Alternative proof of Taylor's formula by only using the linear approximation property So a function $f: E \to F$ between the normed spaces $E,F$ is called differentiable in $x \in E$ if there exists a bounded linear map $Df(x): E \to F$ such that for every $h \in E$ we have $$f(x+h)=f(x)+Df(x)h + o(||h||). \tag{1}$$
If $f$ is differentiable for every $x \in E$ and $Df: x \mapsto Df(x)$ is differentiable for every $x \in E$ too we get analogously
$$Df(x+e)=Df(x)+D^2f(x)e+o(||e||). \tag{2}$$
Then $f$ is called twice differentiable and for every $h\in E$ we have the "Taylor expansion of second degree"
$$f(x+h)=f(x)+Df(x)h+\frac{1}{2}D^2f(x)[h] + o(||h||^2), \tag{3}$$
where $D^2f(x)[h]:=(D^2f(x)h)h$ for better readability.
I have two questions:

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*How can $(3)$ be proven without resorting to the "standard proof" of using integrals? I want to show it by only using the linear approximations  given in $(1)$ and $(2)$. Inserting $(2)$ in $(1)$ doesn't result in something useful though. Can this be done?

*Can $(3)$  be used as an alternative definition off twice-differentiability? Analogously what about  the general case of $n$-times differentiability:
$$ f(x+h) = f(x) + \sum_{j=1}^{n} \frac{1}{j!} D^jf(x)[h] + o(\|h\|^n) \tag{4}$$

The differential Df(x) is a linear map that approximates your function f, according to your relation (1).  And that linear map depends on x (the point around which you consider the Taylor expansion).  The map Df , which associates a vector to a linear map doesn't have to be linear indeed.  The second differential  DDf(x)   is a linear map that associates a vector to a linear map , and so on. Your relation (1) tells us that the differential of f exists (definition). Your relation (2) tells us that the differential of Df exists (definition). Your relation (3) tells us that there is a valid finite second order (degree) Taylor expansion of f at x (also a definition). From (1) and (2) you cannot derive (3). That's not always possible  even for real function , even more so in this abstract setting.  In a valid, finite Taylor expansion,  the remainder must grow slower than the last term of the Taylor expansion.  And no, (3) is not equivalent to second order differentiability of f. 
