Wan we do better than O($K^2$)
Difference between en5 and en6, changed 44 character(s)
Consider↵
↵
$R_k = \sum_{j = 0}^{ j = k - 1} \sqrt\nu_j(\sqrt{k - j} - \sqrt{k - 1 - j})$↵
$\newline$↵
Can we compute $\sum_{j = 1}^{j = K} \nu_kR_k$ in less than O(K^2) ?↵
$\newline$↵
$\nu$ and $R$ are arrays of double size $K$↵
$\newline$↵
The answer is a double.

History

 
 
 
 
Revisions
 
 
  Rev. Lang. By When Δ Comment
en9 English chika10 2022-09-28 12:36:58 2
en8 English chika10 2022-09-28 11:48:09 12
en7 English chika10 2022-09-28 11:47:29 7 Tiny change: 'of double size $K$\' -> 'of double of size $K$\'
en6 English chika10 2022-09-28 11:47:06 44
en5 English chika10 2022-09-28 11:45:47 45
en4 English chika10 2022-09-28 11:45:26 14 Tiny change: '^2) ?\n$\nu$ and R are array' -> '^2) ?\n$\newline$\n$\nu$ and $R$ are array' (published)
en3 English chika10 2022-09-28 11:44:44 51 Tiny change: '1 - j})$\n\newline\nCan we c' -> '1 - j})$\n$\newline$\nCan we c'
en2 English chika10 2022-09-28 11:44:31 120
en1 English chika10 2022-09-28 11:41:59 119 Initial revision (saved to drafts)