# Multivariate Taylor theorem with small o notation (Not continuous, only differentiable)

I want to know whether the following is true.

Let $n\in \mathbb{Z}$, $n\geq 2$, and $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be $k$ times differentiable at a point $p\in\mathbb{R}^n$. Then $f(p+v)=f(p)+Df(p)(v)+\frac{1}{2!}D^2f(p)v^{(2)}+\cdots+\frac{1}{k!}D^kf(p)v^{(k)}+o(||v||^k)$

When $n=1$, I know the above statement is true, and it can be proved using L'Hospital's theorem or other methods(Taylor's Theorem with Peano's Form of Remainder). However, when $n\geq 2$, I have tried to prove but I cannot prove the above statement. Note that the above condition of the function $f$ is NOT $k$ times continuously differentiable, ONLY $k$ times differentiable.
(It can be easily proved when $f$ is $\mathcal{C}^k$, then $f(p+v)=f(p)+Df(p)(v)+\frac{1}{2!}D^2f(p)v^{(2)}+\cdots+\frac{1}{k!}D^kf(p^*)v^{(k)}$ for some $p^*$, where $p^*$ is on the line segment between $p$ and $p+v$, therefore $f(p+v)-f(p)-Df(p)(v)-\frac{1}{2!}D^2f(p)v^{(2)}-\cdots-\frac{1}{k!}D^kf(p)v^{(k)}=\frac{1}{k!}\{D^kf(p^*)-D^kf(p)\}v^{(k)}$

, and the continuity of $D^kf$ now implies $o(||v||^k)$)

Without the continuity of $D^kf$, I guess the above statement is false, but I cannot find any counterexamples. Can someone have some idea and give me any hints?

(c.f. $D^mf:\mathbb{R}^n \rightarrow \mathcal{L}^m(\mathbb{R}^n,\mathbb{R})$ and $\mathcal{L}^m(\mathbb{R}^n,\mathbb{R})$ is the set of $m$-multilinear maps of $\mathbb{R}^n$ to $\mathbb{R}$, so $D^mf(p)v^{(m)}:=D^mf(p)(v,v,\cdots,v).$)

My book uses the following lemma. Define the function $r(v)=f(p+v)-\sum_{i=0}^k D^if(p)v^i$. Check that $r$ is k times differentiable at $0$ and $D^ir(0)=0, \forall 0\le i \le k$. It follows that $lim_{v \rightarrow 0} \frac{r(v)}{||v||^k}=0$.
• Many thanks for your answer. I see what I want to prove is equivalent to show that if $r$ is $k$ times differentiable at 0 and $D^ir(0)=0$ for all $0\leq i \leq k$, then $\displaystyle\lim_{v->0} \dfrac{r(v)}{||v||^k}=0$. Unfortunately, I do not know how to prove. May I know what the title of your book is? Thanks, again! – Hyunsoo Cho Feb 2 '18 at 10:06
I conclude that the statement is true. I think the general spherical coordinate system in an $n$-dimensional Euclidean space which is analogous to the spherical coordinate system defined for $3$-dimensional Euclidean space, in which the coordinates consist of a radial coordinate, $\rho$, and $n-1$ angular coodinates $\phi_1, \phi_2, \cdots, \phi_{n-1}$.(See, https://en.wikipedia.org/wiki/N-sphere#Spherical_coordinates)
According to @lzralbu 's answer, we need to show if $r:\mathbb{R}^n\rightarrow \mathbb{R}$ is $k$ times differentiable at $0$ and $D^ir(0)=0$ for all $0≤i≤k$, then $\displaystyle\lim_{v−>0}\dfrac{r(v)}{||v||^k}=0$. Via reparametrization using the above spherical coordinate system, then we now show $\displaystyle\lim_{\rho−>0}\dfrac{r(\rho,\phi_1,\cdots,\phi_{n-1})}{\rho^k}=0$. Let the standard coordinate functions be $x_1,x_2,\cdots,x_n$,and also let $(x_1,x_2,\cdots,x_n)=(\rho,\phi_1,\cdots,\phi_{n-1})$, then we note that $\dfrac{\partial x_1}{\partial\rho},\cdots,\dfrac{\partial x_n}{\partial\rho}$ are all independent about the variable $\rho$.
Apply L'Hospital's theorem to $\displaystyle\lim_{\rho\rightarrow 0}\dfrac{r(\rho,\phi_1,\cdots,\phi_{n-1})}{\rho^k}=\displaystyle\lim_{\rho\rightarrow 0}\dfrac{\frac{\partial r}{\partial \rho}}{k\rho^{k-1}}=\cdots=\displaystyle\lim_{\rho\rightarrow 0}\dfrac{\frac{\partial^{k-1} r}{\partial \rho^{k-1}}}{k!\rho}$, and the last limit value is zero because $D^{k-1}r$ is differentiable at $0$ and $D^kr(0)\equiv 0$, which implies $\displaystyle\lim_{\rho\rightarrow 0}\dfrac{\frac{\partial^{k-1} r}{\partial \rho^{k-1}}}{\rho}=0$ and we finally conclude that $r(v)$ is $o(||v||^k)$.