3 Nov
2004
3 Nov
'04
7:35 a.m.
On Wed, 3 Nov 2004, Marc Charpentier wrote:
Thank you all for the friendly and helpful explanations - and for your patience.
The solution to my problem is finally
f :: Double -> Double f i = (-1)**i/(2**(10*i)) * (-2^5/(4*i+1)-1/(4*i+3)+2^8/(10*i+1) -2^6/(10*i+3)-2^2/(10*i+5)-2^2/(10*i+7)+1/(10*i+9))
I'm afraid that you will get problems with negative bases if you use (**) which allows fractional exponents. If your function is inherently based on positive integer i's you should stick to (^). Further on you should use one of the signatures f :: Integral a => a -> Double f :: Int -> Double f :: Integer -> Double to assert statically that f will always receive integer values and nothing else.