Given a polynomial $y=C_0+C_1 x+C_2 x^2+C_3 x^3 + \ldots$ of some order $N$, I can easily calculate the polynomial of reduced order $M$ by taking only the first $M+1$ terms. This is equivalent to doing a Taylor series expansion with $M<=N$ around $x=0$.
But what if I want to take the Taylor series expansion around a different point $x_c$. In the end, I want the polynomial coefficients of $y_2=K_0+K_1 x + K_2 x^2 + K_3 x^3 + \ldots$ which represents the Taylor's expansion of $y$ around point $x_c$ such that $y(x_c)=y_2(x_c)$ including the first $M$ derivatives.
So given the coefficients $C_i$ with $i=0 \ldots N$, and a location $x_c$ I want to calculate the coefficients $K_j$ with $j=0 \ldots M$.
Example
Given $y=C_0+C_1 x+C_2 x^2$ ( $N=2$ ) then the tangent line ($M=1$) through $x_c$ is
$ y_2 = (C_0-C_2 x_c^2) + (C_1+2 C_2 x_c) x $
or $K_0 = C_0-C_2 x_c^2$, and $K_1 =C_1+2 C_2 x_c$
There must be a way to construct a ($M+1$ by $N+1$ ) matrix that transforms the coefficients $C_i$ into $K_j$. For the above example this matrix is
$ \begin{bmatrix}K_{0}\\ K_{1}\end{bmatrix}=\begin{bmatrix}1 & 0 & -x_{c}^{2}\\ 0 & 1 & 2\, x_{c}\end{bmatrix}\begin{bmatrix}C_{0}\\ C_{1}\\ C_{2}\end{bmatrix} $
Example #2
The reduction of a $5$-th order polynomial to a $3$-rd order around $x_c$ is
$ \begin{bmatrix}K_{0}\\ K_{1}\\ K_{2}\\ K_{3}\end{bmatrix}=\left[\begin{array}{cccc|cc} 1 & & & & -x_{c}^{4} & -4\, x_{c}^{5}\\ & 1 & & & 4\, x_{c}^{3} & 15\, x_{c}^{4}\\ & & 1 & & -6\, x_{c}^{2} & -20\, x_{c}^{3}\\ & & & 1 & 4\, x_{c} & 10\, x_{c}^{2}\end{array}\right]\begin{bmatrix}C_{0}\\ C_{1}\\ C_{2}\\ C_{3}\\ C_{4}\\ C_{5}\end{bmatrix} $
which is a block matrix, and not an upper diagonal one as some of the answers have indicated.