Show that if $|f(v)|\leq C\|v-v_0\|^{k+1}$ and $f$ is $k+1$ times differentiable, then the Taylor polynomial of $f$ about $v_0$ is $0$ Let $f:\mathbb{R}^n\to\mathbb{R}$ be $k+1$ times differentiable and suppose for some $v_0\in\mathbb{R}^n$ that $|f(v)|\leq C\|v-v_0\|^{k+1}$ in some neighbourhood of $v_0$. Show that the $k$-th order Taylor polynomial of $f$ about $v_0$ is $0$.
So I managed to prove that $f(v_0)=0$ and that the first partial derivatives of $f$ are $0$ at $v_0=0$. How do I proceed?
 A: The result is not true without an absolute value: you need $|f(v)|\leq C\|v-v_0\|^{k+1}$. This is simply induction. 
For simplicity of notation, let me do the case $n=1$; it is not hard to generalize the idea for partial derivatives. So we assume that $|f(v)|\leq|v-v_0|^{k+1}$ in a certain neighbourhood of $v_0$. You have already shown that $f(v_0)=f'(v_0)=0$. If we now look at the Taylor polynomial, for $v$ close to $v_0$ we have
$$
f(v)=f''(v_0)\,(v-v_0)^2+f'''(\xi)\,(v-v_0)^3, 
$$
with $\xi$ between $v_0$ and $v$ (so, in the neighbourhood). Thus
$$
|f''(v_0)|=\left| \frac{f(v)-f'''(\xi)(v-v_0)^3}{(v-v_0)^2}\right|\leq|v-v_0|^{k-1}+D|v-v_0|,
$$
where $D$ is a bound for $f'''$ in the neighbourhood (it exists, because $f'''$ is continuous). As $v$ can be chosen arbitrarily close to $v_0$, we obtain that $f''(v_0)=0$. 
Now,since $f(v_0)=f'(v_0)=f''(v_0)=0$, we can rewrite the Taylor polynomial as 
$$
f(v)=f'''(v_0)(v-v_0)^3+f^{(4)}(\xi)(v-v_0)^4
$$
and repeat the argument as above to obtain $f'''(v_0)=0$, etc. 
