Content deleted Content added
Line 260:
:Since <math>2 \cdot 2 ^n = 2^{n+1}</math>, I think your odd and even cases give the same end result as the binary estimate described in the article. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 15:22, 13 July 2013 (UTC)
::In the example in the article, ''S'' has <math>17=2\cdot8+1</math> binary digits, so the odd case above yields <math>2\cdot2^8=512</math> if I'm not mistaken. It also seems reasonable that 512 should be a better initial guess than 256 if 600 is supposed to be better than 300. [[User:Isheden|Isheden]] ([[User talk:Isheden|talk]]) 15:59, 13 July 2013 (UTC)
::''b'' is one less than the number of digits. S has 17 binary digits because it lies between 2^16 and 2^17 so ''b'' = 16 and your method gives an estimate of 2^8 = 256, the same as the method in the article. The actual square root of S is about 354. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 16:19, 13 July 2013 (UTC)
| |||