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I am trying to calculate analytic solution (or locus) of zeros of a very large multi-variable function which is consisted of thousands of nonlinear trigonometric terms. All the variables are real numbers. The function is not differential.

(The equation is a Singular Value of the constrains matrix of a mechanical system)

Here is an example of what I want:


For this equation: $$ f = y\sin(\theta) - z\cos(\theta) $$ Where $ \theta,y,z \in \mathbb{R} $ (and if it helps: $ 0 \le \theta < 2\pi$).

I want all the sets of real values of $\theta$, $y$ and $z$ that satisfy $f = 0$, which would be the following three set of answers:

$$ \theta = \arctan(\frac{z}{y}) $$

$$ z = 0,\ \theta = \{0,...\},\ y\in \mathbb{R} $$

$$ y = 0,\ \theta = \{\pi/2,...\},\ z\in \mathbb{R} $$

Q: Is there any way in any software that this solution could be achieved?

Any help would be much appreciated! Thanks.

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In the particular case you give, you can make the substitution $t = tan(\theta)/2,$ which transforms your equation into a single polynomial in three variables, of which you want to find all real solutions (you can then back out the $\theta$). If your trig functions depend nicely on the argument, this reduction will always work, at which point (since you say "function") you want to find all real real zeros of a multivariate polynomial. I am not sure what form you want for these solutions (generally there will not be a nice parametrization), but Mathematica's Solve[] or Reduce[] will often give you something useful.

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  • $\begingroup$ Thanks a lot for your answer, I will look into your solution and report back! $\endgroup$
    – p8me
    Commented Jul 7, 2013 at 0:28
  • $\begingroup$ The symbolic expression is extremely large (consists of hundreds of millions of terms) and I found it impossible to simplify, and thus I guess nothing would work in my case. So now I am trying to choose a completely different path. $\endgroup$
    – p8me
    Commented Aug 20, 2013 at 3:28

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