 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Infinitely Continued III (Posted on 2010-02-04) Evaluate:

√(6+2√(7+3√(8+4√(9+5√(10+ ....)))))

 See The Solution Submitted by K Sengupta Rating: 4.5000 (2 votes) Comments: ( Back to comment list | You must be logged in to post comments.) re: Is my reasoning valid? Spoiler | Comment 4 of 7 | (In reply to Is my reasoning valid? by Jer)

I think you�re right Jer; for integers >=0 your inspired function, f(x), is just x in disguise.
To get round the singularity that Steve mentions, and find f(3), I believe we can use the first derivative as follows:

(x - 2) f(x + 1) = f(x)2 - x - 2      (1)  This can be differentiated to give

(x - 2) f�(x + 1) + f(x + 1) = 2 f(x) f�(x) - 1

Using x = 0, 1 and 2, this equation gives

-2 f�(1) + f(1) = 2 f(0) f�(0) - 1                (2)
-f�(2) +f(2) = 2 f(1) f�(1) - 1                (3)
f(3) = 2 f(2) f�(2) - 1                 (4)

We know that f(0)=0, f(1)=1 and f(2)=2 by direct substitution into your formula, so

(2) gives           f�(1) = 1
(3) then gives    f�(2) = 1
(4) then gives    f(3) = 3

x=3 in (1) then gives   f(4) = 32 -3 - 2 = 4, which solves the problem.

_________________

I think this may pave the way to an inductive proof that f(x) = x for all positive integers. I�m not sure what happens between the integers, but my plotting software suggests something very interesting... (not had time to explore yet).

 Posted by Harry on 2010-02-05 08:57:46 Please log in:

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