Timeline for Help with understanding a proof on angle preservation
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 22, 2018 at 4:35 | comment | added | Taliant | Ah okay, thanks so much for the clarification. | |
| Mar 20, 2018 at 22:41 | comment | added | Luc Guyot | What you wrote is the numerator of the square of the left-hand side of the right-most part of $(d)$, which matches the left-hand side of the third to last inequality in my proof. The right-hand side of the latter inequality is the square of the denominator times $1 + \varepsilon$. | |
| Mar 20, 2018 at 20:15 | comment | added | Taliant | This is a very basic question, but could you tell me the inequality you got when you squared (d)? I got $\frac{\| R\textbf{x} \|^2}{\| \textbf{x}\|^2} + \frac{\| R\textbf{w} \|^2}{\| \textbf{w} \|^2} - 2 \frac{\langle R\textbf{x}, R\textbf{w} \rangle}{\| x \| \| w \|}$, which doesn't look like it can be re-written as the third to last inequality. | |
| Mar 20, 2018 at 17:01 | vote | accept | Taliant | ||
| Mar 19, 2018 at 17:49 | comment | added | Luc Guyot | @Taliant By squaring the two sides of the right-most inequality in $(d)$, followed by some simple manipulations (getting rid of the denominator, expanding the squared Euclidean norms, etc.). | |
| Mar 19, 2018 at 17:06 | comment | added | Taliant | How did you derive the third to last inequality? | |
| Mar 19, 2018 at 12:31 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Makes the definition of the Gaussian matrix precise
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| Mar 18, 2018 at 21:52 | history | answered | Luc Guyot | CC BY-SA 3.0 |