Can we always find a complex polynomial which (together with its derivatives) satisfies given points? Let $C=\{ c_1,…,c_m\}$ be a finite subset of $\mathbb{C}$. I wonder if we can always find a polynomial $p$ with complex coefficients that satisfies a group of equations as follows:
$p(c_1)=k_{11},p(c_2)=k_{12},…,p(c_m)=k_{1m}$
$p’(c_1)=k_{21},p’(c_2)=k_{22},…,p’(c_m)=k_{2m}$
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.
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$p^{(n)}(c_1)=k_{n1},p^{(n)}(c_2)=k_{n2},…,p^{(n)}(c_m)=k_{nm}$
where $n$ is an arbitrary number.
 A: This is a fairly bullet-headed approach to show we can always solve this. Certainly nothing like a useful algorithm.
The trick is to first solve for when all but one column of values are all zeros. Then we combine the the resulting polynomials.
This is similar to the approach to showing there is a solution in the $n=0$ case, where you get (when $m=3$):
$$p(x)=a_1(x-c_2)(x-c_3)+a_2(x-c_1)(x-c_3)+a_3(x-c_1)(x-c_2)$$ and solve for $a_1,a_2,a_3.$ But with $n=2,$ say, you get:
$$p(x)=q_1(x)(x-c_2)^3(x-c_3)^3+q_2(x)(x-c_1)^3(x-c_3)^3+q_3(x)(x-c_1)^3(x-c_2)^3$$ where the $q_i$ are polynomials of degree $2.$

Let $f(x)=(x-c_1)\cdots (x-c_m).$ Let $f_i(x)=\frac{f(x)}{x-c_i}.$ Note, $f_i$ is a polynomial - we are just removing one of the roots of $f.$
Let $g_i(x)=f_i^{n+1}(x).$
Then $g_i(c_i)\neq 0.$ When $i\neq j,$ and $k=0,\dots,n$ $g_{i}^{(k)}(c_j)=0.$
We will find a polynomial $q_i(x)$ so that $p_i(x)=q_i(x)g_i(x)$ satisfies $p_{i}^{(j)}(c_i)=k_{ji}.$. Implicitly, $p_i^{(k)}(c_j)=0$ when $i\neq j$ and $0\leq k\leq n.$
To find $q_i(x),$ first note we need $q_i(c_i)=k_{0i}g_{i}(c_i)^{-1}.$
Then we need $q_i'(c_i)g_i(c_i)+q_i(c_i)g_i'(c_i)= k_{1i}.$ We again use $g_i(c_i)\neq 0$ is invertible to get a value for $g_i'(c_j).$
We work inductively.
Knowing $q_i^{(j)}(c_i)$ to $j=0,1,\dots,j',$ we can compute what $q_i^{(j'+1)}(c_i)$ must be.
Once we've got all those values, we can write:
$$q_i(x)=\sum_{j=0}^n \frac{q_i^{(j)}(c_i)}{j!}(x-c_i)^j$$
Then your final polynomial is: $$p(x)=\sum_{i=1}^m q_i(x)g_i(x)$$
Since each $g_i$ is of degree $(n+1)(m-1)$ and $q_i$ is of degree at most $n,$ the result is a polynomial of degree at most $(n+1)m-1.$
