A sufficient condition to ensure a function to be a polynomial with degree at most $3$ Let $f\in C^4(\mathbb{R})$ and satisfy
$$f(x+h)=f(x)+f'(x)h+\frac{1}{2}f''(x+\theta h)h^2,$$
with $\theta\in (0,1)$ is a constant independent of $x,h$. Show that $f$ is a polynomial with degree at most $3$.
I have no idea on it...
 A: Differentiate the equation with respect to $x$ twice and then substitute $h$ by $\theta h$, we obtain
$$f''(x+\theta h)=f''(x)+f^{(3)}(x)\theta h+\frac12f^{(4)}(x+\theta^2 h)\theta^2 h^2$$
Substitute this back into original equation, we get
$$f(x+h) = f(x) + f'(x)h+\frac{h^2}{2}\left[f''(x)+f^{(3)}(x)\theta h+\frac12f^{(4)}(x+\theta^2 h)\theta^2 h^2\right]$$
Since $f\in C^{4}(\mathbb{R})$, we have a Taylor expansion of $f(x+h)$ of the form
$$f(x+h) = f(x) + f'(x)h+\frac12 f''(x)h^2 + \frac{1}{3!} f^{(3)}(x)h^3 + \frac{1}{4!}f^{(4)}(x + \phi_{x,h} h) h^4$$
for some $\phi_{x,h} \in (0,1)$ which in general depends on both $x$ and $h$.
Compare these two expansion, we get
$$\theta f^{(3)}(x) + \frac12 f^{(4)}(x+\theta^2 h)\theta^2 h
= \frac13f^{(3)}(x) + \frac{1}{12} f^{(4)}(x + \phi_{x,h} h)h$$
If $f^{(3)}(x)$ is not identically zero (i.e $f(x)$ not a polynomial of degree $2$),
then we can take limit of $h \to 0$ at suitable $x$ to conclude $\theta = \frac13$. 
This leads to the identity
$$\frac{1}{18} f^{(4)}(x + \frac19 h) = \frac{1}{12} f^{(4)}(x + \phi_{x,h} h)$$
Since $f \in C^4(\mathbb{R})$, we can then take limit $h \to 0$ again to get
$$\frac{1}{18} f^{(4)}(x) = \frac{1}{12} f^{(4)}(x) \quad\implies\quad f^{(4)}(x) = 0 \quad\text{ for all } x$$
and hence $f(x)$ is a polynomial with degree at most $3$.
