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Consider the sequence $\langle x_n \rangle , \: n \geq 0$ defined by the recurrence relation $x_{n+1} = c . x^2_n -2$, where $c > 0$.

Suppose there exists a non-empty, open interval $(a, b)$ such that for all $x_0$ satisfying $a < x_0 < b$, the sequence converges to a limit. The sequence converges to the value?

1. $\frac{1+\sqrt{1+8c}}{2c}$
2. $\frac{1-\sqrt{1+8c}}{2c}$
3. $2$
4. $\frac{2}{2c-1}$

Take $x_0 = 2$, then it con.verges to $2$
It also converges to 2

Convergence, in mathematics, property (exhibited by certain infinite series and functions) of approaching a limit more and more closely as an argument (variable) of the function increases or decreases or as the number of terms of the series increases. For example, the function y = 1/x converges to zero as x increases.

Lets take a look when $c=1$

The value the recurrence converges to must be,  $\dfrac{1\pm\sqrt{1+8\times1}}{2\times 1} =2,{-1}$

However, when we take the positive square root, i.e. when $x_{i}$ supposedly converges to $\dfrac{1+\sqrt{1+8c}}{2c}$,

the convergence does not hold for the neighborhood.

$\quad x_{i}=\lim\limits_{\delta \to 0} 2+\delta$

$x_{i+1}=1\times x^{2}_{i}-2=\lim\limits_{\delta \to 0}\left(2+\delta\right)^{2}-2$

$\qquad=\lim\limits_{\delta \to 0} 4-2+\delta ^{2}+4\delta$

$x_{i+1}=\lim\limits_{\delta \to 0} 2+\delta ^{2}+4\delta$

We can see that $x_{i+1}$ is further than $x_{i}$ from the assumed convergence value of $\lim\limits_{i \to \infty} x_{i}=2$

Similarly, the value does not converge when $x_{i}$ approaches $2$ from the left side of the number line.

When the negative square root is considered, the convergence does hold for neighbors on either side.

$\quad x_{i}=\lim\limits_{\delta \to 0} \left({-1}\right)+\delta$

$x_{i+1}=1\times x^{2}_{i}-2=\lim\limits_{\delta \to 0} \left(\left({-1}\right) +\delta\right)^{2}-2$

$\qquad =\lim\limits_{\delta \to 0} 1-2+\delta ^{2}-2 \delta$

$x_{i+1}=\lim\limits_{\delta \to 0} -1+\delta^{2}-2\delta$

Also,

$\quad x_{i}=\lim\limits_{\delta \to 0} \left({-1}\right)-\delta$

$x_{i+1}=1\times x^{2}_{i}-2=\lim\limits_{\delta \to 0} \left(\left({-1}\right) -\delta\right)^{2}-2$

$\qquad =\lim\limits_{\delta \to 0} 1-2+\delta ^{2}+2 \delta$

$x_{i+1}=\lim\limits_{\delta \to 0} -1+\delta^{2}+2\delta$

Hence, when negative square root is considered, the value oscillates around the convergence point, and actually converges.

Therefore the answer should be only B.

But at -1.0 also the value seems to diverge, though oscillate:$x_{1}=-1.000199, x_{2}=-0.999599, x_{3}=-1.000799, x_{4}=-0.998399$

what is the meaning of converges??
edited by
Ans is both (A) and (B) I guess…

If we take $c=1$ the options (A) and (B) give, $2$ and $-1$ respectively…

If we take $x_0=1$ then $x_1=x_2=x_3=...=-1$

if we take $x_0=2$ then $x_1=x_2=x_3=...=2$

Because in the question it is not defined what is $a$ or $b$. Assuming $a,b \in \mathbb{R}$

and $a=1-\delta$ and $b=2+\beta$ where $\delta$ and $\beta$ are small positive quantities, $x_0=1$ and $x_0=2$ are both possible.

Please correct the flaw in my logic...

Here,

$x_{n+1} = cx_{n}^{2}-2 , c > 0$

For stability, We can write non-linear first order recurrence as :- $x_{n} = f(x_{n-1})$ (or) $x_{n+1} = f(x_{n})$

So, $x_{n+1} = cx_{n}^{2}-2 , c > 0$  becomes $x_{n} = cx_{n}^{2}-2 , c > 0$ (or) For simplicity, to find fixed stable points , we can write it as :- $x = cx^{2}-2 , c > 0$

Now, after solving the given equation , we will get :-

$x= \frac{1\pm \sqrt{1+8c}}{2c}$

Now, here we have $2$ fixed points $x_{1}= \frac{1 + \sqrt{1+8c}}{2c}$ and $x_{2}= \frac{1 - \sqrt{1+8c}}{2c}$

Now, We have to check that for which fixed point the given recurrence converges.

As we know if $f' < 0$ at a point or slope is negative then it means function $f$ is  strictly decreasing and if $f' > 0$ then it means function $f$ is strictly increasing.

So, if we find $f' < 0$ in a given interval , it means function $f$ is decreasing or converses to a fixed point in the given interval.

So, Now, Since here , $f(x) = cx^{2} - x -2$

So, $f'(x) = 2cx - 1$

Now,

$f'(x_{1}) = 2c * \left ( \frac{1+\sqrt{1+8c}}{2c} \right ) - 1$

So, $f'(x_{1}) \nless 0 , c> 0$

Now,

$f'(x_{2}) = 2c * \left ( \frac{1-\sqrt{1+8c}}{2c} \right ) - 1$

So, $f'(x_{2}) < 0 , c> 0$

So, here local stable point is $\frac{1-\sqrt{1+8c}}{2c}$ for which given recurrence converges.

It's the best answer containing all the purely mathematical statements. I think this should be selected as the best answer.
This should be chosen as best answer
This must be the best answer!
The best explanation, this is the best answer.

A sequence converges means at some point $x_{n+1} = x_{n}$

Then,

$x=cx^2 -2$

or

$cx^2 -x -2 = 0$

Solving for $x$:

$x=\frac{1\pm \sqrt{1+8c}}{2c}$

So both (A) and (B) can be the values.

Sir, what will be the final answer both a and b or only b?
I think the answer is only B.
You are solving according to praggy sir solutions?

Let $c = 1, x_0 = 1$

Then,

$\\x_1 \,\,\,= c \cdot x_0^2 - 2\\\\ \indent= 1(1)^2-2\\ \indent = -1\\\\ x_2 \,\,\,= c \cdot x_1^2 - 2\\\\ \indent= 1(-1)^2 - 2\\ \indent = -1$

So, the value converges to $-1$, which is equal to $\frac{1-\sqrt{1+8\times 1}}{2\times 1}$

exactly , only B will be the answer. as all the term of X converges to -1