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Let $a$ be a positive real number and $(x_n)$ be the sequence given by $x_1>0,$ $$x_{n+1}=\dfrac{1}{2}\Big(x_n+\dfrac{a}{x_n}\Big).$$ It is not difficult to prove that $x_n\to\sqrt{a}$ as $n\to\infty.$

My question is : "Is there any explicit formula for $x_n$ ?"

Bumblebee
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  • I have seen some math contest problems concerning series like $$x_{n+1}=\dfrac{1}{2}\Big(x_n+\dfrac{1}{x_n}\Big).$$ but have never seen an explicit formula for $x_n$. – Vim Feb 02 '15 at 10:01
  • Well this case is way too special LOL – Vim Feb 02 '15 at 10:09
  • I had a try for the case where $x_1=2$ and $a=1$, then the first few terms go like this. $x_1=\frac{2}{1},x_2=\frac{5}{4},x_3=\frac{41}{40},x_4=\frac{3281}{3280},x_5=21523361/21523360...$,So it comes to find a formula for the denominator sequence..according to my observation as $n$ goes up it just goes crazily large...so it seems just too hard to me – Vim Feb 02 '15 at 10:30
  • @Vim: If you substitute $$y_n=\dfrac{x_n}{\sqrt{a}},$$ then you can transform given recurrence into your recurrence relation. May be you can try to give an alternative proof. Good Luck. – Bumblebee Mar 30 '15 at 05:04

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since $$x_{n+1}=\dfrac{x^2_{n}+a}{2x_{n}}$$ so we have $$x_{n+1}-\sqrt{a}=\dfrac{(x_{n}-\sqrt{a})^2}{2x_{n}}\tag{1}$$ $$x_{n+1}+\sqrt{a}=\dfrac{(x_{n}+\sqrt{a})^2}{2x_{n}}\tag{2}$$ $(1)/(2)$ we have $$\dfrac{x_{n+1}-\sqrt{a}}{x_{n+1}+\sqrt{a}}=\left(\dfrac{x_{n}-\sqrt{a}}{x_n+\sqrt{a}}\right)^2=\cdots=\left(\dfrac{x_{1}-\sqrt{a}}{x_{1}+\sqrt{a}}\right)^{2^{n}}$$ so we have $$x_{n}=\sqrt{a}\cdot\dfrac{1+\left(\dfrac{x_{1}-\sqrt{a}}{x_{1}+\sqrt{a}}\right)^{2^{n-1}}}{1-\left(\dfrac{x_{1}-\sqrt{a}}{x_{1}+\sqrt{a}}\right)^{2^{n-1}}}$$

Rory Daulton
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math110
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