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Question Number 99568 by Dwaipayan Shikari last updated on 22/Jun/20

$$Findthevalueof\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+}}}}}}...\mathrm{\infty }bycosfunction$$

Commented by bemath last updated on 22/Jun/20

$$x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...}}}}}$$$$x^2=2+x$$$$x^2-x-2=0$$$$(x-2)(x+1)=0$$$$\{\begin{array}{l}x=2solution\\ x=-1rejected\end{array}$$

Commented by Dwaipayan Shikari last updated on 22/Jun/20

$$Thankyousir.Canyouproveit{using}^{\prime }{cos}^{\prime }function?$$

Commented by Dwaipayan Shikari last updated on 22/Jun/20

$$1+cos\frac{\pi }{4}=2{cos}^2\frac{\pi }{8}$$$$\sqrt{\frac{1}{2}(1+\frac{1}{\sqrt{2}})}=cos\frac{\pi }{8}=cos\frac{\pi }{2^3}=\frac{1}{2}\sqrt{2+\sqrt{2}}$$$$Asweprocceed\sqrt{\frac{1}{2}(1+\frac{1}{2}\sqrt{2+\sqrt{2}})}=cos\frac{\pi }{16}=cos\frac{\pi }{2^4}=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2}}}$$$$so\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}}}...nterms=cos\frac{\pi }{2^{n+1}}$$$$\underset{n\to \mathrm{\infty }}{}limcos\frac{\pi }{2^{n+1}}=cos\left(0\right)=1$$$$So\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+}}}}}}}...\mathrm{\infty }=1$$$$so\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+}}}}}}}}....\mathrm{\infty }=2$$$$$$

Commented by Dwaipayan Shikari last updated on 22/Jun/20

$$Sir,ihavefoundthiswaywhileexperimenting.Isittrue?$$

Answered by mahdi last updated on 21/Jun/20

$$\sqrt{2+\sqrt{2+\sqrt{2+...}}}=\mathrm{A}\Rightarrow$$$$\sqrt{2+\mathrm{A}}=\mathrm{A}\Rightarrow {\mathrm{A}}^2=\mathrm{A}+2\Rightarrow {\mathrm{A}}^2-\mathrm{A}-2=0$$$$\Rightarrow \mathrm{A}=\frac{1\pm \sqrt{5}}{2}\stackrel{\mathrm{A}>0}{\Rightarrow }\mathrm{A}=\frac{1+\sqrt{5}}{2}$$

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