# 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?

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!

• This may be of help with little effort you may modify the argument to suit your situation. – Oliver Diaz Sep 21 '20 at 19:40

## 1 Answer

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.