# Most general form of an analytic function vanishing (with some derivatives) at some points.

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

(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. ?

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.
• @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. – Kavi Rama Murthy Sep 3 '19 at 10:07