I consider the following determinant (Hankel-like?) $$ [f_1,f_2,...,f_n]:=\begin{vmatrix} f_1 & f_2 & \cdots & f_{n-1} & f_n\\ n-1 & f_1 & \cdots & f_{n-2}& f_{n-1}\\ 0 & n-2 & \cdots & \cdots & \cdots\\ \cdots & 0 & \cdots & f_1 & f_2\\ 0 & \cdots & 0 & 1 & f_1\end{vmatrix} $$ Can you enlighten me as to evaluate it?
Thank you.
I haven't succeeded in bringing this to conclusion, but I have had progressing in converting this to an equivalent differential equation.
To simplify notation, I'll use $\langle j,k\rangle$ to denote the determinant $[f_j,f_{i+1}\ldots, f_k]$ with $1\leq j \leq k \leq n$. If we expand such a determinant along the first column, we obtain $$ \langle j,k \rangle = f_j \langle j,k-1 \rangle-(k-j)\langle j+1,n \rangle.$$ One approach would be to apply this repeatedly and hope a nice formula appears. Instead, let me introduce a bivariate generating function (BGF) by multiplying both sides by $x^j y^k$ and summing over $j,k$ (with $\langle j,k\rangle$ understood to vanish unless $1\leq j\leq k$). This gives \begin{align} B(x,y)=\sum_{jk} \langle j,k\rangle x^j y^k &=\sum_{jk} f_j \langle j,k-1\rangle x^j y^k -\sum_{jk} (k-j) \langle j+1,k\rangle x^j y^k\\ &=\sum_{jk}f_j\langle j,k\rangle x^j y^{k+1} -\sum_{jk} (k-j+1) \langle j,k\rangle x^{j-1} y^k\\ &=y\cdot \sum_{jk}f_j \langle j,k\rangle x^j y^k-\frac{1}{x}\cdot \sum_{jk}(k-j+1) \langle j,k\rangle x^j y^k \end{align} The second sum we may write as $ \left(x\partial_x - y \partial y+1\right)B(x,y)$ and so we may rearrange our equation to obtain $$ \frac{1}{xy}\left(1+x+x\frac{\partial}{\partial x}-y\frac{\partial}{\partial y}\right) B(x,y) = \sum_{jk} f_j \langle j,k\rangle x^j y^k$$ We have thus converted much of this equation into a differential operator acting on the BGF. Unfortunately, what this doesn't deal with are the $f_j$ coefficients (i.e. the entire interesting part of the problem.) One can consider special cases of $f_j$ where the RHS side can be similarly manipulated. Otherwise, though, I don't know how to go proceed.