使用递归的平方根（牛顿算法）

Question

Can someone explain to me this recursive pseudocode that finds the square root of a number? I find it difficult to understand since I'm not given what n, p, and e inputs stand for. Thanks.

if abs(e^2 - n) < p
    SR(n,p,e) = e     
else
    SR(n,p,e) = SR(n,p,(e+n/e)/2)

(e begins at n)

Answer 1

n is the number you want the square root of, e is an estimate for the square root, and p is the precision you want, ie the error you are willing to tolerate. The algorithm says: if e is "close enough" to the answer, ie e^2 is within p of n, then e is the answer you are looking for; otherwise, try a better estimate, (e+n/e)2. Why is that a better estimate? if e is larger than sqrt(n), then n/e will be smaller than sqrt(n), so sqrt(n) will be between e and n/e, so try the average of e and n/e as your next estimate. (And vice versa if e is less than sqrt(n)).

Hope this helps,

Bruce

Answer 2

There's rather more to Newton's algorithm than just going from one estimate to a "better estimate". There's some detailed maths behind why the better estimate is what it is.

The idea is that to find the solution to ANY equation of the form f(x) = 0 , (apart from a few exceptional cases), if you have an approximation to x , you can get a BETTER approximation, by looking at the rate of change of f(x) , which is often written f'(x) , and using that to work out how much you have to adjust your estimate by, to get a much better estimate of the true solution.

In the case of a square root, that is, we want to find x=sqrt(n) , we can write f(x)=x^2-n and f'(x)=2x , then use Newton's algorithm to find the right x to make f(x)=0 . What this means is that if we have an estimate e , then to work out our next estimate, we look at f(e)=e^2-n , and we ask how much we have to change e to get rid of this error. Since the rate of change of f is f'(x) , which is 2e at the point (e,e^2-n) , we should divide e^2-n by 2e to work out how much we have to adjust e by, to get our next estimate.

That is, our next estimate should be

  e - (e^2-n) / 2e
= e - (e / 2)  + (n / 2e)
= (e + n / e) / 2

More information on Newton's algorithm can be found at http://tutorial.math.lamar.edu/Classes/CalcI/NewtonsMethod.aspx (which has a lovely diagram that explains how it works) and at http://www.math.brown.edu/UTRA/linapprox.html which goes into some of the more technical detail.