So as I answered another question while correcting a wrong definition of differentiability in the post, I wonder what we can deduce about a function if we tweak the formula for defining differentiability.
Let $U$ be an open set of $\Bbb R^m$. Differentiability of a function $f:U\to\Bbb R^n$ at a point $p\in U$ is generically defined by the existence of a linear map $L:\Bbb R^m\to\Bbb R^n$ such that $$\lim\limits_{h\to0}\frac{\lVert f(p+h)-f(p)-L(h)\rVert}{\lVert h\rVert}=0.$$ (The norms are Euclidean norms.) Now, let's say we "accidentally" raised the denominator to a certain power $k\gt1$ for an integer $k$, and still able the compute the limit as $$\lim\limits_{h\to0}\frac{\lVert f(p+h)-f(p)-L(h)\rVert}{\lVert h\rVert^k}=0.$$
Obviously, this limit being zero is no weaker than differentiability.
What more can we deduce about this function $f$ around $p$?
Some examples of functions $f$ for which this holds or does not hold are given:
- $f_1(x,y)=ax+by$
- $f_2(x,y)=x^2+y^2$
- $f_3(x,y)=x^3+y^3$
We try to evaluate the limits at $p=(0,0)$.
The first one satisfies the limit for all $k$ because the numerator is zero already (the derivative is the linear map $L=f_1$ and the numerator is $f_1(h)-L(h)=0$).
The second one does not satisfy the limit when $k=2$. The limit is computed as $$\lim\limits_{(x,y)\to(0,0)}\frac{x^2+y^2}{x^2+y^2}=1.$$
The third one satisfies the limit when $k=2$: $$\lim\limits_{(x,y)\to(0,0)}\frac{\lvert x^3+y^3\rvert}{x^2+y^2}=0.$$ But it does not satisfy the limit when $k=3$: $$\lim\limits_{(x,y)\to(0,0)}\frac{\lvert x^3+y^3\rvert}{(x^2+y^2)\sqrt{x^2+y^2}}\not=0.$$
I think it has something to do with the function's higher derivatives. The second and higher derivatives of $f_1$ is zero, while the second derivative of $f_3$ at $(0,0)$ is also zero. But the second derivative of the second function is the bilinear map represented by the matrix $$\begin{bmatrix}2&0\\0&2\end{bmatrix}.$$ The third derivative of $f_3$ is the trilinear map with components $t_{ijk}=6$ if $i=j=k$ and $t_{ijk}=0$ otherwise.
I subtract the higher derivatives in the numerator, and obtained the following limits:
$$\lim\limits_{(x,y)\to(0,0)}\frac{\left\lVert f_2(x,y)-\frac{1}{2}\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}2&0\\0&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\right\rVert}{\lVert(x,y)\rVert^k}=0$$
$$\lim\limits_{(x_1,x_2)\to(0,0)}\frac{\left\lVert f_3(x_1,x_2)-\frac{1}{6}\sum_{i,j,k=1}^2t_{ijk}x_ix_jx_k\right\rVert}{\lVert(x_1,x_2)\rVert^k}=0$$
for all $k\gt1$.
In general, we may define that a function $f:U\to\Bbb R^n$ has good order $k$ approximation at a point $p\in U$ if there exists a linear map $T_1:\Bbb R^m\to\Bbb R^n$, a bilinear map $T_2:(\Bbb R^m)^2\to\Bbb R^n$, a trilinear map $T_3:(\Bbb R^m)^3\to\Bbb R^n$, ..., and a $k$-linear map $T_k:(\Bbb R^m)^k\to\Bbb R^n$ such that they are all symmetric and
$$\lim\limits_{h\to0}\frac{\lVert f(p+h)-f(p)-T_1(h)-\frac{1}{2}T_2(h,h)-\frac{1}{6}T_3(h,h,h)-\cdots-\frac{1}{k!}T_k(h,\dots,h)\rVert}{\lVert h\rVert^k}=0.$$
The imposed condition that the multilinear functions shall be symmetric is excessive: if a multilinear function $T_i$ (not necessarily symmetric) satisfies the above limit, then its symmetrisation also satisfies the above limit. But imposing that they are symmetric simplifies the upcoming discussion.
