Bryan Burgers wrote:
On 3/31/07, Scott Brown <[EMAIL PROTECTED]> wrote:
It's working now, thank you.
I changed the definition to
> binom n j = div (fac n) ((fac j)*(fac (n - j)))
> bernoulli n p j = fromIntegral(binom n j)*(p ^ j) * ((1 - p)^(n - j))
As a matter of style suggestion, it might make 'binom' more clear if
you use 'div' as an infix operator:
binom n j = (fac n) `div` ( fac j * fac (n - j) )
And as a matter of efficiency, no one would write binom using factorial,
but would rather write at least
binom u k = (product [(u-i+1) | i <- [1..k]]) `div` (product [1..k])
and even better would be to do it this way
-- bb u k = toInteger $ product [ (u-i+1) / i | i <- [1..k]]
but that does not type (for a good reason). The issue is that it is
possible to prove that the above is an integer, but the compiler can't
see that :-(
That can be done as
import Data.Ratio
bb u k = numerator $ product [ (u-i+1) / i | i <- [1..k]]
Of course, the above is fast if and only if the gcd operation in
Data.Ratio has been well optimized.
Jacques
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