Re: [math-fun] fib(1/2)

Subject: [math-fun] fib(1/2)

Binet's and another formula give Sqrt[1/5 - (2 I)/5] and ((1 - I) (Cosh[ArcCsch[2]/2] + I Sinh[ArcCsch[2]/2]))/Sqrt[5] but FullSimplify can't get the former from the latter. But it agrees they're equal. --rwg

Using Maple: converting to exp gives an expression in radicals which 'radnormal' at least simplifies to -1/10*(2*I*5^(1/2)-5^(1/2)-5)/(1/2+1/2*5^(1/2))^(1/2); The package 'gfun' has a command 'algfuntoalgeq' which gives an equation for that, 8000-16000*I-40000*y^2 and solving it gives 1/5*(5-10*I)^(1/2) after picking that root.