Re: arcsin

*To*: mathgroup at christensen.cybernetics.net*Subject*: [mg1576] Re: [mg1550] arcsin*From*: Richard Mercer <richard at seuss.math.wright.edu>*Date*: Sat, 1 Jul 1995 02:58:06 -0400

> Is there a recurrence relation that one can use to compute > an arcsin??? How can one numerically compute the value > of arcsin with a recurrence relation (restated for > clarity)? Can someone direct me to some resource or > formula for numerical solutions of this function... > You don't say whether you want the BEST answer or just AN answer. Here is a simple 2-minute answer. A numerical analyst could certainly do better. Finding arcsin(a) can be seen as solving the equation a = sin(x), or sin(x) - a = 0, with the proviso that the solution should lie within the interval [-pi/2,+pi/2]. This can be done iteratively using Newton's Method with f(x) = sin(x) - a: (i) Select x(0) = 0. (ii) iterate the equation x(n+1) = x(n) - f(x(n))/f'(x(n)), or x(n+1) = x(n) - (sin(x(n)) - a)/cos(x(n)) If you choose a in [-1,1] (a good idea!) this will converge to a root, which may not lie with [-pi/2,pi/2]. By adding a multiple of 2pi, you can easily move the root into [-pi,pi]. Then using the identity sin(pi - x) = sin(x) you can find a root within [-pi/2,pi/2]. This method will have performance problems when a is close to -1 or +1. A more sophisticated method would overcome that. Richard Mercer