Timeline for The real and the imaginary part of a vector
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 16, 2023 at 16:02 | comment | added | Sanae Kochiya | I see. From your definition $Z^q:=Z^{2^q,c^q,\epsilon^q}$ and how you define $Z^{n, c, \epsilon}$, I thought $Z^q$ is a $2^q$ by $2^q$ matrix but now it is clear. | |
| Jul 16, 2023 at 11:09 | comment | added | Iosif Pinelis | @SanaeKochiya : Thank you for your appreciation. As for the matrix $Z^q$, its entries are the $z_{ij}^q$'s with $i$ and $j$ in $K^q$. So, $Z^q$ is a $K^q\times K^q$ matrix, in the sense that its rows and columns are indexed by elements of the set $K^q$. So, I would prefer to leave this as it is. Similarly, we have the $c^q_k$'s with $k\in K^q$; so, we have $c^q\in \mathbb R^{K^q}$. | |
| Jul 16, 2023 at 4:41 | review | Suggested edits | |||
| Jul 16, 2023 at 11:10 | |||||
| Jul 16, 2023 at 3:27 | comment | added | Sanae Kochiya | This is a brilliant example! I am sorry for the lat response and am suggesting some revision to your answers. For instance, given that $Z^q$ is a $2^q$ by $2^q$ matrix, in the first equality in th equation right above (30), the sum should range from $1$ to $2^q$ if we want to have $Z^q c^q$. After replacing $K_q$ by $1\leq k \leq 2^q$ in (40), we still have (40) holds since $Z^q c^q$ is in $\mathbb{R}^{2^q}$ | |
| Jul 16, 2023 at 2:55 | vote | accept | Sanae Kochiya | ||
| Jul 14, 2023 at 19:00 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
deleted 6 characters in body
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| Jul 14, 2023 at 18:55 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |