🧵 Untitled Thread
Anonymous at Mon, 22 Jul 2024 00:58:26 UTC No. 16291695
>use perturbation theory and calculus to find the inverse of a matrix that is close to the identity
I have a math degree and I've never seen this. How to? I will respect /sci/ hw policy and not post the problem unless someone asks. Just looking for some reading materials.
Anonymous at Mon, 22 Jul 2024 01:20:06 UTC No. 16291707
>>16291695
We physicists are on a whole other level...
Anonymous at Mon, 22 Jul 2024 01:39:19 UTC No. 16291715
>>16291707
I very much doubt that. I'm sure I'll be lapping you after I see a few theorems
Anonymous at Mon, 22 Jul 2024 03:18:29 UTC No. 16291755
You can use 1+A+A^2 + ... = (1-A)^-1 for A with frobenius norm less than 1 (frobenius norm is sqrt(sum of squares of entries))
Anonymous at Mon, 22 Jul 2024 03:20:49 UTC No. 16291757
>>16291695
https://en.wikipedia.org/wiki/Neuma
Anonymous at Mon, 22 Jul 2024 03:31:38 UTC No. 16291760
>>16291695
you sure do seem to like talking about yourself on social media
Anonymous at Mon, 22 Jul 2024 11:46:52 UTC No. 16292008
>>16291755
>>16291757
This doesn't seem to help with the problem
>>16291760
?
Anonymous at Mon, 22 Jul 2024 11:52:02 UTC No. 16292015
>>16292008
Try sending the actuakl problem
Anonymous at Mon, 22 Jul 2024 15:10:43 UTC No. 16292204
>>16291695
Is this actually any better than the Penrose inverse?
Anonymous at Mon, 22 Jul 2024 15:58:41 UTC No. 16292254
>>16291695
Just post the problem
Anonymous at Mon, 22 Jul 2024 17:20:22 UTC No. 16292332
Anonymous at Mon, 22 Jul 2024 17:22:43 UTC No. 16292335
>>16292015
>>16292254
>>16292332
Shit, wrong pdf. #2
Anonymous at Mon, 22 Jul 2024 17:48:18 UTC No. 16292369
>>16292335
Take x=10^{-3}. Now take the n^th derivative of either B^2 or B^{-1} and set x=0 to find the parts of those expressions which are proportional to 10^{-3n}
Anonymous at Mon, 22 Jul 2024 18:03:06 UTC No. 16292388
>>16292369
For instance I will find B^2 up to the second order (~10^{-6}) terms
B^2 = C_0 + C_1 10^{-3} + C_2 10^{-6}+...
C_0 = B(0)*B(0)
where B(0) = {{0,0,1,0}, {0,0,0,1},{-1,0,0,0},{0,-1,0,0}}
C_1 = B'(0)*B(0)+ B(0)*B'(0)
where B'(0) = {{-1,0,0,0},{0,0,0,0},{0,0,0,0},{0,
C_2 = (1/2)*( B''(0)*B(0) + 2 B'(0)B'(0) + B(0)*B''(0))
(1/2)*B''(0)= {{{{0,0,0,-1},{0,0,0,0},{0,0,0,-1},
get it?
Anonymous at Mon, 22 Jul 2024 19:30:08 UTC No. 16292542
>>16292388
I think so. I'll get to it when I have pen and paper
Anonymous at Tue, 23 Jul 2024 01:14:11 UTC No. 16292931
>>16292388
>>16292369
Got it. And for the matrix inverse question I use frobenius norm from>>16291755
Thanks so much.