Exercise 2.4.5

Question

[Calculating Square Roots]
  Let $x_{1}=2$, and define
  $$
  x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{2}{x_{n}}\right)
  $$
  \begin{enumerate}[label=(\alph*)] 
  \item Show that $x_{n}^{2}$ is always greater than or equal to 2 , and then use this to prove that $x_{n}-x_{n+1} \geq 0$. Conclude that $\lim x_{n}=\sqrt{2}$.
  \item Modify the sequence $\left(x_{n}\right)$ so that it converges to $\sqrt{c}$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Clearly $x_1^2 \ge 2$, now procede by induction. if $x_{n}^2 \ge 2$ then we have
    $$
    x_{n+1}^2 = \frac 14\left(\frac{x_n^2 + 2}{x_n}ight)^2
    = \frac 14\left(\frac{(x_n^2 + 2)^2}{x_n^2}ight)
    \ge \frac 14\left(\frac{(x_n^2 + 2)^2}{2}ight)
    $$
    Now since $x_n^2 \ge 2$ we have $(x_n^2 + 2)^2 \ge 16$ meaning
    $$
    x_{n+1}^2 = \frac 14\left(\frac{(x_n^2 + 2)^2}{2}ight) \ge 2.
    $$
    Now to show $x_n - x_{n+1} \ge 0$ we use $x_n \ge 0$
    $$
    \begin{aligned}
      x_n - x_{n+1} &= x_n - \frac 12\left(x_n + \frac{2}{x_n}ight) \
                    &= \frac 12 x_n + \frac{1}{x_n} \ge 0
    \end{aligned}
    $$
    Now we know $(x_n) 	o x$ converges by MCT, to show $x^2 = 2$ we equate $x_n = x_{n+1}$ (true in the limit since $|x_n - x_{n+1}|$ becomes arbitrarily small)
    $$
    x = \frac 12 \left(x + \frac{2}{x}ight) \iff x^2 = \frac 12 x^2 + 1 \iff x^2 = 2
    $$
    therefore $x = \pm \sqrt 2$, and since every $x_n$ is positive $x = \sqrt 2$.
  \item Let
    $$
    x_{n+1} = \frac{1}{2}\left(x_n + \frac{c}{x_n}ight)
    $$
    I won't go through the convergence analysis again, but the only fixed point is
    $$
    x = \frac 12\left(x + \frac{c}{x}ight) \implies \frac 12 x^2 = \frac 12 c \implies x^2 = c
    $$
    So if $x_n$ converges, it must converge to $x^2 = c$.
   \end{enumerate}

Exercise 2.4.5

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Exercise 2.4.5

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