most recent 30 from http://mathoverflow.net 2013-05-19T11:28:20Z http://mathoverflow.net/feeds/question/42157 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42157/factorisation-of-analytic-functions peter franek 2010-10-14T14:56:45Z 2010-10-14T20:38:13Z

<p>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</p>

http://mathoverflow.net/questions/42157/factorisation-of-analytic-functions/42159#42159 Denis Serre 2010-10-14T15:20:34Z 2010-10-14T15:20:34Z

<p>The answer is Weirstrass <a href="http://en.wikipedia.org/wiki/Weierstrass_preparation_theorem" rel="nofollow">preparation Theorem</a>.</p>

http://mathoverflow.net/questions/42157/factorisation-of-analytic-functions/42188#42188 Piero D'Ancona 2010-10-14T18:08:13Z 2010-10-14T18:08:13Z

<p>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 <a href="http://www.informaworld.com/smpp/content~db=all~content=a713645852~frm=abslink" rel="nofollow">this</a> paper.</p>

http://mathoverflow.net/questions/42157/factorisation-of-analytic-functions/42206#42206 peter franek 2010-10-14T20:38:13Z 2010-10-14T20:38:13Z

<p>Thank you very much. Does somethink like that hold also over the reals? Peter</p>