Starting by example: If for an analytic function $f$ one has:

$$ f(x_1) = 0, \quad f'(x_1) = 0, \quad f(x_2) = 0 $$

can $f$, in its most general form, be always written as

$$ f(x) = (x-x_1)^2(x-2)g(x) $$

where $g$ is (some) member of the set of all analytic functions?

Generally speaking, if for analytic $f$ one has

$$ f^{(k)}(x_n) = 0 \text{ for all } k<k_n $$

(for somehow chosen $k_n$) can then $f$ be always written as

$$ f(x) = g(x)\times \Pi_{i=1}^{n_{max}} (x-x_i)^{k_i} $$

where $g$ is a general analytic function. ?


(I suppose you are talking about analytic functions on the entire complex plane).

This is true and it follows from the fact that if $f$ has a zero of order $n$ at $z_0$ then $f(z)=(z-z_0)^{n}g(z)$ for some analytic function $g$.

Proof: Let $f(z)=\sum\limits_{k=0}^{\infty} a_k (z-z_0)^{k}$ be the power series expansion of $f$. If $f(z_0)=0$ and $f$ is not identically $0$ then there is a smallest integer $n$ such that $a_0=a_1=...=a_{n-1}=0$ and $a_n \neq 0$. This $n$ is called the order of zero of $f$ at $z_0$. Since the original power series converges so does $\sum\limits_{k=n}^{\infty} a_k(z-z_0)^{k-n}$. [The two series have the same radius of convergence. Let $g(z)$ be the sum of this series. Then $f(z)=(z-z_0)^{n} g(z)$ and $g$ is analytic.

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  • $\begingroup$ I would like to see a more rigorous proof. From (finite) polynomial you can always factor out roots. Bat can you do this with an "infinite" polynomial? $\endgroup$ – F. Jatpil Sep 3 '19 at 9:54
  • $\begingroup$ Yes you can. Due to the analyticity you can really derive and evaluate even in the series expansion without trouble $\endgroup$ – b00n heT Sep 3 '19 at 10:02
  • $\begingroup$ @F.Jatpil I have added a proof of the existence of $g$ in the case of a sinlgle point $x_1$. You just have to repeat this till you exhaust all the $x_i$'s. $\endgroup$ – Kavi Rama Murthy Sep 3 '19 at 10:07

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