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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(4): Solution? | Comment 5 of 12 |
(In reply to re(3): Solution? by Richard)

x=4 appears to be the only solution for any integer n.  I don't know what a non-recursive definition of this sequence would be, but just by observation (using Excel) each of the values of f_n(x) converge rapidly to a limit as n goes to infinity, none of which is equal to 2x.
  Posted by tomarken on 2006-11-19 16:04:04

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