I am looking at Taylor's theorem recently, and I know that there are several different forms of remainders:
Let $ f:[a-\delta,a+\delta]\to\mathbb{R} $ be a $(n+k)$-times differentiable function (we can also weaken this condition such that the $(n+k)$-th derivative, $ f^{(n+k)} $, exists only on $ (a-\delta,a+\delta) $) Then the Taylor series to the order $n$ of $ f $ at point $a$ is of the following form: $$ f(a+h) =f(a)+hf'(a)+\dots+h^{n}\frac{f^{(n)}(a)}{n!}+R_n(h) $$ where $|h|\leq\delta$. The term $ R_n $ is called the remainder of the Taylor series. $$ R_n(h)=o(h^n)$$
Using Cauchy's Mean Value Theorem, we can prove that the remainder can have the following forms: $$ \exists \theta_1 \in (0,1):R_n =h^{n+1}\frac{f^{(n+1)}(a+\theta_1 h)}{(n+1)!}\qquad \text{(Lagrange's form)} $$ $$ \exists \theta_2 \in (0,1):R_n =h^{n+1}(1-\theta_2)^n \frac{f^{(n+1)}(a+\theta_2 h)}{n!}\qquad \text{(Cauchy's form)} $$ Using fundamental theorem of calculus and integral by parts, we can also prove that the remainder has the form: $$ R_n(h)=\int_a^{a+h}{\frac{f^{(n+1)}(t)}{n!}h^{n}\, dt}\qquad\text{(integral form)}$$ Now, my question is, whether the following generalizations hold, and if they do, how can we prove them?
$$ \exists \theta \in (0,1):R_n(h) =h^{n+1}\frac{f^{(n+1)}(a+\theta h)}{(n+1)!}+h^{n+2}\frac{f^{(n+2)}(a+\theta h)}{(n+2)!}\tag1 $$ $$ \exists \theta \in (0,1):R_n(h) =\sum_{r=1}^k{h^{n+r}\frac{f^{(n+r)}(a+\theta h)}{(n+r)!}}\tag2 $$ $$ \exists \theta \in (0,1):R_n(h) =\sum_{r=1}^k{C_{n,r,\theta}h^{n+r}f^{(n+r)}(a+\theta h)} \tag3 $$ where $ C_{n,r,\theta} $ are coefficients depending on $n,r,\theta$.
Thanks in advance.