Timeline for Majorization of cyclic products
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 6, 2016 at 20:45 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
added 2 characters in body
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| Jan 6, 2016 at 20:37 | vote | accept | Wolfgang | ||
| Jan 6, 2016 at 16:59 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
added 9 characters in body
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| Jan 6, 2016 at 16:57 | comment | added | Fedor Petrov | Ah, yes, I should probably rename them | |
| Jan 6, 2016 at 16:11 | comment | added | Wolfgang | In fact, I realize now that your $\alpha_i$'s have nothing to do with mine, Your vector $(\alpha_1,...,\alpha_n)$ is rather the solution of the linear system. For the example and n=4, we get $\alpha_1=3/4,\alpha_2=\alpha_3=0,\alpha_4=1/4$. | |
| Jan 6, 2016 at 10:16 | comment | added | Wolfgang | oh sure, sorry... I misread as $\min_i |\alpha_i+c|\leq 1$... | |
| Jan 6, 2016 at 9:52 | comment | added | Fedor Petrov | maximum of $F(c):=\sum |\alpha_i+c|$ is of course infinite, so if minimum does not exceed 1, there exists $c$ such that $F(c)=1$ as we need. | |
| Jan 6, 2016 at 9:50 | comment | added | Wolfgang | OK now I see. The product over i yields the $\sum \beta_k$. Nice! | |
| Jan 6, 2016 at 9:27 | comment | added | Fedor Petrov | Last = is = since $\sum \beta_k=1$ | |
| Jan 6, 2016 at 8:06 | comment | added | Wolfgang | Very interesting approach! I think it works. As we can divide all the $\alpha_i$ wlog by a common factor, there will always such a c, and it will even be positive, so the sign function isn't needed. I think the last "=" after Jensen must be a "$\le$" since you use $\beta_k\le1$ don't you? And why do you mention $\min_c \sum_i |\alpha_i+c|\leq 1$? max? | |
| Jan 6, 2016 at 0:22 | history | answered | Fedor Petrov | CC BY-SA 3.0 |