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| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 2 years, 6 months |
| seen | May 14 at 16:58 | |
| stats | profile views | 207 |
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May 7 |
revised |
Preconditioning and effects on precision of solution of LSE corrected reference |
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May 7 |
comment |
Preconditioning and effects on precision of solution of LSE I do not think that your answer makes sense. Preconditioning precisely means decreasing the condition number. So there is no 'balancing' necessary between error magnification and convergence rates. In particular, my question regards only the numerical level - it is about the stability of numerical algorithms. |
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May 3 |
asked | Preconditioning and effects on precision of solution of LSE |
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Apr 5 |
comment |
Boundedness of a Solution Operator @user17904: You need to show that for Sobolev space $H$ of sufficient regularity the trace from $H$ to $L^2(\partial\Omega)$ has closed range. Than a bounded right-inverse exists by functional analysis. Typically, you can prove surjectivity, say, by partition of unity, smoothing and an explicit construction. |
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Mar 20 |
revised |
fundamental theorem of linear inequalities added 336 characters in body |
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Mar 7 |
comment |
pseudoinverse under change of norm I have completely rewritten the question. |
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Mar 7 |
revised |
pseudoinverse under change of norm complete rewrite |
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Mar 7 |
accepted | What can be said about the minimizer of $\| b - A x \| ^2+ \| x \|^2$? |
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Mar 7 |
asked | pseudoinverse under change of norm |
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Mar 7 |
asked | What can be said about the minimizer of $\| b - A x \| ^2+ \| x \|^2$? |
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Mar 6 |
asked | Mappings preserving convex polyhedra |
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Mar 4 |
comment |
An variation of “Lk” in simplicial complexes @StefanH.: Yes, thanks. |
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Mar 4 |
revised |
An variation of “Lk” in simplicial complexes edited body |
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Mar 4 |
asked | An variation of “Lk” in simplicial complexes |
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Feb 23 |
awarded | Popular Question |
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Feb 13 |
comment |
How can not-equals be expressed as an inequality for a linear programming model It is intutive it does not work, because polyhedrons are convex and closed, while not-equals correspond to non-convex open regions. |
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Feb 13 |
asked | fundamental theorem of linear inequalities |
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Feb 1 |
accepted | Expansion of $x^{-1/2}$ at $0$ |
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Feb 1 |
revised |
Expansion of $x^{-1/2}$ at $0$ clarification |
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Feb 1 |
asked | Expansion of $x^{-1/2}$ at $0$ |
