为什么ghci和编译代码的数值结果不同？

Question

我有一个计算定义积分的haskell代码。当我从ghci运行它时，我得到：

tol = 1.0e-12 :: 11.83933789116509 :: 0.0

如果第一列是容差，则第二列是整数值，第三列是错误。

但是当我运行绝对相同的代码时，只使用-O0编译，我得到：

tol = 1.0e-12 :: 11.839337891165126 :: 3.552713678800501e-14

为什么会这样？

P.S：我想，代码并不重要。只是很多双算术。

UPD ：这是最小的例子：

type R = Double
type F = R -> R

linearScale :: Fractional a => Int -> a -> a -> [a]
linearScale n x0 x1 = let h = x1 - x0 in
   map (\m -> x0 + h * fromIntegral m / fromIntegral n) [0 .. n]

triangulate :: [[Double]] -> [[Double]]
triangulate [] = []
triangulate (x:xs) = x : (triangulate xs')
    where xs' = map f xs
          f ys
            | head ys == 0 = drop 1 ys
            | otherwise = drop 1 $ zipWith (-) (map (* c) ys) x
            where c = (head x) / (head ys)

solve :: [[Double]] -> [Double] -> [Double]
solve ms bs = f (reverse ms) (reverse bs) []
    where f [] _ zs = zs
          f _ [] zs = zs
          f (x:xs) (y:ys) zs = f xs ys (z:zs)
            where z = (y - (sum $ zipWith (*) (tail x) zs)) / head x

gauss :: [[Double]] -> [Double] -> [Double]
gauss xs ys = solve ms bs
    where xs' = zipWith (\a b -> a ++ [b]) xs ys
          ts = triangulate xs'
          ms = map init ts
          bs = map last ts

integrateWithN :: Int -> F -> (Int -> R -> R -> R) -> R -> R -> R
integrateWithN n f mu a b = sum . zipWith (*) rs $ map f xs
    where xs = linearScale n a b
          mus = [mu s a b | s <- [0 .. n - 1]]
          xss = [map (^ i) xs | i <- [0 .. n - 1]]
          rs = gauss xss mus

main :: IO ()
main = print $ integrateWithN 60 f mu a b
    where
    f x = x
    intp k x = -0.39876 * exp(1.16315 * fromIntegral k) * (16 - 5 * x) ** (3/4)
    mu k a b = intp k b - intp k a
    a = 1.7
    b = 3.2

您将获得运行此代码的不同结果，该代码使用-O0和ghci编译。

UPD： ghc 7.10.3

我的结果：