It may be important (if you're actually doing something besides your
math homework with this
iterative method) to know what kind of error you're generating by using such approximations. Taking a slightly different approach from that of
PiGuy, look at Newton's method as derived from
Taylor Polynomial expansion:
Suppose that f (and its first and second derivative) are continuous on the interval [a,b] Let x' (an arbitrary value on the given interval) be a guess of what p (the root of our function f) might be, with the stipulation that f'(x') != 0, and that the difference between x' and the actual root p is "small" (ahh math terms). The first Taylor Polynomial of f(x) about x' is:
f(x) = f(x') + (x-x')f'(x')+((x-x')²/2!)f''(þ(x))
Where þ(x) is somewhere between x and x'. Since
f(p) = 0 (that's the solution we're looking for, so it is true by definition), evaluating the above equation with x = p gives:
f(p) = 0 = f(x') + (p-x')f'(x')+((p-x')²/2!)f''(þ(p))
Newton's Method assumes that (since the difference between x' and p is small) |p-x'| is small and therefore (p-x')² is a lot smaller. (A number less than one, when
squared, becomes smaller.) Thus
that term is dropped and you end up with:
0 (approximately)= f(x') + (p-x')f'(x)
And you can see the rest of that in
PiGuy's writeup.
This WU is concerned with the term that was dropped. It is, in its entirety:
((p-x')²/2!)f''(þ(x'))
"That's not of much use," you might be thinking, "You don't know what þ(x') is, not to mention x'."
You would be correct in thinking that it is unknown, but incorrect in thinking that it's not useful. Given the function's second derivative (f''(x)) and the interval over which it is to be evaluated, it is possible to find the maximum value of that second derivative on the interval. Even if you can't get the third derivative to do critical point analysis, you could still brute force the numbers out of it computationally to find an approximate maximum. So now we have that part of the term. Finally, there are a couple of ways to find (p-x'), but one is to take the worst-case scenario (half the width of the available interval) and plug that in. You can plug in the minimums of these values as well and find your error bounds.