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Question Number 142725 by mathlove last updated on 04/Jun/21

Commented by Olaf_Thorendsen last updated on 04/Jun/21

$$https://peterjamesthomas.com/category/general/social-media/?hcb=1$$

Commented by MJS_new last updated on 04/Jun/21

$$\mathrm{there}\mathrm{are}\mathrm{about}7900000000\mathrm{people}\mathrm{on}\mathrm{this}\mathrm{planet}$$$$\mathrm{and}\mathrm{probably}\mathrm{less}\mathrm{than}79000000\mathrm{can}\mathrm{solve}$$$$\mathrm{this}:$$$$x^4-8x^3+2x^2+8x+1=0$$$$\mathrm{I}\mathrm{can}-\mathrm{so}\mathrm{what}?!$$

Commented by 1549442205PVT last updated on 05/Jun/21

$$\mathrm{This}\mathrm{equation}\mathrm{is}\mathrm{too}\mathrm{easy}!$$$${\mathrm{x}}^4-{8\mathrm{x}}^3+{2\mathrm{x}}^2+8\mathrm{x}+1=0$$$$\iff ({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2})-8(\mathrm{x}-\frac{1}{\mathrm{x}})+2=0.\mathrm{put}(\mathrm{x}-\frac{1}{\mathrm{x}})=\mathrm{t}$$$$\mathrm{we}\mathrm{get}{\mathrm{t}}^2+2-8\mathrm{t}+2=0\iff {\mathrm{t}}^2-8\mathrm{t}+4=0$$$${\mathrm{\Delta }}^{\prime }=12={\left(2\sqrt{3}\right)}^2\Rightarrow \mathrm{t}=4\pm 2\sqrt{3}$$$$\mathrm{i})\mathrm{x}-\frac{1}{\mathrm{x}}=4+2\sqrt{3}\iff {\mathrm{x}}^2-(4+2\sqrt{3})\mathrm{x}-1=0$$$${\mathrm{\Delta }}^{\prime }={(2+\sqrt{3})}^2+1=8+4\sqrt{3}={\left[\sqrt{2}(\sqrt{3}+1)\right]}^2$$$$\mathrm{x}=2+\sqrt{3}\pm (\sqrt{6}+\sqrt{2})$$$$\mathrm{ii})\mathrm{x}-\frac{1}{\mathrm{x}}=4-2\sqrt{3}\iff {\mathrm{x}}^2-2(2-\sqrt{3})\mathrm{x}-1=0$$$${\mathrm{\Delta }}^{\prime }={(2-\sqrt{3})}^2+1=8-4\sqrt{3}={(\sqrt{6}-\sqrt{2})}^2$$$$\mathrm{x}=2-\sqrt{3}\pm (\sqrt{6}-\sqrt{2})$$

Commented by MJS_new last updated on 05/Jun/21

$$\mathrm{Ok},\mathrm{now}\mathrm{ask}1000\mathrm{randomly}\mathrm{chosen}\mathrm{people}$$$$\mathrm{and}\mathrm{tell}\mathrm{us}\mathrm{the}\mathrm{percentage}\mathrm{of}\mathrm{right}\mathrm{answers}$$

Commented by Dwaipayan Shikari last updated on 05/Jun/21

$$Actuallynoonehasdoneityet.ItisRiemannHypothesis.$$$$0.0000000001\mathrm{\%}candothis$$

Commented by MJS_new last updated on 05/Jun/21

$$\mathrm{I}\mathrm{know}.\mathrm{but}\mathrm{follow}\mathrm{the}\mathrm{link}\mathrm{posted}\mathrm{by}\mathrm{Sir}\mathrm{Olaf}.$$

Commented by Tawa11 last updated on 06/Nov/21

$$\mathrm{Nice}$$

Answered by Dwaipayan Shikari last updated on 05/Jun/21

$$z=w+bi$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^z}=\underset{n=1}{\sum ^{\mathrm{\infty }}}{n}^{-w}{n}^{-bi}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}{n}^{-w}{e}^{-bilog\left(n\right)}=\underset{n=1}{\sum ^{\mathrm{\infty }}}{n}^{-w}cos\left(blog\left(n\right)\right)-{in}^{-w}sin\left(blog\left(n\right)\right)$$$$\mathrm{\Re }\left(\zeta \left(z\right)\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{cos\left(blog\left(n\right)\right)}{n^w}$$$$Now\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{cos\left(blog\left(n\right)\right)}{n^w}-i\frac{sin(blog\left(n\right)}{n^w}=0$$$$Peoplebelievethat\mathrm{\Re }\left(z\right)=\frac{1}{2}$$

Commented by MJS_new last updated on 04/Jun/21

$$\mathrm{welcome}\mathrm{among}\mathrm{the}\mathrm{chosen}\mathrm{few}!$$

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