How to show $(x_n)$ defined $x_1 = a$, $x_{n+1} = 2x_n - x^2_ny$ is increasing if $ay < 2$

Question

$y,a > 0$. Let $x_n = (y^{-1} + \epsilon)$ for $\epsilon \in \mathbb{R}$. Then $x_{n+1} = y^{-1}-\epsilon^2y$, so sup$(x_n) = y^{-1}$. How do i show $(x_n)$ is an increasing sequence so i can conclude $(x_n)$ converges to $y^{-1}$?

Answer 1

Hint: Let $$z_n=x_ny-1$$ with $z_1=ay-1<1$.

Answer 2

Part 1 for $0<ay\leq 1$.

Look at the function $f(x)=2x-yx^2$ such that $x_{n+1}=f(x_n)$.

• Pr 1 $f'(x)=2-2yx\geq 0\iff 1\geq yx$, i.e. function is increasing.
• Pr 2 this function gets it's maximum value at $x_{\max}=\frac{1}{y}$, thus $\color{blue}{f(x)\leq} f\left(\frac{1}{y}\right)=\color{blue}{\frac{1}{y}}$
• Pr 3 and finally for $x\geq0$: $1\geq yx \Rightarrow x\geq yx^2 \Rightarrow -x\leq -yx^2 \Rightarrow \color{blue}{x}=2x-x \color{blue}{\leq} 2x-yx^2=\color{blue}{f(x)}$

As a result, if $0\leq a\leq \frac{1}{y} \overset{\color{red}{Pr3}}{\Rightarrow} 0\leq x_1=a\leq f(a)=x_2 \overset{\color{red}{Pr2}}{\leq} \frac{1}{y}$ and by induction $$0\leq x_{n-1}\leq x_n \leq \frac{1}{y} \overset{\color{red}{Pr1}}{\Rightarrow} 0\leq \color{blue}{x_n}=f(x_{n-1})\color{blue}{\leq} f(x_{n})=\color{blue}{x_{n+1}} \overset{\color{red}{Pr2}}{\leq} \frac{1}{y}$$ The sequence is increasing and bounded by $0$ and $\frac{1}{y}$ thus it has a limit $L$ which is the solution of $L=2L-yL^2 \Rightarrow L=\frac{1}{y}$.

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Part 2 for $1<ay<2$.

Function $f(x)$ has the property that $0\leq f(x)\overset{\color{red}{Pr2}}{\leq}\frac{1}{y}$ for $0\leq x \leq \frac{2}{y}$, because $2x-yx^2 = x(2-yx)\geq 0$, and a 2nd degree polynomial with negative leading coefficient is positive between its roots. Then for $1<ay<2 \Rightarrow 0<a<\frac{2}{y}$ we have $0\leq x_1=f(a)\leq \frac{1}{y}$ or $0\leq x_1y\leq1$ and from this moment on, we are in part 1 of the recursion.

Answer 3

Wstp $x_n < x_{n+1}$ or that $y^{-1} -\epsilon < y^{-1} - \epsilon^2y$ for $\epsilon \in \mathbb{R}^+$. Since $\forall n \in \mathbb{N}$, $x_n > 0$, $\epsilon < y^{-1}$. Then $y < \epsilon^{-1}$, so $y^{-1} - \epsilon^2y > y^{-1} - \epsilon$.