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Another function problem (Posted on 2006-11-19) |
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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
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Submitted by K Sengupta
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Rating: 4.0000 (1 votes)
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Solution:
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(Hide)
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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
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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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