Taking sum both side from 1 to 1995 and some simplifying job we would obtain $ \sum_{i = 0}^{1994}x_i - \frac {1}{x_i} = 0$ ..........(#) Also , simplfying the expression also gives us $ (x_i\cdot x_{i - 1} - 1)(2x_i - x_{i - 1}) = 0$ If $ x_i\cdot x_{i - 1} = 1$ then (#) would gives us $ x_{1994} = \frac {1}{x_{1994}}$ or $ x_{1994} = 1$ . So this means $ x_0 = x_{1995} = \frac {1}{x_{1994}} = 1$ If $ 2x_i = x_{i - 1}$ $ \implies x_i = x_0\left(\frac {1}{2}\right)^i$ . Hence from (#) we get $ \sum_{i = 0}^{1994}x_0\left(\frac {1}{2}\right)^i - \frac {2^i}{x_0} = 0$ simplify to get $ x_0 = 2^{997}$ . So the maximum is $ x_0 = 2^{997}$

" wrote: note that the given equation is equivalent to: $ 2x_i^2 - \left(x_{i - 1} + \frac 2{x_{i - 1}}\right)x_i + 1 = 0$ so either $ x_i = \frac 1{x_{i - 1}}$ or $ x_i = \frac 12x_{i - 1}$. now we claim that $ x_i = 2^{k_i}x_0^{\epsilon_i}$ in which $ \left|k_i\right|\leq i$ and $ \epsilon_i = ( - 1)^{k_i + i}$,this claim can be proved using induction and the fact that $ x_i = \frac 1{x_{i - 1}}\textrm{ or }\frac {x_{i - 1}}2$.so we have: $ x_{1995} = 2^kx_0^\epsilon$ in which $ \epsilon = \epsilon_{1995},k = k_{1995},0\leq |k|\leq 1995,\epsilon = ( - 1)^{1995 + k}$ and $ x_0 = x_{1995}$ now note that if $ k$ was an odd number,then $ \epsilon = 1$ hence $ 2^k = 1$ which is a contradiction,so assume that $ k$ is an even number,then $ \epsilon = - 1$ and $ x_0^2 = 2^k$,now note that $ |k|\leq 1995$ so $ k\leq 1994$ hence $ x_0\leq 2^{997}$. now we'll show that the case where $ x_0 = 2^{997}$ is possible,just let $ x_0 = 2^{997}$,$ x_i = \frac 12x_{i - 1}$ for $ 1\leq i\leq 1994$ and $ x_{1995} = \frac 1{x_{1994}}$ which gives us: $ 2^{997} = x_0 = x_{1995}$,so the maximum value of $ x_0$ is $ 2^{997}$ QED

Could somebody please check this: If we pass the 2x_i to the right side and x_(i-1)' then we work out the sums of fractions, we get that 2xi = x_(i-1).... COULD SOMEBODY PLEASE CHECK THIS????? Thanks

shyong wrote: If $ x_i\cdot x_{i - 1} = 1$ then (#) would gives us $ x_{1994} = \frac {1}{x_{1994}}$ or $ x_{1994} = 1$ . So this means $ x_0 = x_{1995} = \frac {1}{x_{1994}} = 1$ If $ 2x_i = x_{i - 1}$ $ \implies x_i = x_0\left(\frac {1}{2}\right)^i$ . Hence from (#) we get $ \sum_{i = 0}^{1994}x_0\left(\frac {1}{2}\right)^i - \frac {2^i}{x_0} = 0$ simplify to get $ x_0 = 2^{997}$ . So the maximum is $ x_0 = 2^{997}$ I don't understand why there ar only two cases? Why do you assume that for all $i's$ just only one condition is true? Please, explain anyone

Hello, First rewrite the condition as $$(\forall i\in [|1,1995|] x_i=\frac{1}{x_{i-1}} \ or\ x_i=\frac{x_{i-1}}{2} $$ then prove by strong induction that if $x_{2k+1}=x_0$ then $x_0\leq 2^k$ then deduce $x_0\leq 2^{997}$ and give and example of such sequence for $x_0=2^997$ ...

