Timeline for Inequality with Hermite polynomials
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| May 27 at 20:49 | comment | added | Dabed | Yes but (and I should probably not insist much more since is my who is lacking in understanding) is that I fail to see what happens in between the two expressions $\lambda_{N}(z)^{-1}=(\min_{\deg P\leq N-1 ,~P(z)=1}\int_{K}|P(w)|^{2}d\mu(w))^{-1} =...?...= \max_{\alpha_{k}}\frac{|\sum_{k=0}^{N-1}\alpha_{k}P_{k}(z)|^{2}}{\sum_{k=0}^{N-1}|\alpha_{k}|^{2}}$ | |
| May 27 at 18:58 | comment | added | user111 | @Dabed the denominator in the ratio is the L2-norm squared of the polynomial (with respect to $\mu$). The numerator accounts for the normalization $P(z)=1$. | |
| May 27 at 18:17 | comment | added | Dabed | Trying to learn here, I think I got an idea of the proof but not sure how $\lambda_{N}(z)=\min_{\deg P\leq N-1,~P(z)=1}\int_{K}|P(w)|^{2}d\mu(w)$ translates into $\lambda_{N}(z)^{-1}=\max_{\alpha_{k}}\frac{|\sum_{k=0}^{N-1}\alpha_{k}P_{k}(z)|^{2}}{\sum_{k=0}^{N-1}|\alpha_{k}|^{2}}$ if it were argmin instead of min it would look a bit like linear regression ($\hat\beta=\mbox{argmin}||X-\beta Y||\rightarrow \hat\beta=(X^TX)^{-1}XY)$ but still can't make sense of how things go | |
| May 27 at 6:15 | history | edited | user111 | CC BY-SA 4.0 |
deleted 1 character in body
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| May 27 at 6:08 | history | answered | user111 | CC BY-SA 4.0 |