# Multiplicity of zeros and derivative [closed]

Given a function $$f$$ and point $$x$$ we define the multiplicity of its zeroes as $$v_f (x)= \min(k\in \mathbb{N} \mid f^k(x)\neq0)$$ where $$k$$ denotes the $$k$$-th derivative. Given this definition what I wish to ask is this: How do I prove that $$v_{f\cdot g}(x)=v_f(x)+v_g(x)$$ where $$f\cdot g$$ represents the product of two functions.

$$(f\cdot g)^{(n)} (x) = \sum_{k=0}^n \binom{n}{k} f^{(n-k)} (x) g^{(k)} (x)\\$$

In order for this to be nonzero, at least one of the summands must be nonzero. In order for the $$k$$'th summand to be nonzero neither factor can be zero which means that:

$$n - k \geq v_f (x)\\ k \geq v_g (x)\\$$

Adding these two inequalities gives $$n \geq v_f (x) + v_g (x)$$. This shows that $$v_{f \cdot g}(x) \geq v_f (x) + v_g (x)$$.

But if we use $$N = v_f (x) + v_g (x)$$ only one term remains. The one with $$k = v_g (x)$$

$$(f \cdot g)^{(N)} (x) = \binom{N}{v_g (x)} f^{(v_f (x))} (x) g^{(v_g (x))} (x)\\$$

This is a product of nonzero quantities so it is nonzero as well. This shows that $$v_{f \cdot g} (x) \leq N$$.