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Another function problem (Posted on 20061119) 

Define a sequence of functions f_{0}, f_{1}, f_{2}, ......, by
f_{0}(x)= 8, for all real x, and f_{n+1}(x) = sqrt(x^{2} + 6f_{n}(x)); for all real x and all nonnegative integers n.
Solve the equation
f_{n}(x)= 2x

Submitted by K Sengupta

Rating: 4.0000 (1 votes)


Solution:

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At the outset, we note that x = 4 is a solution of f_{1}(x)= 2x
Hence, f_{2}(x) = x^{2}+6f_{1}(x)= 2x, also has a solution x =4
Accordingly, by induction it can be shown that if f_{n}(x) = 2x has a solution x=4, so does the equation f_{n+1}(x)( i.e. possesses a solution of x=4
By induction, it also follows that f_{n}(x)/x decreases as x increases for every n.
Consequently, f_{n}(x)= 2x possesses the unique solution x=4
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In addition, refer to the explanation provided by Paul in this location.
Joel also provides a good methodology here.

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