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Let $f(x)=x^{2}-a$. Show that the Newton Method leads to the recurrence
$x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{a}{x_{n}}\right) .$
Heron of Alexandria (60 CE?) used a pre-algebra version of the above recurrence. It is still at the heart of computer algorithms for finding square roots.
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We have $f(x)=2 x$. The Newton Method therefore leads to the recurrence
$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}=x_{n}-\frac{x_{n}^{2}-a}{2 x_{n}} .$
Bring the expression on the right hand side to the common denominator $2 x_{n}$. We get
$x_{n+1}=\frac{2 x_{n}^{2}-\left(x_{n}^{2}-a\right)}{2 x_{n}}=\frac{x_{n}^{2}+a}{2 x_{n}}=\frac{1}{2}\left(x_{n}+\frac{a}{x_{n}}\right) .$
by Platinum (101k points)

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