Let's take a look at a few values of x.
g(.5) = ⌊.5 ⌊.5⌋⌋ = ⌊.5*0⌋ = ⌊0⌋ = 0
g(1.9) = ⌊1.9⌊1.9⌋⌋ = ⌊1.9*1⌋ = ⌊1.9⌋ = 1
g(2.4) = ⌊2.4⌊2.4⌋⌋ = ⌊2.4*2⌋ = ⌊4.8⌋ = 4
g(2.7) = ⌊2.7⌊2.7⌋⌋ = ⌊2.7*2⌋ = ⌊5.4⌋ = 5
so, for any 0<z<1, g(2+z) = ⌊(2+z)⌊2+z⌋⌋ = ⌊(2+z)*2⌋ = ⌊4+2z⌋
So, g(2+z) = 4 for z < 0.5
g(2+z) = 5 for z >= 0.5
So, I'd guess that for any integer k>1,
g(k+z) = k^2 for z < 1/k
k^2+1 for 1/k <= z < 2/k
...
k^2+(k-1) for (k-1)/k <= z < 1
I haven't checked it thoroughly, but see what you can do with that. It does seem to indicate that there is no solution for g(x) = 3
Hello,
I've been having trouble on the following question:
We let g(x) = ⌊x⌊x⌋⌋ for x >= 0.
i. Find all x >= 0 so that g(x) = 1.
ii. Find all x >= 0 so that g(x) = 3.
iii. Find all x >= 0 so that g(x) = 5.
iv. Find the number of possible values of g(x) for 0 <= x <= 10.
I started on i. using the properties of a floor function and got 1 <= x⌊x⌋ < 2, but I have no idea what to do next. Please help as soon as possible.
Thanks in advance.
3 answers
Thank you for your response. I have solved (a), (b) and (c). Yes, there is no solution for g(x)=3.
Please help on part iv, I’ve no idea where to start.