This seems to be a higher-dimensional generalisation of the following, but we focus our attention to the difference between a function and multilinear functions: Intuition behind two functions being equal up to order $n$ at $a$
With this definition. I have several follow-up questions in mind:
Like the uniqueness of the derivative when it exists, would the $T_i$ be unique if they exist?
If $f$ has good order $k$ approximation at $p$ for some $k\gt1$, would $f$ also be differentiable at $p$? More generally, does having good order $k$ approximation implies having good order $k-1$ approximation?
If $f$ happens to have good order $k$ approximation at every point $p\in U$, would $f$ actually be $k$ times differentiable on $U$?
Conversely, if $f$ is $k$ times differentiable on $U$, would $f$ have good order $k$ approximation, with the $i$-th derivative satisfy the role of $T_i$? If not, what about lower order approximation? And what if we assume $f$ is $k$ times continuously differentiable?
For 1. I mimic a proof of uniqueness of derivative. If another set of multilinear maps $L_i$ satisfies the role of $T_i$, then
$$\begin{align}&\frac{\lVert(T_1-L_1)(h)+\frac{1}{2}(T_2-L_2)(h,h)+\cdots+\frac{1}{k!}(T_k-L_k)(h,\dots,h)\rVert}{\lVert h\rVert^k}\\=&\frac{\lVert(f(p+h)-f(p))-(f(p+h)-f(p))+(T_1-L_1)(h)+\cdots+\frac{1}{k!}(T_k-L_k)(h,\dots,h)\rVert}{\lVert h\rVert^k}\\ \le&\frac{\lVert f(p+h)-f(p)-T_1(h)-\cdots-\frac{1}{k!}T_k(h,\dots,h)\rVert}{\lVert h\rVert^k}\\&+\frac{\lVert f(p+h)-f(p)-L_1(h)-\cdots-\frac{1}{k!}L_k(h,\dots,h)\rVert}{\lVert h\rVert^k}\\ \to&0\end{align}$$ as $h\to0$. So the first expression has limit $0$ as $h\to0$. Now fix a unit vector $h$ and let $x$ be a variable nonzero real number. We have $$\begin{align}0 & =\lim\limits_{x\to0}\frac{\lVert(T_1-L_1)(xh)+\frac{1}{2}(T_2-L_2)(xh,xh)+\cdots+\frac{1}{k!}(T_k-L_k)(xh,\dots,xh)\rVert}{\lVert xh\rVert^k} \\ & =\lim\limits_{x\to0}\frac{\lVert xa_1(h)+\frac{x^2}{2}a_2(h)+\cdots+\frac{x^k}{k!}a_k(h)\rVert}{\lvert x\rvert^k}\end{align}$$ where $a_i(h)=(T_i-L_i)(h,\dots,h)$ (remember they are vectors). I suppose this implies that $a_i(h)=\vec 0$, for all $i=1,\dots,k$ and all vector $h$, but this is the best I could find.
For 2., I did the following. Assume $f$ has good order $k$ approximation, with $T_i$ chosen to satisfy the limit. Then
$$\begin{align}&\frac{\lVert f(p+h)-f(p)-T_1(h)-\cdots-\frac{1}{(k-1)!}T_{k-1}(h,\dots,h)\rVert}{\lVert h\rVert^{k-1}}\\=&\frac{\lVert f(p+h)-f(p)-T_1(h)-\cdots-\frac{1}{(k-1)!}T_{k-1}(h,\dots,h)-\frac{1}{k!}T_{k}(h,\dots,h)+\frac{1}{k!}T_{k}(h,\dots,h)\rVert}{\lVert h\rVert^{k-1}}\\ \le&\frac{\lVert f(p+h)-f(p)-T_1(h)-\cdots-\frac{1}{(k-1)!}T_{k-1}(h,\dots,h)-\frac{1}{k!}T_{k}(h,\dots,h)\rVert}{\lVert h\rVert^{k-1}}+\frac{\lVert\frac{1}{k!}T_{k}(h,\dots,h)\rVert}{\lVert h\rVert^{k-1}}\\ \to&0.\end{align}$$ So $f$ also has a good order $k-1$ approximation.
I have no idea about 3. and 4.