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I have a very specific question about a part of the proof that the sequence that converges to $\sqrt{2}$ given by: $x_1 = 1, x_{n+1} = \frac{1}{2}(x_n + \frac{2}{x_n})$ is monotonically decreasing.

While I do understand "how" it converges and why showing that $x_{n+1}-x_n \leq 0$ proves that the sequence is monotonically decreasing, I don't understand how I get to $x_{n+1} - x_n = \frac{1}{2}(\frac{2}{x_n}-x_n)$ without knowing what $x_n$ looks like. Thanks for any quick hints and sorry if I'm missing the obvious.

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  • $\begingroup$ The formula for $x_{n+1}-x_n$ just came from manipulating the definition in the first paragraph to pull over a single term of $x_n$. Was that what you were asking? $\endgroup$ – Matthew Daly Aug 16 '19 at 12:04
  • $\begingroup$ Yes, it was. Wow, now that I see the solutions I feel stupid. $\endgroup$ – psyph Aug 16 '19 at 12:06
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    $\begingroup$ Absolutely not. Stupid people don't ask questions when they are confused, which is why they stay stupid. ^_^ $\endgroup$ – Matthew Daly Aug 16 '19 at 12:07
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    $\begingroup$ Note that because $1 < \sqrt2$ you have $x_1 < x_2$ so it is only monotonically decreasing for later steps $\endgroup$ – Henry Aug 16 '19 at 12:12
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$$x_{n+1} - x_n = \frac 12 \left(x_n+\frac{2}{x_n}\right) - x_n = \frac 12 \left(x_n+\frac{2}{x_n}\right) - \frac 12 \cdot(2x_n) = \frac 12 \left(x_n+\frac{2}{x_n} - 2 x_n\right)=\frac 12 \left(\frac{2}{x_n}-x_n\right).$$

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  • $\begingroup$ Thanks, I didn't know I could just do that. $\endgroup$ – psyph Aug 16 '19 at 12:07
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Just plug in $x_{n + 1}=\frac12(x_n+\frac{2}{x_n})$, i.e., $x_{n+1}-x_n = \frac12(x_n+\frac{2}{x_n})-x_n$. Now you can simplify things.

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An easy way is that$$x_{n+1}={1\over 2}\left(x_n+{2\over x_n}\right)={\sqrt 2\over 2}\left({x_n\over \sqrt 2}+{\sqrt 2\over x_n}\right)\ge\sqrt 2$$therefore$$x_{n+1}={1\over 2}\left(x_n+{2\over x_n}\right)={1\over 2}x_n+{1\over x_n}\le {1\over 2}x_n+{1\over 2}x_n=x_n$$

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