Is there a closed-form solution for $c$ in the equation $Ac = b$, where $A$, $c$, $b$ respectively are the below matrices and $n \in Z^+$? $$ \begin{bmatrix} 1 & 1 & 1 & 1 & \dots & 1 \\ 0 & 1 & \frac{1}{2} & \frac{1}{3} & \dots & \frac{1}{n} \\ 0 & 1 & (\frac{1}{2})^2 & (\frac{1}{3})^2 & \dots & (\frac{1}{n})^2 \\ 0 & 1 & (\frac{1}{2})^3 & (\frac{1}{3})^3 & \dots & (\frac{1}{n})^3 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & (\frac{1}{2})^n & (\frac{1}{3})^n & \dots & (\frac{1}{n})^n \end{bmatrix}\begin{bmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ \vdots \\ c_n \end{bmatrix} =\begin{bmatrix} f(0) \\ f'(0) \\ f''(0) \\ f^{(3)}(0) \\ \vdots \\ f^{(n)}(0) \end{bmatrix}. $$ I recognize that there's a Vandermonde matrix inside of $A$, but I'm not sure how to use that to get to a solution.
Sorry for the weird formatting, I can't figure out how to put them all on one line in the StackExchange format.
You can partition the matrix, $$M=\begin{bmatrix}A&B\\C&D\end{bmatrix}$$ where $A=1$ and $D$ is the Vandermonde matrix. Then you can express the inverse of $M$ using the formula in here, which requires the inversion of the Schur’s complement, $A-BD^{-1}C$, which is just 1, and the inverse of the Vandermonde matrix $D$, which has this formula.
Note that I would not recommend doing this in floating point arithmetic because of poor conditioning of $D$ and $M$ itself.