Timeline for Bounding a series of nested integrals
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| May 25, 2018 at 19:59 | comment | added | Ludwig | @lcv: matrices $A_1$ and $A_2$ are real and the commutator $[A_1,A_2]$ is (real and) skew-symmetric. | |
| May 25, 2018 at 19:49 | comment | added | lcv | Are your matrices real or complex? Skew symmetric and real or just skew symmetric? | |
| May 25, 2018 at 18:35 | history | edited | Ludwig | CC BY-SA 4.0 |
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| May 25, 2018 at 0:30 | history | edited | Ludwig | CC BY-SA 4.0 |
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| May 24, 2018 at 4:32 | history | edited | Ludwig | CC BY-SA 4.0 |
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| May 23, 2018 at 9:20 | answer | added | user539887 | timeline score: 1 | |
| May 23, 2018 at 2:03 | comment | added | lcv | @AnthonyQuas why not? I mean, that's the question no? | |
| May 22, 2018 at 17:03 | history | edited | Ludwig | CC BY-SA 4.0 |
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| May 22, 2018 at 5:53 | comment | added | Anthony Quas | If you differentiate the expression, you obtain the differential equation $\dot X(t)=f(t)X(t)$, where $X(t)=I+S(t)$. If $A_1$ and $A_2$ don't commute, there is no reason to expect solutions of this differential equation to remain bounded. | |
| May 22, 2018 at 5:11 | history | edited | Ludwig | CC BY-SA 4.0 |
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| May 22, 2018 at 1:56 | history | edited | Ludwig | CC BY-SA 4.0 |
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| May 22, 2018 at 1:19 | history | asked | Ludwig | CC BY-SA 4.0 |