Finding the limit of a recursive sequence $x_{n+1}=\frac{1}{2}(x_{n}+\frac{M}{x_{n}})$

Question

Define the sequence $\{x_n\}_{n\in\mathbb{N}}$ by $$x_0=\frac{M}{2}, \qquad x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{M}{x_{n}}\right)$$ where $M\in\mathbb{R}$, $M\geq 0$. Find $\lim\limits_{x_n\to\infty}x_{n}$ and prove it exists.

My idea was, if I can prove that $x_{n}\geq x_{n+1}$ then I can say that

$$x_{n}\geq x_{n+1}\iff x_{n}\geq\dfrac{1}{2}\left(x_{n}+\dfrac{M}{x_{n}}\right)\iff 2x_{n}\geq x_{n}+\dfrac{M}{x_{n}}\iff$$ $$ x_{n}\geq\frac{M}{x_{n}}\Longleftrightarrow x_{n}^{2}\geq M\Longleftrightarrow x_{n}\geq\sqrt{M}$$

but I couldn't find a way to prove that $x_{n}\geq x_{n+1}$.

Answer 1

Assuming that $M>0$:

First we show that $(x_n)$ is bounded below:

Note $x_n$ satisfies $x_n^2-2x_{n+1}x_n+M=0$. This quadratic has a real root, so $4x_{n+1}^2-4M\ge 0$.

So $x_{n+1}^2\ge M$.

Next, we show that $(x_n)$ is eventually decreasing:

For $n\ge 2$ $$x_n-x_{n+1}=x_n-{1\over2}\Bigl(x_n+{M\over x_n}\Bigr)={1\over2}{x_n^2-M\over x_n}\ge 0$$

Thus, since $(x_n)$ is bounded below by $\sqrt M$ and is eventually decreasing, $(x_n)$ converges to some $L$ with $L\ge \sqrt M>0$. But then $$ x_{n+1}\rightarrow{1\over2}\Bigl( L+{M\over L}\Bigr). $$

We then must have $$ L={1\over2}\Bigl( L+{M\over L}\Bigr); $$ which implies $L=\sqrt M$.

If $M=0$, then $(x_n)$ obviously converges to 0.

Answer 2

For any positive $M$, this method seems to converge and is decreasing towards a solution after the first iteration (for $x_0 > 2$, it is always decreasing, but for $x_0 < 2$, it takes one iteration to increase and then decreases towards the solution. I'll explain that below). To have $x_n \ge x_{n+1}$, you need $$ x_n \ge x_{n+1} = \frac 12 \left( x_n + \frac M{x_n} \right) \qquad \Longleftrightarrow \qquad x_n^2 - M \ge 0 \qquad \Longleftrightarrow \qquad x_n \ge \sqrt M $$ if we assume that $x_0$ is positive to begin with. (Otherwise, one can prove that a convergent sequence following this pattern will be a root of $x^2 - M$ with $M$ negative, which means that the root $x$ would be complex, which is not what you expect I assume). We show further that $x_n \ge x_{n+1} \ge M$. Now we can see that \begin{align} x_n^2 - 2 \sqrt M x_n + M = (x_n-\sqrt M)^2 \ge 0 \quad & \Longrightarrow \quad x_n + \frac{M}{x_n} \ge 2 \sqrt M \\ & \Longrightarrow \quad \frac 12 \left( x_n + \frac {M}{x_n} \right) \ge \sqrt M \\ \end{align} and therefore we know that after the first iteration, $x_n$ will always be above $\sqrt M$, since the only thing I assumed here was that $x_0$ was positive (strict).

Therefore the sequence is converging because it decreases and it is bounded below by $\sqrt M$. To find the limit, we have $$ x_n \to L \qquad \Longrightarrow \qquad x_{n+1} \to L = \frac 12 \left( L + \frac ML \right) \qquad \Longrightarrow \qquad L = \sqrt M $$ by knowing the limit has to be positive (you'll find a quadratic with two roots in there).

To explain the first iteration thing, you have to note that $$ x_0 = \frac M2 \ge \sqrt M \qquad \Longleftrightarrow \qquad (\sqrt M)^2 - 2 \sqrt M \ge 0 $$ which in turn is equivalent to $\sqrt M \ge 2$, which means $M \ge 4$ and $x_0 \ge 2$. Hope that helps,

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{x_{0}\equiv{M \over 2},\quad x_{n+1}\equiv\half\,\pars{x_{n} + {M \over x_{n}}}.\qquad M \geq 0.}$

Let's $\ds{y_{n} = {x_{n} \over \root{M}}\quad\imp\quad y_{n + 1} = \half\pars{y_{n} + {1 \over y_{n}}}\,,\quad y_{0} = {\root{M} \over 2}}$ \begin{align} {y_{n + 1} + 1 \over y_{n + 1} - 1}& ={\half\pars{y_{n} + 1/y_{n}} + 1 \over \half\pars{y_{n} + 1/y_{n}} - 1} ={y_{n}^{2} + 2y_{n} + 1 \over y_{n}^{2} - 2y_{n} + 1} =\pars{y_{n} + 1 \over y_{n} - 1}^{2} \end{align}

\begin{align} {y_{n + 1} + 1 \over y_{n + 1} - 1}& =\pars{y_{n - 1} + 1 \over y_{n - 1} - 1}^{4} =\pars{y_{n - 2} + 1 \over y_{n - 2} - 1}^{8}=\cdots =\pars{y_{0} + 1 \over y_{0} - 1}^{2^{n + 1}} \end{align}

$$ y_{n + 1} = {\pars{y_{0} + 1}^{2^{n + 1}} + \pars{y_{0} - 1}^{2^{n + 1}} \over \pars{y_{0} + 1}^{2^{n + 1}} - \pars{y_{0} - 1}^{2^{n + 1}}} $$ Now, you can take from here.

Answer 4

if the limit $x$ exists, then $x=(1/2)(x+M/x)$ and $x=\pm\sqrt{M}$. see newton's method applied to $f(x)=x^2-M$