1
$\begingroup$

If I have an analytic function in plane $F(x,y)$ that is zero on a curve $y=f(x)$, is it true that $F=(y-f(x))^n h$, where $h$ is nonzero on the curve? More general, can be somethink said about factorisation of analytic functions? How much is it determined by its zero set? Thx

$\endgroup$
  • $\begingroup$ You are using $f$ for two different things... $\endgroup$ – Mariano Suárez-Álvarez Oct 14 '10 at 14:58
  • $\begingroup$ Thank you very much. Does somethink like that hold also over the reals? Peter $\endgroup$ – Peter Franek Oct 14 '10 at 20:38
  • $\begingroup$ Yes it does.... $\endgroup$ – Thierry Zell Oct 14 '10 at 20:45
3
$\begingroup$

The answer is Weirstrass preparation Theorem.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

You need a combination of Weierstrass preparation and Puiseux series expansion to factor the analytic function, but it is indeed possible. Keep in mind that this is a local factorization near a point of your choice, that the factors may be complex valued and singular (=Holder continuous) at the point, but they are analytic outside the point. Better than writing here a lengthy explanation let me point you at a paper where I wrote all the details since I could not find them in the literature, although this stuff must be well known. See Section 2 of this paper.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.