Re: [Haskell-beginners] Performance of Idiomatic lazy Haskell

Hi Markus Whoops, I hadn't read your email properly and wasn't accounting for the epsilon. Here's a version that does, although it is perhaps a little slow...

import Data.List (unfoldr)

leibniz eps = converge eps ser

ser :: [Double] ser = unfoldr phi (True,1) where phi (sig,d) | sig == True = Just (1/d, (False,d+2)) | otherwise = Just (negate (1/d), (True,d+2))

converge :: Double -> [Double] -> Double converge eps xs = step 0 0 xs where step a b (x:xs) = let a' = a + (4*x) in if abs (a'-b) < eps then a' else step a' a xs

demo1 = leibniz 0.00000000025

Best wishes Stephen