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Suppose that $f \in \mathcal{C}^{k+1}$ is a function such that $f(a) = f'(a) = \cdots f^{(k-1)}(a) = 0$ and $f^{(k)}(a) \neq 0$. Can we write $f(x) = (x-a)^kg(x)$ for some $g \in \mathcal{C}^2$?

My attempt:

Using the Taylor expansion formula and using the hypothesis $f(a) = f'(a) = \cdots f^{(k-1)}(a) = 0$ , we have: $$ f(x) = f^{(k)}(a)\frac{(x-a)^k}{k!} + f^{(k+1)}(\theta)\frac{(x-a)^{k+1}}{(k+1)!}$$

where $\theta$ depends on $x$. If we he define: $$ g(x) = \frac{f^{(k)}(a)}{k!} + f^{(k+1)}(\theta)\frac{(x-a)}{(k+1)!}$$ then $f(x) =(x-a)^kg(x)$. But it is not clear if $g\in \mathcal{C}^2$ since $\theta$ depends on $x$. I was not able to give a counterexample that shows that $g\notin \mathcal{C}^2$ either.

I would appreciate any help. Thanks.

Santos
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  • @MartinR : if we consider $k=1$ in that $f$, its second derivative is not continuous, therefore $f \in \mathcal{C}^2$. – Santos Jun 03 '21 at 13:54
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    Try using the integral form of the remainder instead. True or false, it should follow immediately from that. – Klaus Jun 03 '21 at 13:55
  • Can't you just define $f(x) = (x-a)^k g(x)$ where $g$ is differentiable, but $g''(a)$ does not exist? – Martin R Jun 03 '21 at 13:55
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    @MartinR Yes, already $k = 0$ shows that the numbers don't match. $g \in C^1$ should be ok, though (or change $f \in C^{k+1}$ to $f \in C^{k+2}$ instead). – Klaus Jun 03 '21 at 14:04
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    What about the counterexample that was given here https://math.stackexchange.com/a/4151041 in response to your previous question? – Martin R Jun 03 '21 at 15:10

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