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Another function problem (Posted on 2006-11-19) Difficulty: 3 of 5
Define a sequence of functions f0, f1, f2, ......, by

f0(x)= 8, for all real x, and
fn+1(x) = sqrt(x2 + 6fn(x)); for all real x and all non-negative integers n.

Solve the equation

fn(x)= 2x

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(8): Solution? | Comment 9 of 12 |
(In reply to re(7): Solution? by Richard)

I don't like the x=-4 solution either, I only mentioned it because you were looking for other x values that would satisfy the equation for some fixed n.  But I think it is perfectly reasonable to say sqrt(64) is -8.  In other words, would you say x=-4 is one of the roots of:

f_2:  x^4 - 4x^2 - 192 = 0

f_3:  x^8 - 8x^6 + 16x^4 - 576x^2 - 27648 = 0

..,?  If you consider these non-recursive cases of the problem then x=-4 is certainly a solution.


I don't like x=-4 as a solution because:

a) it doesn't hold for f_0(x), and

b) if you assume f_n(x) = -8, then f_n+1(x) is unsolvable.


  Posted by tomarken on 2006-11-20 01:26:34

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