Newton's iteration for square roots

Let g(a)=s-f(a) be the function whose roots we seek to find, where f(a) is an arbitrary function and s is a constant. If we find the roots of g(a)=0, then we also have found the roots of f(a)=s. Newton's iteration for g(a)=0 can be written as g(a_t)=-g'(a_t)(a_t+1-a_t), or

 

s-f(a_t)=f'(a_t)(a_t+1-a_t) (1)

If we now consider f(a)=a², then we can write (1) as:

 

s-a_t²=2a_t(a_t+1-a_t) (2)

which gives Newton's iterative procedure for finding square roots  

$$a_{t+1} = {1\over 2} \left( a_t + {s\over a_t} \right).$$ (3)