The graphs of $y = 2 + x - \sqrt{x^2 + 2}$ and $y = x$ intersect at $x = a$ and $x = b$ as s

Question

The graphs of $y = 2 + x - \sqrt{x^2 + 2}$ and $y = x$ intersect at $x = a$ and $x = b$ as shown below. Find the values of $a$ and $b$, correct to $3$ decimal places. A sequence of real numbers $x_1, x_2, x_3, \dots$ follows the relation $$x_{n+1} = 2 + x_n - \sqrt{{x_n}^2 + 2} \text{ for}\hspace{0.5em} n = 1, 2, 3, \dots$$ Suppose the sequence converges. Explain why the limit must either be $a$ or $b$. Using the graphs of $y = 2 + x - \sqrt{x^2 + 2}$ and $y = x$, show that If $x_n < a$, then $x_{n+1} < x_n$. If $a < x_n < b$, then $x_n < x_{n+1} < b$. If $x_n > b$, then $b < x_{n+1} < x_n$. Use the results in part (c), determine whether the sequence converges as $n \to \infty$ and the value of the limit if applicable in each case. $x_1 = -2$, $x_1 = 0$, $x_1 = 2$.

Accepted Answer

(a) $a=-\sqrt{2}\approx-1.414,\ b=\sqrt{2}\approx1.414$ (b) $L=a$ or $L=b$ (c)(i) $x_{n+1}<x_n$ (c)(ii) $x_n<x_{n+1}<b$ (c)(iii) $b<x_{n+1}<x_n$ (d)(i) Divergent (d)(ii) Converges to $b=\sqrt{2}$ (d)(iii) Converges to $b=\sqrt{2}$

Recurrence Relations and Summation

Recurrence Relations and Summation

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