Define a function f(x) = √((x+2)+(x-2)√((x+3)+(x-1)√((x+4)+x√((x+5)+(x+1)√((x+6)+ ....)))))
We wish to find f(4)
f(x)^2 = (x+2)+(x-2)√((x+3)+(x-1)√((x+4)+n√((x+5)+(x+1)√((x+6)+ ....))))
f(x)^2 = (x+2)+(x-2)*f(x+1)
f(x+1) = (f(x)^2 - x - 2)/(x-2)
This doesnt feel like a valid argument but the above works if f(x)=x
x+1 = (x^2 - x - 2)/(x-2)
x+1 = (x+1)(x-2)/(x-2)
x+1 = x+1
So f(4)=4
Edited on February 4, 2010, 5:00 pm
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Posted by Jer
on 2010-02-04 15:17:51 |