$$a_0 = \max{\{x,1\}}$$ $$a_{n+1} = {\frac{1}{2} (a_n + \frac{x}{a_n})}$$
I really have some trouble understanding this recursive defined sequence. I just don't know how to calculate the n'th term or even finding $a_1$ first.
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1$\begingroup$ I suppose you mean $a_{n+1}$? $\endgroup$– BernardCommented Nov 19, 2017 at 11:59
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$\begingroup$ oh yes I actually meant $a_{n+1}$. Sorry for the typo $\endgroup$– MsmatCommented Nov 19, 2017 at 12:01
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1$\begingroup$ Hint: the sequence is monotonic. $\endgroup$– GribouillisCommented Nov 19, 2017 at 12:10
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1$\begingroup$ The result can be computed explicitly. Step 1: remove x by letting a(n) = Sqrt(x) b(n) $\endgroup$– Dr. Wolfgang HintzeCommented Nov 19, 2017 at 12:23
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2$\begingroup$ Possible duplicate of Find the limit if it exists of $S_{n+1} = \frac{1}{2}(S_n +\frac{A}{S_n})$ $\endgroup$– rtybaseCommented Nov 21, 2017 at 22:37
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1 Answer
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by the $AM-GM$ inequality you can Show that $$\frac{1}{2}\left(a_n+\frac{x}{a_n}\right)\geq \sqrt{a_n\cdot \frac{x}{a_n}}=\sqrt{x}$$
answered Nov 19, 2017 at 12:10
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$\begingroup$ thank you that really helped me!! And It's the first time I hear from the Inequality of arithmetic and geometric means $\endgroup$– MsmatCommented Nov 19, 2017 at 12:52
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1$\begingroup$ it is $$\frac{a+b}{2}\geq \sqrt{ab}$$ for $$a\geq 0,b\geq 0$$ $\endgroup$ Commented Nov 19, 2017 at 12:53