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Question Number 37818 by prof Abdo imad last updated on 17/Jun/18

$$let0<u_0<1and{u}_{n+1}=\sqrt{\frac{1+u_n}{2}}$$ $$studytheconvergenceof{u}_n$$ $$$$

Commented bymath khazana by abdo last updated on 19/Jun/18

$$letf\left(x\right)=\sqrt{\frac{1+x}{2}}withx⩾0$$ $$f\left(x\right)=x\Rightarrow \frac{1+x}{2}=x^2\Rightarrow 1+x=2x^2\Rightarrow$$ $$2x^2-x-1=0\mathrm{\Delta }=1+8=9$$ $$x_1=\frac{1+3}{4}=1and{x}_2=\frac{1-3}{4}=-\frac{1}{2}wehave$$ $$u_{n+1}=f\left(u_n\right)andfiscontinue\Rightarrow {lim}_{n\to +\mathrm{\infty }}{u}_n=1$$ $$letfindanotherformof{u}_{n}weput$$ $$u_0=cos\alpha \Rightarrow u_1=\sqrt{\frac{1+cos\left(\alpha \right)}{2}}=cos\left(\frac{\alpha }{2}\right)$$ $$letprovethat{u}_n=cos\left(\frac{\alpha }{2^n}\right)$$ $$relationtrueforn=0$$ $$letsuppose{u}_n=cos\left(\frac{\alpha }{2^n}\right)\Rightarrow$$ $$u_{n+1}=\sqrt{\frac{1+u_n}{2}}=\sqrt{\frac{1+cos\left(\frac{\alpha }{2^n}\right)}{2}}$$ $$=cos\left(\frac{\alpha }{2^{n+1}}\right).itsnowclearthat$$ $${lim}_{n\to +\mathrm{\infty }}{u}_n=1.$$

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