The taylor expansion of the multiplication of two functions

Given two function $$f,g \in C^n(\mathbb R)$$, $$h = f*g$$

$$h(x):=f(x)*g(x)$$

how can I show that:

$$T_nh(x) = [T_nf(x)*T_ng(x)]_n$$

where $$[P]_n$$ is the "trimmed" series (The Taylor polynomial will be with the degree <= n)

what I've tried:

• Induction: failed proving the induction step.
• Proving from this lemma: $$lim_a (f(x)-T_n(a))/(x-a)^n = 0$$

WITHOUT USING GENERAL LEIBNIZ RULE

• Hint: you can use the Leibnitz Rule for higher derivatives of the product of two functions. See here: en.wikipedia.org/wiki/General_Leibniz_rule – jflipp Mar 23 at 20:07
• @jflipp Hi, thank you but I didn't mentioned that (obviously) this question is part of my home work, and the next question is to prove Leibniz rule using this question so I cannot use it yet XD – MercyDude Mar 23 at 20:12

Proving from this lemma: $$\lim_{x\to a}\dfrac{f(x)-T_{n,a}(x)}{(x-a)^n}=0$$
What you need is the converse of that lemma: if $$f$$ is at least $$n$$ times differentiable at $$a$$, and $$\lim_{x\to a}\dfrac{f(x)-P(x)}{(x-a)^n}=0$$ for some $$n$$th degree polynomial $$P$$, then $$P$$ is the Taylor polynomial of degree $$n$$ centered at $$a$$ $$T_{n,a}(x)$$.
Why is this true? Subtract the two limits: $$0 = \lim_{x\to a}\left(\frac{f(x)-T_{n,a}(x)}{(x-a)^n}-\frac{f(x)-P(x)}{(x-a)^n}\right) = \lim_{x\to a}\frac{P(x)-T_{n,a}(x)}{(x-a)^n}$$ That numerator is a polynomial of degree at most $$n$$. For the fraction to go to zero, that polynomial must be divisible by $$(x-a)^{n+1}$$, which means it has to be identically zero.