Timeline for Inverse of matrix of generalised harmonic numbers
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| May 18, 2014 at 4:00 | comment | added | Peter Dukes | In hopes of an explicit form, have you tried Bernoulli's formula for $r_{n,s}$? I am hoping it leads to a factorization of $R_{n,s}$ in which each factor is easy to invert. | |
| May 12, 2014 at 7:28 | comment | added | dima | @ChristianRemling: Unfortunately, this is not the case. $H^{-1}$ does have negative entries (as does the inverse of the Hilbert matrix), so even for a 2-by-2 case the matrix $H^{-1}R-I$ has $O(1)$ entries. | |
| May 11, 2014 at 22:12 | comment | added | Christian Remling | It seems to me you're getting this (that is, $R=H(1+B)$ with $\| B\|$ small) anyway here, since $H$, $H^{-1}$ have positive entries, so there are no cancellations when working out the matrix elements of $H^{-1}H$, so an extra $O(n^{-1})$ should keep things small. | |
| May 11, 2014 at 21:33 | comment | added | dima | @ChristianRemling: the problem is that one cannot write $R=H(I+O(1/n))$ in this case. Rather, $O(1/n)H$ stands for a matrix whose $i,j$-th entry is $O(1/n)$ times the $i,j$-th entry of $H$. | |
| May 11, 2014 at 18:24 | comment | added | Christian Remling | $R = H(1+O(1/n))$ and matrix inversion is continuous (and defined on an open set), which gives your claims. These facts are an immediate consequence of the Neumann series for $(1+A)^{-1}$ for small $A$. | |
| May 11, 2014 at 10:28 | history | asked | dima | CC BY-SA 3.0 |