[Haskell-cafe] Fibonacci numbers generator in Haskell

[Doug Quale]

Doug Quale <quale1 at charter.net> writes:

> Mathew Mills <mathewmills at mac.com> 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.

Sorry, I suffered brain lock.  The correct expression is parenthesized
differently:

> fib n = truncate(phi^n/sqrt 5 + 0.5)
>    where phi = (1 + sqrt 5)/2