All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
Approximating radicals (Posted on 2015-02-17) Difficulty: 3 of 5
You can approximate irrational square roots with rational numbers using linear interpolation between the integers as follows:

√1 = 1
√2 ≈ 4/3
√3 ≈ 5/3
√4 = 2
√5 ≈ 11/5
√6 ≈ 12/5
√7 ≈ 13/5
√8 ≈ 14/5
√9 = 3
etc...

How good an approximation is this?
For large numbers, might the previous or next fraction be a better approximation?

No Solution Yet Submitted by Jer    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts More thoughts | Comment 3 of 4 |

Interesting problem.

Let y be an integer between (x-1)^2 and x^2, say (x^2-n)      
Then y^(1/2)≈x-n/(2x-1).      
The gap between x^2 and (x-1)^2 = 2x-1, and the function is most inaccurate at mid range, say n=x.      
      
However:      
(x^2-x)^(1/2)-(x-x/(2x-1)), x=10^n, is only out by around 1 part in 10^(n+1)  
x=10: 0.0131 (y between 82 and 100)      
x=100: 0.00125 (y between 9802 and 10000)      
x=1000: 0.000125 (y between 998002 and 1000000)      
etc.      
Hence,  the function becomes very accurate for large y.  


Edited on February 17, 2015, 9:12 pm
  Posted by broll on 2015-02-17 21:01:15

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information