Timeline for Inverse problems for an asymptotic series which depends on a parameter?
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| Jan 6, 2020 at 2:33 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
Removed the (tag-removed) tag (The question has been bumped anyway - by a new answer.)
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| Dec 27, 2019 at 17:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 27, 2019 at 16:45 | answer | added | Max Alekseyev | timeline score: 1 | |
| Oct 23, 2013 at 10:27 | comment | added | Dirk | This is kind of true. In case of the "truncated SVD" the regularization parameter is the "truncation level" below which no singular values are considered. If you keep this constant while sensing $N$ to infinity you should converge to a regularized solution. Carefully adapting the truncation level to $N$ should converge to the "true" solution" if that exists. Note that for large $N$ you probably do not want to compute the SVD anymore, but solving for the Tikhonov regularized solution is still tractable by conjugate gradient. | |
| Oct 23, 2013 at 8:48 | comment | added | Igor Kotelnikov | @Dirk, I thought that SVD pseudoinverse is a sort of regularization method so that Tikhonov-regularization is not needed. | |
| Oct 21, 2013 at 7:25 | comment | added | Dirk | Looks like weakly-* converging to something strange... For further numerical investigation I would suggest not to use the pure pseudo-inverse but some regularization in the solution of the truncated linear system. Probably ordinary Tikhonov-regularization, i.e. solving $(A^*A + \alpha I)a = A^*b$ (where $A$ is the coefficient matrix and $b$ is the right hand side $b_m = 1/m$) with some small $\alpha$, would be helpful to see what the limit may be. Moreover, one may want to couple $\alpha$ to $N$ in some clever way. Also: Is it clear that a solution exists? | |
| Oct 21, 2013 at 5:03 | history | edited | Igor Kotelnikov | CC BY-SA 3.0 |
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| Oct 21, 2013 at 3:42 | history | edited | Igor Kotelnikov | CC BY-SA 3.0 |
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| Oct 20, 2013 at 11:44 | comment | added | Igor Kotelnikov | @j.c: $a_n$ may not depend on $m$. It is assumend that $m$ runs from 1 to $\infty$, so we have infinite number of equations. | |
| Oct 19, 2013 at 12:04 | comment | added | Jon | @j.c.:Indeed, that is the obvious solution: $a_n=\delta_{n,0}$ with $\nu\rightarrow 0$. | |
| Oct 19, 2013 at 7:57 | comment | added | j.c. | Are there some additional conditions on $a_n(\nu)$ or am I misreading something? Why couldn't you just take all $a_n(\nu)$ to be zero except $a_0$. | |
| Oct 19, 2013 at 2:09 | history | edited | Igor Kotelnikov | CC BY-SA 3.0 |
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| Oct 18, 2013 at 14:00 | history | edited | Igor Kotelnikov | CC BY-SA 3.0 |
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| Oct 18, 2013 at 8:59 | history | edited | Igor Kotelnikov | CC BY-SA 3.0 |
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| Oct 18, 2013 at 8:32 | review | First posts | |||
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| Oct 18, 2013 at 8:32 | history | edited | Igor Kotelnikov | CC BY-SA 3.0 |
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| Oct 18, 2013 at 8:27 | history | edited | Igor Kotelnikov | CC BY-SA 3.0 |
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| Oct 18, 2013 at 8:20 | history | edited | Igor Kotelnikov | CC BY-SA 3.0 |
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| Oct 18, 2013 at 8:13 | history | asked | Igor Kotelnikov | CC BY-SA 3.0 |