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Question Number 7532 by Yozzia last updated on 02/Sep/16

$$Find{x}_n(n\in \mathbb{Z})satisfying{x}_0=0,x_1=1and$$$$x_{n+1}=x_n\sqrt{x_{n-1}^2+1}+x_{n-1}\sqrt{x_n^2+1}forn⩾1.$$

Commented by sou1618 last updated on 03/Sep/16

$$whenn=1$$$$x_2=1$$$$whenn=2$$$$x_3=2\sqrt{2}$$$$whenn=3$$$$x_4=2\sqrt{2}\sqrt{2}+\sqrt{8+1}=7$$$$...$$$$whenn>2x_n,x_{n-1}>1$$$$x_{n+1}=x_n\sqrt{x_{n-1}^2+1}+x_{n-1}\sqrt{x_n^2+1}\cdot \cdot \cdot (\ast )$$$$x_n=\frac{1}{2}(y_n-\frac{1}{y_n})(y_n⩾1)$$$$\frac{1}{2}(y_{n+1}-y_{n+1}^{-1})=\frac{1}{2}(y_n-y_n^{-1})\frac{1}{2}(y_{n-1}+y_{n-1}^{-1})+\frac{1}{2}(y_{n-1}-y_{n-1}^{-1})\frac{1}{2}(y_n+y_n^{-1})$$$$2(y_{n+1}-y_{n+1}^{-1})=(y_n-y_n^{-1})(y_{n-1}+y_{n-1}^{-1})+(y_{n-1}-y_{n-1}^{-1})(y_n+y_n^{-1})$$$$2(y_{n+1}-y_{n+1}^{-1})=2y_n{y}_{n-1}-2y_n^{-1}{y}_n^{-1}$$$$y_{n+1}-\frac{1}{y_{n+1}}=y_n{y}_{n-1}-\frac{1}{y_n{y}_{n-1}}$$$$y_{n+1}^2-(y_n{y}_{n-1}-\frac{1}{y_n{y}_{n-1}})y_{n+1}-1=0$$$$(y_{n+1}-y_n{y}_{n-1})(y_{n+1}+\frac{1}{y_n{y}_{n-1}})=0$$$$y_{n+1}=y_n{y}_{n-1}(\because y_n>0)$$$$$$$$x_0=0,y_0^2-0y_0-1=0$$$$y_0=1$$$$x_1=1,y_1^2-2y_1-1=0$$$$y_1=1+\sqrt{2}$$$$y_n=y_{n-1}{y}_{n-2}=y_{n-2}^2{y}_{n-3}=y_{n-3}^3{y}_{n-4}^2=y_{n-4}^5{y}_{n-5}^3=...$$$$y_2=1(1+\sqrt{2}),y_3={(1+\sqrt{2})}^2,y_4={(1+\sqrt{2})}^3,y_5={(1+\sqrt{2})}^5...$$$$1123581321...fibonacci$$$$y_n={(1+\sqrt{2})}^{f\left(n\right)}\{\begin{array}{l}f\left(n\right)=f(n-1)+f(n-2)\\ f\left(0\right)=0,f\left(1\right)=1\end{array}$$$$f\left(n\right)=\frac{1}{\sqrt{5}}\{{\left(\frac{1+\sqrt{5}}{2}\right)}^n-{\left(\frac{1-\sqrt{5}}{2}\right)}^n\}$$$$=\frac{{\varphi }^n-{(-\varphi )}^{-n}}{\sqrt{5}},(\varphi =\frac{1+\sqrt{5}}{2})$$$$$$$$$$$$x_n=\frac{1}{2}\{{(1+\sqrt{2})}^{f\left(n\right)}-{(1+\sqrt{2})}^{-f\left(n\right)}\}$$$$++++++++++++$$$$x_0=\frac{1}{2}\{{(1+\sqrt{2})}^0-{(1+\sqrt{2})}^0\}=0$$$$x_1=\frac{1}{2}\{1+\sqrt{2}-\frac{1}{1+\sqrt{2}}\}=\frac{{(1+\sqrt{2})}^2-1^2}{2(1+\sqrt{2})}=1$$$$x_2=x_1=1$$$$x_3=\frac{1}{2}\{{(1+\sqrt{2})}^2-\frac{1}{{(1+\sqrt{2})}^2}\}=\frac{{\left({(1+\sqrt{2})}^2\right)}^2-1^2}{2{(1+\sqrt{2})}^2}$$$$=\frac{(3+2\sqrt{2}+1)(3+2\sqrt{2}-1)}{2{(1+\sqrt{2})}^2}$$$$=\frac{\left(2\sqrt{2}(1+\sqrt{2})\right)\left(2(1+\sqrt{2})\right)}{2{(1+\sqrt{2})}^2}=2\sqrt{2}$$$$...$$

Commented by sou1618 last updated on 03/Sep/16

$$ifyouneedfibonacci=f\left(n\right)$$$$a_0=0,a_1=1$$$$a_n=a_{n-1}+a_{n-2}(n⩾2)$$$$a_n-{pa}_{n-1}=q(a_{n-1}-{pa}_{n-2})$$$$a_n=(p+q)a_{n-1}-{pqa}_{n-2}$$$$\{\begin{array}{l}p+q=1\\ pq=-1\end{array}$$$$p^2-p+p=0$$$$p=\frac{1\pm \sqrt{5}}{2}$$$$(p,q)=(\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2})$$$$\{\begin{array}{l}{a}_n-{pa}_{n-1}=q(a_{n-1}-{pa}_{n-2})\\ a_n-{qa}_{n-1}=p(a_{n-1}-{qa}_{n-2})\end{array}$$$$whenn=2$$$$\{\begin{array}{l}{a}_1-{pa}_0=1\\ a_1-{qa}_0=1\end{array}$$$$so$$$$\{\begin{array}{l}{a}_n-{pa}_{n-1}=q^{n-1}\cdot \cdot \cdot \left(1\right)\\ a_n-{qa}_{n-1}=p^{n-1}\cdot \cdot \cdot \left(2\right)\end{array}$$$$\left(2\right)-\left(1\right)\Rightarrow$$$$(p-q)a_{n-1}=p^{n-1}-q^{n-1}$$$$a_n=\frac{1}{p-q}(p^n-q^n)$$$$a_n=\frac{1}{\sqrt{5}}\{{\left(\frac{1+\sqrt{5}}{2}\right)}^n-{\left(\frac{1-\sqrt{5}}{2}\right)}^n\}$$$$$$$$f\left(n\right)=a_n$$

Commented by Yozzia last updated on 03/Sep/16

$$Nicely!Whatifyoulet{x}_n=sinh\left\{u\left(n\right)\right\}initially?$$$$$$$$$$

Commented by sou1618 last updated on 03/Sep/16

$${it}^{\prime }sniceidea!!$$$$x_n=sinh\left(u_n\right)$$$$sinh\left(u_{n+1}\right)=sinh\left(u_n\right)cosh\left(u_{n-1}\right)+sinh\left(u_{n-1}\right)cosh\left(u_n\right)$$$$sinh\left(u_{n+1}\right)=sinh(u_n+u_{n-1})$$$$u_{n+1}=u_n+u_{n-1}$$$$u_0=0$$$$u_1=ln(1+\sqrt{2})$$$$so$$$$u_n=ln(1+\sqrt{2})\times f\left(n\right)$$$$$$$$x_n=sinh\left\{ln(1+\sqrt{2})f\left(n\right)\right\}$$$$=\frac{1}{2}\{{(1+\sqrt{2})}^{f\left(n\right)}-{(1+\sqrt{2})}^{-f\left(n\right)}\}$$

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