23 Mar
2007
23 Mar
'07
7:41 a.m.
It is :-
my_sqrt t = last (take 20 (iterate (\n -> n/2 + t/(2 * n)) t))
It is a bit crude though. 20 iterations is a bit arbitrary. I don't suppose there is a easy way to iterate until the results stop changing.
Here's something for you to play with: my_sqrt t = fix (\n -> n/2 + t/(2 * n)) t fix :: (Double -> Double) -> Double -> Double fix f x | x == fx = x | otherwise = f (fix f fx) where fx = f x My numerical analysis is a bit on the dodgy side, but you probably don't want to use == to test for convergence. Maybe you can find a nice way to write fix using some prelude functions? Cheers, Bernie.