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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
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