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

Home > Numbers
Infinitely Continued III (Posted on 2010-02-04) Difficulty: 3 of 5
Evaluate:

√(6+2√(7+3√(8+4√(9+5√(10+ ....)))))

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

Comments: ( Back to comment list | You must be logged in to post comments.)
A couple of things Comment 7 of 7 |
First what if we go backwards?
Using my function f(x) = sqrt((x+2)+(x-2)f(x+1))
f(0) = sqrt(2+-2*f(1)) = sqrt(2+-2) = 0
f(-1) = sqrt(1+-3*f(0)) = sqrt(1+0) = 1
which is promising but
f(-2) = sqrt(0+-4*f(-1)) = sqrt(-4) = 2i
which is less so
f(-3) =sqrt(-1+-5*f(-2)) = sqrt(-1-10i) = 2.12519-2.35052i
and they seem to continue as irrational complex numbers without too much pattern.

Second the recursive proof that f(x)=x for x a positive integer:
We now know f(3)=3
assume f(x)=x to prove f(x+1)=x+1

f(x+1) = (f(x)^2 - (x+2)) / (x-2)
= ((x)^2 - x - 2) / (x-2)
=(x+1)(x-2) / (x-2)
=x+1

[I was very close to this in my other post but it requires x>2 which we didn't have at the time.]

Finally to show this formula works for any real number > 0
We prove f(x+a) = x+a where x is a positive integer and a is a positive real.
x+a = f(x)+a
so we have f(x+a)=f(x)+a and we are done.


  Posted by Jer on 2010-02-06 15:20:25
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 (0)
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