If $f$ is smooth and we write with Taylor's formula $f(x)=f(0)+f'(0)x+...+\frac{1}{k!}f^{(k)}(0)x^k+h(x)x^k$ is then also $h$ a smooth function?

Question

Let $f\colon\mathbb{R}\to\mathbb{R}$ be smooth and write with Taylor's formula $$f(x)=f(0)+f'(0)x+...+\frac{1}{k!}f^{(k)}(0)x^k+h(x)x^k$$ is then also $h$ a smooth function? Obviously $$h(x)=\frac{1}{x^k}\left(f(x)-f(0)-f'(0)x-...-\frac{1}{k!}f^{(k)}(0)x^k\right)$$ is smooth for $x\neq 0$. I can only show that $h$ is once-differentiable at $0$. (Arguing similar as in the solution to the question here. If needed I can provide the details.) Thanks for any ideas!

Answer 1

Taylor's theorem with the integral form of the remainder is $$ f(x)=f(0)+f'(0)x+...+\frac{1}{k!}f^{(k)}(0)x^k+ R_k(x) $$ with $$ R_k(x) = \frac{1}{k!} \int_0^x (x-t)^k f^{(k+1)}(t) \, dt = \frac{x^k}{k!} \int_0^1 (1-s)^k f^{(k+1)}(xs) \, ds $$ so that your $h(x)$ has the representation $$ h(x) = \frac{1}{k!} \int_0^1 (1-s)^k f^{(k+1)}(xs) \, ds \, . $$ It follows that if $f$ is “smooth” (in the sense of “infinitely differentiable”) then the same holds for $h$. More generally, if $f^{(k+1)}$ exists and is continuous then $h$ is continuous, and if $f^{(k+1+l)}$ exists and is continuous then $h$ is $l$ times continuously differentiable.