm{s}\mathrm{solution}$$$$\mathrm{if}D<0:$$$$\frac{p^3}{27}+\frac{q^2}{4}<0\iff -1<\frac{27q^2}{4p^3}<1$$$$\Rightarrow \mathrm{we}\mathrm{can}\mathrm{put}\frac{27q^2}{4p^3}=\sin \theta \mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{convenient}$$$$\mathrm{to}\mathrm{let}\theta =3\alpha$$$$\frac{27q^2}{4p^3}=\sin 3\alpha$$$$\mathrm{now}\mathrm{let}x=az\mathrm{and}a=2\sqrt{\frac{-p}{3}}\Rightarrow \mathrm{equation}\left(1\right)$$$$\mathrm{becomes}$$$$a^3{z}^3+az+q=0$$$$\Rightarrow$$$$4z^3-3z-\frac{3\sqrt{3}q}{2\sqrt{-p^3}}=0$$$$\mathrm{now}{(-\frac{3\sqrt{3}q}{2\sqrt{-p^3}})}^2=\mid \frac{27q^2}{4p^3}\mid =\sin 3\alpha$$$$4z^3-3z+\sin 3\alpha =0$$$$z^3-\frac{3}{4}z+\frac{\sin 3\alpha }{4}=0$$$$\mathrm{we}\mathrm{know}\mathrm{that}\sin 3\alpha =3\sin \alpha -{4\sin }^3\alpha \iff$$$$\iff {\sin }^3\alpha -\frac{3}{4}\sin \alpha +\frac{\sin 3\alpha }{4}=0$$$$\mathrm{we}\mathrm{can}\mathrm{put}z=\sin \alpha \mathrm{and}\mathrm{get}\mathrm{the}\mathrm{solution}$$

Answered by Olaf last updated on 19/Sep/20

$$$$$$x=u+v$$$$3(u^3+v^3)+9uv(u+v)-3(u+v)-\frac{1}{2}=0$$$$3(u^3+v^3)+3(u+v)(3uv-1)-\frac{1}{2}=0$$$$\mathrm{we}\mathrm{choose}u\mathrm{and}v\mathrm{such}\mathrm{as}uv=\frac{1}{3}$$$$\Rightarrow u^3{v}^3=\frac{1}{27}$$$$\mathrm{and}3(u^3+v^3)-\frac{1}{2}=0$$$$u^3+v^3=\frac{1}{6}$$$$\mathrm{Let}\mathrm{U}=u^3,\mathrm{V}=v^3$$$$\mathrm{and}\mathrm{S}=\mathrm{U}+\mathrm{V}=\frac{1}{6},\mathrm{P}=\mathrm{UV}=\frac{1}{27}$$$$\mathrm{We}\mathrm{know}\mathrm{S}\mathrm{and}\mathrm{P}:$$$$z^2-\mathrm{S}z+\mathrm{P}=0$$$$\mathrm{\Delta }={\mathrm{S}}^2-4\mathrm{P}=\frac{1}{36}-\frac{4}{27}=-\frac{13}{108}$$$$\mathrm{U}=\frac{\mathrm{S}-\sqrt{\mathrm{\Delta }}}{2}=\frac{1}{2}(\frac{1}{6}-\frac{1}{6}i\sqrt{\frac{13}{3}})$$$$\mathrm{U}=\frac{\mathrm{S}-\sqrt{\mathrm{\Delta }}}{2}=\frac{1}{12}(1-i\sqrt{\frac{13}{3}})$$$$\mathrm{V}=\frac{\mathrm{S}+\sqrt{\mathrm{\Delta }}}{2}=\frac{1}{12}(1+i\sqrt{\frac{13}{3}})$$$$\mid \mathrm{U}\mid =\mid \mathrm{V}\mid =\frac{1}{12}\times \sqrt{\frac{16}{3}}=\frac{1}{3\sqrt{3}}$$$$\mathrm{Let}\theta =\mathrm{ArgU}=-\mathrm{ArgV}=\arctan \sqrt{\frac{13}{3}}$$$$\mathrm{U}=\frac{1}{3\sqrt{3}}{e}^{i\theta }\mathrm{and}\mathrm{V}=\frac{1}{3\sqrt{3}}{e}^{-i\theta }$$$$u=\stackrel{3}{}\sqrt{\mathrm{U}}=\frac{1}{\sqrt{3}}{e}^{i(\frac{\theta }{3}+\frac{2k\pi }{3})}k=0,1,2$$$$v=\stackrel{3}{}\sqrt{\mathrm{V}}=\frac{1}{\sqrt{3}}{e}^{i(-\frac{\theta }{3}+\frac{2k\pi }{3})}k=0,1,2$$$$\mathrm{for}k=0,u=\frac{1}{\sqrt{3}}{e}^{i\frac{\theta }{3}}\mathrm{and}v=\frac{1}{\sqrt{3}}{e}^{-i\frac{\theta }{3}}$$$$x=u+v=\frac{1}{\sqrt{3}}(e^{i\frac{\theta }{3}}+e^{-i\frac{\theta }{3}})$$$$x=u+v=\frac{2}{\sqrt{3}}\cos \frac{\theta }{3}$$$$x=u+v=\frac{2}{\sqrt{3}}\cos \left(\frac{1}{3}\arctan \sqrt{\frac{13}{3}}\right)$$$$\mathrm{we}\mathrm{have}\mathrm{the}\mathrm{first}\mathrm{root}...$$

Commented by MJS_new last updated on 19/Sep/20

$$\mathrm{yes}\mathrm{but}\mathrm{the}\mathrm{formula}\mathrm{I}\mathrm{posted}\mathrm{is}\mathrm{standard}...$$

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