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 $x_n$ ?"
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 $$x_{n}=\sqrt{a}\cdot\dfrac{1+\left(\dfrac{x_{1}-\sqrt{a}}{x_{1}+\sqrt{a}}\right)^{2^{n-1}}}{1-\left(\dfrac{x_{1}-\sqrt{a}}{x_{1}+\sqrt{a}}\right)^{2^{n-1}}}$$