I am reading a book and the author claims the following:
Let $f(x,t)$ be smooth, then we can apply Taylor's theorem in $t$ and obtain:
$$f(x,t) = f(x,0)+f'(x,0)t+\frac{f''(x,0)}{2}t^2+h(x,t)t^3$$
where prime denote derivatives in $t$ and $h$ is smooth in $x,t$.
I looked at wikipedia and another similar question posted here 4 years ago, but
1.) The Peano form of the remainder should be multiplied by $t^2$ to obtain the remainder, not $t^3$. 2.) In the other thread, it is only shown that the peano form $h \rightarrow 0$ for $x \rightarrow 0^+$, it says nothing about the differentiability of $h$.
Does anyone know how one would prove the statement above?
The integral form of the remainder term ( cf. Wikipedia) is (aside from a constant $1/k!$ that I'll ignore to simplify typesetting) $$R_k(x)= \int_{t=0}^{t=x} (x-t)^{k} f^{k+1}(t) dt$$ Hold $x$ fixed. Making the change of variables $t=x s$ we get $$R_k(x)= x^{k+1}\int_{s=0}^{s=1} (1-s)^k f^{k+1}(x s) ds$$
Note that $h=\int_{s=0}^{s=1} (1-s)^k f^{k+1}(x s) ds$ is a smooth function of $x$.
In your problem, $f$ also depends on a second parameter, but this formula shows that $h$ is also a smooth function of that parameter.