Re: Solve/Approx: th / (1 - cos th) = c

From: edA-qa mort-ora-y (usenet_at_disemia.com)
Date: 09/15/04 

Date: Wed, 15 Sep 2004 13:28:20 +0200

David W. Cantrell wrote:
> realized its importance, based on your experience in the thread
> "Solve: cos(th) / th = c ?" from last July:

Yes, I realize the limits of c and the expected results were quite
important -- I made a mistake not posting them the first time.

I am however quite interested in knowing how these problems are done in
general, and your reply to the main post is quite helpful in that
regards. My desk is quite full of expressions with sin/cos and I
eventually need to approximate all of them (most of them have other
geometric approximations I can use in the meantime).

So, to the specific problem:
        th / (1 - cos th) = c
Solve/approx for th

> Is c perhaps restricted to some finite interval?
        0 <= c < 1

there must be an answer in the range 0 <= th <= pi, with an expected
answer in 0 <= th <= pi/2

> What error would be acceptable in an approximate solution?
> Is that absolute or relative error?
> Is that error in th, or is that actually (based on your desire in the
> previously cited thread) error in c once the approximation for th is
> substituted in th/(1 - cos(th))?

        Substituing calculated th back into equation should result in a c that
is within 0.001 of the input c (relative error wrt c). 0.01 is also
workable so long as the error is continuous and smooth. 

-- 
edA-qa mort-ora-y
Idea Architect
http://disemia.com/