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 Another function problem (Posted on 2006-11-19)
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

 Submitted by K Sengupta Rating: 4.0000 (1 votes) Solution: (Hide) At the outset, we note that x = 4 is a solution of f1(x)= 2x Hence, f2(x) = x2+6f1(x)= 2x, also has a solution x =4 Accordingly, by induction it can be shown that if fn(x) = 2x has a solution x=4, so does the equation fn+1(x)( i.e. possesses a solution of x=4 By induction, it also follows that fn(x)/x decreases as x increases for every n. Consequently, fn(x)= 2x possesses the unique solution x=4**************************************** In addition, refer to the explanation provided by Paul in this location. Joel also provides a good methodology here.

 Subject Author Date re(2): Solution? Joel 2006-11-21 23:22:33 Solution Rajesh 2006-11-21 13:17:42 a complete solution Paul 2006-11-20 10:07:25 re(8): Solution? tomarken 2006-11-20 01:26:34 re(7): Solution? Richard 2006-11-19 20:40:39 re(6): Solution? tomarken 2006-11-19 19:45:48 re(5): Solution? Richard 2006-11-19 18:02:36 re(4): Solution? tomarken 2006-11-19 16:04:04 re(3): Solution? Richard 2006-11-19 12:40:29 re(2): Solution? tomarken 2006-11-19 12:14:56 re: Solution? Richard 2006-11-19 12:01:06 Solution? tomarken 2006-11-19 11:17:32

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