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Is there an analytical formula for the inverse of a complex matrix whose elements are sets of "power series" except the last term is scaled?

Let $0<x_1<x_2<...<x_n$ be monotonically increasing. The matrix $A$ is formed by $$A=\begin{bmatrix}\frac{1}{x_1} & \frac{1}{x^2_1} &\dots & \frac{1}{x^n_1}+I_1\\\frac{1}{x_2} & \frac{1}{x^2_2}&\dots & \frac{1}{x^n_2}+I_2\\\vdots&\vdots&\ddots&\vdots\\\frac{1}{x_n} & \frac{1}{x^2_n}&\dots & \frac{1}{x^n_n}+I_n\end{bmatrix}$$ The last column terms $[A_{in}]$ in the original problem are the sum of the highest power of $x_i^n$ and series of Laplace frequency shifting like this $$[A_{in}]=\frac{1}{x^n_i}+\frac{\int_0^\infty f(t)e^{-(i-1)\epsilon}e^{-xt}dt}{x^n_i}$$ where $\epsilon=x_j-x_i,j=i+1$, and $f(t)$ is a very slowly convergent function like $\mathrm{Sinc}(t)$.

fonini
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MathArt
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    I think this case is similar to Vandermonde matrix. see http://math.stackexchange.com/questions/698254/proof-of-vandermonde-matrix-inverse-formula . also https://proofwiki.org/wiki/Inverse_of_Vandermonde's_Matrix – K.K.McDonald Apr 25 '16 at 13:05
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    This matrix is closely related to the Vandermonde matrix. There is a nice formula for the determinant. There exists a formula for the inverse, but it's not particularly nice. – Ben Grossmann Apr 25 '16 at 13:09
  • As K.K.McDonald and Omnomnomnom point out, $A$ is simply the tranpose of a Vandermonde matrix. Since $(A^{-1})^{T}=(A^{T})^{-1}$, you can use the expression on the page linked to by K.K.McDonald. – parsiad Apr 25 '16 at 13:40
  • (P.s. it might be misleading to liken the elements to power series; consider editing that) – parsiad Apr 25 '16 at 14:57

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As K.K.McDonald and Omnomnomnom point out, $A$ is simply the tranpose of a Vandermonde matrix: $$ A=\begin{bmatrix}\frac{1}{x_{1}} & \frac{1}{x_{1}^{2}} & \dots & \frac{1}{x_{1}^{n}}\\ \frac{1}{x_{2}} & \frac{1}{x_{2}^{2}} & \dots & \frac{1}{x_{2}^{n}}\\ \vdots & \vdots & \ddots & \vdots\\ \frac{1}{x_{n}} & \frac{1}{x_{n}^{2}} & \dots & \frac{1}{x_{n}^{n}} \end{bmatrix}=\begin{bmatrix}\frac{1}{x_{1}} & \frac{1}{x_{2}} & \dots & \frac{1}{x_{n}}\\ \left(\frac{1}{x_{1}}\right)^{2} & \left(\frac{1}{x_{2}}\right)^{2} & \dots & \left(\frac{1}{x_{n}}\right)^{2}\\ \vdots & \vdots & \ddots & \vdots\\ \left(\frac{1}{x_{1}}\right)^{n} & \left(\frac{1}{x_{2}}\right)^{n} & \dots & \left(\frac{1}{x_{n}}\right)^{n} \end{bmatrix}^{\intercal} $$ Since $(A^{-1})^{\intercal}=(A^{\intercal})^{-1}$, we can use the expression on the page linked to by K.K.McDonald to get $A^{-1}=(b_{ij})$ where $$ b_{ij}=\begin{cases} \dfrac{x_{j}\left(-1\right)^{i-1}{\displaystyle \sum_{\substack{1\leq m_{1}<\cdots<m_{n-i}\leq n\\ m_{1},\ldots,m_{n-i}\ne j } }\frac{1}{x_{m_{1}}}\cdots\frac{1}{x_{m_{n-i}}}}}{{\displaystyle \prod_{\substack{1\le m\le n\\ m\ne j } }\left(\frac{1}{x_{m}}-\frac{1}{x_{j}}\right)}} & \text{if }1\le i<n;\\ \dfrac{x_{j}}{{\displaystyle \prod_{\substack{1\le m\le n\\ m\ne j } }\left(\frac{1}{x_{j}}-\frac{1}{x_{m}}\right)}} & \text{if }i=n. \end{cases} $$ You should check to make sure that my $b_{ij}$s are correct.

parsiad
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