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.


1 Answer 1


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

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 $v_{f \cdot g} (x) \leq N$.


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