Answer:
No, not usually. With the inherent inaccuracy of floating point this risks creating a non-terminating loop. This next example shows a while loop.
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 common 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 gets the exact value, it stays fixed at that
value.
Try it:
Say that n == 4 and that our first
appoximation to the square root is x == 1.
Use the formula to get the next approximation:
x' =(1/2)(1 + 4/1)