Function with equal nth derivatives to two functions at two points.

Question

I have a function, $p(x)$, connecting two functions, $a(x)$ and $b(x)$, at two separate points. The first derivatives are also equal at those points. The following equation is a representation of $p(x)$:\begin{align}p\left(x\right)=\frac{\left(f'\left(x_{0}\right)-\frac{f\left(x_{0}\right)}{x_{0}-x_{1}}\right)\left(x-x_{1}\right)^{K}\left(x-x_{0}\right)+\left(g'\left(x_{1}\right)+\frac{g\left(x_{1}\right)}{x_{0}-x_{1}}\right)\left(x-x_{0}\right)^{K}\left(x-x_{1}\right)}{\left(x_{0}-x_{1}\right)^{K}}+\frac{f\left(x_{0}\right)\left(x-x_{1}\right)\cos\left(\frac{\pi\left(x-x_{0}\right)}{2\left(x_{0}-x_{1}\right)}\right)-g\left(x_{1}\right)\left(x-x_{0}\right)\cos\left(\frac{\pi\left(x-x_{1}\right)}{2\left(x_{0}-x_{1}\right)}\right)}{x_{0}-x_{1}}\end{align}\begin{align}K=2n \hspace{1cm} n\in \mathbb{N} \hspace{1cm}K \geq 2\end{align} Is there an easier way to write functions like this (Maybe as a sum) given by the set of equations:\begin{align}a^{[N]}(x_0)=p^{[N]}(x_0)\end{align}\begin{align}b^{[N]}(x_1)=p^{[N]}(x_1)\end{align}\begin{align}N=\{0,1,2,...,c|c \in \mathbb{N}\} \end{align} Edit: The simplest form of this is where c=0:\begin{align}\frac{b(x_1)(x-x_0)^{n+1}-a(x_0)(x-x_1)^{n+1}}{(x_1-x_0)^{n+1}}\end{align}It may help to represent $K_N$ as $K_N = c(n+1)$.