After we solve the quadratic and realize $x_i\in \{1/x_{i-1}, x_{i-1}/2\}$, we can see that $x_{1995}=\frac{x_0 ^ {{(-1)}^b}}{2^m}$. Let $f(x)=1/x, g(x)=x/2$, we know that $x_{i+1}\in\{f(x_i), g(x_i)\}$. Also notice the parity of $m$ is the parity of the number of times g was applied (since f does not affect its parity), and $b$ has the same parity as the number of times $f$ was applied. Lastly, notice that we must apply these funcions 1995 times to get from $x_0$ to $x_{1995}$. So case 1, b even: $x_0=x_{1995}=\frac{x_0}{2^m}\Rightarrow 1=2^{-m}\Rightarrow m$ even, but then both f and g were applied an even number of times, hence they cant be applied 1995 times, ergo contradiction. Case 2, b odd: $x_0=\frac{1}{x_0} \frac{1}{2^m}\Rightarrow x_0^2=2^{-m}$. Now, every time we apply f the absolute value of $m$ does not change, and when we apply $g$ it either increases by one, or decreases by one, so $|m|\le 1994$. We can apply g at most 1994 times because b is odd, hence we must apply f at least one time. So $x_0\le \sqrt{2^{1994}}=2^{997}$. Which is possible by doing $x_{1995}=g(f(f(...f(2^{997})...)))$

orl wrote: Find the maximum value of $ x_{0}$ for which there exists a sequence $ x_{0},x_{1}\cdots ,x_{1995}$ of positive reals with $ x_{0} = x_{1995}$, such that \[ x_{i - 1} + \frac {2}{x_{i - 1}} = 2x_{i} + \frac {1}{x_{i}}, \]for all $ i = 1,\cdots ,1995$. We have $ x_{i - 1} + \frac {2}{x_{i - 1}} = 2x_{i} + \frac {1}{x_{i}}, \iff x_{i}\in {\frac{x_{i-1}}{2},\frac{1}{x_i}}$ We claim the answer is $x_0=2^{997}$, achieved by $(x_1,\dots,x_{1995})=(2^{996}, 2^{995}, \dots, 2, 1, \frac{1}{2},\dots,\frac{1}{2^{996}},\frac{1}{2^{997}},2^{997})$. Assume that $x_0>2^{997}$. Then, define $y_n={\log_2 x_n}$. Then, $y_0>997$ and $y_i\in {y_{i-1}-1,-y_{i-1}}$ for all $i=1,\dots,1995$. So, $(y_{i+1}-y_i+1)(y_{i+1}+y_i)=0$ or $y_{i+1}^2+y_{i+1}=y_{i}^2-y_i$, which when summed cyclically gives $y_1+y_2+\dots+y_{1995}=0$. Claim: $0\in {y_1,\dots,y_{1995}}$ proof. Since $y_o=y_{1995}$ there must exist $j\in{1,\dots 1995}$ such that $y_{j+1}=-y_j$. Now, all possibilities for $(y_{j-1},y_{j+2})$ are $$(y_i+1, y_i); (y_i+1, -y_i-1); (-y_i, y_i); (-y_i, -y_i-1)$$. In all 4 cases, we have generated a new sequence of $y$'s with the $y_i\in {y_{i-1}-1,-y_{i-1}}$ condition and sum $0$ but with $1995-2=1993$ elements. Inductively, we reach a point where the last $y_z$ remains and it has to be $0$ since the sum$=0$ is invariant. $\blacksquare$ Now, $y_z=0$ for some $z\in {1,\dots,1995}$ and since the absolute values of consecutive $y$'s differ by at most $1$, $| |y_z| - |y_0| |>997$ and $| |y_z| - |y_{1995}| |>997$ must imply that $|z-0|>997$ and $|z-1995|>997$, which leads to $998>z>997$, which is a contradiction.