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

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 (which you, perhaps, have buried in the darkest corner of your basement). But, to make the formula plausible, look 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.

### QUESTION 8:

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)
```