Re: [Haskell-cafe] Fibonacci numbers generator in Haskell

Mathew Mills writes:

Is there anything that can be done (easily) to reduce the rounding errors?

The hint that I gave before is one easy way.

fib :: Integer -> Integer fib x = let phi = ( 1 + sqrt 5 ) / 2 in truncate( ( 1 / sqrt 5 ) * ( phi ^ x + 0.5) )

You run out of precision eventually. IEEE Double's give you about 15 decimal digits, so the results become approximate for x > 75.