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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

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

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 re(6): Solution? | Comment 7 of 12 |
(In reply to re(5): Solution? by Richard)

I'll try to state my original idea a little more clearly, and add that x = -4 also should work.  If you set f_1(x) equal to 2x, you get:

2x = sqrt(x^2 + 48)
4x^2 = x^2 + 48
3x^2 = 48
x^2 = 16
x = 4 or -4

Plugging x = 4 back into the equation for f_1(x):

f_1(x) = sqrt(16+48) = 8   (or choose -8, if you plug in x = -4)

But then f_0(x) = f_1(x) = 8, so solving f_2(x) will be exactly the same as f_1(x), and so on for f_3, f_4, etc.  It will always ultimately reduce to:

2x = sqrt(64)

No matter what n you choose x=4 and x=-4 will be the only solutions.

If you want to try to solve them non-recursively, it quickly gets cumbersome (at least if you do it by hand):

f_2:  x^4 - 4x^2 - 192 = 0

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

etc...

 Posted by tomarken on 2006-11-19 19:45:48

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