No, not usually. With the inherent inaccuracy of floating point this risks creating a non-terminating loop. This next example shows a while loop.
Newton's method
is a way to compute the square root of
a number.
Say that n is the number and that x
is an approximation to the square root of n.
Then:
x' =(1/2)(x + n/x)
x' is an even better approximation to the square root.
The reasons for this are buried in your calculus book.
Here is what happens if
the approximation x happens to be exactly the
square root.
In other words, what if x = n0.5.
Then:
x' = (1/2)(x + n/x)
= (1/2)( n0.5 + n/n0.5 )
= (1/2)(n0.5 + n0.5)
= n0.5
If x reaches the exact value, it stays fixed at that
value.
Try it:
Say that n == 4 and that our first
approximation to the square root is x == 1.
Use the formula to get the next approximation:
x' =(1/2)(1 + 4/1)