*** edited to show correct argument of floor function in solution.
F((x+1)
3) = x
3, where F(x) = x - floor(x)
0 <= F(x) < 1, so 0 <= x < 1
F((x+1)3) = x3 + 3x2 + 3x + 1 - floor((x+1)3) = x3
floor((x+1)3) =3x2 + 3x + 1 = 3(x2 + x) + 1, an integer >=0 and < 7 as x < 1
x2 + x = 5/3, x = -1/2+sqrt(23/3)/2
x2 + x = 4/3, x = -1/2+sqrt(19/3)/2
x2 + x = 1, x = -1/2+sqrt(5)/2
x2 + x = 2/3, x = -1/2+sqrt(11/3)/2
x2 + x = 1/3, x = -1/2+sqrt(7/3)/2
x2 + x = 0, x = 0
Edited on September 27, 2016, 6:29 pm
Edited on September 27, 2016, 6:30 pm
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Posted by ken
on 2016-09-27 13:43:49 |
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