If $f: \mathbb{R} \rightarrow \mathbb{R} \in C^{n}$ (class $C^{n}$) and $f^{(i)}(a) = 0, \forall i \in \{0,1,2,\cdots ,n\}$ . Prove that for $k\leq n$ :
$$ f(x)=(x-a)^k g(x)$$
Where $g : \mathbb{R} \rightarrow \mathbb{R} \in C^{n-k}$
My attempt :
By Taylor's theorem: for $x\in \mathbb{R}$ exists $c_{x}$ between $x$ and $a$ such that :
$$f(x)=f(a)+f'(a)(x-a)+\cdot + \dfrac{f^{(k-1)}(a)}{(k-1)!}(x-a)^{k-1}+\dfrac{f^{(k)}(c_x)}{k!}(x-a)^{k} $$
Then :
$$ f(x) = (x-a)^{k} \dfrac{f^{(k)}(c_x)}{k!}$$
My idea is define $g(x)= \dfrac{f^{(k)}(c_x)}{k!}$
I would like to know if that definition is good or not.
The fact that $g$ is $C^{n-k}$ is clear from the fact that $f$ is $C^{n}$
You are complicating things by introducing $c_x$. Why not just define $g(x)$ to be $\frac {f(x)} {(x-a)^{k}}$ for $x \neq a$ and $0$ for $x =a$. Repeated application of L'Hopital's Rule show that $g$ is $C^{n-k}$.