3

I have some function $F(\omega): \mathbb R\to\mathbb C$. The function $F(\omega)$ has both roots and singularities. Fortunately, I can calculate positions of singularities analytically.

So my problem is to find roots inside domain between singularities. Here I have difficulty, because I cannot estimate if there is one, many or none roots inside the domain. Now I scan $\omega$ through the domain for change of simultaneous change of real and imaginary part of $F(\omega)$, what is both inaccurate and ineffective.

If there some method to estimate number of roots inside a domain?

I don't know if this is a right place to post questions about computational math, so don't hesitate to show me the right place to post :)

J. M. ain't a mathematician
  • 75,051
galadog
  • 133
  • If $C$ is a closed contour that does not enclose any singularity then the number of zeroes inside $C$ is given by $$\int_C\frac{F'}{F}d\omega.$$

    If there are poles enclosed by $C$ you have to add them to the result to get the only the number of zeroes.

     – WimC Jun 01 '12 at 16:29
  • Thank you for your answer. I'll try this technique, the problem here is computation of $F'$. Am I right that contour can be just a segment in real axis? – galadog Jun 01 '12 at 16:46
  • I forgot the factor $(2 \pi i)^{-1}$. – WimC Jun 01 '12 at 16:49
  • @galadog no, the contour must be a closed loop in the plane. That integral calculates the number of zeros inside the loop. – Antonio Vargas Jun 01 '12 at 16:59

1 Answers1

2

See Computing the Zeros of Analytic Functions for theory and code.

Despite the title, it handles computing zeros and poles of meromorphic functions.

lhf
  • 216,483
  • Are we sure that the above reference really helps? From looking at the statement of the question, it appears we are talking about a function from the reals into the complex plane, rather than an analytic or meromorphic function. (Unless I have it all wrong). – Old John Jun 01 '12 at 16:48
  • @OldJohn: good point. Could still work, but that would depend a bit on the actual form of $F$. – WimC Jun 01 '12 at 16:52
  • @OldJohn, right, I just assumed the function could be extended to the complex plane. – lhf Jun 01 '12 at 17:00