Binet's formula for the nth Fibonacci numbers is remarkable because the equation converts via a few arithmetic operations an irrational number $\phi$ into an integer sequence. However, using finite precision arithmetic, one would always have some (small) roundoff error. Is there a discussion/description somewhere of how to calculate the Fibonacci sequence using Binet's formula (ie not the recurrence relation) and floating point arithmetic which results in no roundoff error?
Calculating Binet's formula for Fibonacci numbers with arbitrary precision
algorithms;discrete mathematics;floating point;arithmetic;mathematical programming
As the comments show, roundig is a respected method for Fibonacci computations. See also Computation by rounding in the wiki-lemma.You avoid roundoff by computing exactly, but using numbers of the form $a+b\sqrt5$: $(a+b\sqrt5)(c+d\sqrt5) = (ac+5bd)+(ad+bc)\sqrt5$. Technically that is $Z[\sqrt5]$ I believe. But I now realise that you also have to take into account that $a,b$ are fractions.But the recursion is so nice, why not use it? Based on the matrix form you can get the $n$-th Fibo number in a logarithmic number of steps.