Classic square roots procedure

1,006 views

Published on

Published in: Education
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
1,006
On SlideShare
0
From Embeds
0
Number of Embeds
6
Actions
Shares
0
Downloads
0
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

Classic square roots procedure

  1. 1. The Classic iterative method for finding square roots 1 ASn 1 2 Sn Sn - Dave Coulson, 2012
  2. 2. 5 ??? The classic method for determining square roots is an iterative method that might take three or four cycles to get an answer that satisfies you.
  3. 3. 5 2 (First guess) Here’s how it was shown to me back in the stone age: 5 2 2.5 1 2 (2 2.5) 2.25 (Second guess)
  4. 4. 5 2 (First guess) Repeat the process to get a better guess. 5 2 2.5 1 2 (2 2.5) 2.25 (Second guess) 5 2.25 2.22... 1 2 (2.25 2.22 ) 2.235 (Third guess)
  5. 5. Here’s how it would look on a spreadsheet
  6. 6. The technique converges pretty quickly.
  7. 7. The trouble is, though, if you’re going to use aspreadsheet, why not just hit the square root key?
  8. 8. Math technology is everywhere today,so this method is kind of outdated now.
  9. 9. The process I’ve described can begeneralised to this double-step process: A Sn Tn Repeat until frustrated. 1 Sn 1 2 ( S n Tn ) Where Sn is the ‘nth’ guess at the square root Tn is just a temporary number used in the process.
  10. 10. It can be reduced furtherinto a single formula: 1 A Sn 1 2 Sn SnUntil the invention of thecalculator, this was the best wayto obtain a square root, exceptfor simply looking it up in book.
  11. 11. Improvements: See what happens when two steps are combined into one: The formula has been put inside itself , A Sn 1 1 2 Sn making a new faster that proceeds twice Sn as fast 1 1 A A Sn 2 2 2 Sn Sn 1 A 2 Sn Sn
  12. 12. Improvements: 1 A Sn 1 2 Sn Sn A Sn Sn A Sn 2 See how the formula simplifies 4 A Sn Sn
  13. 13. Improvements: f1n f 2n A A A where f 2n 4 f1n f 2n f 1n
  14. 14. Improvements:This is the formula I created (using a different approach)and demonstrated in one of my other presentations. f1n f 2n A A A where f 2n 4 f1n f 2n f 1n
  15. 15. Improvements:For example, 2 2.5 5 5 5 where 2.5 4 2 2.5 2
  16. 16. Improvements:For example, 5 1.125 1.22
  17. 17. Improvements:For example, 5 2.375 Equivalent to two steps of the original formula.
  18. 18. [END]

×