# Is the remainder of first-order Taylor expansion still continuously differentiable?

Let $f: {\mathbb R}^n \to {\mathbb R}^n$ be a continuously differentiable function. Then, we can rewrite its first-order Taylor expansion at $x \in {\mathbb R}^n$ for $h \in {\mathbb R}^n$ that \begin{align} f(x+h) = f(x) + f'(x) h + r(h), \end{align} where $f'(x) \in {\mathbb R}^{n\times n}$. Since $f$ is differentiable, we have \begin{align} r'(0)=\lim_{h\to 0} \frac {r(h)}{h} = \lim_{h\to 0}\frac {f(x+h) - f(x) - f'(x) h}{h} = 0. \end{align}

Now, the question is coming. Is r(h) continuously differentiable for $h$?

From my point of view, r(h) is continuously differentiable for $h$, and the reason is given as follows:

I rewrite the first equation as \begin{align} r(h) = f(x+h) - f(x) - f'(x) h. \end{align}

For any fixed $x \in {\mathbb R}^n$, $f(x+h)$ is continuously differentiable w.r.t. $h$, so is $f'(x) h$. Thus, in my opinion, $r(h)$ is continuously differentiable with $r'(0) = 0$.

However, I am not sure this reason is correct...

• yes, it's correct
– zhw.
Oct 20, 2015 at 5:49

Yes, of course. $r(h) = f(x+h) - f(x) -f'(x)h$ and all terms on the right hand side are continuously differentiable, then so is the left hand side.