Timeline for Error bounds on the expansion of square root of matrix
Current License: CC BY-SA 4.0
12 events
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| Sep 29, 2020 at 16:45 | comment | added | yoshi | But the $n+1$ term bounds the error when Taylor series is within the radius of convergence, right? (The analogous condition holds for the example above) | |
| Sep 29, 2020 at 10:02 | comment | added | Federico Poloni | No, the $n+1$st term does not estimate the error in Taylor series, in general. Case in point: $1/(1-x)$. | |
| Sep 28, 2020 at 15:36 | comment | added | yoshi | My inclination was to think this works like taylor series approximation where the n+1 estimates the error. I suspect I'm wrong somehow though. I thought this might be correct because of the final inequality in the wiki article -- perhaps I could just use a n term taylor appx for $f_k$ and the actual square root for $f_l$: en.wikipedia.org/wiki/… | |
| Sep 28, 2020 at 14:15 | history | edited | YCor | CC BY-SA 4.0 |
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| Sep 28, 2020 at 14:14 | comment | added | Federico Poloni | Why do you expect the first omitted term to bound the error ? This series does not have alternating signs, and as far as I understand this result is not true even for scalar inputs. | |
| Sep 28, 2020 at 13:39 | history | edited | yoshi |
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| Sep 26, 2020 at 19:03 | comment | added | Mahdi - Free Palestine | As another approach, see mathoverflow.net/questions/301272/… | |
| Sep 26, 2020 at 16:37 | comment | added | yoshi | ah yes -- In this post I am considering a normalized $A$ in the expansion -- $A/\|A\|$ - which is 1/4 and 1 on the diagonal. So for this matrix I believe the spectrum is in the disk centered at 1 with radius 1. | |
| Sep 26, 2020 at 15:08 | comment | added | Carlo Beenakker | doesn't it state the requirement that the spectrum of $A$ is contained within a unit disc in the complex plane? | |
| Sep 26, 2020 at 15:03 | comment | added | yoshi | I should have mentioned that the wiki article I was using for the formula is here: en.wikipedia.org/wiki/Square_root_of_a_matrix#Power_series -- the condition they have is $\|(I - A)\|^n \leq 1$ -- which is true for this case. Where are you seeing the eigenvalue condition you are referring too? | |
| Sep 26, 2020 at 14:22 | comment | added | Carlo Beenakker | the Wikipedia formula you cite says that it needs the eigenvalues of $A$ to lie in the interval (0,2), which does not apply to your choice of $A$. | |
| Sep 26, 2020 at 13:59 | history | asked | yoshi | CC BY-SA 4.0 |